Chapter 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.”

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_.

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.

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_.

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.

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.

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.

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_.

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.

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.

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] 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.

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.

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.

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_.

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.

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.

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.

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_).

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.

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:—

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.

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.

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.

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.