Part 3

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] [Illustration: Fig. 35.] If either of the sides of the greater square is parallel to the plane of the picture, as _DG_ in Fig. 35., _DG_ of course must be equal to _AC_, and _DH_ equal to _BC_/2, and the construction is as in Fig. 35.

PROBLEM XIV.

TO DRAW A TRUNCATED CIRCULAR CONE, GIVEN IN POSITION AND MAGNITUDE, THE TRUNCATIONS BEING IN HORIZONTAL PLANES, AND THE AXIS OF THE CONE VERTICAL.

Let _ABCD_, Fig. 37., be the portion of the cone required.

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.