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| committer | nfenwick <nfenwick@pglaf.org> | 2025-01-27 17:16:01 -0800 |
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Anyone seeking to utilize +this eBook outside of the United States should confirm copyright +status under the laws that apply to them. diff --git a/README.md b/README.md new file mode 100644 index 0000000..5e6db16 --- /dev/null +++ b/README.md @@ -0,0 +1,2 @@ +Project Gutenberg (https://www.gutenberg.org) public repository for +eBook #60816 (https://www.gutenberg.org/ebooks/60816) diff --git a/old/60816-0.txt b/old/60816-0.txt deleted file mode 100644 index edb2c66..0000000 --- a/old/60816-0.txt +++ /dev/null @@ -1,3446 +0,0 @@ -The Project Gutenberg EBook of The Elements of Perspective, by John Ruskin - -This eBook is for the use of anyone anywhere in the United States and most -other parts of the world at no cost and with almost no restrictions -whatsoever. You may copy it, give it away or re-use it under the terms of -the Project Gutenberg License included with this eBook or online at -www.gutenberg.org. If you are not located in the United States, you'll have -to check the laws of the country where you are located before using this ebook. - -Title: The Elements of Perspective - arranged for the use of schools and intended to be read - in connection with the first three books of Euclid - -Author: John Ruskin - -Release Date: November 30, 2019 [EBook #60816] - -Language: English - -Character set encoding: UTF-8 - -*** START OF THIS PROJECT GUTENBERG EBOOK THE ELEMENTS OF PERSPECTIVE *** - - - - -Produced by Juliet Sutherland, David Wilson and the Online -Distributed Proofreading Team at http://www.pgdp.net - - - - - - - - - - +--------------------------------------------------------------------+ - | | - | Transcriber’s Note | - | | - | In the original book the author used lowercase italic letters and | - | lowercase small capitals to label the geometric diagrams. These | - | have been here transcribed as lowercase and uppercase italic | - | letters respectively (_aA_). | - | | - | This file should be read using a font that supports the following | - | Unicode characters: | - | ∠ angle, ∶ ratio, ∷ proportion, ∴ therefore, ′ prime, and | - | ″ double prime. | - | | - +--------------------------------------------------------------------+ - - - - - +----------------------------------+ - | | - | Library Edition | - | | - +----------------------------------+ - | | - | | - | THE COMPLETE WORKS | - | OF | - | JOHN RUSKIN | - | | - | | - | ELEMENTS OF DRAWING AND | - | PERSPECTIVE | - | THE TWO PATHS | - | UNTO THIS LAST | - | MUNERA PULVERIS | - | SESAME AND LILIES | - | ETHICS OF THE DUST | - | | - | | - +----------------------------------+ - | NATIONAL LIBRARY ASSOCIATION | - | NEW YORK CHICAGO | - +----------------------------------+ - - - - - THE ELEMENTS OF PERSPECTIVE - - ARRANGED FOR THE USE OF SCHOOLS - - AND INTENDED TO BE READ IN CONNECTION WITH THE - FIRST THREE BOOKS OF EUCLID. - - - - - CONTENTS. - - PAGE - Preface ix - - Introduction 1 - - PROBLEM I. - To fix the Position of a given Point 10 - - PROBLEM II. - To draw a Right Line between two given Points 13 - - PROBLEM III. - To find the Vanishing-Point of a given Horizontal Line 17 - - PROBLEM IV. - To find the Dividing-Points of a given Horizontal Line 23 - - PROBLEM V. - To draw a Horizontal Line, given in Position and Magnitude, - by means of its Sight-Magnitude and Dividing-Points 24 - - PROBLEM VI. - To draw any Triangle, given in Position and Magnitude, in a - Horizontal Plane 27 - - PROBLEM VII. - To draw any Rectilinear Quadrilateral Figure, given in - Position and Magnitude, in a Horizontal Plane 29 - - PROBLEM VIII. - To draw a Square, given in Position and Magnitude, in a - Horizontal Plane 31 - - PROBLEM IX. - To draw a Square Pillar, given in Position and Magnitude, - its Base and Top being in Horizontal Planes 34 - - PROBLEM X. - To draw a Pyramid, given in Position and Magnitude, on a - Square Base in a Horizontal Plane 36 - - PROBLEM XI. - To draw any Curve in a Horizontal or Vertical Plane 38 - - PROBLEM XII. - To divide a Circle drawn in Perspective into any given - Number of Equal Parts 42 - - PROBLEM XIII. - To draw a Square, given in Magnitude, within a larger - Square given in Position and Magnitude; the Sides of the - two Squares being Parallel 45 - - PROBLEM XIV. - To draw a Truncated Circular Cone, given in Position and - Magnitude, the Truncations being in Horizontal Planes, - and the Axis of the Cone vertical 47 - - PROBLEM XV. - To draw an Inclined Line, given in Position and Magnitude 50 - - PROBLEM XVI. - To find the Vanishing-Point of a given Inclined Line 53 - - PROBLEM XVII. - To find the Dividing-Points of a given Inclined Line 55 - - PROBLEM XVIII. - To find the Sight-Line of an Inclined Plane in which Two - Lines are given in Position 57 - - PROBLEM XIX. - To find the Vanishing-Point of Steepest Lines in an Inclined - Plane whose Sight-Line is given 59 - - PROBLEM XX. - To find the Vanishing-Point of Lines perpendicular to the - Surface of a given Inclined Plane 61 - - - APPENDIX. - - I. - Practice and Observations on the preceding Problems 69 - - II. - Demonstrations which could not conveniently be included in - the Text 99 - - - - -PREFACE. - - -For some time back I have felt the want, among Students of Drawing, of a -written code of accurate Perspective Law; the modes of construction in -common use being various, and, for some problems, insufficient. It would -have been desirable to draw up such a code in popular language, so as to -do away with the most repulsive difficulties of the subject; but finding -this popularization would be impossible, without elaborate figures and -long explanations, such as I had no leisure to prepare, I have arranged -the necessary rules in a short mathematical form, which any schoolboy -may read through in a few days, after he has mastered the first three -and the sixth books of Euclid. - -Some awkward compromises have been admitted between the first-attempted -popular explanation, and the severer arrangement, involving irregular -lettering and redundant phraseology; but I cannot for the present do -more, and leave the book therefore to its trial, hoping that, if it be -found by masters of schools to answer its purpose, I may hereafter bring -it into better form.[1] - -An account of practical methods, sufficient for general purposes of -sketching, might indeed have been set down in much less space: but if -the student reads the following pages carefully, he will not only find -himself able, on occasion, to solve perspective problems of a complexity -greater than the ordinary rules will reach, but obtain a clue to many -important laws of pictorial effect, no less than of outline. The subject -thus examined becomes, at least to my mind, very curious and -interesting; but, for students who are unable or unwilling to take it up -in this abstract form, I believe good help will be soon furnished, in a -series of illustrations of practical perspective now in preparation by -Mr. Le Vengeur. I have not seen this essay in an advanced state, but the -illustrations shown to me were very clear and good; and, as the author -has devoted much thought to their arrangement, I hope that his work will -be precisely what is wanted by the general learner. - -Students wishing to pursue the subject into its more extended branches -will find, I believe, Cloquet’s treatise the best hitherto published.[2] - - - [1] Some irregularities of arrangement have been admitted merely for - the sake of convenient reference; the eighth problem, for - instance, ought to have been given as a case of the seventh, but - is separately enunciated on account of its importance. - - Several constructions, which ought to have been given as problems, - are on the contrary given as corollaries, in order to keep the - more directly connected problems in closer sequence; thus the - construction of rectangles and polygons in vertical planes would - appear by the Table of Contents to have been omitted, being given - in the corollary to Problem IX. - - [2] Nouveau Traité Élémentaire de Perspective. Bachelier, 1823. - - - - -THE ELEMENTS OF PERSPECTIVE. - - - - -INTRODUCTION. - - -When you begin to read this book, sit down very near the window, and -shut the window. I hope the view out of it is pretty; but, whatever the -view may be, we shall find enough in it for an illustration of the first -principles of perspective (or, literally, of “looking through”). - -Every pane of your window may be considered, if you choose, as a glass -picture; and what you see through it, as painted on its surface. - -And if, holding your head still, you extend your hand to the glass, you -may, with a brush full of any thick color, trace, roughly, the lines of -the landscape on the glass. - -But, to do this, you must hold your head very still. Not only you must -not move it sideways, nor up and down, but it must not even move -backwards or forwards; for, if you move your head forwards, you will see -_more_ of the landscape through the pane; and, if you move it backwards, -you will see _less_: or considering the pane of glass as a picture, when -you hold your head near it, the objects are painted small, and a great -many of them go into a little space; but, when you hold your head some -distance back, the objects are painted larger upon the pane, and fewer -of them go into the field of it. - -But, besides holding your head still, you must, when you try to trace -the picture on the glass, shut one of your eyes. If you do not, the -point of the brush appears double; and, on farther experiment, you will -observe that each of your eyes sees the object in a different place on -the glass, so that the tracing which is true to the sight of the right -eye is a couple of inches (or more, according to your distance from the -pane,) to the left of that which is true to the sight of the left. - -Thus, it is only possible to draw what you see through the window -rightly on the surface of the glass, by fixing one eye at a given point, -and neither moving it to the right nor left, nor up nor down, nor -backwards nor forwards. Every picture drawn in true perspective may be -considered as an upright piece of glass,[3] on which the objects seen -through it have been thus drawn. Perspective can, therefore, only be -quite right, by being calculated for one fixed position of the eye of -the observer; nor will it ever appear _deceptively_ right unless seen -precisely from the point it is calculated for. Custom, however, enables -us to feel the rightness of the work on using both our eyes, and to be -satisfied with it, even when we stand at some distance from the point it -is designed for. - -Supposing that, instead of a window, an unbroken plate of crystal -extended itself to the right and left of you, and high in front, and -that you had a brush as long as you wanted (a mile long, suppose), and -could paint with such a brush, then the clouds high up, nearly over your -head, and the landscape far away to the right and left, might be traced, -and painted, on this enormous crystal field.[4] But if the field were so -vast (suppose a mile high and a mile wide), certainly, after the picture -was done, you would not stand as near to it, to see it, as you are now -sitting near to your window. In order to trace the upper clouds through -your great glass, you would have had to stretch your neck quite back, -and nobody likes to bend their neck back to see the top of a picture. So -you would walk a long way back to see the great picture—a quarter of a -mile, perhaps,—and then all the perspective would be wrong, and would -look quite distorted, and you would discover that you ought to have -painted it from the greater distance, if you meant to look at it from -that distance. Thus, the distance at which you intend the observer to -stand from a picture, and for which you calculate the perspective, -ought to regulate to a certain degree the size of the picture. If you -place the point of observation near the canvas, you should not make the -picture very large: _vice versâ_, if you place the point of observation -far from the canvas, you should not make it very small; the fixing, -therefore, of this point of observation determines, as a matter of -convenience, within certain limits, the size of your picture. But it -does not determine this size by any perspective law; and it is a mistake -made by many writers on perspective, to connect some of their rules -definitely with the size of the picture. For, suppose that you had what -you now see through your window painted actually upon its surface, it -would be quite optional to cut out any piece you chose, with the piece -of the landscape that was painted on it. You might have only half a -pane, with a single tree; or a whole pane, with two trees and a cottage; -or two panes, with the whole farmyard and pond; or four panes, with -farmyard, pond, and foreground. And any of these pieces, if the -landscape upon them were, as a scene, pleasantly composed, would be -agreeable pictures, though of quite different sizes; and yet they would -be all calculated for the same distance of observation. - -In the following treatise, therefore, I keep the size of the picture -entirely undetermined. I consider the field of canvas as wholly -unlimited, and on that condition determine the perspective laws. After -we know how to apply those laws without limitation, we shall see what -limitations of the size of the picture their results may render -advisable. - -But although the size of the _picture_ is thus independent of the -observer’s distance, the size of the _object represented_ in the picture -is not. On the contrary, that size is fixed by absolute mathematical -law; that is to say, supposing you have to draw a tower a hundred feet -high, and a quarter of a mile distant from you, the height which you -ought to give that tower on your paper depends, with mathematical -precision, on the distance at which you intend your paper to be placed. -So, also, do all the rules for drawing the form of the tower, whatever -it may be. - -Hence, the first thing to be done in beginning a drawing is to fix, at -your choice, this distance of observation, or the distance at which you -mean to stand from your paper. After that is determined, all is -determined, except only the ultimate size of your picture, which you may -make greater, or less, not by altering the size of the things -represented, but by _taking in more, or fewer_ of them. So, then, before -proceeding to apply any practical perspective rule, we must always have -our distance of observation marked, and the most convenient way of -marking it is the following: - - [Illustration: Fig. 1. PLACING OF THE SIGHT-POINT, SIGHT-LINE, - STATION-POINT, AND STATION-LINE.] - - -I. THE SIGHT-POINT.—Let _ABCD_, Fig. 1., be your sheet of paper, the -larger the better, though perhaps we may cut out of it at last only a -small piece for our picture, such as the dotted circle _NOPQ_. This -circle is not intended to limit either the size or shape of our picture: -you may ultimately have it round or oval, horizontal or upright, small -or large, as you choose. I only dot the line to give you an idea of -whereabouts you will probably like to have it; and, as the operations of -perspective are more conveniently performed upon paper underneath the -picture than above it, I put this conjectural circle at the top of the -paper, about the middle of it, leaving plenty of paper on both sides and -at the bottom. Now, as an observer generally stands near the middle of a -picture to look at it, we had better at first, and for simplicity’s -sake, fix the point of observation opposite the middle of our -conjectural picture. So take the point _S_, the center of the circle -_NOPQ_;—or, which will be simpler for you in your own work, take the -point _S_ at random near the top of your paper, and strike the circle -_NOPQ_ round it, any size you like. Then the point _S_ is to represent -the point _opposite_ which you wish the observer of your picture to -place his eye, in looking at it. Call this point the “Sight-Point.” - - -II. THE SIGHT-LINE.—Through the Sight-point, _S_, draw a horizontal -line, _GH_, right across your paper from side to side, and call this -line the “Sight-Line.” - -This line is of great practical use, representing the level of the eye -of the observer all through the picture. You will find hereafter that if -there is a horizon to be represented in your picture, as of distant sea -or plain, this line defines it. - - -III. THE STATION-LINE.—From _S_ let fall a perpendicular line, _SR_, to -the bottom of the paper, and call this line the “Station-Line.” - -This represents the line on which the observer stands, at a greater or -less distance from the picture; and it ought to be _imagined_ as drawn -right out from the paper at the point s. Hold your paper upright in -front of you, and hold your pencil horizontally, with its point against -the point _S_, as if you wanted to run it through the paper there, and -the pencil will represent the direction in which the line _SR_ ought to -be drawn. But as all the measurements which we have to set upon this -line, and operations which we have to perform with it, are just the same -when it is drawn on the paper itself, below _S_, as they would be if it -were represented by a wire in the position of the leveled pencil, and as -they are much more easily performed when it is drawn on the paper, it is -always in practice, so drawn. - - -IV. THE STATION-POINT.—On this line, mark the distance _ST_ at your -pleasure, for the distance at which you wish your picture to be seen, -and call the point T the “Station-Point.” - - [Illustration: Fig. 2.] - -In practice, it is generally advisable to make the distance _ST_ about -as great as the diameter of your intended picture; and it should, for -the most part, be more rather than less; but, as I have just stated, -this is quite arbitrary. However, in this figure, as an approximation to -a generally advisable distance, I make the distance _ST_ equal to the -diameter of the circle _NOPQ_. Now, having fixed this distance, _ST_, -all the dimensions of the objects in our picture are fixed likewise, and -for this reason:— - -Let the upright line _AB_, Fig. 2., represent a pane of glass placed -where our picture is to be placed; but seen at the side of it, -edgeways; let _S_ be the Sight-point; _ST_ the Station-line, which, in -this figure, observe, is in its true position, drawn out from the paper, -not down upon it; and _T_ the Station-point. - -Suppose the Station-line _ST_ to be continued, or in mathematical -language “produced,” through _S_, far beyond the pane of glass, and let -_PQ_ be a tower or other upright object situated on or above this line. - -Now the _apparent_ height of the tower _PQ_ is measured by the angle -_QTP_, between the rays of light which come from the top and bottom of -it to the eye of the observer. But the _actual_ height of the _image_ of -the tower on the pane of glass _AB_, between us and it, is the distance -_P′Q′_ between the points where the rays traverse the glass. - -Evidently, the farther from the point _T_ we place the glass, making -_ST_ longer, the larger will be the image; and the nearer we place it to -_T_, the smaller the image, and that in a fixed ratio. Let the distance -_DT_ be the direct distance from the Station-point to the foot of the -object. Then, if we place the glass _AB_ at one-third of that whole -distance, _P′Q′_ will be one-third of the real height of the object; if -we place the glass at two-thirds of the distance, as at _EF_, _P″Q″_ -(the height of the image at that point) will be two-thirds the height[5] -of the object, and so on. Therefore the mathematical law is that _P′Q′_ -will be to _PQ_ as _ST_ to _DT_. I put this ratio clearly by itself that -you may remember it: - - _P′Q′_ ∶ _PQ_ ∷ _ST_ ∶ _DT_ - -or in words: - - _P_ dash _Q_ dash is to _PQ_ as _ST_ to _DT_ - -In which formula, recollect that _P′Q′_ is the height of the appearance -of the object on the picture; _PQ_ the height of the object itself; _S_ -the Sight-point; _T_ the Station-point; _D_ a point at the direct -distance of the object; though the object is seldom placed actually on -the line _TS_ produced, and may be far to the right or left of it, the -formula is still the same. - -For let _S_, Fig. 3., be the Sight-point, and _AB_ the glass—here seen -looking _down_ on its _upper edge_, not sideways;—then if the tower -(represented now, as on a map, by the dark square), instead of being at -_D_ on the line _ST_ produced, be at _E_, to the right (or left) of the -spectator, still the apparent height of the tower on _AB_ will be as -_S′T_ to _ET_, which is the same ratio as that of _ST_ to _DT_. - - [Illustration: Fig. 3.] - -Now in many perspective problems, the position of an object is more -conveniently expressed by the two measurements _DT_ and _DE_, than by -the single oblique measurement _ET_. - -I shall call _DT_ the “direct distance” of the object at _E_, and _DE_ -its “lateral distance.” It is rather a license to call _DT_ its “direct” -distance, for _ET_ is the more direct of the two; but there is no other -term which would not cause confusion. - -Lastly, in order to complete our knowledge of the position of an object, -the vertical height of some point in it, above or below the eye, must be -given; that is to say, either _DP_ or _DQ_ in Fig. 2.[6]: this I shall -call the “vertical distance” of the point given. In all perspective -problems these three distances, and the dimensions of the object, must -be stated, otherwise the problem is imperfectly given. It ought not to -be required of us merely to draw _a_ room or _a_ church in perspective; -but to draw _this_ room from _this_ corner, and _that_ church on _that_ -spot, in perspective. For want of knowing how to base their drawings on -the measurement and place of the object, I have known practiced students -represent a parish church, certainly in true perspective, but with a -nave about two miles and a half long. - -It is true that in drawing landscapes from nature the sizes and -distances of the objects cannot be accurately known. When, however, we -know how to draw them rightly, if their size were given, we have only to -_assume a rational approximation_ to their size, and the resulting -drawing will be true enough for all intents and purposes. It does not in -the least matter that we represent a distant cottage as eighteen feet -long, when it is in reality only seventeen; but it matters much that we -do not represent it as eighty feet long, as we easily might if we had -not been accustomed to draw from measurement. Therefore, in all the -following problems the measurement of the object is given. - -The student must observe, however, that in order to bring the diagrams -into convenient compass, the measurements assumed are generally very -different from any likely to occur in practice. Thus, in Fig. 3., the -distance _DS_ would be probably in practice half a mile or a mile, and -the distance _TS_, from the eye of the observer to the paper, only two -or three feet. The mathematical law is however precisely the same, -whatever the proportions; and I use such proportions as are best -calculated to make the diagram clear. - -Now, therefore, the conditions of a perspective problem are the -following: - - The Sight-line _GH_ given, Fig. 1.; - The Sight-point _S_ given; - The Station-point _T_ given; and - The three distances of the object,[7] direct, lateral, and vertical, - with its dimensions, given. - -The size of the picture, conjecturally limited by the dotted circle, is -to be determined afterwards at our pleasure. On these conditions I -proceed at once to construction. - - - [3] If the glass were not upright, but sloping, the objects might - still be drawn through it, but their perspective would then be - different. Perspective, as commonly taught, is always calculated - for a vertical plane of picture. - - [4] Supposing it to have no thickness; otherwise the images would be - distorted by refraction. - - [5] I say “height” instead of “magnitude,” for a reason stated in - Appendix I., to which you will soon be referred. Read on here at - present. - - [6] _P_ and _Q_ being points indicative of the place of the tower’s - base and top. In this figure both are above the sight-line; if the - tower were below the spectator both would be below it, and - therefore measured below _D_. - - [7] More accurately, “the three distances of any point, either in the - object itself, or indicative of its distance.” - - - - -PROBLEM I. - -TO FIX THE POSITION OF A GIVEN POINT.[8] - - -Let _P_, Fig. 4., be the given point. - - [Illustration: Fig. 4.] - -Let its direct distance be _DT_; its lateral distance to the left, _DC_; -and vertical distance _beneath_ the eye of the observer, _CP_. - -[Let _GH_ be the Sight-line, _S_ the Sight-point, and _T_ the -Station-point.][9] - -It is required to fix on the plane of the picture the position of the -point P. - -Arrange the three distances of the object on your paper, as in -Fig. 4.[10] - -Join _CT_, cutting _GH_ in _Q_. - -From _Q_ let fall the vertical line _QP′_. - -Join _PT_, cutting _QP_ in _P′_. - -_P′_ is the point required. - -If the point _P_ is _above_ the eye of the observer instead of below it, -_CP_ is to be measured upwards from _C_, and _QP′_ drawn upwards from -_Q_. The construction will be as in Fig. 5. - - [Illustration: Fig. 5.] - -And if the point _P_ is to the right instead of the left of the -observer, _DC_ is to be measured to the right instead of the left. - -The figures 4. and 5., looked at in a mirror, will show the construction -of each, on that supposition. - -Now read very carefully the examples and notes to this problem in -Appendix I. (page 69). I have put them in the Appendix in order to keep -the sequence of following problems more clearly traceable here in the -text; but you must read the first Appendix before going on. - - - [8] More accurately, “To fix on the plane of the picture the apparent - position of a point given in actual position.” In the headings of - all the following problems the words “on the plane of the - picture” are to be understood after the words “to draw.” The - plane of the picture means a surface extended indefinitely in the - direction of the picture. - - [9] The sentence within brackets will not be repeated in succeeding - statements of problems. It is always to be understood. - - [10] In order to be able to do this, you must assume the distances to - be small; as in the case of some object on the table: how large - distances are to be treated you will see presently; the - mathematical principle, being the same for all, is best - illustrated first on a small scale. Suppose, for instance, _P_ to - be the corner of a book on the table, seven inches below the eye, - five inches to the left of it, and a foot and a half in advance - of it, and that you mean to hold your finished drawing at six - inches from the eye; then _TS_ will be six inches, _TD_ a foot - and a half, _DC_ five inches, and _CP_ seven. - - - - -PROBLEM II. - -TO DRAW A RIGHT LINE BETWEEN TWO GIVEN POINTS. - - - [Illustration: Fig. 6.] - -Let _AB_, Fig. 6., be the given right line, joining the given points _A_ -and _B_. - -Let the direct, lateral, and vertical distances of the point _A_ be -_TD_, _DC_, and _CA_. - -Let the direct, lateral, and vertical distances of the point _B_ be -_TD′_, _DC′_, and _C′B_. - -Then, by Problem I., the position of the point _A_ on the plane of the -picture is _a_. - -And similarly, the position of the point _B_ on the plane of the picture -is _b_. - -Join _ab_. - -Then _ab_ is the line required. - - -COROLLARY I. - -If the line _AB_ is in a plane parallel to that of the picture, one end -of the line _AB_ must be at the same direct distance from the eye of the -observer as the other. - -Therefore, in that case, _DT_ is equal to _D′T_. - -Then the construction will be as in Fig. 7.; and the student will find -experimentally that _ab_ is now parallel to _AB_.[11] - - [Illustration: Fig. 7.] - -And that _ab_ is to _AB_ as _TS_ is to _TD_. - -Therefore, to draw any line in a plane parallel to that of the picture, -we have only to fix the position of one of its extremities, _a_ or _b_, -and then to draw from _a_ or _b_ a line parallel to the given line, -bearing the proportion to it that _TS_ bears to _TD_. - - -COROLLARY II. - -If the line _AB_ is in a horizontal plane, the vertical distance of one -of its extremities must be the same as that of the other. - -Therefore, in that case, _AC_ equals _BC′_ (Fig. 6.). - -And the construction is as in Fig. 8. - - [Illustration: Fig. 8.] - -In Fig. 8. produce _ab_ to the sight-line, cutting the sight-line in -_V_; the point _V_, thus determined, is called the VANISHING-POINT of -the line _AB_. - -Join _TV_. Then the student will find experimentally that _TV_ is -parallel to _AB_.[12] - - -COROLLARY III. - -If the line _AB_ produced would pass through some point beneath or above -the station-point, _CD_ is to _DT_ as _C′D′_ is to _D′T_; in which case -the point _c_ coincides with the point _c′_, and the line _ab_ is -vertical. - -Therefore every vertical line in a picture is, or may be, the -perspective representation of a horizontal one which, produced, would -pass beneath the feet or above the head of the spectator.[13] - - - [11] For by the construction _AT_ ∶ _aT_ ∷ _BT_ ∶ _bT_; and therefore - the two triangles _ABT_, _abT_, (having a common angle _ATB_,) - are similar. - - [12] The demonstration is in Appendix II. Article I. - - [13] The reflection in water of any luminous point or isolated object - (such as the sun or moon) is therefore, in perspective, a - vertical line; since such reflection, if produced, would pass - under the feet of the spectator. Many artists (Claude among the - rest) knowing something of optics, but nothing of perspective, - have been led occasionally to draw such reflections towards a - point at the center of the base of the picture. - - - - -PROBLEM III. - -TO FIND THE VANISHING-POINT OF A GIVEN HORIZONTAL LINE. - - - [Illustration: Fig. 9.] - -Let _AB_, Fig. 9., be the given line. - -From _T_, the station-point, draw _TV_ parallel to _AB_, cutting the -sight-line in _V_. - -_V_ is the Vanishing-point required.[14] - - -COROLLARY I. - -As, if the point _b_ is first found, _V_ may be determined by it, so, if -the point _V_ is first found, _b_ may be determined by it. For let _AB_, -Fig. 10., be the given line, constructed upon the paper as in Fig. 8.; -and let it be required to draw the line _ab_ without using the point -_C′_. - - [Illustration: Fig. 10.] - -Find the position of the point _A_ in _a_. (Problem I.) - -Find the vanishing-point of _AB_ in _V_. (Problem III.) - -Join _aV_. - -Join _BT_, cutting _aV_ in _b_. - -Then _ab_ is the line required.[15] - - -COROLLARY II. - -We have hitherto proceeded on the supposition that the given line was -small enough, and near enough, to be actually drawn on our paper of its -real size; as in the example given in Appendix I. We may, however, now -deduce a construction available under all circumstances, whatever may be -the distance and length of the line given. - - [Illustration: Fig. 11.] - -From Fig. 8. remove, for the sake of clearness, the lines _C′D′_, _bV_, -and _TV_; and, taking the figure as here in Fig. 11., draw from _a_, the -line _aR_ parallel to _AB_, cutting _BT_ in _R_. - - Then _aR_ is to _AB_ as _aT_ is to _AT_. - ---- ---- as _cT_ is to _CT_. - ---- ---- as _TS_ is to _TD_. - -That is to say, _aR_ is the sight-magnitude of _AB_.[16] - - [Illustration: Fig. 12.] - -Therefore, when the position of the point _A_ is fixed in _a_, as in -Fig. 12., and _aV_ is drawn to the vanishing-point; if we draw a line -_aR_ from _a_, parallel to _AB_, and make _aR_ equal to the -sight-magnitude of _AB_, and then join _RT_, the line _RT_ will cut _aV_ -in _b_. - -So that, in order to determine the length of _ab_, we need not draw the -long and distant line _AB_, but only _aR_ parallel to it, and of its -sight-magnitude; which is a great gain, for the line _AB_ may be two -miles long, and the line _aR_ perhaps only two inches. - - -COROLLARY III. - -In Fig. 12., altering its proportions a little for the sake of -clearness, and putting it as here in Fig. 13., draw a horizontal line -_aR′_ and make _aR′_ equal to _aR_. - -Through the points _R_ and _b_ draw _R′M_, cutting the sight-line in -_M_. Join _TV_. Now the reader will find experimentally that _VM_ is -equal to _VT_.[17] - - [Illustration: Fig. 13.] - -Hence it follows that, if from the vanishing-point _V_ we lay off on -the sight-line a distance, _VM_, equal to _VT_; then draw through _a_ a -horizontal line _aR′_, make _aR′_ equal to the sight-magnitude of _AB_, -and join _R′M_; the line _R′M_ will cut _aV_ in _b_. And this is in -practice generally the most convenient way of obtaining the length of -_ab_. - - -COROLLARY IV. - -Removing from the preceding figure the unnecessary lines, and retaining -only _R′M_ and _aV_, as in Fig. 14., produce the line _aR′_ to the other -side of _a_, and make _aX_ equal to _aR′_. - -Join _Xb_, and produce _Xb_ to cut the line of sight in _N_. - - [Illustration: Fig. 14.] - -Then as _XR′_ is parallel to _MN_, and _aR′_ is equal to _aX_, _VN_ -must, by similar triangles, be equal to _VM_ (equal to _VT_ in -Fig. 13.). - -Therefore, on whichever side of _V_ we measure the distance _VT_, so as -to obtain either the point _M_, or the point _N_, if we measure the -sight-magnitude _aR′_ or _aX_ on the opposite side of the line _aV_, the -line joining _R′M_ or _XN_ will equally cut _aV_ in _b_. - -The points _M_ and _N_ are called the “DIVIDING-POINTS” of the original -line _AB_ (Fig. 12.), and we resume the results of these corollaries in -the following three problems. - - - [14] The student will observe, in practice, that, his paper lying flat - on the table, he has only to draw the line _TV_ on its horizontal - surface, parallel to the given horizontal line _AB_. In theory, - the paper should be vertical, but the station-line _ST_ - horizontal (see its definition above, page 5); in which case - _TV_, being drawn parallel to _AB_, will be horizontal also, and - still cut the sight-line in _V_. - - The construction will be seen to be founded on the second - Corollary of the preceding problem. - - It is evident that if any other line, as _MN_ in Fig. 9., - parallel to _AB_, occurs in the picture, the line _TV_, drawn - from _T_, parallel to _MN_, to find the vanishing-point of _MN_, - will coincide with the line drawn from _T_, parallel to _AB_, to - find the vanishing-point of _AB_. - - Therefore _AB_ and _MN_ will have the same vanishing-point. - - Therefore all parallel horizontal lines have the same - vanishing-point. - - It will be shown hereafter that all parallel _inclined_ lines - also have the same vanishing-point; the student may here accept - the general conclusion—“_All parallel lines have the same - vanishing-point._” - - It is also evident that if _AB_ is parallel to the plane of the - picture, _TV_ must be drawn parallel to _GH_, and will therefore - never cut _GH_. The line _AB_ has in that case no - vanishing-point: it is to be drawn by the construction given in - Fig. 7. - - It is also evident that if _AB_ is at right angles with the plane - of the picture, _TV_ will coincide with _TS_, and the - vanishing-point of _AB_ will be the sight-point. - - [15] I spare the student the formality of the _reductio ad absurdum_, - which would be necessary to prove this. - - [16] For definition of Sight-Magnitude, see Appendix I. It ought to - have been read before the student comes to this problem; but I - refer to it in case it has not. - - [17] The demonstration is in Appendix II. Article II. p. 101. - - - - -PROBLEM IV. - -TO FIND THE DIVIDING-POINTS OF A GIVEN HORIZONTAL LINE. - - - [Illustration: Fig. 15.] - -Let the horizontal line _AB_ (Fig. 15.) be given in position and -magnitude. It is required to find its dividing-points. - -Find the vanishing-point _V_ of the line _AB_. - -With center _V_ and distance _VT_, describe circle cutting the -sight-line in _M_ and _N_. - -Then _M_ and _N_ are the dividing-points required. - -In general, only one dividing-point is needed for use with any -vanishing-point, namely, the one nearest _S_ (in this case the point -_M_). But its opposite _N_, or both, may be needed under certain -circumstances. - - - - -PROBLEM V. - -TO DRAW A HORIZONTAL LINE, GIVEN IN POSITION AND MAGNITUDE, BY MEANS -OF ITS SIGHT-MAGNITUDE AND DIVIDING-POINTS. - - - [Illustration: Fig. 16.] - -Let _AB_ (Fig. 16.) be the given line. - -Find the position of the point _A_ in _a_. - -Find the vanishing-point _V_, and most convenient dividing-point _M_, -of the line _AB_. - -Join _aV_. - -Through _a_ draw a horizontal line _ab′_ and make _ab′_ equal to the -sight-magnitude of _AB_. Join _b′M_, cutting _aV_ in _b_. - -Then _ab_ is the line required. - - -COROLLARY I. - - [Illustration: Fig. 17.] - -Supposing it were now required to draw a line _AC_ (Fig. 17.) twice as -long as _AB_, it is evident that the sight-magnitude _ac′_ must be -twice as long as the sight-magnitude _ab′_; we have, therefore, merely -to continue the horizontal line _ab′_, make _b′c′_ equal to _ab′_, -join _cM′_, cutting _aV_ in _c_, and _ac_ will be the line required. -Similarly, if we have to draw a line _AD_, three times the length of -_AB_, _ad′_ must be three times the length of _ab′_, and, joining -_d′M_, _ad_ will be the line required. - -The student will observe that the nearer the portions cut off, _bc_, -_cd_, etc., approach the point _V_, the smaller they become; and, -whatever lengths may be added to the line _AD_, and successively cut -off from _aV_, the line _aV_ will never be cut off entirely, but the -portions cut off will become infinitely small, and apparently “vanish” -as they approach the point _V_; hence this point is called the -“vanishing” point. - - -COROLLARY II. - -It is evident that if the line _AD_ had been given originally, and we -had been required to draw it, and divide it into three equal parts, we -should have had only to divide its sight-magnitude, _ad′_, into the -three equal parts, _ab′_, _b′c′_, and _c′d′_, and then, drawing to _M_ -from _b′_ and _c′_, the line _ad_ would have been divided as required -in _b_ and _c_. And supposing the original line _AD_ be divided -_irregularly into any number_ of parts, if the line _ad′_ be divided -into a similar number in the same proportions (by the construction -given in Appendix I.), and, from these points of division, lines are -drawn to _M_, they will divide the line _ad_ in true perspective into -a similar number of proportionate parts. - -The horizontal line drawn through _a_, on which the sight-magnitudes -are measured, is called the “MEASURING-LINE.” - -And the line _ad_, when properly divided in _b_ and _c_, or any other -required points, is said to be divided “IN PERSPECTIVE RATIO” to the -divisions of the original line _AD_. - -If the line _aV_ is above the sight-line instead of beneath it, the -measuring-line is to be drawn above also: and the lines _b′M_, _c′M_, -etc., drawn _down_ to the dividing-point. Turn Fig. 17. upside down, -and it will show the construction. - - - - -PROBLEM VI. - -TO DRAW ANY TRIANGLE, GIVEN IN POSITION AND MAGNITUDE, IN A HORIZONTAL -PLANE. - - - [Illustration: Fig. 18.] - -Let _ABC_ (Fig. 18.) be the triangle. - -As it is given in position and magnitude, one of its sides, at least, -must be given in position and magnitude, and the directions of the two -other sides. - -Let _AB_ be the side given in position and magnitude. - -Then _AB_ is a horizontal line, in a given position, and of a given -length. - -Draw the line _AB_. (Problem V.) - -Let _ab_ be the line so drawn. - -Find _V_ and _V′_, the vanishing-points respectively of the lines _AC_ -and _BC_. (Problem III.) - -From _a_ draw _aV_, and from _b_, draw _bV′_, cutting each other in -_c_. - -Then _abc_ is the triangle required. - -If _AC_ is the line originally given, _ac_ is the line which must be -first drawn, and the line _V′b_ must be drawn from _V′_ to _c_ and -produced to cut _ab_ in _b_. Similarly, if _BC_ is given, _Vc_ must be -drawn to _c_ and produced, and _ab_ from its vanishing-point to _b_, -and produced to cut _ac_ in _a_. - - - - -PROBLEM VII. - -TO DRAW ANY RECTILINEAR QUADRILATERAL FIGURE, GIVEN IN POSITION AND -MAGNITUDE, IN A HORIZONTAL PLANE. - - - [Illustration: Fig. 19.] - -Let _ABCD_ (Fig. 19.) be the given figure. - -Join any two of its opposite angles by the line _BC_. - -Draw first the triangle _ABC_. (Problem VI.) - -And then, from the base _BC_, the two lines _BD_, _CD_, to their -vanishing-points, which will complete the figure. It is unnecessary to -give a diagram of the construction, which is merely that of Fig. 18. -duplicated; another triangle being drawn on the line _AC_ or _BC_. - - -COROLLARY. - -It is evident that by this application of Problem VI. any given -rectilinear figure whatever in a horizontal plane may be drawn, since -any such figure may be divided into a number of triangles, and the -triangles then drawn in succession. - -More convenient methods may, however, be generally found, according -to the form of the figure required, by the use of succeeding problems; -and for the quadrilateral figure which occurs most frequently in -practice, namely, the square, the following construction is more -convenient than that used in the present problem. - - - - -PROBLEM VIII. - -TO DRAW A SQUARE, GIVEN IN POSITION AND MAGNITUDE, IN A HORIZONTAL -PLANE. - - - [Illustration: Fig. 20.] - -Let _ABCD_, Fig. 20., be the square. - -As it is given in position and magnitude, the position and magnitude -of all its sides are given. - -Fix the position of the point _A_ in _a_. - -Find _V_, the vanishing-point of _AB_; and _M_, the dividing-point of -_AB_, nearest _S_. - -Find _V′_, the vanishing-point of _AC_; and _N_, the dividing-point of -_AC_, nearest _S_. - -Draw the measuring-line through _a_, and make _ab′_, _ac′_, each equal -to the sight-magnitude of _AB_. - -(For since _ABCD_ is a square, _AC_ is equal to _AB_.) - -Draw _aV′_ and _c′N_, cutting each other in _c_. - -Draw _aV_, and _b′M_, cutting each other in _b_. - -Then _ac_, _ab_, are the two nearest sides of the square. - -Now, clearing the figure of superfluous lines, we have _ab_, _ac_, -drawn in position, as in Fig. 21. - - [Illustration: Fig. 21.] - -And because _ABCD_ is a square, _CD_ (Fig. 20.) is parallel to _AB_. - -And all parallel lines have the same vanishing-point. (Note to -Problem III.) - -Therefore, _V_ is the vanishing-point of _CD_. - -Similarly, _V′_ is the vanishing-point of _BD_. - -Therefore, from _b_ and _c_ (Fig. 22.) draw _bV′_, _cV_, cutting each -other in _d_. - -Then _abcd_ is the square required. - - -COROLLARY I. - -It is obvious that any rectangle in a horizontal plane may be drawn by -this problem, merely making _ab′_, on the measuring-line, Fig. 20., -equal to the sight-magnitude of one of its sides, and _ac′_ the -sight-magnitude of the other. - - -COROLLARY II. - -Let _abcd_, Fig. 22., be any square drawn in perspective. Draw the -diagonals _ad_ and _bc_, cutting each other in _C_. Then _C_ is the -center of the square. Through _C_, draw _ef_ to the vanishing-point of -_ab_, and _gh_ to the vanishing-point of _ac_, and these lines will -bisect the sides of the square, so that _ag_ is the perspective -representation of half the side _ab_; _ae_ is half _ac_; _ch_ is half -_cd_; and _bf_ is half _bd_. - - [Illustration: Fig. 22.] - - -COROLLARY III. - -Since _ABCD_, Fig. 20., is a square, _BAC_ is a right angle; and as -_TV_ is parallel to _AB_, and _TV′_ to _AC_, _V′TV_ must be a right -angle also. - -As the ground plan of most buildings is rectangular, it constantly -happens in practice that their angles (as the corners of ordinary -houses) throw the lines to the vanishing-points thus at right angles; -and so that this law is observed, and _VTV′_ is kept a right angle, it -does not matter in general practice whether the vanishing-points are -thrown a little more or a little less to the right or left of _S_: but -it matters much that the relation of the vanishing-points should be -accurate. Their position with respect to _S_ merely causes the -spectator to see a little more or less on one side or other of the -house, which may be a matter of chance or choice; but their -rectangular relation determines the rectangular shape of the building, -which is an essential point. - - - - -PROBLEM IX. - -TO DRAW A SQUARE PILLAR, GIVEN IN POSITION AND MAGNITUDE, ITS BASE AND -TOP BEING IN HORIZONTAL PLANES. - - -Let _AH_, Fig. 23., be the square pillar. - -Then, as it is given in position and magnitude, the position and -magnitude of the square it stands upon must be given (that is, the -line _AB_ or _AC_ in position), and the height of its side _AE_. - - [Illustration: Fig. 23.] [Illustration: Fig. 24.] - -Find the sight-magnitudes of _AB_ and _AE_. Draw the two sides _ab_, -_ac_, of the square of the base, by Problem VIII., as in Fig. 24. From -the points _a_, _b_, and _c_, raise vertical lines _ae_, _cf_, _bg_. - -Make _ae_ equal to the sight-magnitude of _AE_. - -Now because the top and base of the pillar are in horizontal planes, -the square of its top, _FG_, is parallel to the square of its base, -_BC_. - -Therefore the line _EF_ is parallel to _AC_, and _EG_ to _AB_. - -Therefore _EF_ has the same vanishing-point as _AC_, and _EG_ the same -vanishing-point as _AB_. - -From _e_ draw _ef_ to the vanishing-point of _ac_, cutting _cf_ in -_f_. - -Similarly draw _eg_ to the vanishing-point of _ab_, cutting _bg_ in -_g_. - -Complete the square _gf_ in _h_, by drawing _gh_ to the -vanishing-point of _ef_, and _fh_ to the vanishing-point of _eg_, -cutting each other in _h_. Then _aghf_ is the square pillar required. - - -COROLLARY. - -It is obvious that if _AE_ is equal to _AC_, the whole figure will be -a cube, and each side, _aefc_ and _aegb_, will be a square in a given -vertical plane. And by making _AB_ or _AC_ longer or shorter in any -given proportion, any form of rectangle may be given to either of the -sides of the pillar. No other rule is therefore needed for drawing -squares or rectangles in vertical planes. - -Also any triangle may be thus drawn in a vertical plane, by inclosing -it in a rectangle and determining, in perspective ratio, on the sides -of the rectangle, the points of their contact with the angles of the -triangle. - -And if any triangle, then any polygon. - -A less complicated construction will, however, be given hereafter.