Part 2

[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.

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.

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.

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]

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.

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.

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′_.

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.

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]

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]

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_.

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.

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.

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.

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.

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.

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.

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.

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_.

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_.

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.

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]

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.)

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.

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.

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_.