An example will be necessary to make this problem clear to the general student.
The nearest corner of a piece of pattern on the carpet is 4½feet beneath the eye, 2feet to our right and 3½feet in direct distance from us. We intend to make a drawing of the pattern which shall be seen properly when held 1½foot from the eye. It is required to fix the position of the corner of the piece of pattern.
[Geometric diagram]
Fig.51.
Let AB, Fig.51., be our sheet of paper, some 3feet wide. Make ST equal to 1½foot. Draw the line of sight through S. Produce TS, and make DS equal to 2feet, therefore TD equal to 3½feet. Draw DC, equal to 2feet; CP, equal to 4feet. Join TC (cutting the sight-line in Q) and TP.
Let fall the vertical QP', then P' is the point required.
If the lines, as in the figure, fall outside of your sheet of paper, in order to draw them, it is necessary to attach other sheets of paper to its edges. This is inconvenient, but must be done [p70] at first that you may see your way clearly; and sometimes afterwards, though there are expedients for doing without such extension in fast sketching.
It is evident, however, that no extension of surface could be of any use to us, if the distance TD, instead of being 3½ feet, were 100feet, or a mile, as it might easily be in a landscape.
It is necessary, therefore, to obtain some other means of construction; to do which we must examine the principle of the problem.
In the analysis of Fig.2., in the introductory remarks, I used the word “height” only of the tower, QP, because it was only to its vertical height that the law deduced from the figure could be applied. For suppose it had been a pyramid, as OQP, Fig.52., then the image of its side, QP, being, like every other magnitude, limited on the glass AB by the lines coming from its extremities, would appear only of the length Q'S; and it is not true that Q'S is to QP as TS is to TP. But if we let fall a vertical QD from Q, so as to get the vertical height of the pyramid, then it is true that Q'S is to QD as TS is to TD.
[Geometric diagram]
Fig.52.
Supposing this figure represented, not a pyramid, but a triangle on the ground, and that QD and QP are horizontal lines, expressing lateral distance from the line TD, still the rule would be false for QP and true for QD. And, similarly, it is true for all lines which are parallel, like QD, to [p71] the plane of the picture AB, and false for all lines which are inclined to it at an angle.
Hence generally. Let PQ (Fig.2. in Introduction, p.6) be any magnitude parallel to the plane of the picture; and P'Q' its image on the picture.
Then always the formula is true which you learned in the Introduction: P'Q' is to PQ as ST is to DT.
Now the magnitude P dash Q dash in this formula I call the “SIGHT-MAGNITUDE” of the line PQ. The student must fix this term, and the meaning of it, well in his mind. The “sight-magnitude” of a line is the magnitude which bears to the real line the same proportion that the distance of the picture bears to the distance of the object. Thus, if a tower be a hundred feet high, and a hundred yards off; and the picture, or piece of glass, is one yard from the spectator, between him and the tower; the distance of picture being then to distance of tower as 1 to 100, the sight-magnitude of the tower’s height will be as 1 to 100; that is to say, one foot. If the tower is two hundred yards distant, the sight-magnitude of its height will be half a foot, and so on.
But farther. It is constantly necessary, in perspective operations, to measure the other dimensions of objects by the sight-magnitude of their vertical lines. Thus, if the tower, which is a hundred feet high, is square, and twenty-five feet broad on each side; if the sight-magnitude of the height is one foot, the measurement of the side, reduced to the same scale, will be the hundredth part of twenty-five feet, or three inches: and, accordingly, I use in this treatise the term “sight-magnitude” indiscriminately for all lines reduced in the same proportion as the vertical lines of the object. If I tell you to find the “sight-magnitude” of any line, I mean, always, find the magnitude which bears to that line the proportion of ST to DT; or, in simpler terms, reduce the line to the scale which you have fixed by the first determination of the length ST.