[18] - - - [18] See page 96 (note), after you have read Problem XVI. - - - - -PROBLEM X. - -TO DRAW A PYRAMID, GIVEN IN POSITION AND MAGNITUDE, ON A SQUARE BASE -IN A HORIZONTAL PLANE. - - - [Illustration: Fig. 25.] - -Let _AB_, Fig. 25., be the four-sided pyramid. As it is given in -position and magnitude, the square base on which it stands must be -given in position and magnitude, and its vertical height, _CD_.[19] - - [Illustration: Fig. 26.] - -Draw a square pillar, _ABGE_, Fig. 26., on the square base of the -pyramid, and make the height of the pillar _AF_ equal to the vertical -height of the pyramid _CD_ (Problem IX.). Draw the diagonals _GF_, -_HI_, on the top of the square pillar, cutting each other in _C_. -Therefore _C_ is the center of the square _FGHI_. (Prob. VIII. -Cor. II.) - - [Illustration: Fig. 27.] - -Join _CE_, _CA_, _CB_. - -Then _ABCE_ is the pyramid required. If the base of the pyramid is -above the eye, as when a square spire is seen on the top of a -church-tower, the construction will be as in Fig. 27. - - - [19] If, instead of the vertical height, the length of _AD_ is - given, the vertical must be deduced from it. See the Exercises - on this Problem in the Appendix, p. 79. - - - - -PROBLEM XI. - -TO DRAW ANY CURVE IN A HORIZONTAL OR VERTICAL PLANE. - - - [Illustration: Fig. 28.] - -Let _AB_, Fig. 28., be the curve. - -Inclose it in a rectangle, _CDEF_. - -Fix the position of the point _C_ or _D_, and draw the rectangle. -(Problem VIII. Coroll. I.)[20] - -Let _CDEF_, Fig. 29., be the rectangle so drawn. - - [Illustration: Fig. 29.] - -If an extremity of the curve, as _A_, is in a side of the rectangle, -divide the side _CE_, Fig. 29., so that _AC_ shall be (in perspective -ratio) to _AE_ as _AC_ is to _AE_ in Fig. 28. (Prob. V. Cor. II.) - -Similarly determine the points of contact of the curve and rectangle -_e_, _f_, _g_. - -If an extremity of the curve, as _B_, is not in a side of the -rectangle, let fall the perpendiculars _Ba_, _Bb_ on the rectangle -sides. Determine the correspondent points _a_ and _b_ in Fig. 29., as -you have already determined _A_, _B_, _e_, and _f_. - -From _b_, Fig. 29., draw _bB_ parallel to _CD_,[21] and from _a_ draw -_aB_ to the vanishing-point of _DF_, cutting each other in _B_. Then -_B_ is the extremity of the curve. - -Determine any other important point in the curve, as _P_, in the same -way, by letting fall _Pq_ and _Pr_ on the rectangle’s sides. - -Any number of points in the curve may be thus determined, and the -curve drawn through the series; in most cases, three or four will be -enough. Practically, complicated curves may be better drawn in -perspective by an experienced eye than by rule, as the fixing of the -various points in haste involves too many chances of error; but it is -well to draw a good many by rule first, in order to give the eye its -experience.[22] - - -COROLLARY. - -If the curve required be a circle, Fig. 30., the rectangle which -incloses it will become a square, and the curve will have four points -of contact, _ABCD_, in the middle of the sides of the square. - - [Illustration: Fig. 30.] - -Draw the square, and as a square may be drawn about a circle in any -position, draw it with its nearest side, _EG_, parallel to the -sight-line. - -Let _EF_, Fig. 31., be the square so drawn. - -Draw its diagonals _EF_, _GH_; and through the center of the square -(determined by their intersection) draw _AB_ to the vanishing-point of -_GF_, and _CD_ parallel to _EG_. Then the points _ABCD_ are the four -points of the circle’s contact. - - [Illustration: Fig. 31.] - -On _EG_ describe a half square, _EL_; draw the semicircle _KAL_; and -from its center, _R_, the diagonals _RE_, _RG_, cutting the circle in -_x_, _y_. - -From the points _x_ _y_, where the circle cuts the diagonals, raise -perpendiculars, _Px_, _Qy_, to _EG_. - -From _P_ and _Q_ draw _PP′_, _QQ′_, to the vanishing-point of _GF_, -cutting the diagonals in _m_, _n_, and _o_, _p_. - -Then _m_, _n_, _o_, _p_ are four other points in the circle. - -Through these eight points the circle may be drawn by the hand -accurately enough for general purposes; but any number of points -required may, of course, be determined, as in Problem XI. - -The distance _EP_ is approximately one-seventh of _EG_, and may be -assumed to be so in quick practice, as the error involved is not -greater than would be incurred in the hasty operation of drawing the -circle and diagonals. - -It may frequently happen that, in consequence of associated -constructions, it may be inconvenient to draw _EG_ parallel to the -sight-line, the square being perhaps first constructed in some oblique -direction. In such cases, _QG_ and _EP_ must be determined in -perspective ratio by the dividing-point, the line _EG_ being used as a -measuring-line. - - [_Obs._ In drawing Fig. 31. the station-point has been taken much - nearer the paper than is usually advisable, in order to show the - character of the curve in a very distinct form. - - If the student turns the book so that _EG_ may be vertical, - Fig. 31. will represent the construction for drawing a circle in a - vertical plane, the sight-line being then of course parallel to - _GL_; and the semicircles _ADB_, _ACB_, on each side of the - diameter _AB_, will represent ordinary semicircular arches seen in - perspective. In that case, if the book be held so that the line - _EH_ is the top of the square, the upper semicircle will represent - a semicircular arch, _above_ the eye, drawn in perspective. But if - the book be held so that the line _GF_ is the top of the square, - the upper semicircle will represent a semicircular arch, _below_ - the eye, drawn in perspective. - - If the book be turned upside down, the figure will represent a - circle drawn on the ceiling, or any other horizontal plane above - the eye; and the construction is, of course, accurate in every - case.] - - - [20] Or if the curve is in a vertical plane, Coroll. to Problem IX. - As a rectangle may be drawn in any position round any given - curve, its position with respect to the curve will in either - case be regulated by convenience. See the Exercises on this - Problem, in the Appendix, p. 85. - - [21] Or to its vanishing-point, if _CD_ has one. - - [22] Of course, by dividing the original rectangle into any number - of equal rectangles, and dividing the perspective rectangle - similarly, the curve may be approximately drawn without any - trouble; but, when accuracy is required, the points should be - fixed, as in the problem. - - - - -PROBLEM XII. - -TO DIVIDE A CIRCLE DRAWN IN PERSPECTIVE INTO ANY GIVEN NUMBER OF EQUAL -PARTS. - - -Let _AB_, Fig. 32., be the circle drawn in perspective. It is required -to divide it into a given number of equal parts; in this case, 20. - -Let _KAL_ be the semicircle used in the construction. Divide the -semicircle _KAL_ into half the number of parts required; in this case, -10. - -Produce the line _EG_ laterally, as far as may be necessary. - -From _O_, the center of the semicircle _KAL_, draw radii through the -points of division of the semicircle, _p_, _q_, _r_, etc., and produce -them to cut the line _EG_ in _P_, _Q_, _R_, etc. - -From the points _PQR_ draw the lines _PP′_, _QQ′_, _RR′_, etc., -through the center of the circle _AB_, each cutting the circle in two -points of its circumference. - -Then these points divide the perspective circle as required. - -If from each of the points _p_, _q_, _r_, a vertical were raised to -the line _EG_, as in Fig. 31., and from the point where it cut _EG_ a -line were drawn to the vanishing-point, as _QQ′_ in Fig. 31., this -line would also determine two of the points of division. - - [Illustration: Fig. 32.] - -If it is required to divide a circle into any number of given -_un_equal parts (as in the points _A_, _B_, and _C_, Fig. 33.), the -shortest way is thus to raise vertical lines from _A_ and _B_ to the -side of the perspective square _XY_, and then draw to the -vanishing-point, cutting the perspective circle in _a_ and _b_, the -points required. Only notice that if any point, as _A_, is on the -nearer side of the circle _ABC_, its representative point, _a_, must -be on the nearer side of the circle _abc_; and if the point _B_ is on -the farther side of the circle _ABC_, _b_ must be on the farther side -of _abc_. If any point, as _C_, is so much in the lateral arc of the -circle as not to be easily determinable by the vertical line, draw the -horizontal _CP_, find the correspondent _p_ in the side of the -perspective square, and draw _pc_ parallel to _XY_, cutting the -perspective circle in _c_. - - [Illustration: Fig. 33.] - - -COROLLARY. - -It is obvious that if the points _P′_, _Q′_, _R_, etc., by which the -circle is divided in Fig. 32., be joined by right lines, the resulting -figure will be a regular equilateral figure of twenty sides inscribed -in the circle. And if the circle be divided into given unequal parts, -and the points of division joined by right lines, the resulting figure -will be an irregular polygon inscribed in the circle with sides of -given length. - -Thus any polygon, regular or irregular, inscribed in a circle, may be -inscribed in position in a perspective circle. - - - - -PROBLEM XIII. - -TO DRAW A SQUARE, GIVEN IN MAGNITUDE, WITHIN A LARGER SQUARE GIVEN IN -POSITION AND MAGNITUDE; THE SIDES OF THE TWO SQUARES BEING PARALLEL. - - - [Illustration: Fig. 34.] - -Let _AB_, Fig. 34., be the sight-magnitude of the side of the smaller -square, and _AC_ that of the side of the larger square. - -Draw the larger square. Let _DEFG_ be the square so drawn. - -Join _EG_ and _DF_. - -On either _DE_ or _DG_ set off, in perspective ratio, _DH_ equal to -one half of _BC_. Through _H_ draw _HK_ to the vanishing-point of -_DE_, cutting _DF_ in _I_ and _EG_ in _K_. Through _I_ and _K_ draw -_IM_, _KL_, to vanishing-point of _DG_, cutting _DF_ in _L_ and _EG_ -in _M_. Join _LM_. - -Then _IKLM_ is the smaller square, inscribed as required.[23] - - -COROLLARY. - - [Illustration: Fig. 36.] - -If, instead of one square within another, it be required to draw one -circle within another, the dimensions of both being given, inclose -each circle in a square. Draw the squares first, and then the circles -within, as in Fig. 36. - - - [23] [Illustration: Fig. 35.] If either of the sides of the greater - square is parallel to the plane of the picture, as _DG_ in - Fig. 35., _DG_ of course must be equal to _AC_, and _DH_ equal - to _BC_/2, and the construction is as in Fig. 35. - - - - -PROBLEM XIV. - -TO DRAW A TRUNCATED CIRCULAR CONE, GIVEN IN POSITION AND MAGNITUDE, -THE TRUNCATIONS BEING IN HORIZONTAL PLANES, AND THE AXIS OF THE CONE -VERTICAL. - - -Let _ABCD_, Fig. 37., be the portion of the cone required. - - [Illustration: Fig. 37.] - -As it is given in magnitude, its diameters must be given at the base -and summit, _AB_ and _CD_; and its vertical height, _CE_.[24] - -And as it is given in position, the center of its base must be given. - - [Illustration: Fig. 38.] - -Draw in position, about this center,[25] the square pillar _afd_, -Fig. 38., making its height, _bg_, equal to _CE_; and its side, _ab_, -equal to _AB_. - -In the square of its base, _abcd_, inscribe a circle, which therefore -is of the diameter of the base of the cone, _AB_. - -In the square of its top, _efgh_, inscribe concentrically a circle -whose diameter shall equal _CD_. (Coroll. Prob. XIII.) - -Join the extremities of the circles by the right lines _kl_, _nm_. -Then _klnm_ is the portion of cone required. - - -COROLLARY I. - -If similar polygons be inscribed in similar positions in the circles -_kn_ and _lm_ (Coroll. Prob. XII.), and the corresponding angles of -the polygons joined by right lines, the resulting figure will be a -portion of a polygonal pyramid. (The dotted lines in Fig. 38., -connecting the extremities of two diameters and one diagonal in the -respective circles, occupy the position of the three nearest angles of -a regular octagonal pyramid, having its angles set on the diagonals -and diameters of the square _ad_, inclosing its base.) - -If the cone or polygonal pyramid is not truncated, its apex will be -the center of the upper square, as in Fig. 26. - - -COROLLARY II. - -If equal circles, or equal and similar polygons, be inscribed in the -upper and lower squares in Fig. 38., the resulting figure will be a -vertical cylinder, or a vertical polygonal pillar, of given height and -diameter, drawn in position. - - -COROLLARY III. - -If the circles in Fig. 38., instead of being inscribed in the squares -_bc_ and _fg_, be inscribed in the sides of the solid figure _be_ and -_df_, those sides being made square, and the line _bd_ of any given -length, the resulting figure will be, according to the constructions -employed, a cone, polygonal pyramid, cylinder, or polygonal pillar, -drawn in position about a horizontal axis parallel to _bd_. - -Similarly, if the circles are drawn in the sides _gd_ and _ec_, the -resulting figures will be described about a horizontal axis parallel -to _ab_. - - - [24] Or if the length of its side, _AC_, is given instead, take - _ae_, Fig. 37., equal to half the excess of _AB_ over _CD_; - from the point _e_ raise the perpendicular _ce_. With center - _a_, and distance _AC_, describe a circle cutting _ce_ in _c_. - Then _ce_ is the vertical height of the portion of cone - required, or _CE_. - - [25] The direction of the side of the square will of course be - regulated by convenience. - - - - -PROBLEM XV. - -TO DRAW AN INCLINED LINE, GIVEN IN POSITION AND MAGNITUDE. - - -We have hitherto been examining the conditions of horizontal and -vertical lines only, or of curves inclosed in rectangles. - - [Illustration: Fig. 39.] [Illustration: Fig. 40.] - -We must, in conclusion, investigate the perspective of inclined lines, -beginning with a single one given in position. For the sake of -completeness of system, I give in Appendix II. Article III. the -development of this problem from the second. But, in practice, the -position of an inclined line may be most conveniently defined by -considering it as the diagonal of a rectangle, as _AB_ in Fig. 39., -and I shall therefore, though at some sacrifice of system, examine it -here under that condition. - -If the sides of the rectangle _AC_ and _AD_ are given, the slope of -the line _AB_ is determined; and then its position will depend on that -of the rectangle. If, as in Fig. 39., the rectangle is parallel to the -picture plane, the line _AB_ must be so also. If, as in Fig. 40., the -rectangle is inclined to the picture plane, the line _AB_ will be so -also. So that, to fix the position of _AB_, the line _AC_ must be -given in position and magnitude, and the height _AD_. - - [Illustration: Fig. 41.] - -If these are given, and it is only required to draw the single line -_AB_ in perspective, the construction is entirely simple; thus:— - -Draw the line _AC_ by Problem I. - -Let _AC_, Fig. 41., be the line so drawn. From _a_ and _c_ raise the -vertical lines _ad_, _cb_. Make _ad_ equal to the sight-magnitude of -_AD_. From _d_ draw _db_ to the vanishing-point of _ac_, cutting _bc_ -in _b_. - -Join _ab_. Then _ab_ is the inclined line required. - - [Illustration: Fig. 42.] - -If the line is inclined in the opposite direction, as _DC_ in -Fig. 42., we have only to join _dc_ instead of _ab_ in Fig. 41., and -_dc_ will be the line required. - -I shall hereafter call the line _AC_, when used to define the position -of an inclined line _AB_ (Fig. 40.), the “relative horizontal” of the -line _AB_. - - -OBSERVATION. - - [Illustration: Fig. 43.] - -In general, inclined lines are most needed for gable roofs, in which, -when the conditions are properly stated, the vertical height of the -gable, _XY_, Fig. 43., is given, and the base line, _AC_, in position. -When these are given, draw _AC_; raise vertical _AD_; make _AD_ equal -to sight-magnitude of _XY_; complete the perspective-rectangle _ADBC_; -join _AB_ and _DC_ (as by dotted lines in figure); and through the -intersection of the dotted lines draw vertical _XY_, cutting _DB_ in -_Y_. Join _AY_, _CY_; and these lines are the sides of the gable. If -the length of the roof _AA′_ is also given, draw in perspective the -complete parallelopiped _A′D′BC_, and from _Y_ draw _YY′_ to the -vanishing-point of _AA′_, cutting _D′B′_ in _Y′_. Join _A′Y_, and you -have the slope of the farther side of the roof. - - [Illustration: Fig. 44.] - -The construction above the eye is as in Fig. 44.; the roof is reversed -in direction merely to familiarize the student with the different -aspects of its lines. - - - - -PROBLEM XVI. - -TO FIND THE VANISHING-POINT OF A GIVEN INCLINED LINE. - - -If, in Fig. 43. or Fig. 44., the lines _AY_ and _A′Y′_ be produced, -the student will find that they meet. - -Let _P_, Fig. 45., be the point at which they meet. - -From _P_ let fall the vertical _PV_ on the sight-line, cutting the -sight-line in _V_. - -Then the student will find experimentally that _V_ is the -vanishing-point of the line _AC_.[26] - -Complete the rectangle of the base _AC′_, by drawing _A′C′_ to _V_, -and _CC′_ to the vanishing-point of _AA′_. - -Join _Y′C′_. - -Now if _YC_ and _Y′C′_ be produced downwards, the student will find -that they meet. - -Let them be produced, and meet in _P′_. - -Produce _PV_, and it will be found to pass through the point _P′_. - -Therefore if _AY_ (or _CY_), Fig. 45., be any inclined line drawn in -perspective by Problem XV., and _AC_ the relative horizontal (_AC_ in -Figs. 39, 40.), also drawn in perspective. - -Through _V_, the vanishing-point of _AV_, draw the vertical _PP′_ -upwards and downwards. - -Produce _AY_ (or _CY_), cutting _PP′_ in _P_ (or _P′_). - -Then _P_ is the vanishing-point of _AY_ (or _P′_ of _CY_). - - [Illustration: Fig. 45.] - -The student will observe that, in order to find the point _P_ by this -method, it is necessary first to draw a portion of the given inclined -line by Problem XV. Practically, it is always necessary to do so, and, -therefore, I give the problem in this form. - -Theoretically, as will be shown in the analysis of the problem, the -point _P_ should be found by drawing a line from the station-point -parallel to the given inclined line: but there is no practical means -of drawing such a line; so that in whatever terms the problem may be -given, a portion of the inclined line (_AY_ or _CY_) must always be -drawn in perspective before P can be found. - - - [26] The demonstration is in Appendix II. Article III. - - - - -PROBLEM XVII. - -TO FIND THE DIVIDING-POINTS OF A GIVEN INCLINED LINE. - - - [Illustration: Fig. 46.] - -Let _P_, Fig. 46., be the vanishing-point of the inclined line, and -_V_ the vanishing-point of the relative horizontal. - -Find the dividing-points of the relative horizontal, _D_ and _D′_. - -Through _P_ draw the horizontal line _XY_. - -With center _P_ and distance _DP_ describe the two arcs _DX_ and -_D′Y_, cutting the line _XY_ in _X_ and _Y_. - -Then _X_ and _Y_ are the dividing-points of the inclined line.[27] - -_Obs._ The dividing-points found by the above rule, used with the -ordinary measuring-line, will lay off distances on the retiring -inclined line, as the ordinary dividing-points lay them off on the -retiring horizontal line. - -Another dividing-point, peculiar in its application, is sometimes -useful, and is to be found as follows:— - - [Illustration: Fig. 47.] - -Let _AB_, Fig. 47., be the given inclined line drawn in perspective, -and _Ac_ the relative horizontal. - -Find the vanishing-points, _V_ and _E_, of _Ac_ and _AB_; _D_, the -dividing-point of _Ac_; and the sight-magnitude of _Ac_ on the -measuring-line, or _AC_. - -From _D_ erect the perpendicular _DF_. - -Join _CB_, and produce it to cut _DE_ in _F_. Join _EF_. - -Then, by similar triangles, _DF_ is equal to _EV_, and _EF_ is -parallel to _DV_. - -Hence it follows that if from _D_, the dividing-point of _Ac_, we -raise a perpendicular and make _DF_ equal to _EV_, a line _CF_, drawn -from any point _C_ on the measuring-line to _F_, will mark the -distance _AB_ on the inclined line, _AB_ being the portion of the -given inclined line which forms the diagonal of the vertical rectangle -of which _AC_ is the base. - - - [27] The demonstration is in Appendix II., p. 104. - - - - -PROBLEM XVIII. - -TO FIND THE SIGHT-LINE OF AN INCLINED PLANE IN WHICH TWO LINES ARE -GIVEN IN POSITION.[28] - - -As in order to fix the position of a line two points in it must be -given, so in order to fix the position of a plane, two lines in it -must be given. - - [Illustration: Fig. 48] - -Let the two lines be _AB_ and _CD_, Fig. 48. - -As they are given in position, the relative horizontals _AE_ and _CF_ -must be given. - -Then by Problem XVI. the vanishing-point of _AB_ is _V_, and of _CD_, -_V′_. - -Join _VV′_ and produce it to cut the sight-line in _X_. - -Then _VX_ is the sight-line of the inclined plane. - -Like the horizontal sight-line, it is of indefinite length; and may be -produced in either direction as occasion requires, crossing the -horizontal line of sight, if the plane continues downward in that -direction. - -_X_ is the vanishing-point of all horizontal lines in the inclined -plane. - - - [28] Read the Article on this problem in the Appendix, p. 97, before - investigating the problem itself. - - - - -PROBLEM XIX. - -TO FIND THE VANISHING-POINT OF STEEPEST LINES IN AN INCLINED PLANE -WHOSE SIGHT-LINE IS GIVEN. - - - [Illustration: Fig. 49.] - -Let _VX_, Fig. 49., be the given sight-line. - -Produce it to cut the horizontal sight-line in _X_. - -Therefore _X_ is the vanishing-point of horizontal lines in the given -inclined plane. (Problem XVIII.) - -Join _TX_, and draw _TY_ at right angles to _TX_. - -Therefore _Y_ is the rectangular vanishing-point corresponding to -_X_.[29] - -From _Y_ erect the vertical _YP_, cutting the sight-line of the -inclined plane in _P_. - -Then _P_ is the vanishing-point of steepest lines in the plane. - -All lines drawn to it, as _QP_, _RP_, _NP_, etc., are the steepest -possible in the plane; and all lines drawn to _X_, as _QX_, _OX_, -etc., are horizontal, and at right angles to the lines _PQ_, _PR_, -etc. - - - [29] That is to say, the vanishing-point of horizontal lines drawn at - right angles to the lines whose vanishing-point is _X_. - - - - -PROBLEM XX. - -TO FIND THE VANISHING-POINT OF LINES PERPENDICULAR TO THE SURFACE OF A -GIVEN INCLINED PLANE. - - - [Illustration: Fig. 50.] - -As the inclined plane is given, one of its steepest lines must be given, -or may be ascertained. - -Let _AB_, Fig. 50., be a portion of a steepest line in the given plane, -and _V_ the vanishing-point of its relative horizontal. - -Through _V_ draw the vertical _GF_ upwards and downwards. - -From _A_ set off any portion of the relative horizontal _AC_, and on -_AC_ describe a semicircle in a vertical plane, _ADC_, cutting _AB_ in -_E_. - -Join _EC_, and produce it to cut _GF_ in _F_. - -Then _F_ is the vanishing-point required. - -For, because _AEC_ is an angle in a semicircle, it is a right angle; -and therefore the line _EF_ is at right angles to the line _AB_; and -similarly all lines drawn to _F_, and therefore parallel to _EF_, are -at right angles with any line which cuts them, drawn to the -vanishing-point of _AB_. - -And because the semicircle _ADC_ is in a vertical plane, and its -diameter _AC_ is at right angles to the horizontal lines traversing -the surface of the inclined plane, the line _EC_, being in this -semicircle, is also at right angles to such traversing lines. And -therefore the line _EC_, being at right angles to the steepest lines -in the plane, and to the horizontal lines in it, is perpendicular to -its surface. - - * * * * * - -The preceding series of constructions, with the examples in the first -Article of the Appendix, put it in the power of the student to draw -any form, however complicated,[30] which does not involve intersection -of curved surfaces. I shall not proceed to the analysis of any of -these more complex problems, as they are entirely useless in the -ordinary practice of artists. For a few words only I must ask the -reader’s further patience, respecting the general placing and scale of -the picture. - -As the horizontal sight-line is drawn through the sight-point, and the -sight-point is opposite the eye, the sight-line is always on a level -with the eye. Above and below the sight-line, the eye comprehends, as -it is raised or depressed while the head is held upright, about an -equal space; and, on each side of the sight-point, about the same -space is easily seen without turning the head; so that if a picture -represented the true field of easy vision, it ought to be circular, -and have the sight-point in its center. But because some parts of any -given view are usually more interesting than others, either the -uninteresting parts are left out, or somewhat more than would -generally be seen of the interesting parts is included, by moving the -field of the picture a little upwards or downwards, so as to throw the -sight-point low or high. The operation will be understood in a moment -by cutting an aperture in a piece of pasteboard, and moving it up and -down in front of the eye, without moving the eye. It will be seen to -embrace sometimes the low, sometimes the high objects, without -altering their perspective, only the eye will be opposite the lower -part of the aperture when it sees the higher objects, and _vice -versâ_. - -There is no reason, in the laws of perspective, why the picture should -not be moved to the right or left of the sight-point, as well as up or -down. But there is this practical reason. The moment the spectator -sees the horizon in a picture high, he tries to hold his head high, -that is, in its right place. When he sees the horizon in a picture -low, he similarly tries to put his head low. But, if the sight-point -is thrown to the left hand or right hand, he does not understand that -he is to step a little to the right or left; and if he places himself, -as usual, in the middle, all the perspective is distorted. Hence it is -generally unadvisable to remove the sight-point laterally, from the -center of the picture. The Dutch painters, however, fearlessly take -the license of placing it to the right or left; and often with good -effect. - -The rectilinear limitation of the sides, top, and base of the picture -is of course quite arbitrary, as the space of a landscape would be -which was seen through a window; less or more being seen at the -spectator’s pleasure, as he retires or advances. - -The distance of the station-point is not so arbitrary. In ordinary -cases it should not be less than the intended greatest dimension -(height or breadth) of the picture. In most works by the great masters -it is more; they not only calculate on their pictures being seen at -considerable distances, but they like breadth of mass in buildings, -and dislike the sharp angles which always result from station-points -at short distances.[31] - -Whenever perspective, done by true rule, looks wrong, it is always -because the station-point is too near. Determine, in the outset, at -what distance the spectator is likely to examine the work, and never -use a station-point within a less distance. - -There is yet another and a very important reason, not only for care in -placing the station-point, but for that accurate calculation of -distance and observance of measurement which have been insisted on -throughout this work. All drawings of objects on a reduced scale are, -if rightly executed, drawings of the appearance of the object at the -distance which in true perspective reduces it to that scale. They are -not _small_ drawings of the object seen near, but drawings the _real -size_ of the object seen far off. Thus if you draw a mountain in a -landscape, three inches high, you do not reduce all the features of -the near mountain so as to come into three inches of paper. You could -not do that. All that you can do is to give the appearance of the -mountain, when it is so far off that three inches of paper would -really hide it from you. It is precisely the same in drawing any other -object. A face can no more be reduced in scale than a mountain can. It -is infinitely delicate already; it can only be quite rightly rendered -on its own scale, or at least on the slightly diminished scale which -would be fixed by placing the plate of glass, supposed to represent -the field of the picture, close to the figures. Correggio and Raphael -were both fond of this slightly subdued magnitude of figure. Colossal -painting, in which Correggio excelled all others, is usually the -enlargement of a small picture (as a colossal sculpture is of a small -statue), in order to permit the subject of it to be discerned at a -distance. The treatment of colossal (as distinguished from ordinary) -paintings will depend therefore, in general, on the principles of -optics more than on those of perspective, though, occasionally, -portions may be represented as if they were the projection of near -objects on a plane behind them. In all points the subject is one of -great difficulty and subtlety; and its examination does not fall -within the compass of this essay. - -Lastly, it will follow from these considerations, and the conclusion -is one of great practical importance, that, though pictures may be -enlarged, they cannot be reduced, in copying them. All attempts to -engrave pictures completely on a reduced scale are, for this reason, -nugatory. The best that can be done is to give the aspect of the -picture at the distance which reduces it in perspective to the size -required; or, in other words, to make a drawing of the distant effect -of the picture. Good painting, like nature’s own work, is infinite, -and unreduceable. - -I wish this book had less tendency towards the infinite and -unreduceable. It has so far exceeded the limits I hoped to give it, -that I doubt not the reader will pardon an abruptness of conclusion, -and be thankful, as I am myself, to get to an end on any terms. - - - [30] As in algebraic science, much depends, in complicated - perspective, on the student’s ready invention of expedients, - and on his quick sight of the shortest way in which the - solution may be accomplished, when there are several ways. - - [31] The greatest masters are also fond of parallel perspective, - that is to say, of having one side of their buildings fronting - them full, and therefore parallel to the picture plane, while - the other side vanishes to the sight-point. This is almost - always done in figure backgrounds, securing simple and balanced - lines. - - - - -APPENDIX. - - -I. - -PRACTICE AND OBSERVATIONS. - - -II. - -DEMONSTRATIONS. - - - - -I. - -PRACTICE AND OBSERVATIONS ON THE PRECEDING PROBLEMS. - - - - -PROBLEM I. - - -An example will be necessary to make this problem clear to the general -student. - -The nearest corner of a piece of pattern on the carpet is 4½ feet -beneath the eye, 2 feet to our right and 3½ feet in direct distance -from us. We intend to make a drawing of the pattern which shall be -seen properly when held 1½ foot from the eye. It is required to fix -the position of the corner of the piece of pattern. - - [Illustration: Fig. 51.] - -Let _AB_, Fig. 51., be our sheet of paper, some 3 feet wide. Make _ST_ -equal to 1½ foot. Draw the line of sight through _S_. Produce _TS_, -and make _DS_ equal to 2 feet, therefore _TD_ equal to 3½ feet. Draw -_DC_, equal to 2 feet; _CP_, equal to 4 feet. Join _TC_ (cutting the -sight-line in _Q_) and _TP_. - -Let fall the vertical _QP′_, then _P′_ is the point required. - -If the lines, as in the figure, fall outside of your sheet of paper, -in order to draw them, it is necessary to attach other sheets of paper -to its edges. This is inconvenient, but must be done at first that -you may see your way clearly; and sometimes afterwards, though there -are expedients for doing without such extension in fast sketching. - -It is evident, however, that no extension of surface could be of any -use to us, if the distance _TD_, instead of being 3½ feet, were 100 -feet, or a mile, as it might easily be in a landscape. - -It is necessary, therefore, to obtain some other means of -construction; to do which we must examine the principle of the -problem. - - -In the analysis of Fig. 2., in the introductory remarks, I used the -word “height” only of the tower, _QP_, because it was only to its -vertical height that the law deduced from the figure could be applied. -For suppose it had been a pyramid, as _OQP_, Fig. 52., then the image -of its side, _QP_, being, like every other magnitude, limited on the -glass _AB_ by the lines coming from its extremities, would appear only -of the length _Q′S_; and it is not true that _Q′S_ is to _QP_ as _TS_ -is to _TP_. But if we let fall a vertical _QD_ from _Q_, so as to get -the vertical height of the pyramid, then it is true that _Q′S_ is to -_QD_ as _TS_ is to _TD_. - - [Illustration: Fig. 52.] - -Supposing this figure represented, not a pyramid, but a triangle on -the ground, and that _QD_ and _QP_ are horizontal lines, expressing -lateral distance from the line _TD_, still the rule would be false for -_QP_ and true for _QD_. And, similarly, it is true for all lines which -are parallel, like _QD_, to the plane of the picture _AB_, and false -for all lines which are inclined to it at an angle. - -Hence generally. Let _PQ_ (Fig. 2. in Introduction, p. 6) be any -magnitude _parallel to the plane of the picture_; and _P′Q′_ its image -on the picture. - -Then always the formula is true which you learned in the Introduction: -_P′Q′_ is to _PQ_ as _ST_ is to _DT_. - -Now the magnitude _P_ dash _Q_ dash in this formula I call the -“SIGHT-MAGNITUDE” of the line _PQ_. The student must fix this term, -and the meaning of it, well in his mind. The “sight-magnitude” of a -line is the magnitude which bears to the real line the same proportion -that the distance of the picture bears to the distance of the object. -Thus, if a tower be a hundred feet high, and a hundred yards off; and -the picture, or piece of glass, is one yard from the spectator, -between him and the tower; the distance of picture being then to -distance of tower as 1 to 100, the sight-magnitude of the tower’s -height will be as 1 to 100; that is to say, one foot. If the tower is -two hundred yards distant, the sight-magnitude of its height will be -half a foot, and so on. - -But farther. It is constantly necessary, in perspective operations, -to measure the other dimensions of objects by the sight-magnitude of -their vertical lines. Thus, if the tower, which is a hundred feet -high, is square, and twenty-five feet broad on each side; if the -sight-magnitude of the height is one foot, the measurement of the -side, reduced to the same scale, will be the hundredth part of -twenty-five feet, or three inches: and, accordingly, I use in this -treatise the term “sight-magnitude” indiscriminately for all lines -reduced in the same proportion as the vertical lines of the object. If -I tell you to find the “sight-magnitude” of any line, I mean, always, -find the magnitude which bears to that line the proportion of _ST_ to -_DT_; or, in simpler terms, reduce the line to the scale which you -have fixed by the first determination of the length _ST_. - -Therefore, you must learn to draw quickly to scale before you do -anything else; for all the measurements of your object must be -reduced to the scale fixed by _ST_ before you can use them in your -diagram. If the object is fifty feet from you, and your paper one -foot, all the lines of the object must be reduced to a scale of one -fiftieth before you can use them; if the object is two thousand feet -from you, and your paper one foot, all your lines must be reduced to -the scale of one two-thousandth before you can use them, and so on. -Only in ultimate practice, the reduction never need be tiresome, for, -in the case of large distances, accuracy is never required. If a -building is three or four miles distant, a hairbreadth of accidental -variation in a touch makes a difference of ten or twenty feet in -height or breadth, if estimated by accurate perspective law. Hence it -is never attempted to apply measurements with precision at such -distances. Measurements are only required within distances of, at the -most, two or three hundred feet. Thus it may be necessary to represent -a cathedral nave precisely as seen from a spot seventy feet in front -of a given pillar; but we shall hardly be required to draw a cathedral -three miles distant precisely as seen from seventy feet in advance of -a given milestone. Of course, if such a thing be required, it can be -done; only the reductions are somewhat long and complicated: in -ordinary cases it is easy to assume the distance _ST_ so as to get at -the reduced dimensions in a moment. Thus, let the pillar of the nave, -in the case supposed, be 42 feet high, and we are required to stand -70 feet from it: assume _ST_ to be equal to 5 feet. Then, as 5 is to -70 so will the sight-magnitude required be to 42; that is to say, the -sight-magnitude of the pillar’s height will be 3 feet. If we make _ST_ -equal to 2½ feet, the pillar’s height will be 1½ foot, and so on. - -And for fine divisions into irregular parts which cannot be measured, -the ninth and tenth problems of the sixth book of Euclid will serve -you: the following construction is, however, I think, more practically -convenient:— - -The line _AB_ (Fig. 53.) is divided by given points, _a_, _b_, _c_, -into a given number of irregularly unequal parts; it is required to -divide any other line, _CD_, into an equal number of parts, bearing -to each other the same proportions as the parts of _AB_, and arranged -in the same order. - -Draw the two lines parallel to each other, as in the figure. - -Join _AC_ and _BD_, and produce the lines _AC_, _BD_, till they meet -in _P_. - -Join _aP_, _bP_, _cP_, cutting _cD_ in _f_, _g_, _h_. - -Then the line _CD_ is divided as required, in _f_, _g_, _h_. - -In the figure the lines _AB_ and _CD_ are accidentally perpendicular -to _AP_. There is no need for their being so. - - [Illustration: Fig. 53.] - -Now, to return to our first problem. - -The construction given in the figure is only the quickest mathematical -way of obtaining, on the picture, the sight-magnitudes of _DC_ and -_PC_, which are both magnitudes parallel with the picture plane. But -if these magnitudes are too great to be thus put on the paper, you -have only to obtain the reduction by scale. Thus, if _TS_ be one foot, -_TD_ eighty feet, _DC_ forty feet, and _CP_ ninety feet, the distance -_QS_ must be made equal to one eightieth of _DC_, or half a foot; and -the distance _QP′_, one eightieth of _CP_, or one eightieth of ninety -feet; that is to say, nine eighths of a foot, or thirteen and a half -inches. The lines _CT_ and _PT_ are thus _practically_ useless, it -being only necessary to measure _QS_ and _QP_, on your paper, of the -due sight-magnitudes. But the mathematical construction, given in -Problem I., is the basis of all succeeding problems, and, if it is -once thoroughly understood and practiced (it can only be thoroughly -understood by practice), all the other problems will follow easily. - -Lastly. Observe that any perspective operation whatever may be -performed with reduced dimensions of every line employed, so as to -bring it conveniently within the limits of your paper. When the -required figure is thus constructed on a small scale, you have only to -enlarge it accurately in the same proportion in which you reduced the -lines of construction, and you will have the figure constructed in -perspective on the scale required for use. - - - - -PROBLEM IX. - - -The drawing of most buildings occurring in ordinary practice will -resolve itself into applications of this problem. In general, any -house, or block of houses, presents itself under the main conditions -assumed here in Fig. 