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

[p13]
PROBLEM II.

TO DRAW A RIGHT LINE BETWEEN TWO GIVEN POINTS.

[Geometric diagram]
Fig. 6.

Let A B, 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 T D, D C, and C A.

Let the direct, lateral, and vertical distances of the point B be T D′, D C′, 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 a b.

Then a b is the line required.

[p14]
COROLLARY I.

If the line A B is in a plane parallel to that of the picture, one end of the line A B must be at the same direct distance from the eye of the observer as the other.

Therefore, in that case, D T is equal to D′ T.

Then the construction will be as in Fig. 7.; and the student will find experimentally that a b is now parallel to A B.[Footnote 11]

[Geometric diagram]
Fig. 7.

And that a b is to A B as T S is to T D.

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 T S bears to T D.

[p15]
COROLLARY II.

If the line A B 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, A C equals B C′ (Fig. 6.).

And the construction is as in Fig. 8.

[Geometric diagram]
Fig. 8.

In Fig. 8. produce a b to the sight-line, cutting the sight-line in V; the point V, thus determined, is called the Vanishing-Point of the line A B.

Join T V. Then the student will find experimentally that T V is parallel to A B.[Footnote 12]

[p16]
COROLLARY III.

If the line A B produced would pass through some point beneath or above the station-point, C D is to D T as C′ D′ is to D′ T; in which case the point c coincides with the point c′, and the line a b 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.[Footnote 13]

[p17]
PROBLEM III.

TO FIND THE VANISHING-POINT OF A GIVEN HORIZONTAL LINE.

[Geometric diagram]
Fig. 9.

Let A B, Fig. 9., be the given line.

From T, the station-point, draw T V parallel to A B, cutting the sight-line in V.

V is the Vanishing-point required.[Footnote 14]

[p18]
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 A B, Fig. 10., be the given line, constructed upon the paper as in Fig. 8.; and let it be required to draw the line a b without using the point C′.

[Geometric diagram]
Fig. 10.

Find the position of the point A in a. (Problem I.)

[p19]
Find the vanishing-point of A B in V. (Problem III.)

Join a V.

Join B T, cutting a V in b.

Then a b is the line required.[Footnote 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.

[Geometric diagram]
Fig. 11.

From Fig. 8. remove, for the sake of clearness, the lines [p20] C′ D′, b V, and T V; and, taking the figure as here in Fig. 11., draw from a, the line a R parallel to A B, cutting B T in R.

Then a R is to A B as a T is to A T.

Then — is to — as c T is to C T.

Then — is to — as T S is to T D.

That is to say, a R is the sight-magnitude of A B.[Footnote 16]

[Geometric diagram]
Fig. 12.

Therefore, when the position of the point A is fixed in a, as in Fig. 12., and a V is drawn to the vanishing-point; if we draw a line a R from a, parallel to A B, and make a R equal to the sight-magnitude of A B, and then join R T, the line R T will cut a V in b.

So that, in order to determine the length of a b, we need not draw the long and distant line A B, but only a R parallel to it, and of its sight-magnitude; which is a great gain, for the line A B may be two miles long, and the line a R perhaps only two inches.

[p21]
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 a R′ and make a R′ equal to a R.

Through the points R and b draw R′ M, cutting the sight-line in M. Join T V. Now the reader will find experimentally that V M is equal to V T.[Footnote 17]

[Geometric diagram]
Fig. 13.

Hence it follows that, if from the vanishing-point V we lay off on the sight-line a distance, V M, equal to V T; then draw through a a horizontal line a R′, make a R′ equal to the sight-magnitude of A B, and join R′ M; the line R′ M will cut a V in b. And this is in practice generally the most convenient way of obtaining the length of a b.

[p22]
COROLLARY IV.

Removing from the preceding figure the unnecessary lines, and retaining only R′ M and a V, as in Fig. 14., produce the line a R′ to the other side of a, and make a X equal to a R′.

Join X b, and produce X b to cut the line of sight in N.

[Geometric diagram]
Fig. 14.