Therefore, you must learn to draw quickly to scale before you do anything else; for all the measurements of your object [p72] must be reduced to the scale fixed by ST before you can use them in your diagram. If the object is fifty feet from you, and your paper one foot, all the lines of the object must be reduced to a scale of one fiftieth before you can use them; if the object is two thousand feet from you, and your paper one foot, all your lines must be reduced to the scale of one two-thousandth before you can use them, and so on. Only in ultimate practice, the reduction never need be tiresome, for, in the case of large distances, accuracy is never required. If a building is three or four miles distant, a hairbreadth of accidental variation in a touch makes a difference of ten or twenty feet in height or breadth, if estimated by accurate perspective law. Hence it is never attempted to apply measurements with precision at such distances. Measurements are only required within distances of, at the most, two or three hundred feet. Thus it may be necessary to represent a cathedral nave precisely as seen from a spot seventy feet in front of a given pillar; but we shall hardly be required to draw a cathedral three miles distant precisely as seen from seventy feet in advance of a given milestone. Of course, if such a thing be required, it can be done; only the reductions are somewhat long and complicated: in ordinary cases it is easy to assume the distance ST so as to get at the reduced dimensions in a moment. Thus, let the pillar of the nave, in the case supposed, be 42feet high, and we are required to stand 70feet from it: assume ST to be equal to 5feet. Then, as 5 is to 70 so will the sight-magnitude required be to 42; that is to say, the sight-magnitude of the pillar’s height will be 3feet. If we make ST equal to 2½feet, the pillar’s height will be 1½foot, and so on.
And for fine divisions into irregular parts which cannot be measured, the ninth and tenth problems of the sixth book of Euclid will serve you: the following construction is, however, I think, more practically convenient:—
The line AB (Fig.53.) is divided by given points, a, b, c, into a given number of irregularly unequal parts; it is required to divide any other line, CD, into an equal number [p73] of parts, bearing to each other the same proportions as the parts of AB, and arranged in the same order.
Draw the two lines parallel to each other, as in the figure.
Join AC and BD, and produce the lines AC, BD, till they meet in P.
Join aP, bP, cP, cutting cD in f, g, h.
Then the line CD is divided as required, in f, g, h.
In the figure the lines AB and CD are accidentally perpendicular to AP. There is no need for their being so.
[Geometric diagram]
Fig.53.
Now, to return to our first problem.
The construction given in the figure is only the quickest mathematical way of obtaining, on the picture, the sight-magnitudes of DC and PC, which are both magnitudes parallel with the picture plane. But if these magnitudes are too great to be thus put on the paper, you have only to obtain the reduction by scale. Thus, if TS be one foot, TD eighty feet, DC forty feet, and CP ninety feet, the distance QS must be made equal to one eightieth of DC, or half a foot; and the distance QP', one eightieth of CP, or one eightieth of ninety feet; that is to say, nine eighths of a foot, or thirteen and a half inches. The lines CT and PT are thus practically useless, it being only necessary to measure QS [p74] and QP, on your paper, of the due sight-magnitudes. But the mathematical construction, given in ProblemI., is the basis of all succeeding problems, and, if it is once thoroughly understood and practiced (it can only be thoroughly understood by practice), all the other problems will follow easily.
Lastly. Observe that any perspective operation whatever may be performed with reduced dimensions of every line employed, so as to bring it conveniently within the limits of your paper. When the required figure is thus constructed on a small scale, you have only to enlarge it accurately in the same proportion in which you reduced the lines of construction, and you will have the figure constructed in perspective on the scale required for use.
[p75]
PROBLEM IX.
The drawing of most buildings occurring in ordinary practice will resolve itself into applications of this problem. In general, any house, or block of houses, presents itself under the main conditions assumed here in Fig.54. There will be an angle or corner somewhere near the spectator, as AB; and the level of the eye will usually be above the base of the building, of which, therefore, the horizontal upper lines will slope down to the vanishing-points, and the base lines rise to them. The following practical directions will, however, meet nearly all cases:—
[Geometric diagram]
Fig.54.