54. There will be an angle or corner somewhere -near the spectator, as _AB_; and the level of the eye will usually be -above the base of the building, of which, therefore, the horizontal -upper lines will slope down to the vanishing-points, and the base -lines rise to them. The following practical directions will, however, -meet nearly all cases:— - - [Illustration: Fig. 54.] - -Let _AB_, Fig. 54., be any important vertical line in the block of -buildings; if it is the side of a street, you may fix upon such a line -at the division between two houses. If its real height, distance, -etc., are given, you will proceed with the accurate construction of -the problem; but usually you will neither know, nor care, exactly how -high the building is, or how far off. In such case draw the line _AB_, -as nearly as you can guess, about the part of the picture it ought to -occupy, and on such a scale as you choose. Divide it into any -convenient number of equal parts, according to the height you presume -it to be. If you suppose it to be twenty feet high, you may divide it -into twenty parts, and let each part stand for a foot; if thirty feet -high, you may divide it into ten parts, and let each part stand for -three feet; if seventy feet high, into fourteen parts, and let each -part stand for five feet; and so on, avoiding thus very minute -divisions till you come to details. Then observe how high your eye -reaches upon this vertical line; suppose, for instance, that it is -thirty feet high and divided into ten parts, and you are standing so -as to raise your head to about six feet above its base, then the -sight-line may be drawn, as in the figure, through the second division -from the ground. If you are standing above the house, draw the -sight-line above _B_; if below the house, below _A_; at such height or -depth as you suppose may be accurate (a yard or two more or less -matters little at ordinary distances, while at great distances -perspective rules become nearly useless, the eye serving you better -than the necessarily imperfect calculation). Then fix your sight-point -and station-point, the latter with proper reference to the scale of -the line _AB_. As you cannot, in all probability, ascertain the exact -direction of the line _AV_ or _BV_, draw the slope _BV_ as it appears -to you, cutting the sight-line in _V_. Thus having fixed one -vanishing-point, the other, and the dividing-points, must be -accurately found by rule; for, as before stated, whether your entire -group of points (vanishing and dividing) falls a little more or less -to the right or left of _S_ does not signify, but the relation of the -points to each other _does_ signify. Then draw the measuring-line -_BG_, either through _A_ or _B_, choosing always the steeper slope of -the two; divide the measuring-line into parts of the same length as -those used on _AB_, and let them stand for the same magnitudes. Thus, -suppose there are two rows of windows in the house front, each window -six feet high by three wide, and separated by intervals of three feet, -both between window and window and between tier and tier; each of the -divisions here standing for three feet, the lines drawn from _BG_ to -the dividing-point _D_ fix the lateral dimensions, and the divisions -on _AB_ the vertical ones. For other magnitudes it would be necessary -to subdivide the parts on the measuring-line, or on _AB_, as required. -The lines which regulate the inner sides or returns of the windows -(_a_, _b_, _c_, etc.) of course are drawn to the vanishing-point of -_BF_ (the other side of the house), if _FBV_ represents a right angle; -if not, their own vanishing-point must be found separately for these -returns. But see Practice on Problem XI. - - [Illustration: Fig. 55.] - -Interior angles, such as _EBC_, Fig. 55. (suppose the corner of a -room), are to be treated in the same way, each side of the room having -its measurements separately carried to it from the measuring-line. It -may sometimes happen in such cases that we have to carry the -measurement _up_ from the corner _B_, and that the sight-magnitudes -are given us from the length of the line _AB_. For instance, suppose -the room is eighteen feet high, and therefore _AB_ is eighteen feet; -and we have to lay off lengths of six feet on the top of the room -wall, _BC_. Find _D_, the dividing-point of _BC_. Draw a -measuring-line, _BF_, from _B_; and another, _gC_, anywhere above. On -_BF_ lay off _BG_ equal to one third of _AB_, or six feet; and draw -from _D_, through _G_ and _B_, the lines _Gg_, _Bb_, to the upper -measuring-line. Then _gb_ is six feet on that measuring-line. Make -_bc_, _ch_, etc., equal to _bg_; and draw _ce_, _hf_, etc., to _D_, -cutting _BC_ in _e_ and _f_, which mark the required lengths of six -feet each at the top of the wall. - - - - -PROBLEM X. - - -This is one of the most important foundational problems in -perspective, and it is necessary that the student should entirely -familiarize himself with its conditions. - -In order to do so, he must first observe these general relations of -magnitude in any pyramid on a square base. - -Let _AGH′_, Fig. 56., be any pyramid on a square base. - - [Illustration: Fig. 56.] - -The best terms in which its magnitude can be given, are the length of -one side of its base, _AH_, and its vertical altitude (_CD_ in -Fig. 25.); for, knowing these, we know all the other magnitudes. But -these are not the terms in which its size will be usually -ascertainable. Generally, we shall have given us, and be able to -ascertain by measurement, one side of its base _AH_, and either _AG_ -the length of one of the lines of its angles, or _BG_ (or _B′G_) the -length of a line drawn from its vertex, _G_, to the middle of the side -of its base. In measuring a real pyramid, _AG_ will usually be the -line most easily found; but in many architectural problems _BG_ is -given, or is most easily ascertainable. - -Observe therefore this general construction. - - [Illustration: Fig. 57.] - -Let _ABDE_, Fig. 57., be the square base of any pyramid. - -Draw its diagonals, _AE_, _BD_, cutting each other in its center, _C_. - -Bisect any side, _AB_, in _F_. - -From _F_ erect vertical _FG_. - -Produce _FB_ to _H_, and make _FH_ equal to _AC_. - -Now if the vertical altitude of the pyramid (_CD_ in Fig. 25.) be -given, make _FG_ equal to this vertical altitude. - -Join _GB_ and _GH_. - -Then _GB_ and _GH_ are the true magnitudes of _GB_ and _GH_ in -Fig. 56. - -If _GB_ is given, and not the vertical altitude, with center _B_, and -distance _GB_, describe circle cutting _FG_ in _G_, and _FG_ is the -vertical altitude. - -If _GH_ is given, describe the circle from _H_, with distance _GH_, -and it will similarly cut _FG_ in _G_. - -It is especially necessary for the student to examine this -construction thoroughly, because in many complicated forms of -ornaments, capitals of columns, etc., the lines _BG_ and _GH_ become -the limits or bases of curves, which are elongated on the longer (or -angle) profile _GH_, and shortened on the shorter (or lateral) profile -_BG_. We will take a simple instance, but must previously note another -construction. - -It is often necessary, when pyramids are the roots of some ornamental -form, to divide them horizontally at a given vertical height. The -shortest way of doing so is in general the following. - - [Illustration: Fig. 58.] - -Let _AEC_, Fig. 58., be any pyramid on a square base _ABC_, and _ADC_ -the square pillar used in its construction. - -Then by construction (Problem X.) _BD_ and _AF_ are both of the -vertical height of the pyramid. - -Of the diagonals, _FE_, _DE_, choose the shortest (in this case _DE_), -and produce it to cut the sight-line in _V_. - -Therefore _V_ is the vanishing-point of _DE_. - -Divide _DB_, as may be required, into the sight-magnitudes of the -given vertical heights at which the pyramid is to be divided. - - [Illustration: Fig. 59.] [Illustration: Fig. 60.] - -From the points of division, 1, 2, 3, etc., draw to the -vanishing-point _V_. The lines so drawn cut the angle line of the -pyramid, _BE_, at the required elevations. Thus, in the figure, it is -required to draw a horizontal black band on the pyramid at three -fifths of its height, and in breadth one twentieth of its height. The -line _BD_ is divided into five parts, of which three are counted from -_B_ upwards. Then the line drawn to _V_ marks the base of the black -band. Then one fourth of one of the five parts is measured, which -similarly gives the breadth of the band. The terminal lines of the -band are then drawn on the sides of the pyramid parallel to _AB_ (or -to its vanishing-point if it has one), and to the vanishing-point of -_BC_. - -If it happens that the vanishing-points of the diagonals are awkwardly -placed for use, bisect the nearest base line of the pyramid in _B_, as -in Fig. 59. - -Erect the vertical _DB_ and join _GB_ and _DG_ (_G_ being the apex of -pyramid). - -Find the vanishing-point of _DG_, and use _DB_ for division, carrying -the measurements to the line _GB_. - -In Fig. 59., if we join _AD_ and _DC_, _ADC_ is the vertical profile -of the whole pyramid, and _BDC_ of the half pyramid, corresponding to -_FGB_ in Fig. 57. - - [Illustration: Fig. 61.] - -We may now proceed to an architectural example. - -Let _AH_, Fig. 60., be the vertical profile of the capital of a -pillar, _AB_ the semi-diameter of its head or abacus, and _FD_ the -semi-diameter of its shaft. - -Let the shaft be circular, and the abacus square, down to the level -_E_. - -Join _BD_, _EF_, and produce them to meet in _G_. - -Therefore _ECG_ is the semi-profile of a reversed pyramid containing -the capital. - -Construct this pyramid, with the square of the abacus, in the required -perspective, as in Fig. 61.; making _AE_ equal to _AE_ in Fig. 60., -and _AK_, the side of the square, equal to twice _AB_ in Fig. 60. Make -_EG_ equal to _CG_, and _ED_ equal to _CD_. Draw _DF_ to the -vanishing-point of the diagonal _DV_ (the figure is too small to -include this vanishing-point), and _F_ is the level of the point _F_ -in Fig. 60., on the side of the pyramid. - -Draw _Fm_, _Fn_, to the vanishing-points of _AH_ and _AK_. Then _Fn_ -and _Fm_ are horizontal lines across the pyramid at the level _F_, -forming at that level two sides of a square. - - [Illustration: Fig. 62.] - -Complete the square, and within it inscribe a circle, as in Fig. 62., -which is left unlettered that its construction may be clear. At the -extremities of this draw vertical lines, which will be the sides of -the shaft in its right place. It will be found to be somewhat smaller -in diameter than the entire shaft in Fig. 60., because at the center -of the square it is more distant than the nearest edge of the square -abacus. The curves of the capital may then be drawn approximately by -the eye. They are not quite accurate in Fig. 62., there being a -subtlety in their junction with the shaft which could not be shown on -so small a scale without confusing the student; the curve on the left -springing from a point a little way round the circle behind the shaft, -and that on the right from a point on this side of the circle a little -way within the edge of the shaft. But for their more accurate -construction see Notes on Problem XIV. - - - - -PROBLEM XI. - - -It is seldom that any complicated curve, except occasionally a spiral, -needs to be drawn in perspective; but the student will do well to -practice for some time any fantastic shapes which he can find drawn on -flat surfaces, as on wall-papers, carpets, etc., in order to accustom -himself to the strange and great changes which perspective causes in -them. - - [Illustration: Fig. 63.] - -The curves most required in architectural drawing, after the circle, -are those of pointed arches; in which, however, all that will be -generally needed is to fix the apex, and two points in the sides. Thus -if we have to draw a range of pointed arches, such as _APB_, Fig. 63., -draw the measured arch to its sight-magnitude first neatly in a -rectangle, _ABCD_; then draw the diagonals _AD_ and _BC_; where they -cut the curve draw a horizontal line (as at the level _E_ in the -figure), and carry it along the range to the vanishing-point, fixing -the points where the arches cut their diagonals all along. If the arch -is cusped, a line should be drawn, at _F_ to mark the height of the -cusps, and verticals raised at _G_ and _H_, to determine the interval -between them. Any other points may be similarly determined, but these -will usually be enough. Figure 63. shows the perspective construction -of a square niche of good Veronese Gothic, with an uncusped arch of -similar size and curve beyond. - - [Illustration: Fig. 64.] - -In Fig. 64. the more distant arch only is lettered, as the -construction of the nearest explains itself more clearly to the eye -without letters. The more distant arch shows the general construction -for all arches seen underneath, as of bridges, cathedral aisles, etc. -The rectangle _ABCD_ is first drawn to contain the outside arch; then -the depth of the arch, _Aa_, is determined by the measuring-line, and -the rectangle, _abcd_, drawn for the inner arch. - -_Aa_, _Bb_, etc., go to one vanishing-point; _AB_, _ab_, etc., to the -opposite one. - -In the nearer arch another narrow rectangle is drawn to determine the -cusp. The parts which would actually come into sight are slightly -shaded. - - - - -PROBLEM XIV. - - -Several exercises will be required on this important problem. - -I. It is required to draw a circular flat-bottomed dish narrower at -the bottom than the top; the vertical depth being given, and the -diameter at the top and bottom. - - [Illustration: Fig. 65.] - -Let _ab_, Fig. 65., be the diameter of the bottom, _ac_ the diameter -of the top, and _ad_ its vertical depth. - -Take _AD_ in position equal to _ac_. - -On _AD_ draw the square _ABCD_, and inscribe in it a circle. - -Therefore, the circle so inscribed has the diameter of the top of the -dish. - -From _A_ and _D_ let fall verticals, _AE_, _DH_, each equal to _ad_. - -Join _EH_, and describe square _EFGH_, which accordingly will be equal -to the square _ABCD_, and be at the depth _ad_ beneath it. - -Within the square _EFGH_ describe a square _IK_, whose diameter shall -be equal to _ab_. - -Describe a circle within the square _IK_. Therefore the circle so -inscribed has its diameter equal to _ab_; and it is in the center of -the square _EFGH_, which is vertically beneath the square _ABCD_. - -Therefore the circle in the square _IK_ represents the bottom of the -dish. - -Now the two circles thus drawn will either intersect one another, or -they will not. - -If they intersect one another, as in the figure, and they are below -the eye, part of the bottom of the dish is seen within it. - - [Illustration: Fig. 66.] - -To avoid confusion, let us take then two intersecting circles without -the inclosing squares, as in Fig. 66. - -Draw right lines, _ab_, _cd_, touching both circles externally. Then -the parts of these lines which connect the circles are the sides of -the dish. They are drawn in Fig. 65. without any prolongations, but -the best way to construct them is as in Fig. 66. - -If the circles do not intersect each other, the smaller must either be -within the larger or not within it. - -If within the larger, the whole of the bottom of the dish is seen from -above, Fig. 67. _a_. - - [Illustration: Fig. 67.] - -If the smaller circle is not within the larger, none of the bottom is -seen inside the dish, _b_. - -If the circles are above instead of beneath the eye, the bottom of the -dish is seen beneath it, _c_. - -If one circle is above and another beneath the eye, neither the bottom -nor top of the dish is seen, _d_. Unless the object be very large, the -circles in this case will have little apparent curvature. - -II. The preceding problem is simple, because the lines of the profile -of the object (_ab_ and _cd_, Fig. 66.) are straight. But if these -lines of profile are curved, the problem becomes much more complex: -once mastered, however, it leaves no farther difficulty in -perspective. - -Let it be required to draw a flattish circular cup or vase, with a -given curve of profile. - -The basis of construction is given in Fig. 68., half of it only being -drawn, in order that the eye may seize its lines easily. - - [Illustration: Fig. 68.] - -Two squares (of the required size) are first drawn, one above the -other, with a given vertical interval, _AC_, between them, and each is -divided into eight parts by its diameters and diagonals. In these -squares two circles are drawn; which are, therefore, of equal size, -and one above the other. Two smaller circles, also of equal size, are -drawn within these larger circles in the construction of the present -problem; more may be necessary in some, none at all in others. - -It will be seen that the portions of the diagonals and diameters of -squares which are cut off between the circles represent radiating -planes, occupying the position of the spokes of a wheel. - -Now let the line _AEB_, Fig. 69., be the profile of the vase or cup to -be drawn. - -Inclose it in the rectangle _CD_, and if any portion of it is not -curved, as _AE_, cut off the curved portion by the vertical line _EF_, -so as to include it in the smaller rectangle _FD_. - -Draw the rectangle _ACBD_ in position, and upon it construct two -squares, as they are constructed on the rectangle _ACD_ in Fig. 68.; -and complete the construction of Fig. 68., making the radius of its -large outer circles equal to _AD_, and of its small inner circles -equal to _AE_. - -The planes which occupy the position of the wheel spokes will then -each represent a rectangle of the size of _FD_. The construction is -shown by the dotted lines in Fig. 69.; _c_ being the center of the -uppermost circle. - - [Illustration: Fig. 69.] - -Within each of the smaller rectangles between the circles, draw the -curve _EB_ in perspective, as in Fig. 69. - -Draw the curve _xy_, touching and inclosing the curves in the -rectangles, and meeting the upper circle at _y_.[32] - -Then _xy_ is the contour of the surface of the cup, and the upper -circle is its lip. - -If the line _xy_ is long, it may be necessary to draw other rectangles -between the eight principal ones; and, if the curve of profile _AB_ is -complex or retorted, there may be several lines corresponding to _XY_, -inclosing the successive waves of the profile; and the outer curve -will then be an undulating or broken one. - - [Illustration: Fig. 70.] - -III. All branched ornamentation, forms of flowers, capitals of -columns, machicolations of round towers, and other such arrangements -of radiating curve, are resolvable by this problem, using more or -fewer interior circles according to the conditions of the curves. -Fig. 70. is an example of the construction of a circular group of -eight trefoils with curved stems. One outer or limiting circle is -drawn within the square _EDCF_, and the extremities of the trefoils -touch it at the extremities of its diagonals and diameters. A smaller -circle is at the vertical distance _BC_ below the larger, and _A_ is -the angle of the square within which the smaller circle is drawn; but -the square is not given, to avoid confusion. The stems of the trefoils -form drooping curves, arranged on the diagonals and diameters of the -smaller circle, which are dotted. But no perspective laws will do work -of this intricate kind so well as the hand and eye of a painter. - -IV. There is one common construction, however, in which, singularly, -the hand and eye of the painter almost always fail, and that is the -fillet of any ordinary capital or base of a circular pillar (or any -similar form). It is rarely necessary in practice to draw such minor -details in perspective; yet the perspective laws which regulate them -should be understood, else the eye does not see their contours rightly -until it is very highly cultivated. - - [Illustration: Fig. 71.] - -Fig. 71. will show the law with sufficient clearness; it represents -the perspective construction of a fillet whose profile is a -semicircle, such as _FH_ in Fig. 60., seen above the eye. Only half -the pillar with half the fillet is drawn, to avoid confusion. - -_Q_ is the center of the shaft. - -_PQ_ the thickness of the fillet, sight-magnitude at the shaft’s -center. - -Round _P_ a horizontal semicircle is drawn on the diameter of the -shaft _ab_. - -Round _Q_ another horizontal semicircle is drawn on diameter _cd_. - -These two semicircles are the upper and lower edges of the fillet. - -Then diagonals and diameters are drawn as in Fig. 68., and, at their -extremities, semicircles in perspective, as in Fig. 69. - -The letters _A_, _B_, _C_, _D_, and _E_, indicate the upper and -exterior angles of the rectangles in which these semicircles are to be -drawn; but the inner vertical line is not dotted in the rectangle at -_C_, as it would have confused itself with other lines. - -Then the visible contour of the fillet is the line which incloses and -touches[33] all the semicircles. It disappears behind the shaft at the -point _H_, but I have drawn it through to the opposite extremity of -the diameter at _d_. - -Turned upside down the figure shows the construction of a basic -fillet. - -The capital of a Greek Doric pillar should be drawn frequently for -exercise on this fourteenth problem, the curve of its echinus being -exquisitely subtle, while the general contour is simple. - - - [32] This point coincides in the figure with the extremity of the - horizontal diameter, but only accidentally. - - [33] The engraving is a little inaccurate; the inclosing line - should touch the dotted semicircles at _A_ and _B_. The student - should draw it on a large scale. - - - - -PROBLEM XVI. - - -It is often possible to shorten other perspective operations -considerably, by finding the vanishing-points of the inclined lines of -the object. Thus, in drawing the gabled roof in Fig. 43., if the gable -_AYC_ be drawn in perspective, and the vanishing-point of _AY_ -determined, it is not necessary to draw the two sides of the -rectangle, _A′D′_ and _D′B′_, in order to determine the point _Y′_; -but merely to draw _YY′_ to the vanishing-point of _AA′_ and _A′Y′_ to -the vanishing-point of _AY_, meeting in _Y′_, the point required. - -Again, if there be a series of gables, or other figures produced by -parallel inclined lines, and retiring to the point _V_, as in -Fig. 72.,[34] it is not necessary to draw each separately, but merely -to determine their breadths on the line _AV_, and draw the slopes of -each to their vanishing-points, as shown in Fig. 72. Or if the gables -are equal in height, and a line be drawn from _Y_ to _V_, the -construction resolves itself into a zigzag drawn alternately to _P_ -and _Q_, between the lines _YV_ and _AV_. - -The student must be very cautious, in finding the vanishing-points of -inclined lines, to notice their relations to the horizontals beneath -them, else he may easily mistake the horizontal to which they belong. - -Thus, let _ABCD_, Fig. 73., be a rectangular inclined plane, and let -it be required to find the vanishing-point of its diagonal _BD_. - -Find _V_, the vanishing-point of _AD_ and _BC_. - -Draw _AE_ to the opposite vanishing-point, so that _DAE_ may represent -a right angle. - -Let fall from _B_ the vertical _BE_, cutting _AE_ in _E_. - -Join _ED_, and produce it to cut the sight-line in _V′_. - - [Illustration: Fig. 72.] - -Then, since the point _E_ is vertically under the point _B_, the -horizontal line _ED_ is vertically under the inclined line _BD_. - - [Illustration: Fig. 73.] - -So that if we now let fall the vertical _V′P_ from _V′_, and produce -_BD_ to cut _V′P_ in _P_, the point _P_ will be the vanishing-point of -_BD_, and of all lines parallel to it.[35] - - - [34] The diagram is inaccurately cut. _YV_ should be a right line. - - [35] The student may perhaps understand this construction better - by completing the rectangle _ADFE_, drawing _DF_ to the - vanishing-point of _AE_, and _EF_ to _V_. The whole figure, - _BF_, may then be conceived as representing half the gable roof - of a house, _AF_ the rectangle of its base, and _AC_ the - rectangle of its sloping side. - - In nearly all picturesque buildings, especially on the - Continent, the slopes of gables are much varied (frequently - unequal on the two sides), and the vanishing-points of their - inclined lines become very important, if accuracy is required - in the intersections of tiling, sides of dormer windows, etc. - - Obviously, also, irregular triangles and polygons in vertical - planes may be more easily constructed by finding the - vanishing-points of their sides, than by the construction given - in the corollary to Problem IX.; and if such triangles or - polygons have others concentrically inscribed within them, as - often in Byzantine mosaics, etc., the use of the - vanishing-points will become essential. - - - - -PROBLEM XVIII. - - -Before examining the last three problems it is necessary that you -should understand accurately what is meant by the position of an -inclined plane. - -Cut a piece of strong white pasteboard into any irregular shape, and -dip it in a sloped position into water. However you hold it, the edge -of the water, of course, will always draw a horizontal line across its -surface. The direction of this horizontal line is the direction of the -inclined plane. (In beds of rock geologists call it their “strike.”) - - [Illustration: Fig. 74.] - -Next, draw a semicircle on the piece of pasteboard; draw its diameter, -_AB_, Fig. 74., and a vertical line from its center, _CD_; and draw -some other lines, _CE_, _CF_, etc., from the center to any points in -the circumference. - -Now dip the piece of pasteboard again into water, and, holding it at -any inclination and in any direction you choose, bring the surface of -the water to the line _AB_. Then the line _CD_ will be the most -steeply inclined of all the lines drawn to the circumference of the -circle; _GC_ and _HC_ will be less steep; and _EC_ and _FC_ less steep -still. The nearer the lines to _CD_, the steeper they will be; and the -nearer to _AB_, the more nearly horizontal. - -When, therefore, the line _AB_ is horizontal (or marks the water -surface), its direction is the direction of the inclined plane, and -the inclination of the line _DC_ is the inclination of the inclined -plane. In beds of rock geologists call the inclination of the line -_DC_ their “dip.” - -To fix the position of an inclined plane, therefore, is to determine -the direction of any two lines in the plane, _AB_ and _CD_, of which -one shall be horizontal and the other at right angles to it. Then any -lines drawn in the inclined plane, parallel to _AB_, will be -horizontal; and lines drawn parallel to _CD_ will be as steep as _CD_, -and are spoken of in the text as the “steepest lines” in the plane. - -But farther, whatever the direction of a plane may be, if it be -extended indefinitely, it will be terminated, to the eye of the -observer, by a boundary line, which, in a horizontal plane, is -horizontal (coinciding nearly with the visible horizon);—in a vertical -plane, is vertical;—and, in an inclined plane, is inclined. - -This line is properly, in each case, called the “sight-line” of such -plane; but it is only properly called the “horizon” in the case of a -horizontal plane: and I have preferred using always the term -“sight-line,” not only because more comprehensive, but more accurate; -for though the curvature of the earth’s surface is so slight that -practically its visible limit always coincides with the sight-line of -a horizontal plane, it does not mathematically coincide with it, and -the two lines ought not to be considered as theoretically identical, -though they are so in practice. - -It is evident that all vanishing-points of lines in any plane must be -found on its sight-line, and, therefore, that the sight-line of any -plane may be found by joining any two of such vanishing-points. Hence -the construction of Problem XVIII. - - - - -II. - -DEMONSTRATIONS WHICH COULD NOT CONVENIENTLY BE INCLUDED IN THE TEXT. - - - - -I. - -THE SECOND COROLLARY, PROBLEM II. - - -In Fig. 8. omit the lines _CD_, _C′D′_, and _DS_; and, as here in -Fig. 75., from _a_ draw _ad_ parallel to _AB_, cutting _BT_ in _d_; -and from _d_ draw _de_ parallel to _BC′_. - - [Illustration: Fig. 75.] - - Now as _ad_ is parallel to _AB_— - _AC_ ∶ _ac_ ∷ _BC′_ ∶ _de_; - but _AC_ is equal to _BC′_— - ∴ _ac_ = _de_. - - Now because the triangles _acV_, _bc′V_, are similar— - _ac_ ∶ _bc′_ ∷ _aV_ ∶ _bV_; - and because the triangles _deT_, _bc′T_ are similar— - _de_ ∶ _bc′_ ∷ _dT_ ∶ _bT_. - - But _ac_ is equal to _de_— - ∴ _aV_ ∶ _bV_ ∷ _dT_ ∶ _bT_; - ∴ the two triangles _abd_, _bTV_, are similar, and their angles - are alternate; - ∴ _TV_ is parallel to _ad_. - - But _ad_ is parallel to _AB_— - ∴ _TV_ is parallel to _AB_. - - - - -II. - -THE THIRD COROLLARY, PROBLEM III. - - -In Fig. 13., since _aR_ is by construction parallel to _AB_ in -Fig. 12., and _TV_ is by construction in Problem III. also parallel to -_AB_— - - ∴ _aR_ is parallel to _TV_, - ∴ _abR_ and _TbV_ are alternate triangles, - ∴ _aR_ ∶ _TV_ ∷ _ab_ ∶ _bV_. - - Again, by the construction of Fig. 13., _aR′_ is parallel to _MV_— - ∴ _abR′_ and _MbV_ are alternate triangles, - ∴ _aR′_ ∶ _MV_ ∷ _ab_ ∶ _bV_. - - And it has just been shown that also - _aR_ ∶ _TV_ ∷ _ab_ ∶ _bV_— - ∴ _aR′_ ∶ _MV_ ∷ _aR_ ∶ _TV_. - - But by construction, _aR′_ = _aR_— - ∴ _MV_ = _TV_. - - - - -III. - -ANALYSIS OF PROBLEM XV. - - -We proceed to take up the general condition of the second problem, -before left unexamined, namely, that in which the vertical distances -_BC′_ and _AC_ (Fig. 6. page 13), as well as the direct distances _TD_ -and _TD′_ are unequal. - -In Fig. 6., here repeated (Fig. 76.), produce _C′B_ downwards, and -make _C′E_ equal to _CA_. - - [Illustration: Fig. 76.] - -Join _AE_. - -Then, by the second Corollary of Problem II., _AE_ is a horizontal -line. - -Draw _TV_ parallel to _AE_, cutting the sight-line in _V_. - - ∴ _V_ is the vanishing-point of _AE_. - -Complete the constructions of Problem II. and its second Corollary. - -Then by Problem II. _ab_ is the line _AB_ drawn in perspective; and by -its Corollary _ae_ is the line _AE_ drawn in perspective. - - -From _V_ erect perpendicular _VP_, and produce _ab_ to cut it in _P_. - -Join _TP_, and from _e_ draw _ef_ parallel to _AE_, and cutting _AT_ -in _f_. - -Now in triangles _EBT_ and _AET_, as _eb_ is parallel to _EB_ and _ef_ -to _AE_;—_eb_ ∶ _ef_ ∷ _EB_ ∶ _AE_. - -But _TV_ is also parallel to _AE_ and _PV_ to _eb_. - -Therefore also in the triangles _aPV_ and _aVT_, - - _eb_ ∶ _ef_ ∷ _PV_ ∶ _VT_. - -Therefore _PV_ ∶ _VT_ ∷ _EB_ ∶ _AE_. - -And, by construction, angle _TPV_ = ∠ _AEB_. - -Therefore the triangles _TVP_, _AEB_, are similar; and _TP_ is -parallel to _AB_. - -Now the construction in this problem is entirely general for any -inclined line _AB_, and a horizontal line _AE_ in the same vertical -plane with it. - -So that if we find the vanishing-point of _AE_ in _V_, and from _V_ -erect a vertical _VP_, and from _T_ draw _TP_ parallel to _AB_, -cutting _VP_ in _P_, _P_ will be the vanishing-point of _AB_, and (by -the same proof as that given at page 17) of all lines parallel to it. - - [Illustration: Fig. 77.] - -Next, to find the dividing-point of the inclined line. - -I remove some unnecessary lines from the last figure and repeat it -here, Fig. 77., adding the measuring-line _aM_, that the student may -observe its position with respect to the other lines before I remove -any more of them. - -Now if the line _AB_ in this diagram represented the length of the -line _AB_ in reality (as _AB_ _does_ in Figs. 10. and 11.), we should -only have to proceed to modify Corollary III. of Problem II. to this -new construction. We shall see presently that _AB_ does not represent -the actual length of the inclined line _AB_ in nature, nevertheless we -shall first proceed as if it did, and modify our result afterwards. - -In Fig. 77. draw _ad_ parallel to _AB_, cutting _BT_ in _d_. - -Therefore _ad_ is the sight-magnitude of _AB_, as _aR_ is of _AB_ in -Fig. 11. - - [Illustration: Fig. 78.] - -Remove again from the figure all lines except _PV_, _VT_, _PT_, _ab_, -_ad_, and the measuring-line. - -Set off on the measuring-line _am_ equal to _ad_. - -Draw _PQ_ parallel to _am_, and through _b_ draw _mQ_, cutting _PQ_ in -_Q_. - -Then, by the proof already given in page 20, _PQ_ = _PT_. - -Therefore if _P_ is the vanishing-point of an inclined line _AB_, and -_QP_ is a horizontal line drawn through it, make _PQ_ equal to _PT_, -and _am_ on the measuring-line equal to the sight-magnitude of the -line _AB_ _in the diagram_, and the line joining _mQ_ will cut _aP_ in -_b_. - - -We have now, therefore, to consider what relation the length of the -line _AB_ in this diagram, Fig. 77., has to the length of the line -_AB_ in reality. - -Now the line _AE_ in Fig. 77. represents the length of _AE_ in -reality. - -But the angle _AEB_, Fig. 77., and the corresponding angle in all the -constructions of the earlier problems, is in reality a right angle, -though in the diagram necessarily represented as obtuse. - - [Illustration: Fig. 79.] - -Therefore, if from _E_ we draw _EC_, as in Fig. 79., at right angles -to _AE_, make _EC_ = _EB_, and join _AC_, _AC_ will be the real length -of the line _AB_. - -Now, therefore, if instead of _am_ in Fig. 78., we take the real -length of _AB_, that real length will be to _am_ as _AC_ to _AB_ in -Fig. 79. - -And then, if the line drawn to the measuring-line _PQ_ is still to cut -_aP_ in _b_, it is evident that the line _PQ_ must be shortened in the -same ratio that _am_ was shortened; and the true dividing-point will -be _Q′_ in Fig. 80., fixed so that _Q′P′_ shall be to _QP_ as _am′_ is -to _am_; _am′_ representing the real length of _AB_. - -But _am′_is therefore to _am_ as _AC_ is to _AB_ in Fig. 79. - -Therefore _PQ′_ must be to _PQ_ as _AC_ is to _AB_. - -But _PQ_ equals _PT_ (Fig. 78.); and _PV_ is to _VT_ (in Fig. 78.) as -_BE_ is to _AE_ (Fig. 79.). - -Hence we have only to substitute _PV_ for _EC_, and _VT_ for _AE_, in -Fig. 79., and the resulting diagonal _AC_ will be the required length -of _PQ′_. - - [Illustration: Fig. 80.] - -It will be seen that the construction given in the text (Fig. 46.) is -the simplest means of obtaining this magnitude, for _VD_ in Fig. 46. -(or _VM_ in Fig. 15.) = _VT_ by construction in Problem IV. It should, -however, be observed, that the distance _PQ′_ or _PX_, in Fig. 46., -may be laid on the sight-line of the inclined plane itself, if the -measuring-line be drawn parallel to that sight-line. And thus any form -may be drawn on an inclined plane as conveniently as on a horizontal -one, with the single exception of the radiation of the verticals, -which have a vanishing-point, as shown in Problem XX. - - -THE END. - - - -Transcriber’s Note - -A handful of unequivocal typographical errors has been corrected. - -For increased clarity, a few diagrams have been shifted from their -original position in the text. - - - - - -End of Project Gutenberg's The Elements of Perspective, by John Ruskin - -*** END OF THIS PROJECT GUTENBERG EBOOK THE ELEMENTS OF PERSPECTIVE *** - -***** This file should be named 60816-0.txt or 60816-0.zip ***** -This and all associated files of various formats will be found in: - http://www.gutenberg.org/6/0/8/1/60816/ - -Produced by Juliet Sutherland, David Wilson and the Online -Distributed Proofreading Team at http://www.pgdp.net - -Updated editions will replace the previous one--the old editions will -be renamed. - -Creating the works from print editions not protected by U.S. copyright -law means that no one owns a United States copyright in these works, -so the Foundation (and you!) can copy and distribute it in the United -States without permission and without paying copyright -royalties. 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} - p.illolt {float: left; } - .ADE { - display: block; - visibility: visible; - } - div.ADE {padding-top: 3em; } - .ADE li { - padding-top: 1.5em; - border-bottom: thin solid silver; - } - .eqnlink { - display: inline; - visibility: visible; - font-size: small; - padding-left: 0.5em; - } - .eqnimage { - text-align: right; - display: inline-block; - text-indent: 0; - width: 4em; - padding-right: 2em; - } -} -@media screen {/* effectively an override for Kindle */ - .ADE, .eqnlink { - display: none; - visibility: hidden; - } -} - - /* XML end ]]>*/ - -</style> -</head> -<body> - - -<pre> - -The Project Gutenberg EBook of The Elements of Perspective, by John Ruskin - -This eBook is for the use of anyone anywhere in the United States and most -other parts of the world at no cost and with almost no restrictions -whatsoever. You may copy it, give it away or re-use it under the terms of -the Project Gutenberg License included with this eBook or online at -www.gutenberg.org. If you are not located in the United States, you'll have -to check the laws of the country where you are located before using this ebook. - -Title: The Elements of Perspective - arranged for the use of schools and intended to be read - in connection with the first three books of Euclid - -Author: John Ruskin - -Release Date: November 30, 2019 [EBook #60816] - -Language: English - -Character set encoding: UTF-8 - -*** START OF THIS PROJECT GUTENBERG EBOOK THE ELEMENTS OF PERSPECTIVE *** - - - - -Produced by Juliet Sutherland, David Wilson and the Online -Distributed Proofreading Team at http://www.pgdp.net - - - - - - -</pre> - - - - -<hr class="ww" /> -<div class="tnotefonts"> -<h2>Transcriber’s Note</h2> -<p>This EBook requires support for the following -Unicode characters: -<span class="maths">∠</span> angle, -<span class="maths">∶</span> ratio, -<span class="maths">∷</span> proportion, -<span class="maths">∴</span> therefore, -<span class="maths">′</span> prime, and -<span class="maths">″</span> double prime.</p> -<!-- note the images at the end of the book for epub, for instances of - ∶ ratio and ∷ proportion, these being the most likely glyphs to be missing --> -<p class="ADE">Roman-numbered footnotes <a title="See image" - href="#eqn1" class="eqnlink">[i]</a> link to images of text where the ratio and proportion symbols occur.</p> -</div> - -<div class="titlepage"> - -<p class="edition"><a name="png.001" id="png.001" href="#png.001"><span class="pagenum"><span - class="ns">[p</span>i<span class="ns">]<br - /></span></span></a>Library Edition</p> - -<hr /> -<p class="fakeh1">THE COMPLETE WORKS<br - /><small><small>OF</small></small><br - /><big>JOHN RUSKIN</big></p> - -<p class="booklist">ELEMENTS OF DRAWING AND<br - />PERSPECTIVE<br - />THE TWO PATHS<br - />UNTO THIS LAST<br - />MUNERA PULVERIS<br - />SESAME AND LILIES<br - />ETHICS OF THE DUST</p> - -<hr /> - -<p class="pub"><big>NATIONAL LIBRARY ASSOCIATION</big><br -/>NEW YORK CHICAGO</p> - -</div> - -<div class="subtitle"> -<h1 title="The Elements of Perspective"><a name="png.003" id="png.003" href="#png.003"><span class="pagenum"><span - class="ns">[p</span>iii<span class="ns">]<br - /></span></span></a><big>THE ELEMENTS OF PERSPECTIVE</big></h1> - -<p id="arranged">ARRANGED FOR THE USE OF SCHOOLS</p> - -<p id="intended"><small><small>AND INTENDED TO BE READ IN CONNECTION WITH THE<br - />FIRST THREE BOOKS OF EUCLID.</small></small></p> -</div> - - -<div class="toc"> - -<h2 class="pr" title="Contents"><a name="png.005" id="png.005" href="#png.005"><span class="pagenum"><span - class="ns">[p</span>v<span class="ns">]<br - /></span></span></a>CONTENTS.</h2> -<hr class="short" /> - -<table summary="Table of Contents"> -<tr><th class="dots"> </th><th class="pg">PAGE</th></tr> - -<tr><td class="dots"><p class="dotz"><span class="text"><a href="#png.009">Preface</a></span></p></td> - <td class="pg"><a href="#png.009">ix</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text"><a href="#png.011">Introduction</a></span></p></td> - <td class="pg"><a href="#png.011">1</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.020">PROBLEM I.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To fix the Position of a given Point</span></p></td> - <td class="pg"><a href="#png.020">10</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.023">PROBLEM II.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To draw a Right Line between two given Points</span></p></td> - <td class="pg"><a href="#png.023">13</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.027">PROBLEM III.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To find the Vanishing-Point of a given Horizontal Line</span></p></td> - <td class="pg"><a href="#png.027">17</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.033">PROBLEM IV.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To find the Dividing-Points of a given Horizontal Line</span></p></td> - <td class="pg"><a href="#png.033">23</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.034">PROBLEM V.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To draw a Horizontal Line, given in Position and Magnitude, -by means of its Sight-Magnitude and Dividing-Points</span></p></td> - <td class="pg"><a href="#png.034">24</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.037">PROBLEM VI.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To draw any Triangle, given in Position and Magnitude, in a -Horizontal Plane</span></p></td> - <td class="pg"><a href="#png.037">27</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.039">PROBLEM VII.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To draw any Rectilinear Quadrilateral Figure, given in -Position and Magnitude, in a Horizontal Plane</span></p></td> - <td class="pg"><a href="#png.039">29</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.041">PROBLEM VIII.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To draw a Square, given in Position and Magnitude, in a -Horizontal Plane</span></p></td> - <td class="pg"><a href="#png.041">31</a></td></tr> - -<tr><td class="chap" colspan="2"><a name="png.006" id="png.006" href="#png.006"><span class="pagenum"><span - class="ns">[p</span>vi<span class="ns">]<br - /></span></span></a><a href="#png.044">PROBLEM IX.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To draw a Square Pillar, given in Position and Magnitude, -its Base and Top being in Horizontal Planes</span></p></td> - <td class="pg"><a href="#png.044">34</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.046">PROBLEM X.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To draw a Pyramid, given in Position and Magnitude, on a -Square Base in a Horizontal Plane</span></p></td> - <td class="pg"><a href="#png.046">36</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.048">PROBLEM XI.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To draw any Curve in a Horizontal or Vertical Plane</span></p></td> - <td class="pg"><a href="#png.048">38</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.052">PROBLEM XII.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To divide a Circle drawn in Perspective into any given -Number of <span class="nw">Equal Parts</span></span></p></td> - <td class="pg"><a href="#png.052">42</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.055">PROBLEM XIII.