The Elements of Perspective arranged for the use of schools and intended to be read in connection with the first three books of Euclid by Ruskin, John

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+----------------------------------+ | | | 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.