Let AB, Fig.54., be any important vertical line in the block of buildings; if it is the side of a street, you may fix upon such a line at the division between two houses. If its real height, distance, etc., are given, you will proceed with [p76] the accurate construction of the problem; but usually you will neither know, nor care, exactly how high the building is, or how far off. In such case draw the line AB, as nearly as you can guess, about the part of the picture it ought to occupy, and on such a scale as you choose. Divide it into any convenient number of equal parts, according to the height you presume it to be. If you suppose it to be twenty feet high, you may divide it into twenty parts, and let each part stand for a foot; if thirty feet high, you may divide it into ten parts, and let each part stand for three feet; if seventy feet high, into fourteen parts, and let each part stand for five feet; and so on, avoiding thus very minute divisions till you come to details. Then observe how high your eye reaches upon this vertical line; suppose, for instance, that it is thirty feet high and divided into ten parts, and you are standing so as to raise your head to about six feet above its base, then the sight-line may be drawn, as in the figure, through the second division from the ground. If you are standing above the house, draw the sight-line above B; if below the house, below A; at such height or depth as you suppose may be accurate (a yard or two more or less matters little at ordinary distances, while at great distances perspective rules become nearly useless, the eye serving you better than the necessarily imperfect calculation). Then fix your sight-point and station-point, the latter with proper reference to the scale of the line AB. As you cannot, in all probability, ascertain the exact direction of the line AV or BV, draw the slope BV as it appears to you, cutting the sight-line in V. Thus having fixed one vanishing-point, the other, and the dividing-points, must be accurately found by rule; for, as before stated, whether your entire group of points (vanishing and dividing) falls a little more or less to the right or left of S does not signify, but the relation of the points to each other does signify. Then draw the measuring-line BG, either through A or B, choosing always the steeper slope of the two; divide the measuring-line into parts of the same length as those used on AB, and let them stand for the [p77] same magnitudes. Thus, suppose there are two rows of windows in the house front, each window six feet high by three wide, and separated by intervals of three feet, both between window and window and between tier and tier; each of the divisions here standing for three feet, the lines drawn from BG to the dividing-point D fix the lateral dimensions, and the divisions on AB the vertical ones. For other magnitudes it would be necessary to subdivide the parts on the measuring-line, or on AB, as required. The lines which regulate the inner sides or returns of the windows (a, b, c, etc.) of course are drawn to the vanishing-point of BF (the other side of the house), if FBV represents a right angle; if not, their own vanishing-point must be found separately for these returns. But see Practice on ProblemXI.
[Geometric diagram]
Fig.55.
Interior angles, such as EBC, Fig.55. (suppose the corner of a room), are to be treated in the same way, each side of the room having its measurements separately carried to it from the measuring-line. It may sometimes happen in such cases that we have to carry the measurement up from the corner B, and that the sight-magnitudes are given us from the length of the line AB. For instance, suppose the room is eighteen feet high, and therefore AB is eighteen feet; and we have to lay off lengths of six feet on the top of the room wall, BC. Find D, the dividing-point of BC. Draw a [p78] measuring-line, BF, from B; and another, gC, anywhere above. On BF lay off BG equal to one third of AB, or six feet; and draw from D, through G and B, the lines Gg, Bb, to the upper measuring-line. Then gb is six feet on that measuring-line. Make bc, ch, etc., equal to bg; and draw ce, hf, etc., to D, cutting BC in e and f, which mark the required lengths of six feet each at the top of the wall.
This is one of the most important foundational problems in perspective, and it is necessary that the student should entirely familiarize himself with its conditions.
In order to do so, he must first observe these general relations of magnitude in any pyramid on a square base.
Let AGH', Fig.56., be any pyramid on a square base.
[Geometric diagram]
Fig.56.
The best terms in which its magnitude can be given, are the length of one side of its base, AH, and its vertical altitude (CD in Fig.25.); for, knowing these, we know all the other magnitudes. But these are not the terms in which its size will be usually ascertainable. Generally, we shall have given us, and be able to ascertain by measurement, one side of its base AH, and either AG the length of one of the lines of its angles, or BG (or B'G) the length of a line drawn from its vertex, G, to the middle of the side of its base. In measuring a real pyramid, AG will usually be the line most easily found; but in many architectural problems BG is given, or is most easily ascertainable.
Observe therefore this general construction.
[Geometric diagram]
Fig.57.
Let ABDE, Fig.57., be the square base of any pyramid.
Draw its diagonals, AE, BD, cutting each other in its center, C.
Bisect any side, AB, in F.
From F erect vertical FG.
Produce FB to H, and make FH equal to AC.
Now if the vertical altitude of the pyramid (CD in Fig.25.) be given, make FG equal to this vertical altitude.
[p80]
Join GB and GH.
Then GB and GH are the true magnitudes of GB and GH in Fig.56.