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To draw a Square, given in Magnitude, within a larger -Square given in Position and Magnitude; the Sides of the -two Squares being Parallel</span></p></td> - <td class="pg"><a href="#png.055">45</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.057">PROBLEM XIV.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To draw a Truncated Circular Cone, given in Position and -Magnitude, the Truncations being in Horizontal Planes, -and the Axis of the Cone vertical</span></p></td> - <td class="pg"><a href="#png.057">47</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.060">PROBLEM XV.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To draw an Inclined Line, given in Position and Magnitude</span></p></td> - <td class="pg"><a href="#png.060">50</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.063">PROBLEM XVI.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To find the Vanishing-Point of a given Inclined Line</span></p></td> - <td class="pg"><a href="#png.063">53</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.065">PROBLEM XVII.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To find the Dividing-Points of a given Inclined Line</span></p></td> - <td class="pg"><a href="#png.065">55</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.067">PROBLEM XVIII.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To find the Sight-Line of an Inclined Plane in which Two -Lines are given in Position</span></p></td> - <td class="pg"><a href="#png.067">57</a></td></tr> - -<tr><td class="chap" colspan="2"><a name="png.007" id="png.007" href="#png.007"><span class="pagenum"><span - class="ns">[p </span>vii<span class="ns">]<br - /></span></span></a><a href="#png.069">PROBLEM XIX.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To find the Vanishing-Point of Steepest Lines in an Inclined -Plane whose Sight-Line <span class="nw">is given</span></span></p></td> - <td class="pg"><a href="#png.069">59</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.071">PROBLEM XX.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">To find the Vanishing-Point of Lines perpendicular to the -Surface of a given Inclined Plane</span></p></td> - <td class="pg"><a href="#png.071">61</a></td></tr> - - -<tr><td class="appen" colspan="2"><hr class="short" />APPENDIX.</td></tr> - -<tr><td class="chap" colspan="2"><a href="#png.079">I.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">Practice and Observations on the preceding Problems</span></p></td> - <td class="pg"><a href="#png.079">69</a></td></tr> - - -<tr><td class="chap" colspan="2"><a href="#png.109">II.</a></td></tr> - -<tr><td class="dots"><p class="dotz"><span class="text">Demonstrations which could not conveniently be included in -<span class="nw">the Text</span></span></p></td> - <td class="pg"><a href="#png.109">99</a></td></tr> -</table> - - -</div> - -<div class="preface"> - -<h2 class="pr" title="Preface"><a name="png.009" id="png.009" href="#png.009"><span class="pagenum"><span - class="ns">[p </span>ix<span class="ns">]<br - /></span></span></a>PREFACE.</h2> - - -<p><span class="smc">For</span> some time back I have felt the want, among Students -of Drawing, of a written code of accurate Perspective Law; -the modes of construction in common use being various, and, -for some problems, insufficient. It would have been desirable -to draw up such a code in popular language, so as to do -away with the most repulsive difficulties of the subject; -but finding this popularization would be impossible, without -elaborate figures and long explanations, such as I had no -leisure to prepare, I have arranged the necessary rules in a -short mathematical form, which any schoolboy may read -through in a few days, after he has mastered the first three -and the sixth books of Euclid.</p> - -<p>Some awkward compromises have been admitted between -the first-attempted popular explanation, and the severer -arrangement, involving irregular lettering and redundant -phraseology; but I cannot for the present do more, and leave -the book therefore to its trial, hoping that, if it be found by -masters of schools to answer its purpose, I may hereafter -bring it into better form.<a name="fn1" id="fn1"></a><a title="Go to footnote 1" - href="#Footnote1" class="fnanchor"><span - class="ns">[Footnote </span>1<span class="ns">] - </span></a></p> - -<p>An account of practical methods, sufficient for general -purposes of sketching, might indeed have been set down in -<a name="png.010" id="png.010" href="#png.010"><span class="pagenum"><span - class="ns">[p</span>x<span class="ns">] - </span></span></a>much less space: but if the student reads the following pages -carefully, he will not only find himself able, on occasion, -to solve perspective problems of a complexity greater than -the ordinary rules will reach, but obtain a clue to many -important laws of pictorial effect, no less than of outline. -The subject thus examined becomes, at least to my mind, -very curious and interesting; but, for students who are -unable or unwilling to take it up in this abstract form, I -believe good help will be soon furnished, in a series of illustrations -of practical perspective now in preparation by Mr. Le Vengeur. I have not seen this essay in an advanced -state, but the illustrations shown to me were very clear and -good; and, as the author has devoted much thought to their -arrangement, I hope that his work will be precisely what -is wanted by the general learner.</p> - -<p>Students wishing to pursue the subject into its more -extended branches will find, I believe, Cloquet’s treatise the -best hitherto published.<a name="fn2" id="fn2"></a><a title="Go to footnote 2" - href="#Footnote2" class="fnanchor"><span - class="ns">[Footnote </span>2<span class="ns">] - </span></a></p> - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote1" id="Footnote1"><span class="ns">[Footnote </span - >1<span class="ns">: </span></a> </span>Some irregularities of arrangement have been admitted merely for -the sake of convenient reference; the eighth problem, for instance, -ought to have been given as a case of the seventh, but is separately -enunciated on account of its importance.</small></p> - -<p class="ctd"><small>Several constructions, which ought to have been given as problems, -are on the contrary given as corollaries, in order to keep the more -directly connected problems in closer sequence; thus the construction -of rectangles and polygons in vertical planes would appear by the <a href="#png.005">Table -of Contents</a> to have been omitted, being given in the <a href="#png.045">corollary to Problem IX</a>.<span class="ns">]</span> - <a title="Return to text" href="#fn1" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote2" id="Footnote2"><span class="ns">[Footnote </span - >2<span class="ns">: </span></a> </span><cite>Nouveau Traité Élémentaire de Perspective. Bachelier, 1823.</cite><span class="ns">]</span> - <a title="Return to text" href="#fn2" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - -</div> - -<div class="chap"> -<h2 class="fakeh1" title="Introduction"><a name="png.011" id="png.011" href="#png.011"><span class="pagenum"><span - class="ns">[p</span>1<span class="ns">]<br - /></span></span></a><small>THE</small><br - />ELEMENTS OF PERSPECTIVE.</h2> - -<hr class="short" /> - - -<p class="fakeh2">INTRODUCTION.</p> - - -<p><span class="smc">When</span> you begin to read this book, sit down very near -the window, and shut the window. I hope the view out of -it is pretty; but, whatever the view may be, we shall find -enough in it for an illustration of the first principles of -perspective (or, literally, of “looking through”).</p> - -<p>Every pane of your window may be considered, if you -choose, as a glass picture; and what you see through it, as -painted on its surface.</p> - -<p>And if, holding your head still, you extend your hand to -the glass, you may, with a brush full of any thick color, trace, -roughly, the lines of the landscape on the glass.</p> - -<p>But, to do this, you must hold your head very still. Not -only you must not move it sideways, nor up and down, but -it must not even move backwards or forwards; for, if you -move your head forwards, you will see <em>more</em> of the landscape -through the pane; and, if you move it backwards, you will -see <em>less</em>: or considering the pane of glass as a picture, when -you hold your head near it, the objects are painted small, -and a great many of them go into a little space; but, when -you hold your head some distance back, the objects are -painted larger upon the pane, and fewer of them go into the -field of it.</p> - -<p>But, besides holding your head still, you must, when you -try to trace the picture on the glass, shut one of your eyes. -If you do not, the point of the brush appears double; and, -<a name="png.012" id="png.012" href="#png.012"><span class="pagenum"><span - class="ns">[p</span>2<span class="ns">] - </span></span></a>on farther experiment, you will observe that each of your -eyes sees the object in a different place on the glass, so that -the tracing which is true to the sight of the right eye is a -couple of inches (or more, according to your distance from -the pane,) to the left of that which is true to the sight of -the left.</p> - -<p>Thus, it is only possible to draw what you see through -the window rightly on the surface of the glass, by fixing one -eye at a given point, and neither moving it to the right nor -left, nor up nor down, nor backwards nor forwards. Every -picture drawn in true perspective may be considered as an -upright piece of glass,<a name="fn3" id="fn3"></a><a title="Go to footnote 3" - href="#Footnote3" class="fnanchor"><span - class="ns">[Footnote </span>3<span class="ns">] - </span></a> on which the objects seen through -it have been thus drawn. Perspective can, therefore, only be -quite right, by being calculated for one fixed position of the -eye of the observer; nor will it ever appear <em>deceptively</em> right -unless seen precisely from the point it is calculated for. -Custom, however, enables us to feel the rightness of the work -on using both our eyes, and to be satisfied with it, even -when we stand at some distance from the point it is -designed for.</p> - -<p>Supposing that, instead of a window, an unbroken plate -of crystal extended itself to the right and left of you, and -high in front, and that you had a brush as long as you -wanted (a mile long, suppose), and could paint with such -a brush, then the clouds high up, nearly over your head, -and the landscape far away to the right and left, might be -traced, and painted, on this enormous crystal field.<a name="fn4" id="fn4"></a><a title="Go to footnote 4" - href="#Footnote4" class="fnanchor"><span - class="ns">[Footnote </span>4<span class="ns">] - </span></a> But -if the field were so vast (suppose a mile high and a mile -wide), certainly, after the picture was done, you would not -stand as near to it, to see it, as you are now sitting near to -your window. In order to trace the upper clouds through -your great glass, you would have had to stretch your neck -<a name="png.013" id="png.013" href="#png.013"><span class="pagenum"><span - class="ns">[p</span>3<span class="ns">] - </span></span></a>quite back, and nobody likes to bend their neck back to see -the top of a picture. So you would walk a long way back -to see the great picture—a quarter of a mile, perhaps,—and -then all the perspective would be wrong, and would -look quite distorted, and you would discover that you ought -to have painted it from the greater distance, if you meant -to look at it from that distance. Thus, the distance at which -you intend the observer to stand from a picture, and for -which you calculate the perspective, ought to regulate to -a certain degree the size of the picture. If you place the -point of observation near the canvas, you should not make -the picture very large: <em>vice versâ<!-- TN: circumflex invisible --></em>, if you place the point of -observation far from the canvas, you should not make it -very small; the fixing, therefore, of this point of observation -determines, as a matter of convenience, within certain limits, -the size of your picture. But it does not determine this size -by any perspective law; and it is a mistake made by many -writers on perspective, to connect some of their rules definitely -with the size of the picture. For, suppose that you -had what you now see through your window painted actually -upon its surface, it would be quite optional to cut out any -piece you chose, with the piece of the landscape that was -painted on it. You might have only half a pane, with a single -tree; or a whole pane, with two trees and a cottage; or two -panes, with the whole farmyard and pond; or four panes, -with farmyard, pond, and foreground. And any of these -pieces, if the landscape upon them were, as a scene, pleasantly -composed, would be agreeable pictures, though of quite -different sizes; and yet they would be all calculated for the -same distance of observation.</p> - -<p>In the following treatise, therefore, I keep the size of -the picture entirely undetermined. I consider the field of -canvas as wholly unlimited, and on that condition determine -the perspective laws. After we know how to apply those -laws without limitation, we shall see what limitations of the -size of the picture their results may render advisable.</p> - -<p>But although the size of the <em>picture</em> is thus independent -<a name="png.014" id="png.014" href="#png.014"><span class="pagenum"><span - class="ns">[p</span>4<span class="ns">] - </span></span></a>of the observer’s distance, the size of the <em>object represented</em> in -the picture is not. On the contrary, that size is fixed by -absolute mathematical law; that is to say, supposing you have -to draw a tower a hundred feet high, and a quarter of a mile -distant from you, the height which you ought to give that -tower on your paper depends, with mathematical precision, -on the distance at which you intend your paper to be placed. -So, also, do all the rules for drawing the form of the tower, -whatever it may be.</p> - -<p>Hence, the first thing to be done in beginning a drawing -is to fix, at your choice, this distance of observation, or the -distance at which you mean to stand from your paper. After -that is determined, all is determined, except only the ultimate -size of your picture, which you may make greater, or -less, not by altering the size of the things represented, but -by <em>taking in more, or fewer</em> of them. So, then, before proceeding -to apply any practical perspective rule, we must -always have our distance of observation marked, and the -most convenient way of marking it is the following:</p> - - -<p class="illo">PLACING OF THE SIGHT-POINT, SIGHT-LINE, STATION-POINT, -AND STATION-LINE.<br - /><img id="f.1" src="images/illus-014.png" alt="[Geometric diagram]" /><br - /><b>Fig. 1.</b></p> - - -<h3 class="ss" title="I. The Sight-Point">I. The Sight-Point.—</h3><p>Let <span class="maths">A B C D</span>, <a href="#f.1">Fig. 1.</a>, be your sheet -<a name="png.015" id="png.015" href="#png.015"><span class="pagenum"><span - class="ns">[p</span>5<span class="ns">] - </span></span></a>of paper, the larger the better, though perhaps we may cut -out of it at last only a small piece for our picture, such as the -dotted circle <span class="maths">N O P Q</span>. This circle is not intended to limit -either the size or shape of our picture: you may ultimately -have it round or oval, horizontal or upright, small or large, -as you choose. I only dot the line to give you an idea of -whereabouts you will probably like to have it; and, as the -operations of perspective are more conveniently performed -upon paper underneath the picture than above it, I put this -conjectural circle at the top of the paper, about the middle of -it, leaving plenty of paper on both sides and at the bottom. -Now, as an observer generally stands near the middle of a -picture to look at it, we had better at first, and for simplicity’s -sake, fix the point of observation opposite the middle of -our conjectural picture. So take the point <span class="maths">S</span>, the center -of the circle <span class="maths">N O P Q</span>;—or, which will be simpler for you in -your own work, take the point <span class="maths">S</span> at random near the top of -your paper, and strike the circle <span class="maths">N O P Q</span> round it, any size -you like. Then the point <span class="maths">S</span><!-- TN: original has normal roman S --> is to represent the point <em>opposite</em> -which you wish the observer of your picture to place his eye, -in looking at it. Call this point the “Sight-Point.”</p><p class="endrunin"> </p> - - -<h3 class="ss" title="II. The Sight-Line">II. The Sight-Line.—</h3><p>Through the Sight-point, <span class="maths">S</span>, draw -a horizontal line, <span class="maths">G H</span>, right across your paper from side to -side, and call this line the “Sight-Line.”</p><p class="endrunin"> </p> - -<p>This line is of great practical use, representing the level -of the eye of the observer all through the picture. You will -find hereafter that if there is a horizon to be represented in -your picture, as of distant sea or plain, this line defines it.</p> - - -<h3 class="ss" title="III. The Station-Line">III. The Station-Line.—</h3><p>From <span class="maths">S</span> let fall a perpendicular -line, <span class="maths">S R</span>, to the bottom of the paper, and call this line the -“Station-Line.”</p><p class="endrunin"> </p> - -<p>This represents the line on which the observer stands, at a -greater or less distance from the picture; and it ought to be -<em>imagined</em> as drawn right out from the paper at the point <span class="maths">S</span>. -Hold your paper upright in front of you, and hold your pencil -horizontally, with its point against the point <span class="maths">S</span>, as if you -<a name="png.016" id="png.016" href="#png.016"><span class="pagenum"><span - class="ns">[p</span>6<span class="ns">] - </span></span></a>wanted to run it through the paper there, and the pencil will -represent the direction in which the line <span class="maths">S R</span> ought to be -drawn. But as all the measurements which we have to set -upon this line, and operations which we have to perform with -it, are just the same when it is drawn on the paper itself, -below <span class="maths">S</span>, as they would be if it were represented by a wire in -the position of the leveled pencil, and as they are much more -easily performed when it is drawn on the paper, it is always -in practice, so drawn.</p> - - -<h3 class="ss" title="IV. The Station-Point">IV. The Station-Point.—</h3><p>On this line, mark the distance -<span class="maths">S T</span> at your pleasure, for the distance at which you wish -your picture to be seen, and call the point <span class="maths">T</span><!-- TN: original has normal T --> the “Station-Point.”</p> - -<p class="illo"><img id="f.2" src="images/illus-016.png" alt="[Geometric diagram]" /><br - /><b>Fig. 2.</b></p> - -<p>In practice, it is generally advisable to make the distance -<span class="maths">S T</span> about as great as the diameter of your intended picture; -and it should, for the most part, be more rather than less; -but, as I have just stated, this is quite arbitrary. However, -in this figure, as an approximation to a generally advisable -distance, I make the distance <span class="maths">S T</span> equal to the diameter of the -circle <span class="maths">N O P Q</span>. Now, having fixed this distance, <span class="maths">S T</span>, all the -dimensions of the objects in our picture are fixed likewise, -and for this <span class="nw">reason:—</span></p> - -<p>Let the upright line <span class="maths">A B</span>, <a href="#f.2">Fig. 2.</a>, represent a pane of glass -placed where our picture is to be placed; but seen at the side -<a name="png.017" id="png.017" href="#png.017"><span class="pagenum"><span - class="ns">[p</span>7<span class="ns">] - </span></span></a>of it, edgeways; let <span class="maths">S</span> be the Sight-point; <span class="maths">S T</span> the Station-line, -which, in this figure, observe, is in its true position, drawn -out from the paper, not down upon it; and <span class="maths">T</span> the Station-point.</p> - -<p>Suppose the Station-line <span class="maths">S T</span> to be continued, or in mathematical -language “produced,” through <span class="maths">S</span>, far beyond the pane -of glass, and let <span class="maths">P Q</span> be a tower or other upright object situated -on or above this line.</p> - -<p>Now the <em>apparent</em> height of the tower <span class="maths">P Q</span> is measured by -the angle <span class="maths">Q T P</span>, between the rays of light which come from -the top and bottom of it to the eye of the observer. But the -<em>actual</em> height of the <em>image</em> of the tower on the pane of glass -<span class="maths">A B</span>, between us and it, is the distance <span class="maths">P′ Q′</span> between the -points where the rays traverse the glass.</p> - -<p>Evidently, the farther from the point <span class="maths">T</span> we place the glass, -making <span class="maths">S T</span> longer, the larger will be the image; and the -nearer we place it to <span class="maths">T</span>, the smaller the image, and that in -a fixed ratio. Let the distance <span class="maths">D T</span> be the direct distance from -the Station-point to the foot of the object. Then, if we place -the glass <span class="maths">A B</span> at one-third of that whole distance, <span class="maths">P′ Q′</span> will be -one-third of the real height of the object; if we place the -glass at two-thirds of the distance, as at <span class="maths">E F</span>, <span class="maths">P″ Q″</span> (the -height of the image at that point) will be two-thirds the -height<a name="fn5" id="fn5"></a><a title="Go to footnote 5" - href="#Footnote5" class="fnanchor"><span - class="ns">[Footnote </span>5<span class="ns">] - </span></a> of the object, and so on. Therefore the mathematical -law is that <span class="maths">P′ Q′</span> will be to <span class="maths">P Q</span> as <span class="maths">S T</span> to <span class="maths">D T</span>. I put -this ratio clearly by itself that you may remember it:</p> - -<div class="displaymath"> -<a name="eq1" id="eq1"></a><span class="maths">P′ Q′ ∶ P Q ∷ S T ∶ D T</span><a title="See image" - href="#eqn1" class="eqnlink">[<span class="ns">eqn </span>i]</a> -<p>or in words:</p> -<span class="maths">P</span> dash <span class="maths">Q</span> dash is to <span class="maths">P Q</span> as <span class="maths">S T</span> to <span class="maths">D T</span> -</div> - -<p class="noindent">In which formula, recollect that <span class="maths">P′ Q′</span> is the height of the -appearance of the object on the picture; <span class="maths">P Q</span> the height of the -object itself; <span class="maths">S</span> the Sight-point;<!-- TN: original has colon --> <span class="maths">T</span> the Station-point; <span class="maths">D</span> a point -at the direct distance of the object; though the object is -<a name="png.018" id="png.018" href="#png.018"><span class="pagenum"><span - class="ns">[p</span>8<span class="ns">] - </span></span></a>seldom placed actually on the line <span class="maths">T S</span> produced, and may be -far to the right or left of it, the formula is still the same.</p> - -<p>For let <span class="maths">S</span>, <a href="#f.3">Fig. 3.</a>, be the Sight-point, and <span class="maths">A B</span> the glass—here -seen looking <em>down</em> on its <em>upper edge</em>, not sideways;—then -if the tower (represented now, as on a map, by the dark -square), instead of being at <span class="maths">D</span> on the line <span class="maths">S T</span> produced, be at -<span class="maths">E</span>, to the right (or left) of the spectator, -still the apparent height of -the tower on <span class="maths">A B</span> will be as <span class="maths">S′ T</span> to -<span class="maths">E T</span>, which is the same ratio as that -of <span class="maths">S T</span> to <span class="maths">D T</span>.</p> - -<p class="illolt"><img id="f.3" src="images/illus-018.png" alt="[Geometric diagram]" /><br - /><b>Fig. 3.</b></p> - -<p>Now in many perspective problems, -the position of an object is -more conveniently expressed by the -two measurements <span class="maths">D T</span> and <span class="maths">D E</span>, than -by the single oblique measurement -<span class="maths">E T</span>.</p> - -<p>I shall call <span class="maths">D T</span> the “direct distance” -of the object at <span class="maths">E</span>, and <span class="maths">D E</span> -its “lateral distance.” It is rather -a license to call <span class="maths">D T</span> its “direct” -distance, for <span class="maths">E T</span> is the more direct -of the two; but there is no other term which would not cause -confusion.</p> - -<p>Lastly, in order to complete our knowledge of the position -of an object, the vertical height of some point in it, above or -below the eye, must be given; that is to say, either <span class="maths">D P</span> or <span class="maths">D Q</span> -in <a href="#f.2">Fig. 2.</a><a name="fn6" id="fn6"></a><a title="Go to footnote 6" - href="#Footnote6" class="fnanchor"><span - class="ns">[Footnote </span>6<span class="ns">] - </span></a>: this I shall call the “vertical distance” of the -point given. In all perspective problems these three distances, -and the dimensions of the object, must be stated, -otherwise the problem is imperfectly given. It ought not to -be required of us merely to draw <em>a</em> room or <em>a</em> church in perspective; -but to draw <em>this</em> room from <em>this</em> corner, and <em>that</em> -church on <em>that</em> spot, in perspective. For want of knowing -<a name="png.019" id="png.019" href="#png.019"><span class="pagenum"><span - class="ns">[p</span>9<span class="ns">] - </span></span></a>how to base their drawings on the measurement and place -of the object, I have known practiced students represent a -parish church, certainly in true perspective, but with a nave -about two miles and a half long.</p> - -<p>It is true that in drawing landscapes from nature the sizes -and distances of the objects cannot be accurately known. -When, however, we know how to draw them rightly, if their -size were given, we have only to <em>assume a rational approximation</em> -to their size, and the resulting drawing will be true -enough for all intents and purposes. It does not in the least -matter that we represent a distant cottage as eighteen feet -long, when it is in reality only seventeen; but it matters much -that we do not represent it as eighty feet long, as we easily -might if we had not been accustomed to draw from measurement. -Therefore, in all the following problems the measurement -of the object is given.</p> - -<p>The student must observe, however, that in order to bring -the diagrams into convenient compass, the measurements -assumed are generally very different from any likely to occur -in practice. Thus, in <a href="#f.3">Fig. 3.</a>, the distance <span class="maths">D S</span> would be probably -in practice half a mile or a mile, and the distance <span class="maths">T S</span>, -from the eye of the observer to the paper, only two or three -feet. The mathematical law is however precisely the same, -whatever the proportions; and I use such proportions as are -best calculated to make the diagram clear.</p> - -<p>Now, therefore, the conditions of a perspective problem -are the following:</p> - -<ul> -<li>The Sight-line <span class="maths">G H</span> given, <a href="#f.1">Fig. 1.</a>;</li> -<li>The Sight-point <span class="maths">S</span> given;</li> -<li>The Station-point <span class="maths">T</span> given; and</li> -<li>The three distances of the object,<a name="fn7" id="fn7"></a><a title="Go to footnote 7" - href="#Footnote7" class="fnanchor"><span - class="ns">[Footnote </span>7<span class="ns">] - </span></a> direct, lateral, and - vertical, with its dimensions, given.</li> -</ul> - -<p>The size of the picture, conjecturally limited by the dotted -circle, is to be determined afterwards at our pleasure. On -these conditions I proceed at once to construction.</p> - - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote3" id="Footnote3"><span class="ns">[Footnote </span - >3<span class="ns">: </span></a> </span>If the glass were not upright, but sloping, the objects might still -be drawn through it, but their perspective would then be different. -Perspective, as commonly taught, is always calculated for a vertical -plane of picture.<span class="ns">]</span> - <a title="Return to text" href="#fn3" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote4" id="Footnote4"><span class="ns">[Footnote </span - >4<span class="ns">: </span></a> </span>Supposing it to have no thickness; otherwise the images would -be distorted by refraction.<span class="ns">]</span> - <a title="Return to text" href="#fn4" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote5" id="Footnote5"><span class="ns">[Footnote </span - >5<span class="ns">: </span></a> </span>I say “height” instead of “magnitude,” for a reason stated in -<a href="#png.081">Appendix I.</a>, to which you will soon be referred. Read on here at -present.<span class="ns">]</span> - <a title="Return to text" href="#fn5" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote6" id="Footnote6"><span class="ns">[Footnote </span - >6<span class="ns">: </span></a> </span><span class="maths">P</span> and <span class="maths">Q</span> being points indicative of the place of the tower’s base and -top. In this figure both are above the sight-line; if the tower were -below the spectator both would be below it, and therefore measured -below <span class="maths">D</span>.<span class="ns">]</span> - <a title="Return to text" href="#fn6" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote7" id="Footnote7"><span class="ns">[Footnote </span - >7<span class="ns">: </span></a> </span>More accurately, “the three distances of any point, either in the -object itself, or indicative of its distance.”<span class="ns">]</span> - <a title="Return to text" href="#fn7" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> -</div> - -<div class="chap"> -<h2 class="pr" title="PROBLEM I"><a name="png.020" id="png.020" href="#png.020"><span class="pagenum"><span - class="ns">[p</span>10<span class="ns">]<br - /></span></span></a><a name="pr.i" id="pr.i">PROBLEM I.</a></h2> - -<h3 class="pr" title="To fix the position of a given point">TO FIX THE POSITION OF <span class="nw">A GIVEN POINT.<a name="fn8" id="fn8"></a><a title="Go to footnote 8" - href="#Footnote8" class="fnanchor"><span - class="ns">[Footnote </span>8<span class="ns">] - </span></a></span></h3> - - -<p><span class="smc">Let</span> <span class="maths">P</span>, <a href="#f.4">Fig. 4.</a>, be the given point.</p> - -<p class="illo"><img id="f.4" src="images/illus-020.png" alt="[Geometric diagram]" /><br - /><b>Fig. 4.</b></p> - -<p>Let its direct distance be <span class="maths">D T</span>; its lateral distance to the left, -<span class="maths">D C</span>; and vertical distance <em>beneath</em> the eye of the observer, <span class="maths">C P</span>.</p> - -<p>[Let <span class="maths">G H</span> be the Sight-line, <span class="maths">S</span> the Sight-point, and <span class="maths">T</span> the -Station-point.]<a name="fn9" id="fn9"></a><a title="Go to footnote 9" - href="#Footnote9" class="fnanchor"><span - class="ns">[Footnote </span>9<span class="ns">] - </span></a></p> - -<p>It is required to fix on the plane of the picture the position -of the point <span class="maths">P</span>.</p> - -<p><a name="png.021" id="png.021" href="#png.021"><span class="pagenum"><span - class="ns">[p</span>11<span class="ns">]<br - /></span></span></a>Arrange the three distances of the object on your paper, -as in <a href="#f.4">Fig. 4.</a><a name="fn10" id="fn10"></a><a title="Go to footnote 10" - href="#Footnote10" class="fnanchor"><span - class="ns">[Footnote </span>10<span class="ns">] - </span></a></p> - -<p>Join <span class="maths">C T</span>, cutting <span class="maths">G H</span> in <span class="maths">Q</span>.</p> - -<p>From <span class="maths">Q</span> let fall the vertical line <span class="maths">Q P′</span>.</p> - -<p>Join <span class="maths">P T</span>, cutting <span class="maths">Q P</span> in <span class="maths">P′</span>.</p> - -<p><span class="maths">P′</span> is the point required.</p> - -<p>If the point <span class="maths">P</span> is <em>above</em> the eye of the observer instead of -below it, <span class="maths">C P</span> is to be measured upwards from <span class="maths">C</span>, and <span class="maths">Q P′</span> -drawn upwards from <span class="maths">Q</span>. The construction will be as in <a href="#f.5">Fig. 5.</a></p> - -<p class="illo"><img id="f.5" src="images/illus-021.png" alt="[Geometric diagram]" /><br - /><b>Fig. 5.</b></p> - -<p>And if the point <span class="maths">P</span> is to the right instead of the left of the -<a name="png.022" id="png.022" href="#png.022"><span class="pagenum"><span - class="ns">[p</span>12<span class="ns">] - </span></span></a>observer, <span class="maths">D C</span> is to be measured to the right instead of the -left.</p> - -<p>The figures <a href="#f.4">4.</a> and <a href="#f.5">5.</a>, looked at in a mirror, will show the -construction of each, on that supposition.</p> - -<p>Now read very carefully the examples and notes to this -problem in Appendix I. (<a href="#png.079">page 69</a>). I have put them in the -Appendix in order to keep the sequence of following problems -more clearly traceable here in the text; but you must -read the first Appendix before going on.</p> - - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote8" id="Footnote8"><span class="ns">[Footnote </span - >8<span class="ns">: </span></a> </span>More accurately, “To fix on the plane of the picture the apparent -position of a point given in actual position.” In the headings of all -the following problems the words “on the plane of the picture” are to -be understood after the words “to draw.” The plane of the picture -means a surface extended indefinitely in the direction of the picture.<span class="ns">]</span> - <a title="Return to text" href="#fn8" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote9" id="Footnote9"><span class="ns">[Footnote </span - >9<span class="ns">: </span></a> </span>The sentence within brackets will not be repeated in succeeding -statements of problems. It is always to be understood.<span class="ns">]</span> - <a title="Return to text" href="#fn9" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote10" id="Footnote10"><span class="ns">[Footnote </span - >10<span class="ns">: </span></a> </span>In order to be able to do this, you must assume the distances to be -small; as in the case of some object on the table: how large distances -are to be treated you will see presently; the mathematical principle, -being the same for all, is best illustrated first on a small scale. Suppose, -for instance, <span class="maths">P</span> to be the corner of a book on the table, seven -inches below the eye, five inches to the left of it, and a foot and a -half in advance of it, and that you mean to hold your finished drawing -at six inches from the eye; then <span class="maths">T S</span> will be six inches, <span class="maths">T D</span> a foot and -a half, <span class="maths">D C</span> five inches, and <span class="maths">C P</span> seven.<span class="ns">]</span> - <a title="Return to text" href="#fn10" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM II"><a name="png.023" id="png.023" href="#png.023"><span class="pagenum"><span - class="ns">[p</span>13<span class="ns">]<br - /></span></span></a><a name="pr.ii" id="pr.ii">PROBLEM II.</a></h2> - -<h3 class="pr" title="To draw a right line between two given points">TO DRAW A RIGHT LINE BETWEEN <span class="nw">TWO GIVEN POINTS</span>.</h3> - - -<p class="illo"><img id="f.6" src="images/illus-023.png" alt="[Geometric diagram]" /><br - /><b>Fig. 6.</b></p> - -<p><span class="smc">Let</span> <span class="maths">A B</span>, <a href="#f.6">Fig. 6.</a>, be the given right line, joining the given -points <span class="maths">A</span> and <span class="maths">B</span>.</p> - -<p>Let the direct, lateral, and vertical distances of the point -<span class="maths">A</span> be <span class="maths">T D</span>, <span class="maths">D C</span>, and <span class="maths">C A</span>.</p> - -<p>Let the direct, lateral, and vertical distances of the point -<span class="maths">B</span> be <span class="maths">T D′</span>, <span class="maths">D C′</span>, and <span class="maths">C′ B</span>.</p> - -<p>Then, by <a href="#pr.i">Problem I.</a>, the position of the point <span class="maths">A</span> on the -plane of the picture is <var>a</var>.</p> - -<p>And similarly, the position of the point <span class="maths">B</span> on the plane of -the picture is <var>b</var>.</p> - -<p>Join <var>a b</var>.</p> - -<p>Then <var>a b</var> is the line required.</p> - - -<h3 class="cor" title="Corollary I"><a name="png.024" id="png.024" href="#png.024"><span class="pagenum"><span - class="ns">[p</span>14<span class="ns">]<br - /></span></span></a><a name="cor.ii.i" id="cor.ii.i">COROLLARY I.</a></h3> - -<p>If the line <span class="maths">A B</span> is in a plane parallel to that of the picture, -one end of the line <span class="maths">A B</span> must be at the same direct distance -from the eye of the observer as the other.</p> - -<p>Therefore, in that case, <span class="maths">D T</span> is equal to <span class="maths">D′ T</span>.</p> - -<p>Then the construction will be as in <a href="#f.7">Fig. 7.</a>; and the student -will find experimentally that <var>a b</var> is now parallel to <span class="maths">A B</span>.<a name="fn11" id="fn11"></a><a title="Go to footnote 11" - href="#Footnote11" class="fnanchor"><span - class="ns">[Footnote </span>11<span class="ns">] - </span></a></p> - -<p class="illo"><img id="f.7" src="images/illus-024.png" alt="[Geometric diagram]" /><br - /><b>Fig. 7.</b></p> - -<p>And that <var>a b</var> is to <span class="maths">A B</span> as <span class="maths">T S</span> is to <span class="maths">T D</span>.</p> - -<p>Therefore, to draw any line in a plane parallel to that of -the picture, we have only to fix the position of one of its -extremities, <var>a</var> or <var>b</var>, and then to draw from <var>a</var> or <var>b</var> a line parallel -to the given line, bearing the proportion to it that <span class="maths">T S</span> -bears to <span class="maths">T D</span>.</p> - - -<h3 class="cor" title="Corollary II"><a name="png.025" id="png.025" href="#png.025"><span class="pagenum"><span - class="ns">[p</span>15<span class="ns">]<br - /></span></span></a><a name="cor.ii.ii">COROLLARY II.</a></h3> - -<p>If the line <span class="maths">A B</span> is in a horizontal plane, the vertical distance -of one of its extremities must be the same as that of the other.</p> - -<p>Therefore, in that case, <span class="maths">A C</span> equals <span class="maths">B C′</span> (<a href="#f.6">Fig. 6.</a>).</p> - -<p>And the construction is as in <a href="#f.8">Fig. 8.</a></p> - -<p class="illo"><img id="f.8" src="images/illus-025.png" alt="[Geometric diagram]" /><br - /><b>Fig. 8.</b></p> - -<p>In <a href="#f.8">Fig. 8.</a> produce <var>a b</var> to the sight-line, cutting the sight-line -in <span class="maths">V</span>; the point <span class="maths">V</span>, thus determined, is called the <span class="smc">Vanishing-Point</span> -of the line <span class="maths">A B</span>.</p> - -<p>Join <span class="maths">T V</span>. Then the student will find experimentally that -<span class="maths">T V</span> is parallel to <span class="maths">A B</span>.<a name="fn12" id="fn12"></a><a title="Go to footnote 12" - href="#Footnote12" class="fnanchor"><span - class="ns">[Footnote </span>12<span class="ns">] - </span></a></p> - - -<h3 class="cor" title="Corollary III"><a name="png.026" id="png.026" href="#png.026"><span class="pagenum"><span - class="ns">[p</span>16<span class="ns">]<br - /></span></span></a><a name="cor.ii.iii">COROLLARY III.</a></h3> - -<p>If the line <span class="maths">A B</span> produced would pass through some point -beneath or above the station-point, <span class="maths">C D</span> is to <span class="maths">D T</span> as <span class="maths">C′ D′</span> is to -<span class="maths">D′ T</span>; in which case the point <var>c</var> coincides with the point <var>c′</var>, -and the line <var>a b</var> is vertical.</p> - -<p>Therefore every vertical line in a picture is, or may be, the -perspective representation of a horizontal one which, produced, -would pass beneath the feet or above the head of the -spectator.<a name="fn13" id="fn13"></a><a title="Go to footnote 13" - href="#Footnote13" class="fnanchor"><span - class="ns">[Footnote </span>13<span class="ns">] - </span></a></p> - - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote11" id="Footnote11"><span class="ns">[Footnote </span - >11<span class="ns">: </span></a> </span>For by the construction<a name="eq2" id="eq2"></a> - <span class="maths">A T ∶ <var>a</var> T ∷ B T ∶ <var>b</var> T</span>;<a title="See image" - href="#eqn2" class="eqnlink">[<span class="ns">eqn </span>ii]</a> and therefore the two -triangles <span class="maths">A B T</span>, <span class="maths"><var>a b</var> T</span>, (having a common angle <span class="maths">A T B</span>,) are similar.<span class="ns">]</span> - <a title="Return to text" href="#fn11" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote12" id="Footnote12"><span class="ns">[Footnote </span - >12<span class="ns">: </span></a> </span>The demonstration is in <a href="#png.111">Appendix II. Article I</a>.<span class="ns">]</span> - <a title="Return to text" href="#fn12" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote13" id="Footnote13"><span class="ns">[Footnote </span - >13<span class="ns">: </span></a> </span>The reflection in water of any luminous point or isolated object -(such as the sun or moon) is therefore, in perspective, a vertical line; -since such reflection, if produced, would pass under the feet of the -spectator. Many artists (Claude among the rest) knowing something -of optics, but nothing of perspective, have been led occasionally to draw -such reflections towards a point at the center of the base of the picture.<span class="ns">]</span> - <a title="Return to text" href="#fn13" class="fnreturn" - ><i>Return to text</i></a></small></p> - -</div> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM III"><a name="png.027" id="png.027" href="#png.027"><span class="pagenum"><span - class="ns">[p</span>17<span class="ns">]<br - /></span></span></a><a name="pr.iii" id="pr.iii">PROBLEM III.</a></h2> - -<h3 class="pr" title="To find the vanishing-point of a given horizontal line">TO FIND THE VANISHING-POINT OF A GIVEN <span class="nw">HORIZONTAL LINE</span>.</h3> - - -<p class="illo"><img id="f.9" src="images/illus-027.png" alt="[Geometric diagram]" /><br - /><b>Fig. 9.</b></p> - -<p><span class="smc">Let</span> <span class="maths">A B</span>, <a href="#f.9">Fig. 9.