Then as X R′ is parallel to M N, and a R′ is equal to a X, V N must, by similar triangles, be equal to V M (equal to V T in Fig. 13.).

Therefore, on whichever side of V we measure the distance V T, so as to obtain either the point M, or the point N, if we measure the sight-magnitude a R′ or a X on the opposite side of the line a V, the line joining R′ M or X N will equally cut a V in b.

The points M and N are called the “Dividing-Points” of the original line A B (Fig. 12.), and we resume the results of these corollaries in the following three problems.

[p23]
PROBLEM IV.

TO FIND THE DIVIDING-POINTS OF A GIVEN HORIZONTAL LINE.

[Geometric diagram]
Fig. 15.

Let the horizontal line A B (Fig. 15.) be given in position and magnitude. It is required to find its dividing-points.

Find the vanishing-point V of the line A B.

With center V and distance V T, 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.

[p24]
PROBLEM V.

TO DRAW A HORIZONTAL LINE, GIVEN IN POSITION AND MAGNITUDE, BY MEANS OF ITS SIGHT-MAGNITUDE AND DIVIDING-POINTS.

[Geometric diagram]
Fig. 16.

Let A B (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 A B.

Join a V.

Through a draw a horizontal line a b′ and make a b′ equal to the sight-magnitude of A B. Join b′ M, cutting a V in b.

Then a b is the line required.

[p25]
COROLLARY I.

[Geometric diagram]
Fig. 17.

Supposing it were now required to draw a line A C (Fig. 17.) twice as long as A B, it is evident that the sight-magnitude a c′ must be twice as long as the sight-magnitude a b′; we have, therefore, merely to continue the horizontal line a b′, make b′ c′ equal to a b′, join c M′, cutting a V in c, and a c will be the line required. Similarly, if we have to draw a line A D, three times the length of A B, a d′ must be three times the length of a b′, and, joining d′ M, a d will be the line required.

The student will observe that the nearer the portions cut off, b c, c d, etc., approach the point V, the smaller they become; and, whatever lengths may be added to the line A D, and successively cut off from a V, the line a V 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.

[p26]
COROLLARY II.

It is evident that if the line A D 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, a d′, into the three equal parts, a b′, b′ c′, and c′ d′, and then, drawing to M from b′ and c′, the line a d would have been divided as required in b and c. And supposing the original line A D be divided irregularly into any number of parts, if the line a d′ 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 a d 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 a d, 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 A D.

If the line a V 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.

[p27]
PROBLEM VI.

TO DRAW ANY TRIANGLE, GIVEN IN POSITION AND MAGNITUDE, IN A HORIZONTAL PLANE.

[Geometric diagram]
Fig. 18.

Let A B C (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 A B be the side given in position and magnitude.

Then A B is a horizontal line, in a given position, and of a given length.

Draw the line A B. (Problem V.)

Let a b be the line so drawn.

Find V and V′, the vanishing-points respectively of the lines A C and B C. (Problem III.)

[p28]
From a draw a V, and from b, draw b V′, cutting each other in c.

Then a b c is the triangle required.

If A C is the line originally given, a c is the line which must be first drawn, and the line V′ b must be drawn from V′ to c and produced to cut a b in b. Similarly, if B C is given, V c must be drawn to c and produced, and a b from its vanishing-point to b, and produced to cut a c in a.

[p29]
PROBLEM VII.

TO DRAW ANY RECTILINEAR QUADRILATERAL FIGURE, GIVEN IN POSITION AND MAGNITUDE, IN A HORIZONTAL PLANE.

[Geometric diagram]
Fig. 19.

Let A B C D (Fig. 19.) be the given figure.

Join any two of its opposite angles by the line B C.

Draw first the triangle A B C. (Problem VI.)

And then, from the base B C, the two lines B D, C D, 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 A C or B C.

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 [p30] 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.

[p31]
PROBLEM VIII.

TO DRAW A SQUARE, GIVEN IN POSITION AND MAGNITUDE, IN A HORIZONTAL PLANE.

[Geometric diagram]
Fig. 20.