If GB is given, and not the vertical altitude, with center B, and distance GB, describe circle cutting FG in G, and FG is the vertical altitude.
If GH is given, describe the circle from H, with distance GH, and it will similarly cut FG in G.
It is especially necessary for the student to examine this construction thoroughly, because in many complicated forms of ornaments, capitals of columns, etc., the lines BG and GH become the limits or bases of curves, which are elongated on the longer (or angle) profile GH, and shortened on the shorter (or lateral) profile BG. We will take a simple instance, but must previously note another construction.
It is often necessary, when pyramids are the roots of some ornamental form, to divide them horizontally at a given vertical height. The shortest way of doing so is in general the following.
[Geometric diagram]
Fig.58.
Let AEC, Fig.58., be any pyramid on a square base ABC, and ADC the square pillar used in its construction.
[p81]
Then by construction (ProblemX.) BD and AF are both of the vertical height of the pyramid.
Of the diagonals, FE, DE, choose the shortest (in this case DE), and produce it to cut the sight-line in V.
Therefore V is the vanishing-point of DE.
Divide DB, as may be required, into the sight-magnitudes of the given vertical heights at which the pyramid is to be divided.
[Geometric diagram]
Fig.59. [Geometric diagram]
Fig.60.
From the points of division, 1, 2, 3, etc., draw to the vanishing-point V. The lines so drawn cut the angle line of the pyramid, BE, at the required elevations. Thus, in the figure, it is required to draw a horizontal black band on the pyramid at three fifths of its height, and in breadth one twentieth of its height. The line BD is divided into five parts, of which three are counted from B upwards. Then the line drawn to V marks the base of the black band. Then one fourth of one of the five parts is measured, which similarly gives the breadth of the band. The terminal lines of the band are then drawn on the sides of the pyramid parallel to AB (or to its vanishing-point if it has one), and to the vanishing-point of BC.
[p82]
If it happens that the vanishing-points of the diagonals are awkwardly placed for use, bisect the nearest base line of the pyramid in B, as in Fig.59.
Erect the vertical DB and join GB and DG (G being the apex of pyramid).
Find the vanishing-point of DG, and use DB for division, carrying the measurements to the line GB.
In Fig.59., if we join AD and DC, ADC is the vertical profile of the whole pyramid, and BDC of the half pyramid, corresponding to FGB in Fig.57.
[Geometric diagram]
Fig.61.
We may now proceed to an architectural example.
Let AH, Fig.60., be the vertical profile of the capital of a pillar, AB the semi-diameter of its head or abacus, and FD the semi-diameter of its shaft.
Let the shaft be circular, and the abacus square, down to the level E.
Join BD, EF, and produce them to meet in G.
Therefore ECG is the semi-profile of a reversed pyramid containing the capital.
[p83]
Construct this pyramid, with the square of the abacus, in the required perspective, as in Fig.61.; making AE equal to AE in Fig.60., and AK, the side of the square, equal to twice AB in Fig.60. Make EG equal to CG, and ED equal to CD. Draw DF to the vanishing-point of the diagonal DV (the figure is too small to include this vanishing-point), and F is the level of the point F in Fig.60., on the side of the pyramid.
Draw Fm, Fn, to the vanishing-points of AH and AK. Then Fn and Fm are horizontal lines across the pyramid at the level F, forming at that level two sides of a square.
[Geometric diagram]
Fig.62.
Complete the square, and within it inscribe a circle, as in Fig.62., which is left unlettered that its construction may be clear. At the extremities of this draw vertical lines, which will be the sides of the shaft in its right place. It will be found to be somewhat smaller in diameter than the entire shaft in Fig.60., because at the center of the square it is more distant than the nearest edge of the square abacus. The curves of the capital may then be drawn approximately by the eye. They are not quite accurate in Fig.62., there [p84] being a subtlety in their junction with the shaft which could not be shown on so small a scale without confusing the student; the curve on the left springing from a point a little way round the circle behind the shaft, and that on the right from a point on this side of the circle a little way within the edge of the shaft. But for their more accurate construction see Notes on ProblemXIV.
[p85]
PROBLEM XI.
It is seldom that any complicated curve, except occasionally a spiral, needs to be drawn in perspective; but the student will do well to practice for some time any fantastic shapes which he can find drawn on flat surfaces, as on wall-papers, carpets, etc., in order to accustom himself to the strange and great changes which perspective causes in them.