</a>, be the given line.</p> - -<p>From <span class="maths">T</span>, the station-point, draw <span class="maths">T V</span> parallel to <span class="maths">A B</span>, cutting -the sight-line in <span class="maths">V</span>.</p> - -<p><span class="maths">V</span> is the Vanishing-point required.<a name="fn14" id="fn14"></a><a title="Go to footnote 14" - href="#Footnote14" class="fnanchor"><span - class="ns">[Footnote </span>14<span class="ns">] - </span></a></p> - -<h3 class="cor" title="Corollary I"><a name="png.028" id="png.028" href="#png.028"><span class="pagenum"><span - class="ns">[p</span>18<span class="ns">]<br - /></span></span></a><a name="cor.iii.i" id="cor.iii.i">COROLLARY I.</a></h3> - -<p>As, if the point <var>b</var> is first found, <span class="maths">V</span> may be determined by -it, so, if the point <span class="maths">V</span> is first found, <var>b</var> may be determined by it. -For let <span class="maths">A B</span>, <a href="#f.10">Fig. 10.</a>, be the given line, constructed upon the -paper as in <a href="#f.8">Fig. 8.</a>; and let it be required to draw the line -<var>a b</var> without using the point <span class="maths">C′</span>.</p> - -<p class="illo"><img id="f.10" src="images/illus-028.png" alt="[Geometric diagram]" /><br - /><b>Fig. 10.</b></p> - -<p>Find the position of the point <span class="maths">A</span> in <var>a</var>. (<a href="#pr.i">Problem I</a>.)</p> - -<p><a name="png.029" id="png.029" href="#png.029"><span class="pagenum"><span - class="ns">[p</span>19<span class="ns">]<br - /></span></span></a>Find the vanishing-point of <span class="maths">A B</span> in <span class="maths">V</span>. (<a href="#pr.iii">Problem III</a>.)</p> - -<p>Join <span class="maths"><var>a</var> V</span>.</p> - -<p>Join <span class="maths">B T</span>, cutting <span class="maths"><var>a</var> V</span> in <var>b</var>.</p> - -<p>Then <var>a b</var> is the line required.<a name="fn15" id="fn15"></a><a title="Go to footnote 15" - href="#Footnote15" class="fnanchor"><span - class="ns">[Footnote </span>15<span class="ns">] - </span></a></p> - - -<h3 class="cor" title="Corollary II"><a name="cor.iii.ii" id="cor.iii.ii">COROLLARY II.</a></h3> - -<p>We have hitherto proceeded on the supposition that the -given line was small enough, and near enough, to be actually -drawn on our paper of its real size; as in the example given -in Appendix I. We may, however, now deduce a construction -available under all circumstances, whatever may be the -distance and length of the line given.</p> - -<p class="illo"><img id="f.11" src="images/illus-029.png" alt="[Geometric diagram]" /><br - /><b>Fig. 11.</b></p> - -<p>From <a href="#f.8">Fig. 8.</a> remove, for the sake of clearness, the lines -<a name="png.030" id="png.030" href="#png.030"><span class="pagenum"><span - class="ns">[p</span>20<span class="ns">] - </span></span></a><span class="maths">C′ D′</span>, <span class="maths"><var>b</var> V</span>, and <span class="maths">T V</span>; and, taking the figure as here in <a href="#f.11">Fig. 11.</a>, -draw from <var>a</var>, the line <span class="maths"><var>a</var> R</span> parallel to <span class="maths">A B</span>, cutting <span class="maths">B T</span> in <span class="maths">R</span>.</p> - -<p>Then <span class="maths"><var>a</var> R</span> is to <span class="maths">A B</span> as <span class="maths"><var>a</var> T</span> is to <span class="maths">A T</span>.</p> -<p><span class="phantom">Then </span>—<span class="phantom"> is to </span>— as <span class="maths"><var>c</var> T</span> is to <span class="maths">C T</span>.</p> -<p><span class="phantom">Then </span>—<span class="phantom"> is to </span>— as <span class="maths">T S</span> is to <span class="maths">T D</span>.</p> - -<p>That is to say, <span class="maths"><var>a</var> R</span> is the sight-magnitude of <span class="maths">A B</span>.<a name="fn16" id="fn16"></a><a title="Go to footnote 16" - href="#Footnote16" class="fnanchor"><span - class="ns">[Footnote </span>16<span class="ns">] - </span></a></p> - -<p class="illo"><img id="f.12" src="images/illus-030.png" alt="[Geometric diagram]" /><br - /><b>Fig. 12.</b></p> - -<p>Therefore, when the position of the point <span class="maths">A</span> is fixed in <var>a</var>, -as in <a href="#f.12">Fig. 12.</a>, and <span class="maths"><var>a</var> V</span> is drawn to the vanishing-point; if we -draw a line <span class="maths"><var>a</var> R</span> from <var>a</var>, parallel to <span class="maths">A B</span>, and make <span class="maths"><var>a</var> R</span> equal -to the sight-magnitude of <span class="maths">A B</span>, and then join <span class="maths">R T</span>, the line <span class="maths">R T</span> -will cut <span class="maths"><var>a</var> V</span> in <var>b</var>.</p> - -<p>So that, in order to determine the length of <var>a b</var>, we need -not draw the long and distant line <span class="maths">A B</span>, but only <span class="maths"><var>a</var> R</span> parallel -to it, and of its sight-magnitude; which is a great gain, for -the line <span class="maths">A B</span> may be two miles long, and the line <span class="maths"><var>a</var> R</span> perhaps -only two inches.</p> - -<h3 class="cor" title="Corollary III"><a name="png.031" id="png.031" href="#png.031"><span class="pagenum"><span - class="ns">[p</span>21<span class="ns">]<br - /></span></span></a><a name="cor.iii.iii" id="cor.iii.iii">COROLLARY III.</a></h3> - -<p>In <a href="#f.12">Fig. 12.</a>, altering its proportions a little for the sake -of clearness, and putting it as here in <a href="#f.13">Fig. 13.</a>, draw a horizontal -line <span class="maths"><var>a</var> R′</span> and make <span class="maths"><var>a</var> R′</span> equal to <span class="maths"><var>a</var> R</span>.</p> - -<p>Through the points <span class="maths">R</span> and <var>b</var> draw <span class="maths">R′ M</span>, cutting the sight-line -in <span class="maths">M</span>. Join <span class="maths">T V</span>. Now the reader will find experimentally -that <span class="maths">V M</span> is equal to <span class="maths">V T</span>.<a name="fn17" id="fn17"></a><a title="Go to footnote 17" - href="#Footnote17" class="fnanchor"><span - class="ns">[Footnote </span>17<span class="ns">] - </span></a></p> - -<p class="illo"><img id="f.13" src="images/illus-031.png" alt="[Geometric diagram]" /><br - /><b>Fig. 13.</b></p> - -<p>Hence it follows that, if from the vanishing-point <span class="maths">V</span> we -lay off on the sight-line a distance, <span class="maths">V M</span>, equal to <span class="maths">V T</span>; then -draw through <var>a</var> a horizontal line <span class="maths"><var>a</var> R′</span>, make <span class="maths"><var>a</var> R′</span> equal to the -sight-magnitude of <span class="maths">A B</span>, and join <span class="maths">R′ M</span>; the line <span class="maths">R′ M</span> will cut -<span class="maths"><var>a</var> V</span> in <var>b</var>. And this is in practice generally the most convenient -way of obtaining the length of <var>a b</var>.</p> - -<h3 class="cor" title="Corollary IV"><a name="png.032" id="png.032" href="#png.032"><span class="pagenum"><span - class="ns">[p</span>22<span class="ns">]<br - /></span></span></a><a name="cor.iii.iv" id="cor.iii.iv">COROLLARY IV.</a></h3> - -<p>Removing from the preceding figure the unnecessary lines, -and retaining only <span class="maths">R′ M</span> and <span class="maths"><var>a</var> V</span>, as in <a href="#f.14">Fig. 14.</a>, produce the -line <span class="maths"><var>a</var> R′</span> to the other side of <var>a</var>, and make <span class="maths"><var>a</var> X</span> equal to <span class="maths"><var>a</var> R′</span>.</p> - -<p>Join <span class="maths">X <var>b</var></span>, and produce <span class="maths">X <var>b</var></span> to cut the line of sight in <span class="maths">N</span>.</p> - -<p class="illo"><img id="f.14" src="images/illus-032.png" alt="[Geometric diagram]" /><br - /><b>Fig. 14.</b></p> - -<p>Then as <span class="maths">X R′</span> is parallel to <span class="maths">M N</span>, and <span class="maths"><var>a</var> R′</span> is equal to <span class="maths"><var>a</var> X</span>, -<span class="maths">V N</span> must, by similar triangles, be equal to <span class="maths">V M</span> (equal to <span class="maths">V T</span> -in <a href="#f.13">Fig. 13.</a>).</p> - -<p>Therefore, on whichever side of <span class="maths">V</span> we measure the distance -<span class="maths">V T</span>, so as to obtain either the point <span class="maths">M</span>, or the point <span class="maths">N</span>, if we -measure the sight-magnitude <span class="maths"><var>a</var> R′</span> or <span class="maths"><var>a</var> X</span> on the opposite side -of the line <span class="maths"><var>a</var> V</span>, the line joining <span class="maths">R′ M</span> or <span class="maths">X N</span> will equally cut -<span class="maths"><var>a</var> V</span> in <var>b</var>.</p> - -<p>The points <span class="maths">M</span> and <span class="maths">N</span> are called the “<span class="smc">Dividing-Points</span>” of -the original line <span class="maths">A B</span> (<a href="#f.12">Fig. 12.</a>), and we resume the results of -these corollaries in the following three problems.</p> - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote14" id="Footnote14"><span class="ns">[Footnote </span - >14<span class="ns">: </span></a> </span>The student will observe, in practice, that, his paper lying flat -on the table, he has only to draw the line <span class="maths">T V</span> on its horizontal surface, -parallel to the given horizontal line <span class="maths">A B</span>. In theory, the paper should -be vertical, but the station-line <span class="maths">S T</span> horizontal (see its definition above, -<a href="#png.015">page 5</a>); in which case <span class="maths">T V</span>, being drawn parallel to <span class="maths">A B</span>, will be horizontal -also, and still cut the sight-line in <span class="maths">V</span>.</small></p> - -<p class="ctd"><small>The construction will be seen to be founded on the second <a href="#cor.ii.ii">Corollary</a> of -the preceding problem.</small></p> - -<p class="ctd"><small>It is evident that if any other line, as <span class="maths">M N</span> in <a href="#f.9">Fig. 9.</a>, parallel to <span class="maths">A B</span>, -occurs in the picture, the line <span class="maths">T V</span>, drawn from <span class="maths">T</span>, parallel to <span class="maths">M N</span>, to -find the vanishing-point of <span class="maths">M N</span>, will coincide with the line drawn from -<span class="maths">T</span>, parallel to <span class="maths">A B</span>, to find the vanishing-point of <span class="maths">A B</span>.</small></p> - -<p class="ctd"><small>Therefore <span class="maths">A B</span> and <span class="maths">M N</span> will have the same vanishing-point.</small></p> - -<p class="ctd"><small>Therefore all parallel horizontal lines have the same vanishing-point.</small></p> - -<p class="ctd"><small>It will be shown hereafter that all parallel <em>inclined</em> lines also have -the same vanishing-point; the student may here accept the general -conclusion—“<em>All parallel lines have the same vanishing-point.</em>”</small></p> - -<p class="ctd"><small>It is also evident that if <span class="maths">A B</span> is parallel to the plane of the picture, <span class="maths">T V</span> -must be drawn parallel to <span class="maths">G H</span>, and will therefore never cut <span class="maths">G H</span>. The -line <span class="maths">A B</span> has in that case no vanishing-point: it is to be drawn by the -construction given in <a href="#f.7">Fig. 7.</a></small></p> - -<p class="ctd"><small>It is also evident that if <span class="maths">A B</span> is at right angles with the plane of the -picture, <span class="maths">T V</span> will coincide with <span class="maths">T S</span>, and the vanishing-point of <span class="maths">A B</span> will -be the sight-point.<span class="ns">]</span> - <a title="Return to text" href="#fn14" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote15" id="Footnote15"><span class="ns">[Footnote </span - >15<span class="ns">: </span></a> </span>I spare the student the formality of the <em>reductio ad absurdum</em>, which -would be necessary to prove this.<span class="ns">]</span> - <a title="Return to text" href="#fn15" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote16" id="Footnote16"><span class="ns">[Footnote </span - >16<span class="ns">: </span></a> </span>For definition of Sight-Magnitude, see <a href="#png.081">Appendix I</a>. It ought to -have been read before the student comes to this problem; but I refer -to it in case it has not.<span class="ns">]</span> - <a title="Return to text" href="#fn16" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote17" id="Footnote17"><span class="ns">[Footnote </span - >17<span class="ns">: </span></a> </span>The demonstration is in Appendix II. Article II. <a href="#png.111">p. 101.</a><span class="ns">]</span> - <a title="Return to text" href="#fn17" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM IV"><a name="png.033" id="png.033" href="#png.033"><span class="pagenum"><span - class="ns">[p</span>23<span class="ns">]<br - /></span></span></a><a name="pr.iv" id="pr.iv">PROBLEM IV.</a></h2> - -<h3 class="pr" title="To find the dividing-points of a given horizontal line">TO FIND THE DIVIDING-POINTS OF A GIVEN <span class="nw">HORIZONTAL LINE</span>.</h3> - - -<p class="illo"><img id="f.15" src="images/illus-033.png" alt="[Geometric diagram]" /><br - /><b>Fig. 15.</b></p> - -<p><span class="smc">Let</span> the horizontal line <span class="maths">A B</span> (<a href="#f.15">Fig. 15.</a>) be given in position -and magnitude. It is required to find its dividing-points.</p> - -<p>Find the vanishing-point <span class="maths">V</span> of the line <span class="maths">A B</span>.</p> - -<p>With center <span class="maths">V</span> and distance <span class="maths">V T</span>, describe circle cutting the -sight-line in <span class="maths">M</span> and <span class="maths">N</span>.</p> - -<p>Then <span class="maths">M</span> and <span class="maths">N</span> are the dividing-points required.</p> - -<p>In general, only one dividing-point is needed for use with -any vanishing-point, namely, the one nearest <span class="maths">S</span> (in this case -the point <span class="maths">M</span>). But its opposite <span class="maths">N</span>, or both, may be needed -under certain circumstances.</p> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM V"><a name="png.034" id="png.034" href="#png.034"><span class="pagenum"><span - class="ns">[p</span>24<span class="ns">]<br - /></span></span></a><a name="pr.v" id="pr.v">PROBLEM V.</a></h2> - -<h3 class="pr" title="To draw a horizontal line, given in position and magnitude, -by means of its sight-magnitude and dividing-points">TO DRAW A HORIZONTAL LINE, GIVEN IN POSITION AND MAGNITUDE, -BY MEANS OF ITS SIGHT-MAGNITUDE AND <span class="nw">DIVIDING-POINTS</span>.</h3> - - -<p class="illo"><img id="f.16" src="images/illus-034.png" alt="[Geometric diagram]" /><br - /><b>Fig. 16.</b></p> - -<p><span class="smc">Let</span> <span class="maths">A B</span> (<a href="#f.16">Fig. 16.</a>) be the given line.</p> - -<p>Find the position of the point <span class="maths">A</span> in <var>a</var>.</p> - -<p>Find the vanishing-point <span class="maths">V</span>, and most convenient dividing-point -<span class="maths">M</span>, of the line <span class="maths">A B</span>.</p> - -<p>Join <span class="maths"><var>a</var> V</span>.</p> - -<p>Through <var>a</var> draw a horizontal line <var>a b′</var> and make <var>a b′</var> equal -to the sight-magnitude of <span class="maths">A B</span>. Join <span class="maths"><var>b′</var> M</span>, cutting <span class="maths"><var>a</var> V</span> in <var>b</var>.</p> - -<p>Then <var>a b</var> is the line required.</p> - - -<h3 class="cor" title="Corollary I"><a name="png.035" id="png.035" href="#png.035"><span class="pagenum"><span - class="ns">[p</span>25<span class="ns">]<br - /></span></span></a><a name="cor.v.i" id="cor.v.i">COROLLARY I.</a></h3> - -<p class="illo"><img id="f.17" src="images/illus-035.png" alt="[Geometric diagram]" /><br - /><b>Fig. 17.</b></p> - -<p>Supposing it were now required to draw a line <span class="maths">A C</span> (<a href="#f.17">Fig. 17.</a>) twice as long as <span class="maths">A B</span>, it is evident that the sight-magnitude -<var>a c′</var> must be twice as long as the sight-magnitude <var>a b′</var>; -we have, therefore, merely to continue the horizontal line -<var>a b′</var>, make <var>b′ c′</var> equal to <var>a b′</var>, join <span class="maths"><var>c</var> M′</span>, cutting <span class="maths"><var>a</var> V</span> in <var>c</var>, and -<var>a c</var> will be the line required. Similarly, if we have to draw -a line <span class="maths">A D</span>, three times the length of <span class="maths">A B</span>, <var>a d′</var> must be three -times the length of <var>a b′</var>, and, joining <span class="maths"><var>d′</var> M</span>, <var>a d</var> will be the line -required.</p> - -<p>The student will observe that the nearer the portions cut -off, <var>b c</var>, <var>c d</var>, etc., approach the point <span class="maths">V</span>, the smaller they -become; and, whatever lengths may be added to the line <span class="maths">A D</span>, -and successively cut off from <span class="maths"><var>a</var> V</span>, the line <span class="maths"><var>a</var> V</span> will never be -cut off entirely, but the portions cut off will become infinitely -small, and apparently “vanish” as they approach the point -<span class="maths">V</span>; hence this point is called the “vanishing” point.</p> - - -<h3 class="cor" title="Corollary II"><a name="png.036" id="png.036" href="#png.036"><span class="pagenum"><span - class="ns">[p</span>26<span class="ns">]<br - /></span></span></a><a name="cor.v.ii" id="cor.v.ii">COROLLARY II.</a></h3> - -<p>It is evident that if the line <span class="maths">A D</span> had been given originally, -and we had been required to draw it, and divide it into three -equal parts, we should have had only to divide its sight-magnitude, -<var>a d′</var>, into the three equal parts, <var>a b′</var>, <var>b′ c′</var>, and -<var>c′ d′</var>, and then, drawing to <span class="maths">M</span> from <var>b′</var> and <var>c′</var>, the line <var>a d</var> would -have been divided as required in <var>b</var> and <var>c</var>. And supposing the -original line <span class="maths">A D</span> be divided <em>irregularly into any number</em> of -parts, if the line <var>a d′</var> be divided into a similar number in the -same proportions (by the construction given in Appendix I.), -and, from these points of division, lines are drawn to <span class="maths">M</span>, -they will divide the line <var>a d</var> in true perspective into a similar -number of proportionate parts.</p> - -<p>The horizontal line drawn through <var>a</var>, on which the sight-magnitudes -are measured, is called the “<span class="smc">Measuring-line</span>.”</p> - -<p>And the line <var>a d</var>, when properly divided in <var>b</var> and <var>c</var>, or any -other required points, is said to be divided “<span class="allsc">IN PERSPECTIVE -RATIO</span>” to the divisions of the original line <span class="maths">A D</span>.</p> - -<p>If the line <span class="maths"><var>a</var> V</span> is above the sight-line instead of beneath it, -the measuring-line is to be drawn above also: and the lines -<span class="maths"><var>b′</var> M</span>, <span class="maths"><var>c′</var> M</span>, etc., drawn <em>down</em> to the dividing-point. Turn <a href="#f.17">Fig. 17.</a> upside down, and it will show the construction.</p> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM VI"><a name="png.037" id="png.037" href="#png.037"><span class="pagenum"><span - class="ns">[p</span>27<span class="ns">]<br - /></span></span></a><a name="pr.vi" id="pr.vi">PROBLEM VI.</a></h2> - -<h3 class="pr" title="To draw any triangle, given in position and magnitude, -in a horizontal plane">TO DRAW ANY TRIANGLE, GIVEN IN POSITION AND MAGNITUDE, -IN A <span class="nw">HORIZONTAL PLANE</span>.</h3> - - -<p class="illo"><img id="f.18" src="images/illus-037.png" alt="[Geometric diagram]" /><br - /><b>Fig. 18.</b></p> - -<p><span class="smc">Let</span> <span class="maths">A B C</span> (<a href="#f.18">Fig. 18.</a>) be the triangle.</p> - -<p>As it is given in position and magnitude, one of its sides, -at least, must be given in position and magnitude, and the -directions of the two other sides.</p> - -<p>Let <span class="maths">A B</span> be the side given in position and magnitude.</p> - -<p>Then <span class="maths">A B</span> is a horizontal line, in a given position, and of -a given length.</p> - -<p>Draw the line <span class="maths">A B</span>. (<a href="#pr.v">Problem V.</a>)</p> - -<p>Let <var>a b</var> be the line so drawn.</p> - -<p>Find <span class="maths">V</span> and <span class="maths">V′</span>, the vanishing-points respectively of the -lines <span class="maths">A C</span> and <span class="maths">B C</span>. (<a href="#pr.iii">Problem III.</a>)</p> - -<p><a name="png.038" id="png.038" href="#png.038"><span class="pagenum"><span - class="ns">[p</span>28<span class="ns">]<br - /></span></span></a>From <var>a</var> draw <span class="maths"><var>a</var> V</span>, and from <var>b</var>, draw <span class="maths"><var>b</var> V′</span>, cutting each -other in <var>c</var>.</p> - -<p>Then <var>a b c</var> is the triangle required.</p> - -<p>If <span class="maths">A C</span> is the line originally given, <var>a c</var> is the line which must -be first drawn, and the line <span class="maths">V′ <var>b</var></span> must be drawn from <span class="maths">V′</span> to <var>c</var> -and produced to cut <var>a b</var> in <var>b</var>. Similarly, if <span class="maths">B C</span> is given, <span class="maths">V <var>c</var></span> -must be drawn to <var>c</var> and produced, and <var>a b</var> from its vanishing-point -to <var>b</var>, and produced to cut <var>a c</var> in <var>a</var>.</p> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM VII"><a name="png.039" id="png.039" href="#png.039"><span class="pagenum"><span - class="ns">[p</span>29<span class="ns">]<br - /></span></span></a><a name="pr.vii" id="pr.vii">PROBLEM VII.</a></h2> - -<h3 class="pr" title="To draw any rectilinear quadrilateral figure, given -in position and magnitude, in a horizontal plane">TO DRAW ANY RECTILINEAR QUADRILATERAL FIGURE, GIVEN -IN POSITION AND MAGNITUDE, IN A <span class="nw">HORIZONTAL PLANE</span>.</h3> - - -<p class="illo"><img id="f.19" src="images/illus-039.png" alt="[Geometric diagram]" /><br - /><b>Fig. 19.</b></p> - -<p><span class="smc">Let</span> <span class="maths">A B C D</span> (<a href="#f.19">Fig. 19.</a>) be the given figure.</p> - -<p>Join any two of its opposite angles by the line <span class="maths">B C</span>.</p> - -<p>Draw first the triangle <span class="maths">A B C</span>. (<a href="#pr.vi">Problem VI.</a>)</p> - -<p>And then, from the base <span class="maths">B C</span>, the two lines <span class="maths">B D</span>, <span class="maths">C D</span>, to their -vanishing-points, which will complete the figure. It is unnecessary -to give a diagram of the construction, which is -merely that of <a href="#f.18">Fig. 18.</a> duplicated; another triangle being -drawn on the line <span class="maths">A C</span> or <span class="maths">B C</span>.</p> - - -<h3 class="cor" title="Corollary"><a name="cor.vii.i" id="cor.vii.i">COROLLARY.</a></h3> - -<p>It is evident that by this application of <a href="#pr.vi">Problem VI.</a> any -given rectilinear figure whatever in a horizontal plane may -be drawn, since any such figure may be divided into a number -of triangles, and the triangles then drawn in succession.</p> - -<p>More convenient methods may, however, be generally -<a name="png.040" id="png.040" href="#png.040"><span class="pagenum"><span - class="ns">[p</span>30<span class="ns">] - </span></span></a>found, according to the form of the figure required, by the -use of succeeding problems; and for the quadrilateral figure -which occurs most frequently in practice, namely, the square, -the following construction is more convenient than that used -in the present problem.</p> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM VIII"><a name="png.041" id="png.041" href="#png.041"><span class="pagenum"><span - class="ns">[p</span>31<span class="ns">]<br - /></span></span></a><a name="pr.viii" id="pr.viii">PROBLEM VIII.</a></h2> - -<h3 class="pr" title="To draw a square, given in position and magnitude, in -a horizontal plane">TO DRAW A SQUARE, GIVEN IN POSITION AND MAGNITUDE, IN -A <span class="nw">HORIZONTAL PLANE</span>.</h3> - - -<p class="illo"><img id="f.20" src="images/illus-041.png" alt="[Geometric diagram]" /><br - /><b>Fig. 20.</b></p> - -<p><span class="smc">Let</span> <span class="maths">A B C D</span>, <a href="#f.20">Fig. 20.</a>, be the square.</p> - -<p>As it is given in position and magnitude, the position and -magnitude of all its sides are given.</p> - -<p>Fix the position of the point <span class="maths">A</span> in <var>a</var>.</p> - -<p>Find <span class="maths">V</span>, the vanishing-point of <span class="maths">A B</span>; and <span class="maths">M</span>, the dividing-point -of <span class="maths">A B</span>, nearest <span class="maths">S</span>.</p> - -<p>Find <span class="maths">V′</span>, the vanishing-point of <span class="maths">A C</span>; and <span class="maths">N</span>, the dividing-point -of <span class="maths">A C</span>, nearest <span class="maths">S</span>.</p> - -<p><a name="png.042" id="png.042" href="#png.042"><span class="pagenum"><span - class="ns">[p</span>32<span class="ns">]<br - /></span></span></a>Draw the measuring-line through <var>a</var>, and make <var>a b′</var>, <var>a c′</var>, -each equal to the sight-magnitude of <span class="maths">A B</span>.</p> - -<p>(For since <span class="maths">A B C D</span> is a square, <span class="maths">A C</span> is equal to <span class="maths">A B</span>.)</p> - -<p>Draw <span class="maths"><var>a</var> V′</span> and <span class="maths"><var>c′</var> N</span>, cutting each other in <var>c</var>.</p> - -<p>Draw <span class="maths"><var>a</var> V</span>, and <span class="maths"><var>b′</var> M</span>, cutting each other in <var>b</var>.</p> - -<p>Then <var>a c</var>, <var>a b</var>, are the two nearest sides of the square.</p> - -<p>Now, clearing the figure of superfluous lines, we have -<var>a b</var>, <var>a c</var>, drawn in position, as in <a href="#f.21">Fig. 21.</a></p> - -<p class="illo"><img id="f.21" src="images/illus-042.png" alt="[Geometric diagram]" /><br - /><b>Fig. 21.</b></p> - -<p>And because <span class="maths">A B C D</span> is a square, <span class="maths">C D</span> (<a href="#f.20">Fig. 20.</a>) is parallel -to <span class="maths">A B</span>.</p> - -<p>And all parallel lines have the same vanishing-point. -(Note to <a href="#pr.iii">Problem III.</a>)</p> - -<p>Therefore, <span class="maths">V</span> is the vanishing-point of <span class="maths">C D</span>.</p> - -<p>Similarly, <span class="maths">V′</span> is the vanishing-point of <span class="maths">B D</span>.</p> - -<p>Therefore, from <var>b</var> and <var>c</var> (<a href="#f.22">Fig. 22.</a>) draw <span class="maths"><var>b</var> V′</span>, <span class="maths"><var>c</var> V</span>, cutting -each other in <var>d</var>.</p> - -<p>Then <var>a b c d</var> is the square required.</p> - - -<h3 class="cor" title="Corollary I"><a name="cor.viii.i" id="cor.viii.i">COROLLARY I.</a></h3> - -<p>It is obvious that any rectangle in a horizontal plane may -be drawn by this problem, merely making <var>a b′</var>, on the measuring-line, -<a href="#f.20">Fig. 20.</a>, equal to the sight-magnitude of one of its -sides, and <var>a c′</var> the sight-magnitude of the other.</p> - - -<h3 class="cor" title="Corollary II"><a name="png.043" id="png.043" href="#png.043"><span class="pagenum"><span - class="ns">[p</span>33<span class="ns">]<br - /></span></span></a><a name="cor.viii.ii" id="cor.viii.ii">COROLLARY II.</a></h3> - -<p>Let <var>a b c d</var>, <a href="#f.22">Fig. 22.</a>, be any square drawn in perspective. -Draw the diagonals <var>a d</var> and <var>b c</var>, cutting each other in <span class="maths">C</span>. -Then <span class="maths">C</span> is the center of the square. Through <span class="maths">C</span>, draw <var>e f</var> to -the vanishing-point of <var>a b</var>, and <var>g h</var> to the vanishing-point of -<var>a c</var>, and these lines will bisect the sides of the square, so that -<var>a g</var> is the perspective representation of half the side <var>a b</var>; <var>a e</var> -is half <var>a c</var>; <var>c h</var> is half <var>c d</var>; and <var>b f</var> is half <var>b d</var>.</p> - -<p class="illo"><img id="f.22" src="images/illus-043.png" alt="[Geometric diagram]" /><br - /><b>Fig. 22.</b></p> - - -<h3 class="cor" title="Corollary III"><a name="cor.viii.iii" id="cor.viii.iii">COROLLARY III.</a></h3> - -<p>Since <span class="maths">A B C D</span>, <a href="#f.20">Fig. 20.</a>, is a square, <span class="maths">B A C</span> is a right angle; -and as <span class="maths">T V</span> is parallel to <span class="maths">A B</span>, and <span class="maths">T V′</span> to <span class="maths">A C</span>, <span class="maths">V′ T V</span> must be a -right angle also.</p> - -<p>As the ground plan of most buildings is rectangular, it -constantly happens in practice that their angles (as the corners -of ordinary houses) throw the lines to the vanishing-points -thus at right angles; and so that this law is observed, -and <span class="maths">V T V′</span> is kept a right angle, it does not matter in general -practice whether the vanishing-points are thrown a little -more or a little less to the right or left of <span class="maths">S</span>: but it matters -much that the relation of the vanishing-points should be -accurate. Their position with respect to <span class="maths">S</span> merely causes the -spectator to see a little more or less on one side or other of -the house, which may be a matter of chance or choice; but -their rectangular relation determines the rectangular shape -of the building, which is an essential point.</p> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM IX"><a name="png.044" id="png.044" href="#png.044"><span class="pagenum"><span - class="ns">[p</span>34<span class="ns">]<br - /></span></span></a><a name="pr.ix" id="pr.ix">PROBLEM IX.</a></h2> - -<h3 class="pr" title="To draw a square pillar, given in position and magnitude, -its base and top being in horizontal planes">TO DRAW A SQUARE PILLAR, GIVEN IN POSITION AND MAGNITUDE, -ITS BASE AND TOP BEING IN <span class="nw">HORIZONTAL PLANES</span>.</h3> - - - -<p><span class="smc">Let</span> <span class="maths">A H</span>, <a href="#f.23">Fig. 23.</a>, be the square pillar.</p> - -<p>Then, as it is given in position and magnitude, the position -and magnitude of the square it stands upon must be given -(that is, the line <span class="maths">A B</span> or <span class="maths">A C</span> in -position), and the height of its -side <span class="maths">A E</span>.</p> - -<p class="illolt"><span><img id="f.23" src="images/illus-044a.png" alt="[Geometric diagram]" /><br - /><b>Fig. 23.</b></span><span><img id="f.24" src="images/illus-044b.png" alt="[Geometric diagram]" /><br - /><b>Fig. 24.</b></span></p> - -<p>Find the sight-magnitudes of -<span class="maths">A B</span> and <span class="maths">A E</span>. Draw the two sides -<var>a b</var>, <var>a c</var>, of the square of the base, -by <a href="#pr.viii">Problem VIII.</a>, as in <a href="#f.24">Fig. 24.</a> -From the points <var>a</var>, <var>b</var>, and <var>c</var>, raise -vertical lines <var>a e</var>, <var>c f</var>, <var>b g</var>.</p> - -<p>Make <var>a e</var> equal to the sight-magnitude -of <span class="maths">A E</span>.</p> - -<p>Now because the top and base of the pillar are in horizontal -planes, the square of its top, <span class="maths">F G</span>, is parallel to the -square of its base, <span class="maths">B C</span>.</p> - -<p>Therefore the line <span class="maths">E F</span> is parallel to <span class="maths">A C</span>, and <span class="maths">E G</span> to <span class="maths">A B</span>.</p> - -<p>Therefore <span class="maths">E F</span> has the same vanishing-point as <span class="maths">A C</span>, and <span class="maths">E G</span> -the same vanishing-point as <span class="maths">A B</span>.</p> - -<p>From <var>e</var> draw <var>e f</var> to the vanishing-point of <var>a c</var>, cutting <var>c f</var> -in <var>f</var>.</p> - -<p>Similarly draw <var>e g</var> to the vanishing-point of <var>a b</var>, cutting -<var>b g</var> in <var>g</var>.</p> - -<p>Complete the square <var>g f</var> in <var>h</var>, by drawing <var>g h</var> to the vanishing-point -of <var>e f</var>, and <var>f h</var> to the vanishing-point of <var>e g</var>, cutting -each other in <var>h</var>. Then <var>a g h f</var> is the square pillar required.</p> - - -<h3 class="cor" title="Corollary"><a name="png.045" id="png.045" href="#png.045"><span class="pagenum"><span - class="ns">[p</span>35<span class="ns">]<br - /></span></span></a><a name="cor.ix.i" id="cor.ix.i">COROLLARY.</a></h3> - -<p>It is obvious that if <span class="maths">A E</span> is equal to <span class="maths">A C</span>, the whole figure -will be a cube, and each side, <var>a e f c</var> and <var>a e g b</var>, will be a -square in a given vertical plane. And by making <span class="maths">A B</span> or -<span class="maths">A C</span> longer or shorter in any given proportion, any form of -rectangle may be given to either of the sides of the pillar. -No other rule is therefore needed for drawing squares or -rectangles in vertical planes.</p> - -<p>Also any triangle may be thus drawn in a vertical plane, -by inclosing it in a rectangle and determining, in perspective -ratio, on the sides of the rectangle, the points of their contact -with the angles of the triangle.</p> - -<p>And if any triangle, then any polygon.</p> - -<p>A less complicated construction will, however, be given -hereafter.<a name="fn18" id="fn18"></a><a title="Go to footnote 18" - href="#Footnote18" class="fnanchor"><span - class="ns">[Footnote </span>18<span class="ns">] - </span></a></p> - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote18" id="Footnote18"><span class="ns">[Footnote </span - >18<span class="ns">: </span></a> </span> -See page 96 (<a href="#fn35">note</a>), after you have read <a href="#pr.xvi">Problem XVI</a>.<span class="ns">]</span> - <a title="Return to text" href="#fn18" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - -</div> - -<div class="chap"> -<h2 class="pr" title="PROBLEM X"><a name="png.046" id="png.046" href="#png.046"><span class="pagenum"><span - class="ns">[p</span>36<span class="ns">]<br - /></span></span></a><a name="pr.x" id="pr.x">PROBLEM X.</a></h2> - -<h3 class="pr" title="To draw a pyramid, given in position and magnitude, on -a square base in a horizontal plane">TO DRAW A PYRAMID, GIVEN IN POSITION AND MAGNITUDE, ON -A SQUARE BASE IN A <span class="nw">HORIZONTAL PLANE</span>.</h3> - - -<p class="illo"><img id="f.25" src="images/illus-046a.png" alt="[Geometric diagram]" /><br - /><b>Fig. 25.</b></p> - -<p><span class="smc">Let</span> <span class="maths">A B</span>, <a href="#f.25">Fig. 25.</a>, be the four-sided pyramid. As it is -given in position and magnitude, the square base on which -it stands must be given in position and magnitude, and its -vertical height, <span class="maths">C D</span>.<a name="fn19" id="fn19"></a><a title="Go to footnote 19" - href="#Footnote19" class="fnanchor"><span - class="ns">[Footnote </span>19<span class="ns">] - </span></a></p> - -<p class="illo"><img id="f.26" src="images/illus-046b.png" alt="[Geometric diagram]" /><br - /><b>Fig. 26.</b></p> - -<p>Draw a square pillar, <span class="maths">A B G E</span>, <a href="#f.26">Fig. 26.</a>, on the square base -of the pyramid, and make the height of the pillar <span class="maths">A F</span> equal -<a name="png.047" id="png.047" href="#png.047"><span class="pagenum"><span - class="ns">[p</span>36<span class="ns">] - </span></span></a>to the vertical height of the pyramid <span class="maths">C D</span> (<a href="#png.044">Problem IX.</a>). -Draw the diagonals <span class="maths">G F</span>, <span class="maths">H I</span>, on the top of the square pillar, -cutting each other in <span class="maths">C</span>. Therefore <span class="maths">C</span> is the center of the -square <span class="maths">F G H I</span>. (Prob. VIII. <a href="#cor.viii.ii">Cor. II.</a>)</p> - -<p class="illo"><img id="f.27" src="images/illus-047.png" alt="[Geometric diagram]" /><br - /><b>Fig. 27.</b></p> - -<p>Join <span class="maths">C E</span>, <span class="maths">C A</span>, <span class="maths">C B</span>.</p> - -<p>Then <span class="maths">A B C E</span> is the pyramid required. If the base of the -pyramid is above the eye, as when a square spire is seen on -the top of a church-tower, the construction will be as in -<a href="#f.27">Fig. 27</a>.</p> - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote19" id="Footnote19"><span class="ns">[Footnote </span - >19<span class="ns">: </span></a> </span>If, instead of the vertical height, the length of <span class="maths">A D</span> is given, the -vertical must be deduced from it. See the Exercises on this Problem -in the Appendix, <a href="#png.089">p. 79</a>.<span class="ns">]</span> - <a title="Return to text" href="#fn19" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM XI"><a name="png.048" id="png.048" href="#png.048"><span class="pagenum"><span - class="ns">[p</span>38<span class="ns">]<br - /></span></span></a><a name="pr.xi" id="pr.xi">PROBLEM XI.</a></h2> - -<h3 class="pr" title="To draw any curve in a horizontal or vertical plane">TO DRAW ANY CURVE IN A HORIZONTAL OR <span class="nw">VERTICAL PLANE</span>.</h3> - - -<p class="illo"><img id="f.28" src="images/illus-048a.png" alt="[Geometric diagram]" /><br - /><b>Fig. 28.</b></p> - -<p><span class="smc">Let</span> <span class="maths">A B</span>, <a href="#f.28">Fig. 28.</a>, be the curve.</p> - -<p>Inclose it in a rectangle, <span class="maths">C D E F</span>.</p> - -<p>Fix the position of the point <span class="maths">C</span> or <span class="maths">D</span>, and draw the rectangle. -(Problem VIII. <a href="#cor.viii.i">Coroll. I.</a>)<a name="fn20" id="fn20"></a><a title="Go to footnote 20" - href="#Footnote20" class="fnanchor"><span - class="ns">[Footnote </span>20<span class="ns">] - </span></a></p> - -<p>Let <span class="maths">C D E F</span>, <a href="#f.29">Fig. 29.</a>, be the rectangle so drawn.</p> - -<p class="illolt"><img id="f.29" src="images/illus-048b.png" alt="[Geometric diagram]" /><br - /><b>Fig. 29.</b></p> - -<p>If an extremity of the curve, as <span class="maths">A</span>, is in a side of the rectangle, -divide the side <span class="maths">C E</span>, <a href="#f.29">Fig. 29.</a>, -so that <span class="maths">A C</span> shall be (in perspective -ratio) to <span class="maths">A E</span> as <span class="maths">A C</span> is to <span class="maths">A E</span> in <a href="#f.28">Fig. 28.</a> (Prob. V. <a href="#cor.v.ii">Cor. II.</a>)</p> - -<p>Similarly determine the points of -contact of the curve and rectangle -<var>e</var>, <var>f</var>, <var>g</var>.</p> - -<p>If an extremity of the curve, as <span class="maths">B</span>, -is not in a side of the rectangle, let -<a name="png.049" id="png.049" href="#png.049"><span class="pagenum"><span - class="ns">[p</span>39<span class="ns">] - </span></span></a>fall the perpendiculars <span class="maths">B <var>a</var></span>, <span class="maths">B <var>b</var></span> on the rectangle sides. Determine -the correspondent points <var>a</var> and <var>b</var> in <a href="#f.29">Fig. 29.</a>, as you have -already determined <span class="maths">A</span>, <span class="maths">B</span>, <var>e</var>, and <var>f</var>.</p> - -<p>From <var>b</var>, <a href="#f.29">Fig. 29.</a>, draw <span class="maths"><var>b</var> B</span> parallel to <span class="maths">C D</span>,<a name="fn21" id="fn21"></a><a title="Go to footnote 21" - href="#Footnote21" class="fnanchor"><span - class="ns">[Footnote </span>21<span class="ns">] - </span></a> and from <var>a</var> -draw <span class="maths"><var>a</var> B</span> to the vanishing-point of <span class="maths">D F</span>, cutting each other in -<span class="maths">B</span>. Then <span class="maths">B</span> is the extremity of the curve.</p> - -<p>Determine any other important point in the curve, as <span class="maths">P</span>, in -the same way, by letting fall <span class="maths">P <var>q</var></span> and <span class="maths">P <var>r</var></span> on the rectangle’s -sides.</p> - -<p>Any number of points in the curve may be thus determined, -and the curve drawn through the series; in most -cases, three or four will be enough. Practically, complicated -curves may be better drawn in perspective by an experienced -eye than by rule, as the fixing of the various points -in haste involves too many chances of error; but it is well to -draw a good many by rule first, in order to give the eye its -experience.<a name="fn22" id="fn22"></a><a title="Go to footnote 22" - href="#Footnote22" class="fnanchor"><span - class="ns">[Footnote </span>22<span class="ns">] - </span></a></p> - - -<h3 class="cor" title="Corollary">COROLLARY.</h3> - -<p>If the curve required be a circle, <a href="#f.30">Fig. 30.</a>, the rectangle -which incloses it will become a square, -and the curve will have four points of -contact, <span class="maths">A B C D</span>, in the middle of the -sides of the square.</p> - -<p class="illort"><img id="f.30" src="images/illus-049.png" alt="[Geometric diagram]" /><br - /><b>Fig. 30.</b></p> - -<p>Draw the square, and as a square -may be drawn about a circle in any -position, draw it with its nearest side, -<span class="maths">E G</span>, parallel to the sight-line.</p> - -<p>Let <span class="maths">E F</span>, <a href="#f.31">Fig. 31.</a>, be the square so -drawn.</p> - -<p><a name="png.050" id="png.050" href="#png.050"><span class="pagenum"><span - class="ns">[p</span>40<span class="ns">]<br - /></span></span></a>Draw its diagonals <span class="maths">E F</span>, <span class="maths">G H</span>; and through the center of the -square (determined by their intersection) draw <span class="maths">A B</span> to the -vanishing-point of <span class="maths">G F</span>, and <span class="maths">C D</span> parallel to <span class="maths">E G</span>. Then the -points <span class="maths">A B C D</span> are the four points of the circle’s contact.</p> - -<p class="illo"><img id="f.31" src="images/illus-050.png" alt="[Geometric diagram]" /><br - /><b>Fig. 31.</b></p> - -<p>On <span class="maths">E G</span> describe a half square, <span class="maths">E L</span>; draw the semicircle -<span class="maths">K A L</span>; and from its center, <span class="maths">R</span>, the diagonals <span class="maths">R E</span>, <span class="maths">R G</span>, cutting -the circle in <var>x</var>, <var>y</var>.</p> - -<p>From the points <var>x</var> <var>y</var>, where the circle cuts the diagonals, -raise perpendiculars, <span class="maths">P <var>x</var></span>, <span class="maths">Q <var>y</var></span>, to <span class="maths">E G</span>.