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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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