Let A B C D, 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 A B; and M, the dividing-point of A B, nearest S.

Find V′, the vanishing-point of A C; and N, the dividing-point of A C, nearest S.

[p32]
Draw the measuring-line through a, and make a b′, a c′, each equal to the sight-magnitude of A B.

(For since A B C D is a square, A C is equal to A B.)

Draw a V′ and c′ N, cutting each other in c.

Draw a V, and b′ M, cutting each other in b.

Then a c, a b, are the two nearest sides of the square.

Now, clearing the figure of superfluous lines, we have a b, a c, drawn in position, as in Fig. 21.

[Geometric diagram]
Fig. 21.

And because A B C D is a square, C D (Fig. 20.) is parallel to A B.

And all parallel lines have the same vanishing-point. (Note to Problem III.)

Therefore, V is the vanishing-point of C D.

Similarly, V′ is the vanishing-point of B D.

Therefore, from b and c (Fig. 22.) draw b V′, c V, cutting each other in d.

Then a b c d is the square required.

COROLLARY I.

It is obvious that any rectangle in a horizontal plane may be drawn by this problem, merely making a b′, on the measuring-line, Fig. 20., equal to the sight-magnitude of one of its sides, and a c′ the sight-magnitude of the other.

[p33]
COROLLARY II.

Let a b c d, Fig. 22., be any square drawn in perspective. Draw the diagonals a d and b c, cutting each other in C. Then C is the center of the square. Through C, draw e f to the vanishing-point of a b, and g h to the vanishing-point of a c, and these lines will bisect the sides of the square, so that a g is the perspective representation of half the side a b; a e is half a c; c h is half c d; and b f is half b d.

[Geometric diagram]
Fig. 22.

COROLLARY III.

Since A B C D, Fig. 20., is a square, B A C is a right angle; and as T V is parallel to A B, and T V′ to A C, V′ T V 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 V T V′ 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.

[p34]
PROBLEM IX.

TO DRAW A SQUARE PILLAR, GIVEN IN POSITION AND MAGNITUDE, ITS BASE AND TOP BEING IN HORIZONTAL PLANES.

Let A H, 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 A B or A C in position), and the height of its side A E.

[Geometric diagram]
Fig. 23.[Geometric diagram]
Fig. 24.

Find the sight-magnitudes of A B and A E. Draw the two sides a b, a c, of the square of the base, by Problem VIII., as in Fig. 24. From the points a, b, and c, raise vertical lines a e, c f, b g.

Make a e equal to the sight-magnitude of A E.

Now because the top and base of the pillar are in horizontal planes, the square of its top, F G, is parallel to the square of its base, B C.

Therefore the line E F is parallel to A C, and E G to A B.

Therefore E F has the same vanishing-point as A C, and E G the same vanishing-point as A B.

From e draw e f to the vanishing-point of a c, cutting c f in f.

Similarly draw e g to the vanishing-point of a b, cutting b g in g.

Complete the square g f in h, by drawing g h to the vanishing-point of e f, and f h to the vanishing-point of e g, cutting each other in h. Then a g h f is the square pillar required.

[p35]
COROLLARY.

It is obvious that if A E is equal to A C, the whole figure will be a cube, and each side, a e f c and a e g b, will be a square in a given vertical plane. And by making A B or A C 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.[Footnote 18]

[p36]
PROBLEM X.

TO DRAW A PYRAMID, GIVEN IN POSITION AND MAGNITUDE, ON A SQUARE BASE IN A HORIZONTAL PLANE.

[Geometric diagram]
Fig. 25.

Let A B, 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, C D.[Footnote 19]

[Geometric diagram]
Fig. 26.

Draw a square pillar, A B G E, Fig. 26., on the square base of the pyramid, and make the height of the pillar A F equal [p36] to the vertical height of the pyramid C D (Problem IX.). Draw the diagonals G F, H I, on the top of the square pillar, cutting each other in C. Therefore C is the center of the square F G H I. (Prob. VIII. Cor. II.)

[Geometric diagram]
Fig. 27.

Join C E, C A, C B.

Then A B C E 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.