</p> - -<p>From <span class="maths">P</span> and <span class="maths">Q</span> draw <span class="maths">P P′</span>, <span class="maths">Q Q′</span>, to the vanishing-point of -<span class="maths">G F</span>, cutting the diagonals in <var>m</var>, <var>n</var>, and <var>o</var>, <var>p</var>.</p> - -<p>Then <var>m</var>, <var>n</var>, <var>o</var>, <var>p</var> are four other points in the circle.</p> - -<p>Through these eight points the circle may be drawn by the -hand accurately enough for general purposes; but any number -of points required may, of course, be determined, as in -<a href="#pr.xi">Problem XI.</a></p> - -<p>The distance <span class="maths">E P</span> is approximately one-seventh of <span class="maths">E G</span>, and -may be assumed to be so in quick practice, as the error involved -is not greater than would be incurred in the hasty -operation of drawing the circle and diagonals.</p> - -<p>It may frequently happen that, in consequence of associated -<a name="png.051" id="png.051" href="#png.051"><span class="pagenum"><span - class="ns">[p</span>41<span class="ns">] - </span></span></a>constructions, it may be inconvenient to draw <span class="maths">E G</span> parallel -to the sight-line, the square being perhaps first constructed -in some oblique direction. In such cases, <span class="maths">Q G</span> and -<span class="maths">E P</span> must be determined in perspective ratio by the dividing-point, -the line <span class="maths">E G</span> being used as a measuring-line.</p> - -<div class="aside"> -<p>[<i>Obs.</i> In drawing <a href="#f.31">Fig. 31.</a> the station-point has been taken much -nearer the paper than is usually advisable, in order to show the character -of the curve in a very distinct form.</p> - -<p>If the student turns the book so that <span class="maths">E G</span> may be vertical, <a href="#f.31">Fig. 31.</a> -will represent the construction for drawing a circle in a vertical plane, -the sight-line being then of course parallel to <span class="maths">G L</span>; and the semicircles -<span class="maths">A D B</span>, <span class="maths">A C B</span>, on each side of the diameter <span class="maths">A B</span>, will represent ordinary -semicircular arches seen in perspective. In that case, if the book be -held so that the line <span class="maths">E H</span> is the top of the square, the upper semicircle -will represent a semicircular arch, <em>above</em> the eye, drawn in perspective. -But if the book be held so that the line <span class="maths">G F</span> is the top of the square, the -upper semicircle will represent a semicircular arch, <em>below</em> the eye, -drawn in perspective.</p> - -<p>If the book be turned upside down, the figure will represent a circle -drawn on the ceiling, or any other horizontal plane above the eye; -and the construction is, of course, accurate in every case.]</p> -</div> - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote20" id="Footnote20"><span class="ns">[Footnote </span - >20<span class="ns">: </span></a> </span>Or if the curve is in a vertical plane, <a href="#png.045">Coroll. to Problem IX</a>. As -a rectangle may be drawn in any position round any given curve, its -position with respect to the curve will in either case be regulated by -convenience. See the Exercises on this Problem, in the Appendix, <a href="#png.095">p. 85</a>.<span class="ns">]</span> - <a title="Return to text" href="#fn20" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote21" id="Footnote21"><span class="ns">[Footnote </span - >21<span class="ns">: </span></a> </span>Or to its vanishing-point, if <span class="maths">C D</span> has one.<span class="ns">]</span> - <a title="Return to text" href="#fn21" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote22" id="Footnote22"><span class="ns">[Footnote </span - >22<span class="ns">: </span></a> </span>Of course, by dividing the original rectangle into any number of -equal rectangles, and dividing the perspective rectangle similarly, the -curve may be approximately drawn without any trouble; but, when -accuracy is required, the points should be fixed, as in the problem.<span class="ns">]</span> - <a title="Return to text" href="#fn22" class="fnreturn" - ><i>Return to text</i></a></small></p> - -</div> -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM XII"><a name="png.052" id="png.052" href="#png.052"><span class="pagenum"><span - class="ns">[p</span>42<span class="ns">]<br - /></span></span></a><a name="pr.xii" id="pr.xii">PROBLEM XII.</a></h2> - -<h3 class="pr" title="To divide a circle drawn in perspective into any -given number of equal parts">TO DIVIDE A CIRCLE DRAWN IN PERSPECTIVE INTO ANY -GIVEN NUMBER <span class="nw">OF EQUAL PARTS</span>.</h3> - - -<p><span class="smc">Let</span> <span class="maths">A B</span>, <a href="#f.32">Fig. 32.</a>, be the circle drawn in perspective. It -is required to divide it into a given number of equal parts; -in this case, 20.</p> - -<p>Let <span class="maths">K A L</span> be the semicircle used in the construction. -Divide the semicircle <span class="maths">K A L</span> into half the number of parts -required; in this case, 10.</p> - -<p>Produce the line <span class="maths">E G</span> laterally, as far as may be necessary.</p> - -<p>From <span class="maths">O</span>, the center of the semicircle <span class="maths">K A L</span>, draw radii -through the points of division of the semicircle, <var>p</var>, <var>q</var>, <var>r</var>, etc., -and produce them to cut the line <span class="maths">E G</span> in <span class="maths">P</span>, <span class="maths">Q</span>, <span class="maths">R</span>, etc.</p> - -<p>From the points <span class="maths">P Q R</span> draw the lines <span class="maths">P P′</span>, <span class="maths">Q Q′</span>, <span class="maths">R R′</span>, etc., -through the center of the circle <span class="maths">A B</span>, each cutting the circle -in two points of its circumference.</p> - -<p>Then these points divide the perspective circle as required.</p> - -<p>If from each of the points <var>p</var>, <var>q</var>, <var>r</var>, a vertical were raised to -the line <span class="maths">E G</span>, as in <a href="#f.31">Fig. 31.</a>, and from the point where it cut -<span class="maths">E G</span> a line were drawn to the vanishing-point, as <span class="maths">Q Q′</span> in <a href="#f.31">Fig. 31.</a>, this line would also determine two of the points of -division.</p> - -<p class="illo"><a name="png.053" id="png.053" href="#png.053"><span class="pagenum"><span - class="ns">[p</span>43<span class="ns">] - </span></span></a><img class="nopr" id="f.32" src="images/illus-053.png" alt="[Geometric diagram]" /><img class="pronly" id="f.32pr" src="images/illus-053pr.png" alt="[Geometric diagram]" /><br - /><b>Fig. 32.</b></p> - -<p>If it is required to divide a circle into any number of -given <em>un</em>equal parts (as in the points <span class="maths">A</span>, <span class="maths">B</span>, and <span class="maths">C</span>, <a href="#f.33">Fig. 33.</a>), -the shortest way is thus to raise vertical lines from <span class="maths">A</span> and <span class="maths">B</span> -to the side of the perspective square <span class="maths">X Y</span>, and then draw to -the vanishing-point, cutting the perspective circle in <var>a</var> and <var>b</var>, -the points required. Only notice that if any point, as <span class="maths">A</span>, is -on the nearer side of the circle <span class="maths">A B C</span>, its representative point, -<var>a</var>, must be on the nearer side of the circle <var>a b c</var>; and if the -point <span class="maths">B</span> is on the farther side of the circle <span class="maths">A B C</span>, <var>b</var> must be -<a name="png.054" id="png.054" href="#png.054"><span class="pagenum"><span - class="ns">[p</span>44<span class="ns">] - </span></span></a>on the farther side of <var>a b c</var>. If any point, as <span class="maths">C</span>, is so much -in the lateral arc of the circle as not to be easily determinable -by the vertical line, draw the horizontal <span class="maths">C P</span>, find the correspondent -<var>p</var> in the side of the perspective square, and draw -<var>p c</var> parallel to <span class="maths">X Y</span>, cutting the perspective circle in <var>c</var>.</p> - -<p class="illo"><img id="f.33" src="images/illus-054.png" alt="[Geometric diagram]" /><br - /><b>Fig. 33.</b></p> - - -<h3 class="cor" title="Corollary" id="cor.xii.i">COROLLARY.</h3> - -<p>It is obvious that if the points <span class="maths">P′</span>, <span class="maths">Q′</span>, <span class="maths">R</span>, etc., by which the -circle is divided in <a href="#f.32">Fig. 32.</a>, be joined by right lines, the resulting -figure will be a regular equilateral figure of twenty -sides inscribed in the circle. And if the circle be divided -into given unequal parts, and the points of division joined -by right lines, the resulting figure will be an irregular polygon -inscribed in the circle with sides of given length.</p> - -<p>Thus any polygon, regular or irregular, inscribed in a circle, -may be inscribed in position in a perspective circle.</p> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM XIII"><a name="png.055" id="png.055" href="#png.055"><span class="pagenum"><span - class="ns">[p</span>45<span class="ns">]<br - /></span></span></a><a name="pr.xiii" id="pr.xiii">PROBLEM XIII.</a></h2> - -<h3 class="pr" title="To draw a square, given in magnitude, within a larger -square given in position and magnitude; the sides of -the two squares being parallel">TO DRAW A SQUARE, GIVEN IN MAGNITUDE, WITHIN A LARGER -SQUARE GIVEN IN POSITION AND MAGNITUDE; THE SIDES OF -THE TWO SQUARES <span class="nw">BEING PARALLEL</span>.</h3> - - -<p class="illo"><img id="f.34" src="images/illus-055a.png" alt="[Geometric diagram]" /><br - /><b>Fig. 34.</b></p> - -<p><span class="smc">Let</span> <span class="maths">A B</span>, <a href="#f.34">Fig. 34.</a>, be the sight-magnitude of the side of -the smaller square, and <span class="maths">A C</span> that of the side of the larger -square.</p> - -<p>Draw the larger square. Let <span class="maths">D E F G</span> be the square so -drawn.</p> - -<p>Join <span class="maths">E G</span> and <span class="maths">D F</span>.</p> - -<p>On either <span class="maths">D E</span> or <span class="maths">D G</span> set off, in perspective ratio, <span class="maths">D H</span> equal -to one half of <span class="maths">B C</span>. Through <span class="maths">H</span> draw <span class="maths">H K</span> to the vanishing-point -of <span class="maths">D E</span>, cutting <span class="maths">D F</span> in <span class="maths">I</span> and <span class="maths">E G</span> in <span class="maths">K</span>. Through <span class="maths">I</span> and -<span class="maths">K</span> draw <span class="maths">I M</span>, <span class="maths">K L</span>, to vanishing-point of <span class="maths">D G</span>, cutting <span class="maths">D F</span> in <span class="maths">L</span> -and <span class="maths">E G</span> in <span class="maths">M</span>. Join <span class="maths">L M</span>.</p> - -<p>Then <span class="maths">I K L M</span> is the smaller square, inscribed as required.<a name="fn23" id="fn23"></a><a title="Go to footnote 23" - href="#Footnote23" class="fnanchor"><span - class="ns">[Footnote </span>23<span class="ns">] - </span></a></p> - - - -<h3 class="cor" title="Corollary"><a name="png.056" id="png.056" href="#png.056"><span class="pagenum"><span - class="ns">[p</span>46<span class="ns">] - </span></span></a><a name="cor.xiii.i" id="cor.xiii.i">COROLLARY.</a></h3> - -<p class="illolt"><img id="f.36" src="images/illus-056.png" alt="[Geometric diagram]" /><br - /><b>Fig. 36.</b></p> - -<p>If, instead of one square within another, it be required to -draw one circle within another, -the dimensions of both -being given, inclose each circle -in a square. Draw the -squares first, and then the -circles within, as in <a href="#f.36">Fig. 36.</a></p> - - - -<div class="footnotes"> -<p class="illort"><img id="f.35" src="images/illus-055b.png" alt="[Geometric diagram]" /><br - /><b>Fig. 35.</b></p> -<p id="fn.f.35"><small><span class="unjust"><a name="Footnote23" id="Footnote23"><span class="ns">[Footnote </span - >23<span class="ns">: </span></a> </span>If either of the sides of the greater -square is parallel to the plane of the -picture, as <span class="maths">D G</span> in <a href="#f.35">Fig. 35.</a>, <span class="maths">D G</span> of -course must be equal to <span class="maths">A C</span>, and -<span class="maths">D H</span> equal to <span class="maths">B C</span>/2, and the construction -is as in <a href="#f.35">Fig. 35.</a><span class="ns">]</span> - <a title="Return to text" href="#fn23" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM XIV"><a name="png.057" id="png.057" href="#png.057"><span class="pagenum"><span - class="ns">[p</span>47<span class="ns">]<br - /></span></span></a><a name="pr.xiv" id="pr.xiv">PROBLEM XIV.</a></h2> - -<h3 class="pr" title="To draw a truncated circular cone, given in position -and magnitude, the truncations being in horizontal -planes, and the axis of the cone vertical">TO DRAW A TRUNCATED CIRCULAR CONE, GIVEN IN POSITION -AND MAGNITUDE, THE TRUNCATIONS BEING IN HORIZONTAL -PLANES, AND THE AXIS OF <span class="nw">THE CONE VERTICAL</span>.</h3> - - -<p><span class="smc">Let</span> <span class="maths">A B C D</span>, <a href="#f.37">Fig. 37.</a>, be the portion of the cone required.</p> - -<p class="illo"><img id="f.37" src="images/illus-057.png" alt="[Geometric diagram]" /><br - /><b>Fig. 37.</b></p> - -<p>As it is given in magnitude, its diameters must be given -at the base and summit, <span class="maths">A B</span> and <span class="maths">C D</span>; and its vertical height, -<span class="maths">C E</span>.<a name="fn24" id="fn24"></a><a title="Go to footnote 24" - href="#Footnote24" class="fnanchor"><span - class="ns">[Footnote </span>24<span class="ns">] - </span></a></p> - -<p>And as it is given in position, the center of its base must -be given.</p> - -<p class="illort"><img id="f.38" src="images/illus-058.png" alt="[Geometric diagram]" /><br - /><b>Fig. 38.</b></p> - -<p>Draw in position, about this center,<a name="fn25" id="fn25"></a><a title="Go to footnote 25" - href="#Footnote25" class="fnanchor"><span - class="ns">[Footnote </span>25<span class="ns">] - </span></a> the square pillar -<a name="png.058" id="png.058" href="#png.058"><span class="pagenum"><span - class="ns">[p</span>48<span class="ns">] - </span></span></a><var>a f d</var>, <a href="#f.38">Fig. 38.</a>, making its height, <var>b g</var>, equal to <span class="maths">C E</span>; and its side, -<var>a b</var>, equal to <span class="maths">A B</span>.</p> - -<p>In the square of its base, <var>a b c -d</var>, inscribe a circle, which therefore -is of the diameter of the base -of the cone, <span class="maths">A B</span>.</p> - -<p>In the square of its top, <var>e f g -h</var>, inscribe concentrically a circle -whose diameter shall equal <span class="maths">C D</span>. -(<a href="#cor.xiii.i">Coroll. Prob. XIII</a>.)</p> - -<p>Join the extremities of the circles -by the right lines <var>k l</var>, <var>n m</var>. -Then <var>k l n m</var> is the portion of cone required.</p> - - -<h3 class="cor" title="Corollary I">COROLLARY I.</h3> - -<p>If similar polygons be inscribed in similar positions in the -circles <var>k n</var> and <var>l m</var> (<a href="#cor.xii.i">Coroll. Prob. XII.</a>), and the corresponding -angles of the polygons joined by right lines, the resulting -figure will be a portion of a polygonal pyramid. (The dotted -lines in <a href="#f.38">Fig. 38.</a>, connecting the extremities of two diameters -and one diagonal in the respective circles, occupy the position -of the three nearest angles of a regular octagonal pyramid, -having its angles set on the diagonals and diameters of the -square <var>a d</var>, inclosing its base.)</p> - -<p>If the cone or polygonal pyramid is not truncated, its apex -will be the center of the upper square, as in <a href="#f.26">Fig. 26.</a></p> - - -<h3 class="cor" title="Corollary II">COROLLARY II.</h3> - -<p>If equal circles, or equal and similar polygons, be inscribed -in the upper and lower squares in <a href="#f.38">Fig. 38.</a>, the resulting -figure will be a vertical cylinder, or a vertical polygonal -pillar, of given height and diameter, drawn in position. -<a name="png.059" id="png.059" href="#png.059"><span class="pagenum"><span - class="ns">[p</span>49<span class="ns">] - </span></span></a></p> - - -<h3 class="cor" title="Corollary III">COROLLARY III.</h3> - -<p>If the circles in <a href="#f.38">Fig. 38.</a>, instead of being inscribed in the -squares <var>b c</var> and <var>f g</var>, be inscribed in the sides of the solid -figure <var>b e</var> and <var>d f</var>, those sides being made square, and the line -<var>b d</var> of any given length, the resulting figure will be, according -to the constructions employed, a cone, polygonal pyramid, -cylinder, or polygonal pillar, drawn in position about a -horizontal axis parallel to <var>b d</var>.</p> - -<p>Similarly, if the circles are drawn in the sides <var>g d</var> and -<var>e c</var>, the resulting figures will be described about a horizontal -axis parallel to <var>a b</var>.</p> - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote24" id="Footnote24"><span class="ns">[Footnote </span - >24<span class="ns">: </span></a> </span> -Or if the length of its side, <span class="maths">A C</span>, is given instead, take <var>a e</var>, <a href="#f.37">Fig. 37.</a>, -equal to half the excess of <span class="maths">A B</span> over <span class="maths">C D</span>; from the point <var>e</var> raise the -perpendicular <var>c e</var>. With center <var>a</var>, and distance <span class="maths">A C</span>, describe a circle -cutting <var>c e</var> in <var>c</var>. Then <var>c e</var> is the vertical height of the portion of cone -required, or <span class="maths">C E</span>.<span class="ns">]</span> - <a title="Return to text" href="#fn24" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote25" id="Footnote25"><span class="ns">[Footnote </span - >25<span class="ns">: </span></a> </span> -The direction of the side of the square will of course be regulated -by convenience.<span class="ns">]</span> - <a title="Return to text" href="#fn25" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM XV"><a name="png.060" id="png.060" href="#png.060"><span class="pagenum"><span - class="ns">[p</span>50<span class="ns">]<br - /></span></span></a><a name="pr.xv" id="pr.xv">PROBLEM XV.</a></h2> - -<h3 class="pr" title="To draw an inclined line, given in position and - magnitude">TO DRAW AN INCLINED LINE, GIVEN IN POSITION <span class="nw">AND -MAGNITUDE</span>.</h3> - - -<p><span class="smc">We</span> have hitherto been examining the conditions of horizontal -and vertical lines only, or of curves inclosed in rectangles.</p> - -<p class="illo"><span class="lt"><img id="f.39" src="images/illus-060a.png" alt="[Geometric diagram]" /><br - /><b>Fig. 39.</b></span> -<span class="rt"><img id="f.40" src="images/illus-060b.png" alt="[Geometric diagram]" /><br - /><b>Fig. 40.</b></span></p> - -<p>We must, in conclusion, investigate the perspective of inclined -lines, beginning with a single one given in position. -For the sake of completeness of system, I give in Appendix II. Article III. the development of this problem from the -second. But, in practice, the position of an inclined line -may be most conveniently defined by considering it as the -diagonal of a rectangle, as <span class="maths">A B</span> in <a href="#f.39">Fig. 39.</a>, and I shall therefore, -though at some sacrifice of system, examine it here -under that condition.</p> - -<p>If the sides of the rectangle <span class="maths">A C</span> and <span class="maths">A D</span> are given, the -slope of the line <span class="maths">A B</span> is determined; and then its position will -depend on that of the rectangle. If, as in <a href="#f.39">Fig. 39.</a>, the rectangle -is parallel to the picture plane, the line <span class="maths">A B</span> must be -so also. If, as in <a href="#f.40">Fig. 40.</a>, the rectangle is inclined to the -<a name="png.061" id="png.061" href="#png.061"><span class="pagenum"><span - class="ns">[p</span>51<span class="ns">] - </span></span></a>picture plane, the line <span class="maths">A B</span> will be so also. So that, to fix -the position of <span class="maths">A B</span>, the line <span class="maths">A C</span> must be given in position and -magnitude, and the height <span class="maths">A D</span>.</p> - -<p class="illort"><img id="f.41" src="images/illus-061a.png" alt="[Geometric diagram]" /><br - /><b>Fig. 41.</b></p> - -<p>If these are given, and it is only required to draw the -single line <span class="maths">A B</span> in perspective, the construction is entirely -simple; <span class="nw">thus:—</span></p> - -<p>Draw the line <span class="maths">A C</span> by <a href="#pr.i">Problem I</a>.</p> - -<p>Let <span class="maths">A C</span>, <a href="#f.41">Fig. 41.</a>, be the line so drawn. From -<var>a</var> and <var>c</var> raise the vertical lines <var>a d</var>, <var>c b</var>. Make <var>a d</var> -equal to the sight-magnitude of <span class="maths">A D</span>. From <var>d</var> draw -<var>d b</var> to the vanishing-point of <var>a c</var>, cutting <var>b c</var> <span class="nw">in <var>b</var>.</span></p> - -<p>Join <var>a b</var>. Then <var>a b</var> is the inclined line required.</p> - -<p class="illortclear"><img id="f.42" src="images/illus-061b.png" alt="[Geometric diagram]" /><br - /><b>Fig. 42.</b></p> - -<p>If the line is inclined in the opposite direction, -as <span class="maths">D C</span> in <a href="#f.42">Fig. 42.</a>, we have only to join <var>d c</var> instead of <var>a b</var> in -Fig. 41., and <var>d c</var> will be the line required.</p> - -<p>I shall hereafter call the line <span class="maths">A C</span>, when used -to define the position of an inclined line <span class="maths">A B</span> -(<a href="#f.40">Fig. 40.</a>), the “relative horizontal” of the -line <span class="maths">A B</span>.</p> - - -<h4>Observation.</h4> - -<p class="illort"><img id="f.43" src="images/illus-061c.png" alt="[Geometric diagram]" /><br - /><b>Fig. 43.</b></p> - -<p>In general, inclined lines are most needed for gable roofs, -in which, when the conditions are properly stated, the vertical -height of the gable, <span class="maths">X Y</span>, <a href="#f.43">Fig. 43.</a>, is given, and the base line, -<span class="maths">A C</span>, in position. When these are -given, draw <span class="maths">A C</span>; raise vertical -<span class="maths">A D</span>; make <span class="maths">A D</span> equal to sight-magnitude -of <span class="maths">X Y</span>;<!-- TN: original lacks semicolon --> complete the -perspective-rectangle <span class="maths">A D B C</span>; join -<span class="maths">A B</span> and <span class="maths">D C</span> (as by dotted lines in -figure); and through the intersection -of the dotted lines draw vertical <span class="maths">X Y</span>, cutting <span class="maths">D B</span> in <span class="maths">Y</span>. -Join <span class="maths">A Y</span>, <span class="maths">C Y</span>; and these lines are the sides of the gable. If -<a name="png.062" id="png.062" href="#png.062"><span class="pagenum"><span - class="ns">[p</span>52<span class="ns">] - </span></span></a>the length of the roof <span class="maths">A A′</span> is also given, draw in perspective -the complete parallelopiped<!-- TN: OED lists as valid alternate spelling --> <span class="maths">A′ D′ B C</span>, and from <span class="maths">Y</span> draw <span class="maths">Y Y′</span> -to the vanishing-point of <span class="maths">A A′</span>, cutting <span class="maths">D′ B′</span> in <span class="maths">Y′</span>. Join <span class="maths">A′ Y</span>, -and you have the slope of the farther side of the roof.</p> - -<p class="illo"><img id="f.44" src="images/illus-062.png" alt="[Geometric diagram]" /><br - /><b>Fig. 44.</b></p> - -<p>The construction above the eye is as in <a href="#f.44">Fig. 44.</a>; the roof -is reversed in direction merely to familiarize the student with -the different aspects of its lines.</p> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM XVI"><a name="png.063" id="png.063" href="#png.063"><span class="pagenum"><span - class="ns">[p</span>53<span class="ns">]<br - /></span></span></a><a name="pr.xvi" id="pr.xvi">PROBLEM XVI.</a></h2> - -<h3 class="pr" title="To find the vanishing-point of a given inclined line">TO FIND THE VANISHING-POINT OF A GIVEN <span class="nw">INCLINED LINE</span>.</h3> - - -<p><span class="smc">If</span>, in <a href="#f.43">Fig. 43.</a> or <a href="#f.44">Fig. 44.</a>, the lines <span class="maths">A Y</span> and <span class="maths">A′ Y′</span> be produced, -the student will find that they meet.</p> - -<p>Let <span class="maths">P</span>, <a href="#f.45">Fig. 45.</a>, be the point at which they meet.</p> - -<p>From <span class="maths">P</span> let fall the vertical <span class="maths">P V</span> on the sight-line, cutting -the sight-line in <span class="maths">V</span>.</p> - -<p>Then the student will find experimentally that <span class="maths">V</span> is the -vanishing-point of the line <span class="maths">A C</span>.<a name="fn26" id="fn26"></a><a title="Go to footnote 26" - href="#Footnote26" class="fnanchor"><span - class="ns">[Footnote </span>26<span class="ns">] - </span></a></p> - -<p>Complete the rectangle of the base <span class="maths">A C′</span>, by drawing <span class="maths">A′ C′</span> -to <span class="maths">V</span>, and <span class="maths">C C′</span> to the vanishing-point of <span class="maths">A A′</span>.</p> - -<p>Join <span class="maths">Y′ C′</span>.</p> - -<p>Now if <span class="maths">Y C</span> and <span class="maths">Y′ C′</span> be produced downwards, the student -will find that they meet.</p> - -<p>Let them be produced, and meet in <span class="maths">P′</span>.</p> - -<p>Produce <span class="maths">P V</span>, and it will be found to pass through the -point <span class="maths">P′</span>.</p> - -<p>Therefore if <span class="maths">A Y</span> (or <span class="maths">C Y</span>), <a href="#f.45">Fig. 45.</a>, be any inclined line -drawn in perspective by <a href="#pr.xv">Problem XV.</a>, and <span class="maths">A C</span> the relative -horizontal (<span class="maths">A C</span> in <a href="#f.39">Figs. 39</a>, <a href="#f.40">40.</a>), also drawn in perspective.</p> - -<p>Through <span class="maths">V</span>, the vanishing-point of <span class="maths">A V</span>, draw the vertical -<span class="maths">P P′</span> upwards and downwards.</p> - -<p>Produce <span class="maths">A Y</span> (or <span class="maths">C Y</span>), cutting <span class="maths">P P′</span> in <span class="maths">P</span> (or <span class="maths">P′</span>).</p> - -<p>Then <span class="maths">P</span> is the vanishing-point of <span class="maths">A Y</span> (or <span class="maths">P′</span> of <span class="maths">C Y</span>).</p> - -<p class="illo"><img id="f.45" src="images/illus-064.png" alt="[Geometric diagram]" /><br - /><b>Fig. 45.</b></p> - -<p>The student will observe that, in order to find the point <span class="maths">P</span> -by this method, it is necessary first to draw a portion of the -given inclined line by <a href="#pr.xv">Problem XV</a>. Practically, it is always -necessary to do so, and, therefore, I give the problem -in this form.</p> - -<p><a name="png.064" id="png.064" href="#png.064"><span class="pagenum"><span - class="ns">[p</span>54<span class="ns">]<br - /></span></span></a>Theoretically, as will be shown in the analysis of the problem, -the point <span class="maths">P</span> should be found by drawing a line from the -station-point parallel to the given inclined line: but there is -no practical means of drawing such a line; so that in whatever -terms the problem may be given, a portion of the inclined -line (<span class="maths">A Y</span> or <span class="maths">C Y</span>) must always be drawn in perspective -before <span class="maths">P</span> can be found.</p> - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote26" id="Footnote26"><span class="ns">[Footnote </span - >26<span class="ns">: </span></a> </span> -The demonstration is in <a href="#png.112">Appendix II. Article III</a>.<span class="ns">]</span> - <a title="Return to text" href="#fn26" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM XVII"><a name="png.065" id="png.065" href="#png.065"><span class="pagenum"><span - class="ns">[p</span>55<span class="ns">]<br - /></span></span></a><a name="pr.xvii" id="pr.xvii">PROBLEM XVII.</a></h2> - -<h3 class="pr" title="To find the dividing-points of a given inclined line">TO FIND THE DIVIDING-POINTS OF A GIVEN <span class="nw">INCLINED LINE</span>.</h3> - - -<p class="illo"><img id="f.46" src="images/illus-065.png" alt="[Geometric diagram]" /><br - /><b>Fig. 46.</b></p> - -<p><span class="smc">Let</span> <span class="maths">P</span>, <a href="#f.46">Fig. 46.</a>, be the vanishing-point of the inclined line, -and <span class="maths">V</span> the vanishing-point of the relative horizontal.</p> - -<p>Find the dividing-points of the relative horizontal, <span class="maths">D</span> -and <span class="maths">D′</span>.</p> - -<p>Through <span class="maths">P</span> draw the horizontal line <span class="maths">X Y</span>.</p> - -<p>With center <span class="maths">P</span> and distance <span class="maths">D P</span> describe the two arcs <span class="maths">D X</span> -and <span class="maths">D′ Y</span>, cutting the line <span class="maths">X Y</span> in <span class="maths">X</span> and <span class="maths">Y</span>.</p> - -<p>Then <span class="maths">X</span> and <span class="maths">Y</span> are the dividing-points of the inclined line.<a name="fn27" id="fn27"></a><a title="Go to footnote 27" - href="#Footnote27" class="fnanchor"><span - class="ns">[Footnote </span>27<span class="ns">] - </span></a></p> - -<p><i>Obs.</i> The dividing-points found by the above rule, used -with the ordinary measuring-line, will lay off distances on -the retiring inclined line, as the ordinary dividing-points lay -them off on the retiring horizontal line.</p> - -<p>Another dividing-point, peculiar in its application, is -sometimes useful, and is to be found as <span class="nw">follows:—</span></p> - -<p class="illo"><a name="png.066" id="png.066" href="#png.066"><span class="pagenum"><span - class="ns">[p</span>56<span class="ns">] - </span></span></a><img id="f.47" src="images/illus-066.png" alt="[Geometric diagram]" /><br - /><b>Fig. 47.</b></p> - -<p>Let <span class="maths">A B</span>, <a href="#f.47">Fig. 47.</a>, be the given inclined line drawn in perspective, -and <span class="maths">A <var>c</var></span> the relative horizontal.</p> - -<p>Find the vanishing-points, <span class="maths">V</span> and <span class="maths">E</span>, of <span class="maths">A <var>c</var></span> and <span class="maths">A B</span>; <span class="maths">D</span>, the -dividing-point of <span class="maths">A <var>c</var></span>; and the sight-magnitude of <span class="maths">A <var>c</var></span> on the -measuring-line, or <span class="maths">A C</span>.</p> - -<p>From <span class="maths">D</span> erect the perpendicular <span class="maths">D F</span>.</p> - -<p>Join <span class="maths">C B</span>, and produce it to cut <span class="maths">D E</span> in <span class="maths">F</span>. Join <span class="maths">E F</span>.</p> - -<p>Then, by similar triangles, <span class="maths">D F</span> is equal to <span class="maths">E V</span>, and <span class="maths">E F</span> is -parallel to <span class="maths">D V</span>.</p> - -<p>Hence it follows that if from <span class="maths">D</span>, the dividing-point of <span class="maths">A <var>c</var></span>, -we raise a perpendicular and make <span class="maths">D F</span> equal to <span class="maths">E V</span>, a line -<span class="maths">C F</span>, drawn from any point <span class="maths">C</span> on the measuring-line to <span class="maths">F</span>, will -mark the distance <span class="maths">A B</span> on the inclined line, <span class="maths">A B</span> being the portion -of the given inclined line which forms the diagonal of -the vertical rectangle of which <span class="maths">A C</span> is the base.</p> - - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote27" id="Footnote27"><span class="ns">[Footnote </span - >27<span class="ns">: </span></a> </span> -The demonstration is in Appendix II., <a href="#png.114">p. 104</a>.<span class="ns">]</span> - <a title="Return to text" href="#fn27" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM XVIII"><a name="png.067" id="png.067" href="#png.067"><span class="pagenum"><span - class="ns">[p</span>57<span class="ns">]<br - /></span></span></a><a name="pr.xviii" id="pr.xviii">PROBLEM XVIII.</a></h2> - -<h3 class="pr" title="To find the sight-line of an inclined plane in which -two lines are given in position">TO FIND THE SIGHT-LINE OF AN INCLINED PLANE IN WHICH -TWO LINES ARE <span class="nw">GIVEN IN POSITION</span>.<a name="fn28" id="fn28"></a><a title="Go to footnote 28" - href="#Footnote28" class="fnanchor"><span - class="ns">[Footnote </span>28<span class="ns">] - </span></a></h3> - -<p><span class="smc">As</span> in order to fix the position of a line two points in it -must be given, so in order to fix the position of a plane, two -lines in it must be given.</p> - -<p class="illo"><img id="f.48" src="images/illus-067.png" alt="[Geometric diagram]" /><br - /><b>Fig. 48.</b></p> - -<p>Let the two lines be <span class="maths">A B</span> and <span class="maths">C D</span>, <a href="#f.48">Fig. 48.</a></p> - -<p><a name="png.068" id="png.068" href="#png.068"><span class="pagenum"><span - class="ns">[p</span>58<span class="ns">]<br - /></span></span></a>As they are given in position, the relative horizontals <span class="maths">A E</span> -and <span class="maths">C F</span> must be given.</p> - -<p>Then by <a href="#pr.xvi">Problem XVI.</a> the vanishing-point of <span class="maths">A B</span> is <span class="maths">V</span>, -and of <span class="maths">C D</span>, <span class="maths">V′</span>.</p> - -<p>Join <span class="maths">V V′</span> and produce it to cut the sight-line in <span class="maths">X</span>.</p> - -<p>Then <span class="maths">V X</span> is the sight-line of the inclined plane.</p> - -<p>Like the horizontal sight-line, it is of indefinite length; -and may be produced in either direction as occasion requires, -crossing the horizontal line of sight, if the plane continues -downward in that direction.</p> - -<p><span class="maths">X</span> is the vanishing-point of all horizontal lines in the -inclined plane.</p> - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote28" id="Footnote28"><span class="ns">[Footnote </span - >28<span class="ns">: </span></a> </span> -Read the Article on this problem in the Appendix, <a href="#png.107">p. 97</a>, before -investigating the problem itself.<span class="ns">]</span> - <a title="Return to text" href="#fn28" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM XIX"><a name="png.069" id="png.069" href="#png.069"><span class="pagenum"><span - class="ns">[p</span>59<span class="ns">]<br - /></span></span></a><a name="pr.xix" id="pr.xix">PROBLEM XIX.</a></h2> - -<h3 class="pr" title="To find the vanishing-point of steepest lines in an -inclined plane whose sight-line is given">TO FIND THE VANISHING-POINT OF STEEPEST LINES IN AN -INCLINED PLANE WHOSE <span class="nw">SIGHT-LINE IS GIVEN</span>.</h3> - - -<p class="illo"><img id="f.49" src="images/illus-069.png" alt="[Geometric diagram]" /><br - /><b>Fig. 49.</b></p> - -<p><span class="smc">Let</span> <span class="maths">V X</span>, <a href="#f.49">Fig. 49.</a>, be the given sight-line.</p> - -<p>Produce it to cut the horizontal sight-line in <span class="maths">X</span>.</p> - -<p>Therefore <span class="maths">X</span> is the vanishing-point of horizontal lines in -the given inclined plane. (<a href="#pr.xviii">Problem XVIII.</a>)</p> - -<p>Join <span class="maths">T X</span>, and draw <span class="maths">T Y</span> at right angles to <span class="maths">T X</span>.</p> - -<p>Therefore <span class="maths">Y</span> is the rectangular vanishing-point corresponding -to <span class="maths">X</span>.<a name="fn29" id="fn29"></a><a title="Go to footnote 29" - href="#Footnote29" class="fnanchor"><span - class="ns">[Footnote </span>29<span class="ns">] - </span></a></p> - -<p>From <span class="maths">Y</span> erect the vertical <span class="maths">Y P</span>, cutting the sight-line of the -inclined plane in <span class="maths">P</span>.</p> - -<p><a name="png.070" id="png.070" href="#png.070"><span class="pagenum"><span - class="ns">[p</span>60<span class="ns">]<br - /></span></span></a>Then <span class="maths">P</span> is the vanishing-point of steepest lines in the plane.</p> - -<p>All lines drawn to it, as <span class="maths">Q P</span>, <span class="maths">R P</span>, <span class="maths">N P</span>, etc., are the steepest -possible in the plane; and all lines drawn to <span class="maths">X</span>, as <span class="maths">Q X</span>, <span class="maths">O X</span>, -etc., are horizontal, and at right angles to the lines <span class="maths">P Q</span>, <span class="maths">P R</span>, -etc.</p> - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote29" id="Footnote29"><span class="ns">[Footnote </span - >29<span class="ns">: </span></a> </span>That is to say, the vanishing-point of horizontal lines drawn at -right angles to the lines whose vanishing-point is <span class="maths">X</span>.<span class="ns">]</span> - <a title="Return to text" href="#fn29" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - -</div> - -<div class="chap"> - -<h2 class="pr" title="PROBLEM XX"><a name="png.071" id="png.071" href="#png.071"><span class="pagenum"><span - class="ns">[p</span>61<span class="ns">]<br - /></span></span></a><a name="pr.xx" id="pr.xx">PROBLEM XX.</a></h2> - -<h3 class="pr" title="To find the vanishing-point of lines perpendicular to -the surface of a given inclined plane">TO FIND THE VANISHING-POINT OF LINES PERPENDICULAR TO -THE SURFACE OF A GIVEN <span class="nw">INCLINED PLANE</span>.</h3> - - -<p class="illo"><img id="f.50" src="images/illus-071.png" alt="[Geometric diagram]" /><br - /><b>Fig. 50.</b></p> - -<p><span class="smc">As</span> the inclined plane is given, one of its steepest lines -must be given, or may be ascertained.</p> - -<p>Let <span class="maths">A B</span>, <a href="#f.50">Fig. 50.</a>, be a portion of a steepest line in the -<a name="png.072" id="png.072" href="#png.072"><span class="pagenum"><span - class="ns">[p</span>62<span class="ns">] - </span></span></a>given plane, and <span class="maths">V</span> the vanishing-point of its relative horizontal.</p> - -<p>Through <span class="maths">V</span> draw the vertical <span class="maths">G F</span> upwards and downwards.</p> - -<p>From <span class="maths">A</span> set off any portion of the relative horizontal <span class="maths">A C</span>, -and on <span class="maths">A C</span> describe a semicircle in a vertical plane, <span class="maths">A D C</span>, -cutting <span class="maths">A B</span> in <span class="maths">E</span>.</p> - -<p>Join <span class="maths">E C</span>, and produce it to cut <span class="maths">G F</span> in <span class="maths">F</span>.</p> - -<p>Then <span class="maths">F</span> is the vanishing-point required.</p> - -<p>For, because <span class="maths">A E C</span> is an angle in a semicircle, it is a right -angle; and therefore the line <span class="maths">E F</span> is at right angles to the line -<span class="maths">A B</span>; and similarly all lines drawn to <span class="maths">F</span>, and therefore parallel -to <span class="maths">E F</span>, are at right angles with any line which cuts them, -drawn to the vanishing-point of <span class="maths">A B</span>.</p> - -<p>And because the semicircle <span class="maths">A D C</span> is in a vertical plane, and -its diameter <span class="maths">A C</span> is at right angles to the horizontal lines traversing -the surface of the inclined plane, the line <span class="maths">E C</span>, being -in this semicircle, is also at right angles to such traversing -lines. And therefore the line <span class="maths">E C</span>, being at right angles to -the steepest lines in the plane, and to the horizontal lines in -it, is perpendicular to its surface.</p> - -<hr class="tb" /> - -<p><a name="png.073" id="png.073" href="#png.073"><span class="pagenum"><span - class="ns">[p</span>63<span class="ns">]<br - /></span></span></a><span class="smc">The</span> preceding series of constructions, with the examples -in the first Article of the Appendix, put it in the power of -the student to draw any form, however complicated,<a name="fn30" id="fn30"></a><a title="Go to footnote 30" - href="#Footnote30" class="fnanchor"><span - class="ns">[Footnote </span>30<span class="ns">] - </span></a> which -does not involve intersection of curved surfaces. I shall not -proceed to the analysis of any of these more complex problems, -as they are entirely useless in the ordinary practice of -artists. For a few words only I must ask the reader’s -further patience, respecting the general placing and scale of -the picture.</p> - -<p>As the horizontal sight-line is drawn through the sight-point, -and the sight-point is opposite the eye, the sight-line -is always on a level with the eye. Above and below the sight-line, -the eye comprehends, as it is raised or depressed while -the head is held upright, about an equal space; and, on each -side of the sight-point, about the same space is easily seen -without turning the head; so that if a picture represented -the true field of easy vision, it ought to be circular, and have -the sight-point in its center. But because some parts of any -given view are usually more interesting than others, either -the uninteresting parts are left out, or somewhat more than -would generally be seen of the interesting parts is included, -by moving the field of the picture a little upwards or downwards, -so as to throw the sight-point low or high. The operation -will be understood in a moment by cutting an aperture -in a piece of pasteboard, and moving it up and down in front -of the eye, without moving the eye. It will be seen to embrace -sometimes the low, sometimes the high objects, without -<a name="png.074" id="png.074" href="#png.074"><span class="pagenum"><span - class="ns">[p</span>64<span class="ns">] - </span></span></a>altering their perspective, only the eye will be opposite the -lower part of the aperture when it sees the higher objects, -and <em>vice versâ</em>.</p> - -<p>There is no reason, in the laws of perspective, why the -picture should not be moved to the right or left of the sight-point, -as well as up or down. But there is this practical -reason. The moment the spectator sees the horizon in a picture -high, he tries to hold his head high, that is, in its right -place. When he sees the horizon in a picture low, he similarly -tries to put his head low. But, if the sight-point is -thrown to the left hand or right hand, he does not understand -that he is to step a little to the right or left; and if he places -himself, as usual, in the middle, all the perspective is distorted. -Hence it is generally unadvisable to remove the -sight-point laterally, from the center of the picture. The -Dutch painters, however, fearlessly take the license of -placing it to the right or left; and often with good effect.</p> - -<p>The rectilinear limitation of the sides, top, and base of -the picture is of course quite arbitrary, as the space of a landscape -would be which was seen through a window; less or -more being seen at the spectator’s pleasure, as he retires or -advances.</p> - -<p>The distance of the station-point is not so arbitrary. In -ordinary cases it should not be less than the intended greatest -dimension (height or breadth) of the picture. In most -works by the great masters it is more; they not only calculate -on their pictures being seen at considerable distances, -but they like breadth of mass in buildings, and dislike the -sharp angles which always result from station-points at short -distances.<a name="fn31" id="fn31"></a><a title="Go to footnote 31" - href="#Footnote31" class="fnanchor"><span - class="ns">[Footnote </span>31<span class="ns">] - </span></a></p> - -<p>Whenever perspective, done by true rule, looks wrong, it -is always because the station-point is too near. Determine, -<a name="png.075" id="png.075" href="#png.075"><span class="pagenum"><span - class="ns">[p</span>65<span class="ns">] - </span></span></a>in the outset, at what distance the spectator is likely to examine -the work, and never use a station-point within a less -distance.</p> - -<p>There is yet another and a very important reason, not only -for care in placing the station-point, but for that accurate -calculation of distance and observance of measurement which -have been insisted on throughout this work. All drawings -of objects on a reduced scale are, if rightly executed, drawings -of the appearance of the object at the distance which in -true perspective reduces it to that scale. They are not <em>small</em> -drawings of the object seen near, but drawings the <em>real size</em> -of the object seen far off. Thus if you draw a mountain in -a landscape, three inches high, you do not reduce all the features -of the near mountain so as to come into three inches of -paper. You could not do that. All that you can do is to -give the appearance of the mountain, when it is so far off -that three inches of paper would really hide it from you. -It is precisely the same in drawing any other object. A face -can no more be reduced in scale than a mountain can. It is -infinitely delicate already; it can only be quite rightly rendered -on its own scale, or at least on the slightly diminished -scale which would be fixed by placing the plate of glass, supposed -to represent the field of the picture, close to the -figures. Correggio and Raphael were both fond of this -slightly subdued magnitude of figure. Colossal painting, in -which Correggio excelled all others, is usually the enlargement -of a small picture (as a colossal sculpture is of a small -statue), in order to permit the subject of it to be discerned at -a distance. The treatment of colossal (as distinguished from -ordinary) paintings will depend therefore, in general, on -the principles of optics more than on those of perspective, -though, occasionally, portions may be represented as if they -were the projection of near objects on a plane behind them. -In all points the subject is one of great difficulty and subtlety; -and its examination does not fall within the compass -of this essay.</p> - -<p>Lastly, it will follow from these considerations, and the -<a name="png.076" id="png.076" href="#png.076"><span class="pagenum"><span - class="ns">[p</span>66<span class="ns">] - </span></span></a>conclusion is one of great practical importance, that, though -pictures may be enlarged, they cannot be reduced, in copying -them. All attempts to engrave pictures completely on a reduced -scale are, for this reason, nugatory. The best that can -be done is to give the aspect of the picture at the distance -which reduces it in perspective to the size required; or, in -other words, to make a drawing of the distant effect of the -picture. Good painting, like nature’s own work, is infinite, -and unreduceable.</p> - -<p>I wish this book had less tendency towards the infinite and -unreduceable. It has so far exceeded the limits I hoped to -give it, that I doubt not the reader will pardon an abruptness -of conclusion, and be thankful, as I am myself, to get to an -end on any terms.</p> - -<div class="footnotes"> -<p><small><span class="unjust"><a name="Footnote30" id="Footnote30"><span class="ns">[Footnote </span - >30<span class="ns">: </span></a> </span>As in algebraic science, much depends, in complicated perspective, -on the student’s ready invention of expedients, and on his quick -sight of the shortest way in which the solution may be accomplished, -when there are several ways.<span class="ns">]</span> - <a title="Return to text" href="#fn30" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote31" id="Footnote31"><span class="ns">[Footnote </span - >31<span class="ns">: </span></a> </span>The greatest masters are also fond of parallel perspective, that is -to say, of having one side of their buildings fronting them full, and -therefore parallel to the picture plane, while the other side vanishes -to the sight-point. This is almost always done in figure backgrounds, -securing simple and balanced lines.<span class="ns">]</span> - <a title="Return to text" href="#fn31" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - - -</div> - -<div class="apptitle"> - -<h2 class="appendix" title="APPENDIX"><a name="png.077" id="png.077" href="#png.077"><span class="pagenum"><span - class="ns">[p</span>67<span class="ns">]<br - /></span></span></a>APPENDIX.</h2> - -<hr class="short" /> - -<p id="appendix">I.<br - />PRACTICE AND OBSERVATIONS.<br - />II.<br - />DEMONSTRATIONS.</p> - -</div> - -<div class="chap"> - -<h2 class="pr" title="I. Practice and observations on the -preceding problems"><a name="png.079" id="png.079" href="#png.079"><span class="pagenum"><span - class="ns">[p</span>69<span class="ns">]<br - /></span></span></a>I.</h2> -<h3 class="prbig" title="">PRACTICE AND OBSERVATIONS ON THE -<span class="nw">PRECEDING PROBLEMS</span>.</h3> -<hr class="short" /> - - - -<h3 class="cor" title="Problem I"><span class="smc">Problem I.</span></h3> - - -<p><span class="smc">An</span> example will be necessary to make this problem clear -to the general student.</p> - -<p>The nearest corner of a piece of pattern on the carpet is -4½ feet beneath the eye, 2 feet to our right and 3½ feet in -direct distance from us. We intend to make a drawing of -the pattern which shall be seen properly when held 1½ foot -from the eye. It is required to fix the position of the corner -of the piece of pattern.</p> - -<p class="illort"><img id="f.51" src="images/illus-079.png" alt="[Geometric diagram]" /><br - /><b>Fig. 51.</b></p> - -<p>Let <span class="maths">A B</span>, <a href="#f.51">Fig. 51.</a>, be our sheet of -paper, some 3 feet wide. Make <span class="maths">S T</span> -equal to 1½ foot. Draw the line of -sight through <span class="maths">S</span>. Produce <span class="maths">T S</span>, and -make <span class="maths">D S</span> equal to 2 feet, therefore <span class="maths">T D</span> -equal to 3½ feet. Draw <span class="maths">D C</span>, equal to -2 feet; <span class="maths">C P</span>, equal to 4 feet. Join -<span class="maths">T C</span> (cutting the sight-line in <span class="maths">Q</span>) -and <span class="maths">T P</span>.</p> - -<p>Let fall the vertical <span class="maths">Q P′</span>, then <span class="maths">P′</span> -is the point required.</p> - -<p>If the lines, as in the figure, fall -outside of your sheet of paper, in -order to draw them, it is necessary to attach other sheets of -paper to its edges. This is inconvenient, but must be done -<a name="png.080" id="png.080" href="#png.080"><span class="pagenum"><span - class="ns">[p</span>70<span class="ns">] - </span></span></a>at first that you may see your way clearly; and sometimes -afterwards, though there are expedients for doing without -such extension in fast sketching.</p> - -<p>It is evident, however, that no extension of surface could -be of any use to us, if the distance <span class="maths">T D</span>, instead of being 3½ -feet, were 100 feet, or a mile, as it might easily be in a -landscape.</p> - -<p>It is necessary, therefore, to obtain some other means of -construction; to do which we must examine the principle of -the problem.</p> - - -<p class="tb">In the analysis of <a href="#f.2">Fig. 2.</a>, in the introductory remarks, I -used the word “height” only of the tower, <span class="maths">Q P</span>, because it -was only to its vertical height that the law deduced from the -figure could be applied. For suppose it had been a pyramid, -as <span class="maths">O Q P</span>, <a href="#f.52">Fig. 52.</a>, then the image of its side, <span class="maths">Q P</span>, being, -like every other magnitude, limited on the glass <span class="maths">A B</span> by the -lines coming from its extremities, would appear only of the -length <span class="maths">Q′ S</span>; and it is not true that <span class="maths">Q′ S</span> is to <span class="maths">Q P</span> as <span class="maths">T S</span> is to -<span class="maths">T P</span>. But if we let fall a vertical <span class="maths">Q D</span> from <span class="maths">Q</span>, so as to get -the vertical height of the pyramid, then it is true that <span class="maths">Q′ S</span> is -to <span class="maths">Q D</span> as <span class="maths">T S</span> is to <span class="maths">T D</span>.</p> - -<p class="illo"><img id="f.52" src="images/illus-080.png" alt="[Geometric diagram]" /><br - /><b>Fig. 52.</b></p> - -<p>Supposing this figure represented, not a pyramid, but a -triangle on the ground, and that <span class="maths">Q D</span> and <span class="maths">Q P</span> are horizontal -lines, expressing lateral distance from the line <span class="maths">T D</span>, still the -rule would be false for <span class="maths">Q P</span> and true for <span class="maths">Q D</span>. And, similarly, -it is true for all lines which are parallel, like <span class="maths">Q D</span>, to -<a name="png.081" id="png.081" href="#png.081"><span class="pagenum"><span - class="ns">[p</span>71<span class="ns">] - </span></span></a>the plane of the picture <span class="maths">A B</span>, and false for all lines which are -inclined to it at an angle.</p> - -<p>Hence generally. Let <span class="maths">P Q</span> (<a href="#f.2">Fig. 2.</a> in Introduction, <a href="#png.016">p. 6</a>) -be any magnitude <em>parallel to the plane of the picture</em>; and -<span class="maths">P′ Q′</span> its image on the picture.</p> - -<p>Then always the formula is true which you learned in -the Introduction: <span class="maths">P′ Q′</span> is to <span class="maths">P Q</span> as <span class="maths">S T</span> is to <span class="maths">D T</span>.</p> - -<p>Now the magnitude <span class="maths">P</span> dash <span class="maths">Q</span> dash in this formula I call -the “<span class="allsc">SIGHT-MAGNITUDE</span>” of the line <span class="maths">P Q</span>. The student must -fix this term, and the meaning of it, well in his mind. The -“sight-magnitude” of a line is the magnitude which bears -to the real line the same proportion that the distance of the -picture bears to the distance of the object. Thus, if a tower -be a hundred feet high, and a hundred yards off; and the -picture, or piece of glass, is one yard from the spectator, between -him and the tower; the distance of picture being then -to distance of tower as 1 to 100, the sight-magnitude of the -tower’s height will be as 1 to 100; that is to say, one foot. -If the tower is two hundred yards distant, the sight-magnitude -of its height will be half a foot, and so on.</p> - -<p>But farther. It is constantly necessary, in perspective -operations, to measure the other dimensions of objects by -the sight-magnitude of their vertical lines. Thus, if the -tower, which is a hundred feet high, is square, and twenty-five -feet broad on each side; if the sight-magnitude of the -height is one foot, the measurement of the side, reduced to -the same scale, will be the hundredth part of twenty-five feet, -or three inches: and, accordingly, I use in this treatise the -term “sight-magnitude” indiscriminately for all lines reduced -in the same proportion as the vertical lines of the object. -If I tell you to find the “sight-magnitude” of any -line, I mean, always, find the magnitude which bears to that -line the proportion of <span class="maths">S T</span> to <span class="maths">D T</span>; or, in simpler terms, reduce -the line to the scale which you have fixed by the first -determination of the length <span class="maths">S T</span>.</p> - -<p>Therefore, you must learn to draw quickly to scale before -you do anything else; for all the measurements of your object -<a name="png.082" id="png.082" href="#png.082"><span class="pagenum"><span - class="ns">[p</span>72<span class="ns">] - </span></span></a>must be reduced to the scale fixed by <span class="maths">S T</span> before you can -use them in your diagram. If the object is fifty feet from -you, and your paper one foot, all the lines of the object must -be reduced to a scale of one fiftieth before you can use them; -if the object is two thousand feet from you, and your paper -one foot, all your lines must be reduced to the scale of one -two-thousandth before you can use them, and so on. Only -in ultimate practice, the reduction never need be tiresome, -for, in the case of large distances, accuracy is never required. -If a building is three or four miles distant, a hairbreadth of -accidental variation in a touch makes a difference of ten or -twenty feet in height or breadth, if estimated by accurate -perspective law. Hence it is never attempted to apply -measurements with precision at such distances. Measurements -are only required within distances of, at the most, two -or three hundred feet. Thus it may be necessary to represent -a cathedral nave precisely as seen from a spot seventy -feet in front of a given pillar; but we shall hardly be required -to draw a cathedral three miles distant precisely as seen from -seventy feet in advance of a given milestone. Of course, if -such a thing be required, it can be done; only the reductions -are somewhat long and complicated: in ordinary cases it is -easy to assume the distance <span class="maths">S T</span> so as to get at the reduced -dimensions in a moment. Thus, let the pillar of the nave, -in the case supposed, be 42 feet high, and we are required -to stand 70 feet from it: assume <span class="maths">S T</span> to be equal to 5 feet. -Then, as 5 is to 70 so will the sight-magnitude required be -to 42; that is to say, the sight-magnitude of the pillar’s -height will be 3 feet. If we make <span class="maths">S T</span> equal to 2½ feet, the -pillar’s height will be 1½ foot, and so on.</p> - -<p>And for fine divisions into irregular parts which cannot -be measured, the ninth and tenth problems of the sixth book -of Euclid will serve you: the following construction is, -however, I think, more practically <span class="nw">convenient:—</span></p> - -<p>The line <span class="maths">A B</span> (<a href="#f.53">Fig. 53.</a>) is divided by given points, <var>a</var>, <var>b</var>, <var>c</var>, -into a given number of irregularly unequal parts; it is required -to divide any other line, <span class="maths">C D</span>, into an equal number -<a name="png.083" id="png.083" href="#png.083"><span class="pagenum"><span - class="ns">[p</span>73<span class="ns">] - </span></span></a>of parts, bearing to each other the same proportions as the -parts of <span class="maths">A B</span>, and arranged in the same order.</p> - -<p>Draw the two lines parallel to each other, as in the figure.</p> - -<p>Join <span class="maths">A C</span> and <span class="maths">B D</span>, and produce the lines <span class="maths">A C</span>, <span class="maths">B D</span>, till they -meet in <span class="maths">P</span>.</p> - -<p>Join <span class="maths"><var>a</var> P</span>, <span class="maths"><var>b</var> P</span>, <span class="maths"><var>c</var> P</span>, cutting <span class="maths"><var>c</var> D</span> in <var>f</var>, <var>g</var>, <var>h</var>.</p> - -<p>Then the line <span class="maths">C D</span> is divided as required, in <var>f</var>, <var>g</var>, <var>h</var>.</p> - -<p>In the figure the lines <span class="maths">A B</span> and <span class="maths">C D</span> are accidentally perpendicular -to <span class="maths">A P</span>. There is no need for their being so.</p> - -<p class="illo"><img id="f.53" src="images/illus-083.png" alt="[Geometric diagram]" /><br - /><b>Fig. 53.</b></p> - -<p>Now, to return to our first problem.</p> - -<p>The construction given in the figure is only the quickest -mathematical way of obtaining, on the picture, the sight-magnitudes -of <span class="maths">D C</span> and <span class="maths">P C</span>, which are both magnitudes parallel -with the picture plane. But if these magnitudes are -too great to be thus put on the paper, you have only to obtain -the reduction by scale. Thus, if <span class="maths">T S</span> be one foot, <span class="maths">T D</span> -eighty feet, <span class="maths">D C</span> forty feet, and <span class="maths">C P</span> ninety feet, the distance -<span class="maths">Q S</span> must be made equal to one eightieth of <span class="maths">D C</span>, or half a -foot; and the distance <span class="maths">Q P′</span>, one eightieth of <span class="maths">C P</span>, or one eightieth -of ninety feet; that is to say, nine eighths of a foot, or -thirteen and a half inches. The lines <span class="maths">C T</span> and <span class="maths">P T</span> are thus -<em>practically</em> useless, it being only necessary to measure <span class="maths">Q S</span> -<a name="png.084" id="png.084" href="#png.084"><span class="pagenum"><span - class="ns">[p</span>74<span class="ns">] - </span></span></a>and <span class="maths">Q P</span>, on your paper, of the due sight-magnitudes. But -the mathematical construction, given in <a href="#pr.i">Problem I.</a>, is the -basis of all succeeding problems, and, if it is once thoroughly -understood and practiced (it can only be thoroughly understood -by practice), all the other problems will follow easily.</p> - -<p>Lastly. Observe that any perspective operation whatever -may be performed with reduced dimensions of every line employed, -so as to bring it conveniently within the limits of -your paper. When the required figure is thus constructed -on a small scale, you have only to enlarge it accurately -in the same proportion in which you reduced the lines of -construction, and you will have the figure constructed in -perspective on the scale required for use.</p> - - -</div> - -<div class="chap"> - -<h3 class="app" title="Problem IX"><a name="png.085" id="png.085" href="#png.085"><span class="pagenum"><span - class="ns">[p</span>75<span class="ns">]<br - /></span></span></a>PROBLEM IX.</h3> - - -<p><span class="smc">The</span> drawing of most buildings occurring in ordinary -practice will resolve itself into applications of this problem. -In general, any house, or block of houses, presents itself -under the main conditions assumed here in <a href="#f.54">Fig. 54.</a> There -will be an angle or corner somewhere near the spectator, as -<span class="maths">A B</span>; and the level of the eye will usually be above the base -of the building, of which, therefore, the horizontal upper -lines will slope down to the vanishing-points, and the base -lines rise to them. The following practical directions will, -however, meet nearly all <span class="nw">cases:—</span></p> - -<p class="illo"><img id="f.54" src="images/illus-085.png" alt="[Geometric diagram]" /><br - /><b>Fig. 54.</b></p> - -<p>Let <span class="maths">A B</span>, <a href="#f.54">Fig. 54.</a>, be any important vertical line in the -block of buildings; if it is the side of a street, you may fix -upon such a line at the division between two houses. If its -real height, distance, etc., are given, you will proceed with -<a name="png.086" id="png.086" href="#png.086"><span class="pagenum"><span - class="ns">[p</span>76<span class="ns">] - </span></span></a>the accurate construction of the problem; but usually you -will neither know, nor care, exactly how high the building -is, or how far off. In such case draw the line <span class="maths">A B</span>, as nearly -as you can guess, about the part of the picture it ought to -occupy, and on such a scale as you choose. Divide it into -any convenient number of equal parts, according to the -height you presume it to be. If you suppose it to be twenty -feet high, you may divide it into twenty parts, and let each -part stand for a foot; if thirty feet high, you may divide it -into ten parts, and let each part stand for three feet; if -seventy feet high, into fourteen parts, and let each part stand -for five feet; and so on, avoiding thus very minute divisions -till you come to details. Then observe how high your eye -reaches upon this vertical line; suppose, for instance, that it -is thirty feet high and divided into ten parts, and you are -standing so as to raise your head to about six feet above its -base, then the sight-line may be drawn, as in the figure, -through the second division from the ground. If you are -standing above the house, draw the sight-line above <span class="maths">B</span>; if below -the house, below <span class="maths">A</span>; at such height or depth as you suppose -may be accurate (a yard or two more or less matters -little at ordinary distances, while at great distances perspective -rules become nearly useless, the eye serving you better -than the necessarily imperfect calculation). Then fix your -sight-point and station-point, the latter with proper reference -to the scale of the line <span class="maths">A B</span>. As you cannot, in all probability, -ascertain the exact direction of the line <span class="maths">A V</span> or <span class="maths">B V</span>, -draw the slope <span class="maths">B V</span> as it appears to you, cutting the sight-line -in <span class="maths">V</span>. Thus having fixed one vanishing-point, the other, -and the dividing-points, must be accurately found by rule; -for, as before stated, whether your entire group of points -(vanishing and dividing) falls a little more or less to the -right or left of <span class="maths">S</span> does not signify, but the relation of the -points to each other <em>does</em> signify. Then draw the measuring-line -<span class="maths">B G</span>, either through <span class="maths">A</span> or <span class="maths">B</span>, choosing always the steeper -slope of the two; divide the measuring-line into parts of the -same length as those used on <span class="maths">A B</span>, and let them stand for the -<a name="png.087" id="png.087" href="#png.087"><span class="pagenum"><span - class="ns">[p</span>77<span class="ns">] - </span></span></a>same magnitudes. Thus, suppose there are two rows of windows -in the house front, each window six feet high by three -wide, and separated by intervals of three feet, both between -window and window and between tier and tier; each of the -divisions here standing for three feet, the lines drawn from -<span class="maths">B G</span> to the dividing-point <span class="maths">D</span> fix the lateral dimensions, and the -divisions on <span class="maths">A B</span> the vertical ones. For other magnitudes it -would be necessary to subdivide the parts on the measuring-line, -or on <span class="maths">A B</span>, as required. The lines which regulate the -inner sides or returns of the windows (<var>a</var>, <var>b</var>, <var>c</var>, etc.) of course -are drawn to the vanishing-point of <span class="maths">B F</span> (the other side of the -house), if <span class="maths">F B V</span> represents a right angle; if not, their own -vanishing-point must be found separately for these returns. -But see <a href="#png.095">Practice on Problem XI</a>.</p> - -<p class="illo"><img id="f.55" src="images/illus-087.png" alt="[Geometric diagram]" /><br - /><b>Fig. 55.</b></p> - -<p>Interior angles, such as <span class="maths">E B C</span>, <a href="#f.55">Fig. 55.</a> (suppose the corner -of a room), are to be treated in the same way, each side of -the room having its measurements separately carried to it -from the measuring-line. It may sometimes happen in such -cases that we have to carry the measurement <em>up</em> from the -corner <span class="maths">B</span>, and that the sight-magnitudes are given us from -the length of the line <span class="maths">A B</span>. For instance, suppose the room -is eighteen feet high, and therefore <span class="maths">A B</span> is eighteen feet; and -we have to lay off lengths of six feet on the top of the room -wall, <span class="maths">B C</span>. Find <span class="maths">D</span>, the dividing-point of <span class="maths">B C</span>. Draw a -<a name="png.088" id="png.088" href="#png.088"><span class="pagenum"><span - class="ns">[p</span>78<span class="ns">] - </span></span></a>measuring-line, <span class="maths">B F</span>, from <span class="maths">B</span>; and another, <span class="maths"><var>g</var> C</span>, anywhere -above. On <span class="maths">B F</span> lay off <span class="maths">B G</span> equal to one third of <span class="maths">A B</span>, or six -feet; and draw from <span class="maths">D</span>, through <span class="maths">G</span> and <span class="maths">B</span>, the lines <span class="maths">G <var>g</var></span>, <span class="maths">B <var>b</var></span>, -to the upper measuring-line. Then <var>g b</var> is six feet on that -measuring-line. Make <var>b c</var>, <var>c h</var>, etc., equal to <var>b g</var>; and draw -<var>c e</var>, <var>h f</var>, etc., to <span class="maths">D</span>, cutting <span class="maths">B C</span> in <var>e</var> and <var>f</var>, which mark the -required lengths of six feet each at the top of the wall.</p> - - -</div> - -<div class="chap"> - -<h3 class="app" title="Problem X"><a name="png.089" id="png.089" href="#png.089"><span class="pagenum"><span - class="ns">[p</span>79<span class="ns">]<br - /></span></span></a>PROBLEM X.</h3> - -<p><span class="smc">This</span> is one of the most important foundational problems -in perspective, and it is necessary that the student should -entirely familiarize himself with its conditions.</p> - -<p>In order to do so, he must first observe these general -relations of magnitude in any pyramid on a square base.</p> - -<p>Let <span class="maths">A G H′</span>, <a href="#f.56">Fig. 56.</a>, be any pyramid on a square base.</p> - -<p class="illolt"><img id="f.56" src="images/illus-089a.png" alt="[Geometric diagram]" /><br - /><b>Fig. 56.</b></p> - -<p>The best terms in which its magnitude -can be given, are the length of one side of -its base, <span class="maths">A H</span>, and its vertical altitude (<span class="maths">C D</span> -in <a href="#f.25">Fig. 25.</a>); for, knowing these, we know -all the other magnitudes. But these are -not the terms in which its size will be -usually ascertainable. Generally, we shall -have given us, and be able to ascertain by -measurement, one side of its base <span class="maths">A H</span>, and either <span class="maths">A G</span> the -length of one of the lines of its angles, or <span class="maths">B G</span> (or <span class="maths">B′ G</span>) the -length of a line drawn from its vertex, <span class="maths">G</span>, to the middle of -the side of its base. In measuring a real pyramid, <span class="maths">A G</span> will -usually be the line most easily found; but in many architectural -problems <span class="maths">B G</span> is given, or is most easily ascertainable.</p> - -<p>Observe therefore this general construction.</p> - -<p class="illort"><img id="f.57" src="images/illus-089b.png" alt="[Geometric diagram]" /><br - /><b>Fig. 57.</b></p> - -<p>Let <span class="maths">A B D E</span>, <a href="#f.57">Fig. 57.</a>, be the square base -of any pyramid.</p> - -<p>Draw its diagonals, <span class="maths">A E</span>, <span class="maths">B D</span>, cutting -each other in its center, <span class="maths">C</span>.</p> - -<p>Bisect any side, <span class="maths">A B</span>, in <span class="maths">F</span>.</p> - -<p>From <span class="maths">F</span> erect vertical <span class="maths">F G</span>.</p> - -<p>Produce <span class="maths">F B</span> to <span class="maths">H</span>, and make <span class="maths">F H</span> equal -to <span class="maths">A C</span>.</p> - -<p>Now if the vertical altitude of the -pyramid (<span class="maths">C D</span> in <a href="#f.25">Fig. 25.</a>) be given, -make <span class="maths">F G</span> equal to this vertical altitude.</p> - -<p><a name="png.090" id="png.090" href="#png.090"><span class="pagenum"><span - class="ns">[p</span>80<span class="ns">]<br - /></span></span></a>Join <span class="maths">G B</span> and <span class="maths">G H</span>.</p> - -<p>Then <span class="maths">G B</span> and <span class="maths">G H</span> are the true magnitudes of <span class="maths">G B</span> and <span class="maths">G H</span> -in <a href="#f.56">Fig. 56.</a></p> - -<p>If <span class="maths">G B</span> is given, and not the vertical altitude, with center -<span class="maths">B</span>, and distance <span class="maths">G B</span>, describe circle cutting <span class="maths">F G</span> in <span class="maths">G</span>, and <span class="maths">F G</span> -is the vertical altitude.</p> - -<p>If <span class="maths">G H</span> is given, describe the circle from <span class="maths">H</span>, with distance -<span class="maths">G H</span>, and it will similarly cut <span class="maths">F G</span> in <span class="maths">G</span>.</p> - -<p>It is especially necessary for the student to examine this -construction thoroughly, because in many complicated forms -of ornaments, capitals of columns, etc., the lines <span class="maths">B G</span> and -<span class="maths">G H</span> become the limits or bases of curves, which are elongated -on the longer (or angle) profile <span class="maths">G H</span>, and shortened on the -shorter (or lateral) profile <span class="maths">B G</span>. We will take a simple instance, -but must previously note another construction.</p> - -<p>It is often necessary, when pyramids are the roots of some -ornamental form, to divide them horizontally at a given vertical -height. The shortest way of doing so is in general the -following.</p> - -<p class="illo"><img id="f.58" src="images/illus-090.png" alt="[Geometric diagram]" /><br - /><b>Fig. 58.</b></p> - -<p>Let <span class="maths">A E C</span>, <a href="#f.58">Fig. 58.</a>, be any pyramid on a square base <span class="maths">A B C</span>, and <span class="maths">A D C</span> the square pillar used in its construction.</p> - -<p><a name="png.091" id="png.091" href="#png.091"><span class="pagenum"><span - class="ns">[p</span>81<span class="ns">]<br - /></span></span></a>Then by construction (<a href="#pr.x">Problem X.</a>) <span class="maths">B D</span> and <span class="maths">A F</span> are both -of the vertical height of the pyramid.</p> - -<p>Of the diagonals, <span class="maths">F E</span>, <span class="maths">D E</span>, choose the shortest (in this case -<span class="maths">D E</span>), and produce it to cut the sight-line in <span class="maths">V</span>.</p> - -<p>Therefore <span class="maths">V</span> is the vanishing-point of <span class="maths">D E</span>.</p> - -<p>Divide <span class="maths">D B</span>, as may be required, into the sight-magnitudes -of the given vertical heights at which the pyramid is to be -divided.</p> -<p class="illo"><span class="lt"><img id="f.59" src="images/illus-091a.png" alt="[Geometric diagram]" /><br - /><b> Fig. 59. </b></span> -<span class="rt"><img id="f.60" src="images/illus-091b.png" alt="[Geometric diagram]" /><br - /><b> Fig. 60. </b></span></p> - -<p>From the points of division, 1, 2, 3, etc., draw to the -vanishing-point <span class="maths">V</span>. The lines so drawn cut the angle line of -the pyramid, <span class="maths">B E</span>, at the required elevations. Thus, in the -figure, it is required to draw a horizontal black band on the -pyramid at three fifths of its height, and in breadth one -twentieth of its height. The line <span class="maths">B D</span> is divided into five -parts, of which three are counted from <span class="maths">B</span> upwards. Then -the line drawn to <span class="maths">V</span> marks the base of the black band. Then -one fourth of one of the five parts is measured, which similarly -gives the breadth of the band. The terminal lines of -the band are then drawn on the sides of the pyramid parallel -to <span class="maths">A B</span> (or to its vanishing-point if it has one), and to the -vanishing-point of <span class="maths">B C</span>.</p> - -<p><a name="png.092" id="png.092" href="#png.092"><span class="pagenum"><span - class="ns">[p</span>82<span class="ns">]<br - /></span></span></a>If it happens that the vanishing-points of the diagonals -are awkwardly placed for use, bisect the nearest base line of -the pyramid in <span class="maths">B</span>, as in <a href="#f.59">Fig. 59.</a></p> - -<p>Erect the vertical <span class="maths">D B</span> and join <span class="maths">G B</span> and <span class="maths">D G</span> (<span class="maths">G</span> being the -apex of pyramid).</p> - -<p>Find the vanishing-point of <span class="maths">D G</span>, and use <span class="maths">D B</span> for division, -carrying the measurements to the line <span class="maths">G B</span>.</p> - -<p>In <a href="#f.59">Fig. 59.</a>, if we join <span class="maths">A D</span> and <span class="maths">D C</span>, <span class="maths">A D C</span> is the vertical -profile of the whole pyramid, and <span class="maths">B D C</span> of the half pyramid, -corresponding to <span class="maths">F G B</span> in <a href="#f.57">Fig. 57.</a></p> - -<p class="illo"><img id="f.61" src="images/illus-092.png" alt="[Geometric diagram]" /><br - /><b>Fig. 61.</b></p> - - -<p>We may now proceed to an architectural example.</p> - -<p>Let <span class="maths">A H</span>, <a href="#f.60">Fig. 60.</a>, be the vertical profile of the capital of -a pillar, <span class="maths">A B</span> the semi-diameter of its head or abacus, and <span class="maths">F D</span> -the semi-diameter of its shaft.</p> - -<p>Let the shaft be circular, and the abacus square, down to -the level <span class="maths">E</span>.</p> - -<p>Join <span class="maths">B D</span>, <span class="maths">E F</span>, and produce them to meet in <span class="maths">G</span>.</p> - -<p>Therefore <span class="maths">E C G</span> is the semi-profile of a reversed pyramid -containing the capital.</p> - -<p><a name="png.093" id="png.093" href="#png.093"><span class="pagenum"><span - class="ns">[p</span>83<span class="ns">]<br - /></span></span></a>Construct this pyramid, with the square of the abacus, in -the required perspective, as in <a href="#f.61">Fig. 61.</a>; making <span class="maths">A E</span> equal -to <span class="maths">A E</span> in <a href="#f.60">Fig. 60.</a>, and <span class="maths">A K</span>, the side of the square, equal to -twice <span class="maths">A B</span> in <a href="#f.60">Fig. 60.</a> Make <span class="maths">E G</span> equal to <span class="maths">C G</span>, and <span class="maths">E D</span> equal -to <span class="maths">C D</span>. Draw <span class="maths">D F</span> to the vanishing-point of the diagonal <span class="maths">D V</span> -(the figure is too small to include this vanishing-point), and -<span class="maths">F</span> is the level of the point <span class="maths">F</span> in <a href="#f.60">Fig. 60.</a>, on the side of the -pyramid.</p> - -<p>Draw <span class="maths">F <var>m</var></span>, <span class="maths">F <var>n</var></span>, to the vanishing-points of <span class="maths">A H</span> and <span class="maths">A K</span>. -Then <span class="maths">F <var>n</var></span> and <span class="maths">F <var>m</var></span> are horizontal lines across the pyramid at -the level <span class="maths">F</span>, forming at that level two sides of a square.</p> - -<p class="illo"><img id="f.62" src="images/illus-093.png" alt="[Geometric diagram]" /><br - /><b>Fig. 62.</b></p> - -<p>Complete the square, and within it inscribe a circle, as in -Fig. 62., which is left unlettered that its construction may -be clear. At the extremities of this draw vertical lines, -which will be the sides of the shaft in its right place. It -will be found to be somewhat smaller in diameter than the -entire shaft in <a href="#f.60">Fig. 60.</a>, because at the center of the square -it is more distant than the nearest edge of the square abacus. -The curves of the capital may then be drawn approximately -by the eye. They are not quite accurate in <a href="#f.62">Fig. 62.</a>, there -<a name="png.094" id="png.094" href="#png.094"><span class="pagenum"><span - class="ns">[p</span>84<span class="ns">] - </span></span></a>being a subtlety in their junction with the shaft which could -not be shown on so small a scale without confusing the student; -the curve on the left springing from a point a little -way round the circle behind the shaft, and that on the right -from a point on this side of the circle a little way within the -edge of the shaft. But for their more accurate construction -see <a href="#png.097">Notes on Problem XIV</a>.</p> - - - - -</div> - -<div class="chap"> -<h3 class="app" title="Problem XI"><a name="png.095" id="png.095" href="#png.095"><span class="pagenum"><span - class="ns">[p</span>85<span class="ns">]<br - /></span></span></a>PROBLEM XI.</h3> - - -<p><span class="smc">It</span> is seldom that any complicated curve, except occasionally -a spiral, needs to be drawn in perspective; but the student -will do well to practice for some time any fantastic -shapes which he can find drawn on flat surfaces, as on wall-papers, -carpets, etc., in order to accustom himself to the -strange and great changes which perspective causes in them.</p> - -<p class="illo"><img id="f.63" src="images/illus-095.png" alt="[Geometric diagram]" /><br - /><b>Fig. 63.</b></p> - -<p>The curves most required in architectural drawing, after -the circle, are those of pointed arches; in which, however, all -that will be generally needed is to fix the apex, and two -points in the sides. Thus if we have to draw a range of -pointed arches, such as <span class="maths">A P B</span>, <a href="#f.63">Fig. 63.</a>, draw the measured -arch to its sight-magnitude first neatly in a rectangle, <span class="maths">A B C D</span>; then draw the diagonals <span class="maths">A D</span> and <span class="maths">B C</span>; where they cut -the curve draw a horizontal line (as at the level <span class="maths">E</span> in the -figure), and carry it along the range to the vanishing-point, -fixing the points where the arches cut their diagonals all -along. If the arch is cusped, a line should be drawn, at <span class="maths">F</span> to -mark the height of the cusps, and verticals raised at <span class="maths">G</span> and <span class="maths">H</span>, -to determine the interval between them. Any other points -<a name="png.096" id="png.096" href="#png.096"><span class="pagenum"><span - class="ns">[p</span>86<span class="ns">] - </span></span></a>may be similarly determined, but these will usually be enough. -<a href="#f.63">Figure 63.</a> shows the perspective construction of a square -niche of good Veronese Gothic, with an uncusped arch of -similar size and curve beyond.</p> - -<p class="illo"><img id="f.64" src="images/illus-096.png" alt="[Geometric diagram]" /><br - /><b>Fig. 64.</b></p> - -<p>In <a href="#f.64">Fig. 64.</a> the more distant arch only is lettered, as the -construction of the nearest explains itself more clearly to the -eye without letters. The more distant arch shows the general -construction for all arches seen underneath, as of -bridges, cathedral aisles, etc. The rectangle <span class="maths">A B C D</span> is first -drawn to contain the outside arch; then the depth of the -arch, <span class="maths">A <var>a</var></span>, is determined by the measuring-line, and the rectangle, -<var>a b c d</var>, drawn for the inner arch.</p> - -<p><span class="maths">A <var>a</var></span>, <span class="maths">B <var>b</var></span>, etc., go to one vanishing-point; <span class="maths">A B</span>, <var>a b</var>, etc., to -the opposite one.</p> - -<p>In the nearer arch another narrow rectangle is drawn to -determine the cusp. The parts which would actually come -into sight are slightly shaded.</p> - - -</div> - -<div class="chap"> - -<h3 class="app" title="Problem XIV"><a name="png.097" id="png.097" href="#png.097"><span class="pagenum"><span - class="ns">[p</span>87<span class="ns">]<br - /></span></span></a>PROBLEM XIV.</h3> - - -<p><span class="smc">Several</span> exercises will be required on this important -problem.</p> - -<p>I. It is required to draw a circular flat-bottomed dish narrower -at the bottom than the top; the vertical depth being -given, and the diameter at the top and bottom.</p> - -<p class="illo"><img id="f.65" src="images/illus-097.png" alt="[Geometric diagram]" /><br - /><b>Fig. 65.</b></p> - -<p>Let <var>a b</var>, <a href="#f.65">Fig. 65.</a>, be the diameter of the bottom, <var>a c</var> the -diameter of the top, and <var>a d</var> its vertical depth.</p> - -<p>Take <span class="maths">A D</span> in position equal to <var>a c</var>.</p> - -<p>On <span class="maths">A D</span> draw the square <span class="maths">A B C D</span>, and inscribe in it a circle.</p> - -<p>Therefore, the circle so inscribed has the diameter of the -top of the dish.</p> - -<p>From <span class="maths">A</span> and <span class="maths">D</span> let fall verticals, <span class="maths">A E</span>, <span class="maths">D H</span>, each equal to <var>a d</var>.</p> - -<p>Join <span class="maths">E H</span>, and describe square <span class="maths">E F G H</span>, which accordingly -will be equal to the square <span class="maths">A B C D</span>, and be at the depth <var>a d</var> -beneath it.</p> - -<p>Within the square <span class="maths">E F G H</span> describe a square <span class="maths">I K</span>, whose -diameter shall be equal to <var>a b</var>.</p> - -<p>Describe a circle within the square <span class="maths">I K</span>. Therefore the -circle so inscribed has its diameter equal to <var>a b</var>; and it is -<a name="png.098" id="png.098" href="#png.098"><span class="pagenum"><span - class="ns">[p</span>88<span class="ns">] - </span></span></a>in the center of the square <span class="maths">E F G H</span>, which is vertically beneath -the square <span class="maths">A B C D</span>.</p> - -<p>Therefore the circle in the square <span class="maths">I K</span> represents the bottom -of the dish.</p> - -<p>Now the two circles thus drawn will either intersect one -another, or they will not.</p> - -<p>If they intersect one another, as in the figure, and they -are below the eye, part of the bottom of the dish is seen -within it.</p> - -<p class="illo"><img id="f.66" src="images/illus-098a.png" alt="[Geometric diagram]" /><br - /><b>Fig. 66.</b></p> - -<p>To avoid confusion, let us take then two intersecting circles -without the inclosing squares, as in <a href="#f.66">Fig. 66.</a></p> - -<p>Draw right lines, <var>a b</var>, <var>c d</var>, touching both circles externally. -Then the parts of these lines which connect the circles are -the sides of the dish. They are drawn in <a href="#f.65">Fig. 65.</a> without -any prolongations, but the best way to construct them is as -in <a href="#f.66">Fig. 66.</a></p> - -<p>If the circles do not intersect each other, the smaller must -either be within the larger or not within it.</p> - -<p>If within the larger, the whole of the bottom of the dish -is seen from above, <a href="#f.67">Fig. 67.</a> <var>a</var>.</p> - -<p class="illolt"><img id="f.67" src="images/illus-098b.png" alt="[Geometric diagram]" /><br - /><b>Fig. 67.</b></p> - -<p>If the smaller circle is not within -the larger, none of the bottom is seen -inside the dish, <var>b</var>.</p> - -<p>If the circles are above instead of -beneath the eye, the bottom of the -dish is seen beneath it, <var>c</var>.</p> - -<p>If one circle is above and another -beneath the eye, neither the bottom -nor top of the dish is seen, <var>d</var>. Unless -the object be very large, the circles in -this case will have little apparent -curvature.</p> - -<p>II. The preceding problem is simple, -<a name="png.099" id="png.099" href="#png.099"><span class="pagenum"><span - class="ns">[p</span>89<span class="ns">] - </span></span></a>because the lines of the profile of the object (<var>a b</var> and -<var>c d</var>, <a href="#f.66">Fig. 66.</a>) are straight. But if these lines of profile are -curved, the problem becomes much more complex: once mastered, -however, it leaves no farther difficulty in perspective.</p> - -<p>Let it be required to draw a flattish circular cup or vase, -with a given curve of profile.</p> - -<p>The basis of construction is given in <a href="#f.68">Fig. 68.</a>, half of it -only being drawn, in order that the eye may seize its lines -easily.</p> - -<p class="illo"><img id="f.68" src="images/illus-099.png" alt="[Geometric diagram]" /><br - /><b>Fig. 68.</b></p> - -<p>Two squares (of the required size) are first drawn, one -above the other, with a given vertical interval, <span class="maths">A C</span>, between -them, and each is divided into eight parts by its diameters -and diagonals. In these squares two circles are drawn; -which are, therefore, of equal size, and one above the other. -Two smaller circles, also of equal size, are drawn within -these larger circles in the construction of the present problem; -more may be necessary in some, none at all in others.</p> - -<p>It will be seen that the portions of the diagonals and diameters -of squares which are cut off between the circles -represent radiating planes, occupying the position of the -spokes of a wheel.</p> - -<p>Now let the line <span class="maths">A E B</span>, <a href="#f.69">Fig. 69.</a>, be the profile of the vase -or cup to be drawn.</p> - -<p>Inclose it in the rectangle <span class="maths">C D</span>, and if any portion of it is -not curved, as <span class="maths">A E</span>, cut off the curved portion by the vertical -line <span class="maths">E F</span>, so as to include it in the smaller rectangle <span class="maths">F D</span>.</p> - -<p><a name="png.100" id="png.100" href="#png.100"><span class="pagenum"><span - class="ns">[p</span>90<span class="ns">]<br - /></span></span></a>Draw the rectangle <span class="maths">A C B D</span> in position, and upon it construct -two squares, as they are constructed on the rectangle -<span class="maths">A C D</span> in <a href="#f.68">Fig. 68.</a>; and complete the construction of <a href="#f.68">Fig. 68.</a>, -making the radius of its large outer circles equal to <span class="maths">A D</span>, and -of its small inner circles equal to <span class="maths">A E</span>.</p> - -<p>The planes which occupy the position of the wheel spokes -will then each represent a rectangle of the size of <span class="maths">F D</span>. The -construction is shown by the dotted lines in <a href="#f.69">Fig. 69.</a>; <var>c</var> being -the center of the uppermost circle.</p> - -<p class="illo"><img id="f.69" src="images/illus-100.png" alt="[Geometric diagram]" /><br - /><b>Fig. 69.</b></p> - -<p>Within each of the smaller rectangles between the circles, -draw the curve <span class="maths">E B</span> in perspective, as in <a href="#f.69">Fig. 69.</a></p> - -<p>Draw the curve <var>x y</var>, touching and inclosing the curves in -the rectangles, and meeting the upper circle at <var>y</var>.<a name="fn32" id="fn32"></a><a title="Go to footnote 32" - href="#Footnote32" class="fnanchor"><span - class="ns">[Footnote </span>32<span class="ns">] - </span></a></p> - -<p>Then <var>x y</var> is the contour of the surface of the cup, and the -upper circle is its lip.</p> - -<p>If the line <var>x y</var> is long, it may be necessary to draw other -rectangles between the eight principal ones; and, if the curve -of profile <span class="maths">A B</span> is complex or retorted, there may be several -lines corresponding to <span class="maths">X Y</span>, inclosing the successive waves of -the profile; and the outer curve will then be an undulating or -broken one.</p> - -<p class="illo"><a name="png.101" id="png.101" href="#png.101"><span class="pagenum"><span - class="ns">[p</span>91<span class="ns">]<br - /></span></span></a><img class="nopr" id="f.70" src="images/illus-101.png" alt="[Geometric diagram]" /><img class="pronly" id="f.70pr" src="images/illus-101pr.png" alt="[Geometric diagram]" /><br - /><b>Fig. 70.</b></p> - -<p>III. All branched ornamentation, forms of flowers, capitals -of columns, machicolations of round towers, and other such -arrangements of radiating curve, are resolvable by this problem, -using more or fewer interior circles according to the -conditions of the curves. <a href="#f.70">Fig. 70.</a> is an example of the construction -of a circular group of eight trefoils with curved -stems. One outer or limiting circle is drawn within the -square <span class="maths">E D C F</span>, and the extremities of the trefoils touch it -at the extremities of its diagonals and diameters. A -<a name="png.102" id="png.102" href="#png.102"><span class="pagenum"><span - class="ns">[p</span>92<span class="ns">] - </span></span></a>smaller circle is at the vertical distance <span class="maths">B C</span> below the larger, -and <span class="maths">A</span> is the angle of the square within which the smaller -circle is drawn; but the square is not given, to avoid confusion. -The stems of the trefoils form drooping curves, arranged -on the diagonals and diameters of the smaller circle, -which are dotted. But no perspective laws will do work of -this intricate kind so well as the hand and eye of a painter.</p> - -<p>IV. There is one common construction, however, in which, -singularly, the hand and eye of the painter almost always -fail, and that is the fillet of any ordinary capital or base of -a circular pillar (or any similar form). It is rarely necessary -in practice to draw such minor details in perspective; -yet the perspective laws which regulate them should be understood, -else the eye does not see their contours rightly -until it is very highly cultivated.</p> - -<p class="illo"><img id="f.71" src="images/illus-102.png" alt="[Geometric diagram]" /><br - /><b>Fig. 71.</b></p> - -<p>Fig. 71. will show the law with sufficient clearness; it -represents the perspective construction of a fillet whose profile -is a semicircle, such as <span class="maths">F H</span> in <a href="#f.60">Fig. 60.</a>, seen above the -eye. Only half the pillar with half the fillet is drawn, to -avoid confusion.</p> - -<p><a name="png.103" id="png.103" href="#png.103"><span class="pagenum"><span - class="ns">[p</span>93<span class="ns">]<br - /></span></span></a><span class="maths">Q</span> is the center of the shaft.</p> - -<p><span class="maths">P Q</span> the thickness of the fillet, sight-magnitude at the shaft’s -center.</p> - -<p>Round <span class="maths">P</span> a horizontal semicircle is drawn on the diameter -of the shaft <var>a b</var>.</p> - -<p>Round <span class="maths">Q</span> another horizontal semicircle is drawn on diameter -<var>c d</var>.</p> - -<p>These two semicircles are the upper and lower edges of the -fillet.</p> - -<p>Then diagonals and diameters are drawn as in <a href="#f.68">Fig. 68.</a>, -and, at their extremities, semicircles in perspective, as in -<a href="#f.69">Fig. 69</a>.</p> - -<p>The letters <span class="maths">A</span>, <span class="maths">B</span>, <span class="maths">C</span>, <span class="maths">D</span>, and <span class="maths">E</span>, indicate the upper and exterior -angles of the rectangles in which these semicircles are -to be drawn; but the inner vertical line is not dotted in the -rectangle at <span class="maths">C</span>, as it would have confused itself with other -lines.</p> - -<p>Then the visible contour of the fillet is the line which incloses -and touches<a name="fn33" id="fn33"></a><a title="Go to footnote 33" - href="#Footnote33" class="fnanchor"><span - class="ns">[Footnote </span>33<span class="ns">] - </span></a> all the semicircles. It disappears behind -the shaft at the point <span class="maths">H</span>, but I have drawn it through -to the opposite extremity of the diameter at <var>d</var>.</p> - -<p>Turned upside down the figure shows the construction of -a basic fillet.</p> - -<p>The capital of a Greek Doric pillar should be drawn frequently -for exercise on this fourteenth problem, the curve -of its echinus being exquisitely subtle, while the general -contour is simple.</p> - -<div class="footnotes"> - -<p><small><span class="unjust"><a name="Footnote32" id="Footnote32"><span class="ns">[Footnote </span - >32<span class="ns">: </span></a> </span>This point coincides in the figure with the extremity of the horizontal -diameter, but only accidentally.<span class="ns">]</span> - <a title="Return to text" href="#fn32" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust"><a name="Footnote33" id="Footnote33"><span class="ns">[Footnote </span - >33<span class="ns">: </span></a> </span>The engraving is a little inaccurate; the inclosing line should -touch the dotted semicircles at <span class="maths">A</span> and <span class="maths">B</span>. The student should draw it -on a large scale.<span class="ns">]</span> - <a title="Return to text" href="#fn33" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - -</div> - -<div class="chap"> - -<h3 class="app" title="Problem XVI"><a name="png.104" id="png.104" href="#png.104"><span class="pagenum"><span - class="ns">[p</span>94<span class="ns">]<br - /></span></span></a>PROBLEM XVI.</h3> - -<p><span class="smc">It</span> is often possible to shorten other perspective operations -considerably, by finding the vanishing-points of the inclined -lines of the object. Thus, in drawing the gabled roof in -Fig. 43., if the gable <span class="maths">A Y C</span> be drawn in perspective, and the -vanishing-point of <span class="maths">A Y</span> determined, it is not necessary to draw -the two sides of the rectangle, <span class="maths">A′ D′</span> and <span class="maths">D′ B′</span>, in order to determine -the point <span class="maths">Y′</span>; but merely to draw <span class="maths">Y Y′</span> to the vanishing-point -of <span class="maths">A A′</span> and <span class="maths">A′ Y′</span> to the vanishing-point of <span class="maths">A Y</span>, -meeting in <span class="maths">Y′</span>, the point required.</p> - -<p>Again, if there be a series of gables, or other figures produced -by parallel inclined lines, and retiring to the point <span class="maths">V</span>, -as in <a href="#f.72">Fig. 72.</a>,<a name="fn34" id="fn34"></a><a title="Go to footnote 34" - href="#Footnote34" class="fnanchor"><span - class="ns">[Footnote </span>34<span class="ns">] - </span></a> it is not necessary to draw each separately, -but merely to determine their breadths on the line <span class="maths">A V</span>, and -draw the slopes of each to their vanishing-points, as shown -in <a href="#f.72">Fig. 72.</a> Or if the gables are equal in height, and a line -be drawn from <span class="maths">Y</span> to <span class="maths">V</span>, the construction resolves itself into a -zigzag drawn alternately to <span class="maths">P</span> and <span class="maths">Q</span>, between the lines <span class="maths">Y V</span> -and <span class="maths">A V</span>.</p> - -<p>The student must be very cautious, in finding the vanishing-points -of inclined lines, to notice their relations to the -horizontals beneath them, else he may easily mistake the -horizontal to which they belong.</p> - -<p>Thus, let <span class="maths">A B C D</span>, <a href="#f.73">Fig. 73.</a>, be a rectangular inclined plane, -and let it be required to find the vanishing-point of its diagonal -<span class="maths">B D</span>.</p> - -<p>Find <span class="maths">V</span>, the vanishing-point of <span class="maths">A D</span> and <span class="maths">B C</span>.</p> - -<p>Draw <span class="maths">A E</span> to the opposite vanishing-point, so that <span class="maths">D A E</span> -may represent a right angle.</p> - -<p>Let fall from <span class="maths">B</span> the vertical <span class="maths">B E</span>, cutting <span class="maths">A E</span> in <span class="maths">E</span>.</p> - -<p>Join <span class="maths">E D</span>, and produce it to cut the sight-line in <span class="maths">V′</span>.</p> - -<p class="illo"><a name="png.105" id="png.105" href="#png.105"><span class="pagenum"><span - class="ns">[p</span>95<span class="ns">]<br - /></span></span></a><img id="f.72" src="images/illus-105.png" alt="[Geometric diagram]" /><br - /><b>Fig. 72.</b></p> - -<p><a name="png.106" id="png.106" href="#png.106"><span class="pagenum"><span - class="ns">[p</span>96<span class="ns">]<br - /></span></span></a>Then, since the point <span class="maths">E</span> is vertically under the point <span class="maths">B</span>, the -horizontal line <span class="maths">E D</span> is vertically under the inclined line <span class="maths">B D</span>.</p> - -<p class="illo"><img id="f.73" src="images/illus-106.png" alt="[Geometric diagram]" /><br - /><b>Fig. 73.</b></p> - -<p class="noindent">So that if we now let fall the vertical <span class="maths">V′ P</span> from <span class="maths">V′</span>, and produce -<span class="maths">B D</span> to cut <span class="maths">V′ P</span> in <span class="maths">P</span>, the point <span class="maths">P</span> will be the vanishing-point -of <span class="maths">B D</span>, and of all lines parallel to it.<a name="fn35" id="fn35"></a><a title="Go to footnote 35" - href="#Footnote35" class="fnanchor"><span - class="ns">[Footnote </span>35<span class="ns">] - </span></a></p> - -<div class="footnotes"> - -<p><small><span class="unjust2"><a name="Footnote34" id="Footnote34"><span class="ns">[Footnote </span - >34<span class="ns">: </span></a> </span>The diagram is inaccurately cut. <span class="maths">Y V</span> should be a right line.<span class="ns">]</span> -<a title="Return to text" href="#fn34" class="fnreturn" - ><i>Return to text</i></a></small></p> - -<p><small><span class="unjust2"><a name="Footnote35" id="Footnote35"><span class="ns">[Footnote </span - >35<span class="ns">:</span></a> </span>The student may perhaps understand this construction better by -completing the rectangle <span class="maths">A D F E</span>, drawing <span class="maths">D F</span> to the vanishing-point -of <span class="maths">A E</span>, and <span class="maths">E F</span> to <span class="maths">V</span>. The whole figure, <span class="maths">B F</span>, may then be conceived as -representing half the gable roof of a house, <span class="maths">A F</span> the rectangle of its base, -and <span class="maths">A C</span> the rectangle of its sloping side.</small></p> - -<p class="ctd"><small>In nearly all picturesque buildings, especially on the Continent, the -slopes of gables are much varied (frequently unequal on the two sides), -and the vanishing-points of their inclined lines become very important, -if accuracy is required in the intersections of tiling, sides of dormer -windows, etc.</small></p> - -<p class="ctd"><small>Obviously, also, irregular triangles and polygons in vertical planes -may be more easily constructed by finding the vanishing-points of -their sides, than by the construction given in the <a href="#cor.ix.i">corollary to Problem IX.</a>; and if such triangles or polygons have others concentrically inscribed -within them, as often in Byzantine mosaics, etc., the use of -the vanishing-points will become essential.<span class="ns">]</span> -<a title="Return to text" href="#fn35" class="fnreturn" - ><i>Return to text</i></a></small></p> -</div> - -</div> - -<div class="chap"> - -<h3 class="app" title="Problem XVIII"><a name="png.107" id="png.107" href="#png.107"><span class="pagenum"><span - class="ns">[p</span>97<span class="ns">]<br - /></span></span></a>PROBLEM XVIII.</h3> - -<p><span class="smc">Before</span> examining the last three problems it is necessary -that you should understand accurately what is meant by the -position of an inclined plane.</p> - -<p>Cut a piece of strong white pasteboard into any irregular -shape, and dip it in a sloped position into water. However -you hold it, the edge of the water, of course, will always -draw a horizontal line across its surface. The direction of -this horizontal line is the direction of the inclined plane. -(In beds of rock geologists call it their “strike.”)</p> - -<p class="illo"><img id="f.74" src="images/illus-107.png" alt="[Geometric diagram]" /><br - /><b>Fig. 74.</b></p> - -<p>Next, draw a semicircle on the piece of pasteboard; draw -its diameter, <span class="maths">A B</span>, <a href="#f.74">Fig. 74.</a>, and a vertical line from its center, -<span class="maths">C D</span>; and draw some other lines, <span class="maths">C E</span>, <span class="maths">C F</span>, etc., from the center -to any points in the circumference.</p> - -<p>Now dip the piece of pasteboard again into water, and, -holding it at any inclination and in any direction you choose, -bring the surface of the water to the line <span class="maths">A B</span>. Then the -line <span class="maths">C D</span> will be the most steeply inclined of all the lines -drawn to the circumference of the circle; <span class="maths">G C</span> and <span class="maths">H C</span> will -be less steep; and <span class="maths">E C</span> and <span class="maths">F C</span> less steep still. The nearer -the lines to <span class="maths">C D</span>, the steeper they will be; and the nearer to -<span class="maths">A B</span>, the more nearly horizontal.</p> - -<p><a name="png.108" id="png.108" href="#png.108"><span class="pagenum"><span - class="ns">[p</span>98<span class="ns">]<br - /></span></span></a>When, therefore, the line <span class="maths">A B</span> is horizontal (or marks the -water surface), its direction is the direction of the inclined -plane, and the inclination of the line <span class="maths">D C</span> is the inclination -of the inclined plane. In beds of rock geologists call the -inclination of the line <span class="maths">D C</span> their “dip.”</p> - -<p>To fix the position of an inclined plane, therefore, is to -determine the direction of any two lines in the plane, <span class="maths">A B</span> -and <span class="maths">C D</span>, of which one shall be horizontal and the other at -right angles to it. Then any lines drawn in the inclined -plane, parallel to <span class="maths">A B</span>, will be horizontal; and lines drawn -parallel to <span class="maths">C D</span> will be as steep as <span class="maths">C D</span>, and are spoken of in -the text as the “steepest lines” in the plane.</p> - -<p>But farther, whatever the direction of a plane may be, if -it be extended indefinitely, it will be terminated, to the eye -of the observer, by a boundary line, which, in a horizontal -plane, is horizontal (coinciding nearly with the visible horizon);—in -a vertical plane, is vertical;—and, in an inclined -plane, is inclined.</p> - -<p>This line is properly, in each case, called the “sight-line” -of such plane; but it is only properly called the “horizon” -in the case of a horizontal plane: and I have preferred using -always the term “sight-line,” not only because more comprehensive, -but more accurate; for though the curvature of the -earth’s surface is so slight that practically its visible limit -always coincides with the sight-line of a horizontal plane, it -does not mathematically coincide with it, and the two lines -ought not to be considered as theoretically identical, though -they are so in practice.</p> - -<p>It is evident that all vanishing-points of lines in any plane -must be found on its sight-line, and, therefore, that the sight-line -of any plane may be found by joining any two of such -vanishing-points. Hence the construction of <a href="#pr.xviii">Problem XVIII</a>.</p> - - -</div> - -<div class="chap"> - - -<h2 class="pr" title="II. Demonstrations which could not conveniently -be included in the text"><a name="png.109" id="png.109" href="#png.109"><span class="pagenum"><span - class="ns">[p</span>99<span class="ns">]<br - /></span></span></a>II.</h2> -<h3 class="prbig" title="">DEMONSTRATIONS WHICH COULD NOT CONVENIENTLY -BE INCLUDED <span class="nw">IN THE TEXT</span>.</h3> - -<hr class="short" /> - -<h3 class="cor" title="I. The second corollary, Problem II">I.</h3> -<h3 class="pr" title="">THE SECOND COROLLARY, <span class="nw">PROBLEM II</span>.</h3> - - -<p><span class="smc">In</span> <a href="#f.8">Fig. 8.</a> omit the lines <span class="maths">C D</span>, <span class="maths">C′ D′</span>, and <span class="maths">D S</span>; and, as here -in <a href="#f.75">Fig. 75.</a>, from <var>a</var> draw <var>a d</var> parallel to <span class="maths">A B</span>, cutting <span class="maths">B T</span> in -<var>d</var>; and from <var>d</var> draw <var>d e</var> parallel to <span class="maths">B C′<!-- TN: original reads "BC" --></span>.</p> - -<p class="illo"><img id="f.75" src="images/illus-109.png" alt="[Geometric diagram]" /><br - /><b>Fig. 75.</b></p> - -<p>Now as <var>a d</var> is parallel to <span class="nw"><span class="maths">A B</span>—</span></p> - -<div class="displaymath"> -<a name="eq3" id="eq3"></a><span class="maths">A C ∶ <var>a c</var> ∷ B C′<!-- TN: original reads "BC" --> ∶ <var>d e</var></span>;<a title="See image" - href="#eqn3" class="eqnlink">[<span class="ns">eqn </span>iii]</a> - -<p>but <span class="maths">A C</span> is equal to <span class="nw"><span class="maths">B C′</span>—</span></p> - -<span class="maths">∴ <var>a c</var> = <var>d e</var></span>. -</div> - -<p class="gap"><a name="png.110" id="png.110" href="#png.110"><span class="pagenum"><span - class="ns">[p</span>100<span class="ns">]<br - /></span></span></a>Now because the triangles <span class="maths"><var>a c</var> V</span>, <span class="maths"><var>b c′</var> V</span>, are <span class="nw">similar—</span></p> - -<div class="displaymath"> -<a name="eq4" id="eq4"></a><span class="maths"><var>a c</var> ∶ <var>b c′</var> ∷ <var>a</var> V ∶ <var>b</var> V</span>;<a title="See image" - href="#eqn4" class="eqnlink">[<span class="ns">eqn </span>iv]</a> - -<p>and because the triangles <span class="maths"><var>d e</var> T</span>, <span class="maths"><var>b c′</var> T</span> are <span class="nw">similar—</span></p> - -<a name="eq5" id="eq5"></a><span class="maths"><var>d e</var> ∶ <var>b c′</var> ∷ <var>d</var> T ∶ <var>b</var> T</span>.<a title="See image" - href="#eqn5" class="eqnlink">[<span class="ns">eqn </span>v]</a> -</div> - -<p class="gap">But <var>a c</var> is equal to <span class="nw"><var>d e</var>—</span></p> - -<div class="displaymath"> -<a name="eq6" id="eq6"></a><span class="maths">∴ <var>a</var> V ∶ <var>b</var> V ∷ <var>d</var> T ∶ <var>b</var> T</span><a title="See image" - href="#eqn6" class="eqnlink">[<span class="ns">eqn </span>vi]</a>; - -<p><span class="maths">∴</span> the two triangles <var>a b d</var>, <span class="maths"><var>b</var> T V</span>, are similar, and their angles -are alternate;</p> - -<span class="maths">∴ T V</span> is parallel to <var>a d</var>. -</div> - -<p class="gap">But <var>a d</var> is parallel to <span class="nw"><span class="maths">A B</span>—</span></p> - -<div class="displaymath"> -<span class="maths">∴ T V</span> is parallel to <span class="maths">A B</span>.</div> - - -</div> - -<div class="chap"> - -<h3 class="app" title="II. The third corollary, Problem III"><a name="png.111" id="png.111" href="#png.111"><span class="pagenum"><span - class="ns">[p</span>101<span class="ns">]<br - /></span></span></a>II.</h3> -<h3 class="pr" title="">THE THIRD COROLLARY, <span class="nw">PROBLEM III</span>.</h3> - - -<p><span class="smc">In</span> <a href="#f.13">Fig. 13.</a>, since <span class="maths"><var>a</var> R</span> is by construction parallel to <span class="maths">A B</span> in -<a href="#f.12">Fig. 12.</a>, and <span class="maths">T V</span> is by construction in <a href="#pr.iii">Problem III.</a> also -parallel to <span class="nw"><span class="maths">A B</span>—</span></p> - -<div class="displaymath"> -<span class="maths">∴ <var>a</var> R</span> is parallel to <span class="maths">T V</span>,<br - /><span class="maths">∴ <var>a b</var> R</span> and <span class="maths">T <var>b</var> V</span> are alternate triangles,<br - /><a name="eq7" id="eq7"></a><span class="maths">∴ <var>a</var> R ∶ T V ∷ <var>a b</var> ∶ <var>b</var> V</span><a title="See image" - href="#eqn7" class="eqnlink">[<span class="ns">eqn </span>vii]</a>. -</div> - -<p class="gap">Again, by the construction of <a href="#f.13">Fig. 13.</a>, <span class="maths"><var>a</var> R′</span> is parallel to -<span class="nw"><span class="maths">M V</span>—</span></p> - - -<div class="displaymath"> -<span class="maths">∴ <var>a b</var> R′</span> and <span class="maths">M <var>b</var> V</span> are alternate triangles,<br - /><a name="eq8" id="eq8"></a><span class="maths">∴ <var>a</var> R′ ∶ M V ∷ <var>a b</var> ∶ <var>b</var> V</span><a title="See image" - href="#eqn8" class="eqnlink">[<span class="ns">eqn </span>viii]</a>. -</div> - -<p class="gapnoindent">And it has just been shown that also</p> - -<div class="displaymath"> -<a name="eq9" id="eq9"></a><span class="maths"><var>a</var> R ∶ T V ∷ <var>a b</var> ∶ <var>b</var> V</span>—<br - /><span class="maths">∴ <var>a</var> R′ ∶ M V ∷ <var>a</var> R ∶ T V</span><a title="See image" - href="#eqn9" class="eqnlink">[<span class="ns">eqn </span>ix]</a>. -</div> - -<p class="gapnoindent">But by construction, <span class="nw"><span class="maths"><var>a</var> R′</span> = <span class="maths"><var>a</var> R</span>—</span></p> - -<div class="displaymath"> -<span class="maths">∴ M V = T V</span>. -</div> - - - -</div> - -<div class="chap"> - -<h3 class="app" title="III. Analysis of Problem XV"><a name="png.112" id="png.112" href="#png.112"><span class="pagenum"><span - class="ns">[p</span>102<span class="ns">]<br - /></span></span></a>III.</h3> -<h3 class="pr" title="">ANALYSIS OF <span class="nw">PROBLEM XV</span>.</h3> - - -<p><span class="smc">We</span> proceed to take up the general condition of the second -problem, before left unexamined, namely, that in which the -vertical distances <span class="maths">B C′</span> and <span class="maths">A C</span> (<a href="#f.6">Fig. 6.</a> <a href="#png.023">page 13</a>), as well as -the direct distances <span class="maths">T D</span> and <span class="maths">T D′</span> are unequal.</p> - -<p>In <a href="#f.6">Fig. 6.</a>, here repeated (<a href="#f.76">Fig. 76.</a>), produce <span class="maths">C′ B</span> downwards, -and make <span class="maths">C′ E</span> equal to <span class="maths">C A</span>.</p> - -<p class="illo"><img id="f.76" src="images/illus-112.png" alt="[Geometric diagram]" /><br - /><b>Fig. 76.</b></p> - -<p>Join <span class="maths">A E</span>.</p> - -<p>Then, by the second <a href="#cor.ii.ii">Corollary</a> of <a href="#pr.ii">Problem II.</a>, <span class="maths">A E</span> is a -horizontal line.</p> - -<p class="gap">Draw <span class="maths">T V</span> parallel to <span class="maths">A E</span>, cutting the sight-line in <span class="maths">V</span>.</p> - -<div class="displaymath"> -<span class="maths">∴ V</span> is the vanishing-point of <span class="maths">A E</span>. -</div> - -<p><a name="png.113" id="png.113" href="#png.113"><span class="pagenum"><span - class="ns">[p</span>103<span class="ns">]<br - /></span></span></a>Complete the constructions of <a href="#pr.ii">Problem II.</a> and its second -<a href="#cor.ii.ii">Corollary.</a></p> - -<p>Then by <a href="#pr.ii">Problem II.</a> <var>a b</var> is the line <span class="maths">A B</span> drawn in perspective; -and by its <a href="#cor.ii.ii">Corollary</a> <var>a e</var> is the line <span class="maths">A E</span> drawn in -perspective.</p> - -<p class="gap">From <span class="maths">V</span> erect perpendicular <span class="maths">V P</span>, and produce <var>a b</var> to cut -it in <span class="maths">P</span>.</p> - -<p>Join <span class="maths">T P</span>, and from <var>e</var> draw <var>e f</var> parallel to <span class="maths">A E</span>, and cutting -<span class="maths">A T</span> in <var>f</var>.</p> - -<p>Now in triangles <span class="maths">E B T</span> and <span class="maths">A E T</span>, as <var>e b</var> is parallel to <span class="maths">E B</span> -and <var>e f</var> to <span class="nw"><span class="maths">A E</span>;—</span><a name="eq10" id="eq10"></a><span class="nw"><span class="maths"><var>e b</var> ∶ <var>e f</var> ∷ E B ∶ A E</span></span><a title="See image" - href="#eqn10" class="eqnlink">[<span class="ns">eqn </span>x]</a>.</p> - -<p>But <span class="maths">T V</span> is also parallel to <span class="maths">A E</span> and <span class="maths">P V</span> to <var>e b</var>.</p> - -<p>Therefore also in the triangles <span class="maths"><var>a</var> P V</span> and <span class="maths"><var>a</var> V T</span>,</p> - -<div class="displaymath"> -<a name="eq11" id="eq11"></a><span class="maths"><var>e b</var> ∶ <var>e f</var> ∷ P V ∶ V T</span><a title="See image" - href="#eqn11" class="eqnlink">[<span class="ns">eqn </span>xi]</a>. -</div> - -<p>Therefore <a name="eq12" id="eq12"></a><span class="nw"><span class="maths">P V ∶ V T ∷ E B ∶ A E</span></span><a title="See image" - href="#eqn12" class="eqnlink">[<span class="ns">eqn </span>xii]</a>.</p> - -<p>And, by construction, angle <span class="nw"><span class="maths">T P V = ∠ A E B</span>.</span></p> - -<p>Therefore the triangles <span class="maths">T V P</span>, <span class="maths">A E B</span>, are similar; and <span class="maths">T P</span> -is parallel to <span class="maths">A B</span>.</p> - -<p class="gap"><a name="png.114" id="png.114" href="#png.114"><span class="pagenum"><span - class="ns">[p</span>104<span class="ns">]<br - /></span></span></a>Now the construction in this problem is entirely general -for any inclined line <span class="maths">A B</span>, and a horizontal line <span class="maths">A E</span> in the -same vertical plane with it.</p> - -<p>So that if we find the vanishing-point of <span class="maths">A E</span> in <span class="maths">V</span>, and -from <span class="maths">V</span> erect a vertical <span class="maths">V P</span>, and from <span class="maths">T</span> draw <span class="maths">T P</span> parallel -to <span class="maths">A B</span>, cutting <span class="maths">V P</span> in <span class="maths">P</span>, <span class="maths">P</span> will be the vanishing-point of <span class="maths">A B</span>, -and (by the same proof as that given at <a href="#png.027">page 17</a>) of all lines -parallel to it.</p> - -<p class="illo"><img id="f.77" src="images/illus-113.png" alt="[Geometric diagram]" /><br - /><b>Fig. 77.</b></p> - -<p>Next, to find the dividing-point of the inclined line.</p> - -<p>I remove some unnecessary lines from <a href="#f.76">the last figure</a> and -repeat it here, <a href="#f.77">Fig. 77.</a>, adding the measuring-line <span class="maths"><var>a</var> M</span>, that -the student may observe its position with respect to the other -lines before I remove any more of them.</p> - -<p>Now if the line <span class="maths">A B</span> in this diagram represented the length -of the line <span class="maths">A B</span> in reality (as <span class="maths">A B</span> <em>does</em> in <a href="#f.10">Figs. 10.</a> and <a href="#f.11">11.</a>), -we should only have to proceed to modify <a href="#cor.ii.iii">Corollary III.</a> of -<a href="#pr.ii">Problem II.</a> to this new construction. We shall see presently -that <span class="maths">A B</span> does not represent the actual length of the inclined -line <span class="maths">A B</span> in nature, nevertheless we shall first proceed as if it -did, and modify our result afterwards.</p> - -<p><a name="png.115" id="png.115" href="#png.115"><span class="pagenum"><span - class="ns">[p</span>105<span class="ns">]<br - /></span></span></a>In <a href="#f.77">Fig. 77.</a> draw <var>a d</var> parallel to <span class="maths">A B</span>, cutting <span class="maths">B T</span> in <var>d</var>.</p> - -<p>Therefore <var>a d</var> is the sight-magnitude of <span class="maths">A B</span>, as <span class="maths"><var>a</var> R</span> is of -<span class="maths">A B</span> in <a href="#f.11">Fig. 11.</a></p> - -<p class="illo"><img id="f.78" src="images/illus-114.png" alt="[Geometric diagram]" /><br - /><b>Fig. 78.</b></p> - -<p>Remove again from the figure all lines except <span class="maths">P V</span>, <span class="maths">V T</span>, <span class="maths">P T</span>, -<var>a b</var>, <var>a d</var>, and the measuring-line.</p> - -<p>Set off on the measuring-line <var>a m</var> equal to <var>a d</var>.</p> - -<p>Draw <span class="maths">P Q</span> parallel to <var>a m</var>, and through <var>b</var> draw <span class="maths"><var>m</var> Q</span>, cutting -<span class="maths">P Q</span> in <span class="maths">Q</span>.</p> - -<p>Then, by the proof already given in <a href="#png.030">page 20</a>, <span class="nw"><span class="maths">P Q</span> = <span class="maths">P T</span>.</span></p> - -<p>Therefore if <span class="maths">P</span> is the vanishing-point of an inclined line -<span class="maths">A B</span>, and <span class="maths">Q P</span> is a horizontal line drawn through it, make <span class="maths">P Q</span> -equal to <span class="maths">P T</span>, and <var>a m</var> on the measuring-line equal to the -sight-magnitude of the line <span class="maths">A B</span> <em>in the diagram</em>, and the line -joining <span class="maths"><var>m</var> Q</span> will cut <span class="maths"><var>a</var> P</span> in <var>b</var>.</p> - -<p class="gap">We have now, therefore, to consider what relation the -length of the line <span class="maths">A B</span> in this diagram, <a href="#f.77">Fig. 77.</a>, has to the -length of the line <span class="maths">A B</span> in reality.</p> - -<p>Now the line <span class="maths">A E</span> in <a href="#f.77">Fig. 77.</a> represents the length of <span class="maths">A E</span> -in reality.</p> - -<p>But the angle <span class="maths">A E B</span>, <a href="#f.77">Fig. 77.</a>, and the corresponding angle -in all the constructions of the earlier problems, is in reality a -right angle, though in the diagram necessarily represented as -obtuse.</p> - -<p class="illort"><img id="f.79" src="images/illus-115.png" alt="[Geometric diagram]" /><br - /><b>Fig. 79.</b></p> - -<p>Therefore, if from <span class="maths">E</span> we draw <span class="maths">E C</span>, as in <a href="#f.79">Fig. 79.</a>, at right -angles to <span class="maths">A E</span>, make <span class="maths">E C</span> <span class="nw">= <span class="maths">E B</span></span>, and join -<span class="maths">A C</span>, <span class="maths">A C</span> will be the real length of the line -<span class="maths">A B</span>.</p> - -<p>Now, therefore, if instead of <var>a m</var> in <a href="#f.78">Fig. 78.</a>, we take the real length of <span class="maths">A B</span>, that real -length will be to <var>a m</var> as <span class="maths">A C</span> to <span class="maths">A B</span> in <a href="#f.79">Fig. 79.</a></p> - -<p>And then, if the line drawn to the measuring-line -<span class="maths">P Q</span> is still to cut <span class="maths"><var>a</var> P</span> in <var>b</var>, it is evident that the -line <span class="maths">P Q</span> must be shortened in the same ratio that <var>a m</var> was -shortened; and the true dividing-point will be <span class="maths">Q′</span> in <a href="#f.80">Fig. 80.</a>, -fixed so that <span class="maths">Q′ P</span><!-- TN: original reads "P′" --> shall be to <span class="maths">Q P</span> as <var>a m′</var> is to <var>a m</var>; <var>a m′</var> representing -the real length of <span class="maths">A B</span>.</p> - -<p><a name="png.116" id="png.116" href="#png.116"><span class="pagenum"><span - class="ns">[p</span>106<span class="ns">]<br - /></span></span></a>But <var>a m′</var> is therefore to <var>a m</var> as <span class="maths">A C</span> is to <span class="maths">A B</span> in <a href="#f.79">Fig. 79.</a></p> - -<p>Therefore <span class="maths">P Q′</span> must be to <span class="maths">P Q</span> as <span class="maths">A C</span> is to <span class="maths">A B</span>.</p> - -<p>But <span class="maths">P Q</span> equals <span class="maths">P T</span> (<a href="#f.78">Fig. 78.</a>); and <span class="maths">P V</span> is to <span class="maths">V T</span> (in <a href="#f.78">Fig. 78.</a>) as <span class="maths">B E</span> is to <span class="maths">A E</span> (<a href="#f.79">Fig. 79.</a>).</p> - -<p>Hence we have only to substitute <span class="maths">P V</span> for <span class="maths">E C</span>, and <span class="maths">V T</span> for -<span class="maths">A E</span>, in <a href="#f.79">Fig. 79.</a>, and the resulting diagonal <span class="maths">A C</span> will be the -required length of <span class="maths">P Q′</span>.</p> - -<p class="illo"><img id="f.80" src="images/illus-116.png" alt="[Geometric diagram]" /><br - /><b>Fig. 80.</b></p> - -<p>It will be seen that the construction given in the text -(<a href="#f.46">Fig. 46.</a>) is the simplest means of obtaining this magnitude, -for <span class="maths">V D</span> in <a href="#f.46">Fig. 46.</a> (or <span class="maths">V M</span> in <a href="#f.15">Fig. 15.</a>) = <span class="maths">V T</span> by construction -in <a href="#pr.iv">Problem IV</a>. It should, however, be observed, -that the distance <span class="maths">P Q′</span> or <span class="maths">P X</span>, in <a href="#f.46">Fig. 46.</a>, may be laid on the -sight-line of the inclined plane itself, if the measuring-line -be drawn parallel to that sight-line. And thus any form may -be drawn on an inclined plane as conveniently as on a horizontal -one, with the single exception of the radiation of the -verticals, which have a vanishing-point, as shown in <a href="#pr.xx">Problem XX</a>.</p> - -<p class="finis">THE END.</p> - -</div> - -<div class="tnote"> -<h2>Transcriber’s Note</h2> - -<p>A handful of unequivocal typographical errors has been corrected.</p> - -<p>For increased clarity, a few diagrams have been shifted from their original -position in the text.</p> - -<div class="ADE"> -<p>Images for sections of the text where the <span class="maths">∶</span> ratio -and <span class="maths">∷</span> proportion symbols occur.</p> - -<ul> -<li><span class="eqnimage">[i]</span> <a name="eqn1" id="eqn1"><img id="eq1160" src="images/eq1160.png" - alt="P′ Q′ ∶ P Q ∷ S T ∶ D T" /></a><a title="Return to text" href="#eq1" class="fnreturn" - ><small><i>Return to text</i></small></a></li> - -<li><span class="eqnimage">[ii]</span> <a name="eqn2" id="eqn2"><img id="eq1507" src="images/eq1507.png" - alt="A T ∶ a T ∷ B T ∶ b T" /></a><a title="Return to text" href="#eq2" class="fnreturn" - ><small><i>Return to text</i></small></a></li> - -<li><span class="eqnimage">[iii]</span> <a name="eqn3" id="eqn3"><img id="eq4245" src="images/eq4245.png" - alt="A C ∶ a c ∷ B C′ ∶ d e" /></a><a title="Return to text" href="#eq3" class="fnreturn" - ><small><i>Return to text</i></small></a></li> - -<li><span class="eqnimage">[iv]</span> <a name="eqn4" id="eqn4"><img id="eq4257" src="images/eq4257.png" - alt="a c ∶ b c′ ∷ a V ∶ b V" /></a><a title="Return to text" href="#eq4" class="fnreturn" - ><small><i>Return to text</i></small></a></li> - -<li><span class="eqnimage">[v]</span> <a name="eqn5" id="eqn5"><img id="eq4261" src="images/eq4261.png" - alt="d e ∶ b c′ ∷ d T ∶ b T" /></a><a title="Return to text" href="#eq5" class="fnreturn" - ><small><i>Return to text</i></small></a></li> - -<li><span class="eqnimage">[vi]</span> <a name="eqn6" id="eqn6"><img id="eq4267" src="images/eq4267.png" - alt="∴ a V ∶ b V ∷ d T ∶ b T" /></a><a title="Return to text" href="#eq6" class="fnreturn" - ><small><i>Return to text</i></small></a></li> - -<li><span class="eqnimage">[vii]</span> <a name="eqn7" id="eqn7"><img id="eq4298" src="images/eq4298.png" - alt="∴ a R ∶ T V ∷ a b ∶ b V" /></a><a title="Return to text" href="#eq7" class="fnreturn" - ><small><i>Return to text</i></small></a></li> - -<li><span class="eqnimage">[viii]</span> <a name="eqn8" id="eqn8"><img id="eq4307" src="images/eq4307.png" - alt="∴ a R′ ∶ M V ∷ a b ∶ b V" /></a><a title="Return to text" href="#eq8" class="fnreturn" - ><small><i>Return to text</i></small></a></li> - -<li><span class="eqnimage">[ix]</span> <a name="eqn9" id="eqn9"><img id="eq4314" src="images/eq4313_4.png" - alt="a R ∶ T V ∷ a b ∶ b V— - ∴ a R′ ∶ M V ∷ a R ∶ T V" /></a><a title="Return to text" href="#eq9" class="fnreturn" - ><small><i>Return to text</i></small></a></li> - -<li><span class="eqnimage">[x]</span> <a name="eqn10" id="eqn10"><img id="eq4373" src="images/eq4373.png" - alt="e b ∶ e f ∷ E B ∶ A E" /></a><a title="Return to text" href="#eq10" class="fnreturn" - ><small><i>Return to text</i></small></a></li> - -<li><span class="eqnimage">[xi]</span> <a name="eqn11" id="eqn11"><img id="eq4380" src="images/eq4380.png" - alt="e b ∶ e f ∷ P V ∶ V T" /></a><a title="Return to text" href="#eq11" class="fnreturn" - ><small><i>Return to text</i></small></a></li> - -<li><span class="eqnimage">[xii]</span> <a name="eqn12" id="eqn12"><img id="eq4383" src="images/eq4383.png" - alt="P V ∶ V T ∷ E B ∶ A E" /></a><a title="Return to text" href="#eq12" class="fnreturn" - ><small><i>Return to text</i></small></a></li> -</ul> - -</div> -</div> - - -<hr class="ww" /> - - - - - - - - - - - -<pre> - - - - - -End of Project Gutenberg's The Elements of Perspective, by John Ruskin - -*** END OF THIS PROJECT GUTENBERG EBOOK THE ELEMENTS OF PERSPECTIVE *** - -***** This file should be named 60816-h.htm or 60816-h.zip ***** -This and all associated files of various formats will be found in: - http://www.gutenberg.org/6/0/8/1/60816/ - -Produced by Juliet Sutherland, David Wilson and the Online -Distributed Proofreading Team at http://www.pgdp.net - -Updated editions will replace the previous one--the old editions will -be renamed. - -Creating the works from print editions not protected by U.S. copyright -law means that no one owns a United States copyright in these works, -so the Foundation (and you!) can copy and distribute it in the United -States without permission and without paying copyright -royalties. 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