[Geometric diagram]
Fig.63.
The curves most required in architectural drawing, after the circle, are those of pointed arches; in which, however, all that will be generally needed is to fix the apex, and two points in the sides. Thus if we have to draw a range of pointed arches, such as APB, Fig.63., draw the measured arch to its sight-magnitude first neatly in a rectangle, ABCD; then draw the diagonals AD and BC; where they cut the curve draw a horizontal line (as at the level E in the figure), and carry it along the range to the vanishing-point, fixing the points where the arches cut their diagonals all along. If the arch is cusped, a line should be drawn, at F to mark the height of the cusps, and verticals raised at G and H, to determine the interval between them. Any other points [p86] may be similarly determined, but these will usually be enough. Figure 63. shows the perspective construction of a square niche of good Veronese Gothic, with an uncusped arch of similar size and curve beyond.
[Geometric diagram]
Fig.64.
In Fig.64. the more distant arch only is lettered, as the construction of the nearest explains itself more clearly to the eye without letters. The more distant arch shows the general construction for all arches seen underneath, as of bridges, cathedral aisles, etc. The rectangle ABCD is first drawn to contain the outside arch; then the depth of the arch, Aa, is determined by the measuring-line, and the rectangle, abcd, drawn for the inner arch.
Aa, Bb, etc., go to one vanishing-point; AB, ab, etc., to the opposite one.
In the nearer arch another narrow rectangle is drawn to determine the cusp. The parts which would actually come into sight are slightly shaded.
Several exercises will be required on this important problem.
I. It is required to draw a circular flat-bottomed dish narrower at the bottom than the top; the vertical depth being given, and the diameter at the top and bottom.
[Geometric diagram]
Fig.65.
Let ab, Fig.65., be the diameter of the bottom, ac the diameter of the top, and ad its vertical depth.
Take AD in position equal to ac.
On AD draw the square ABCD, and inscribe in it a circle.
Therefore, the circle so inscribed has the diameter of the top of the dish.
From A and D let fall verticals, AE, DH, each equal to ad.
Join EH, and describe square EFGH, which accordingly will be equal to the square ABCD, and be at the depth ad beneath it.
Within the square EFGH describe a square IK, whose diameter shall be equal to ab.
Describe a circle within the square IK. Therefore the circle so inscribed has its diameter equal to ab; and it is [p88] in the center of the square EFGH, which is vertically beneath the square ABCD.
Therefore the circle in the square IK represents the bottom of the dish.
Now the two circles thus drawn will either intersect one another, or they will not.
If they intersect one another, as in the figure, and they are below the eye, part of the bottom of the dish is seen within it.
[Geometric diagram]
Fig.66.
To avoid confusion, let us take then two intersecting circles without the inclosing squares, as in Fig.66.
Draw right lines, ab, cd, touching both circles externally. Then the parts of these lines which connect the circles are the sides of the dish. They are drawn in Fig.65. without any prolongations, but the best way to construct them is as in Fig.66.
If the circles do not intersect each other, the smaller must either be within the larger or not within it.
If within the larger, the whole of the bottom of the dish is seen from above, Fig.67. a.
[Geometric diagram]
Fig.67.
If the smaller circle is not within the larger, none of the bottom is seen inside the dish, b.
If the circles are above instead of beneath the eye, the bottom of the dish is seen beneath it, c.
If one circle is above and another beneath the eye, neither the bottom nor top of the dish is seen, d. Unless the object be very large, the circles in this case will have little apparent curvature.
II. The preceding problem is simple, [p89] because the lines of the profile of the object (ab and cd, Fig.66.) are straight. But if these lines of profile are curved, the problem becomes much more complex: once mastered, however, it leaves no farther difficulty in perspective.
Let it be required to draw a flattish circular cup or vase, with a given curve of profile.
The basis of construction is given in Fig.68., half of it only being drawn, in order that the eye may seize its lines easily.
[Geometric diagram]
Fig.68.
Two squares (of the required size) are first drawn, one above the other, with a given vertical interval, AC, between them, and each is divided into eight parts by its diameters and diagonals. In these squares two circles are drawn; which are, therefore, of equal size, and one above the other. Two smaller circles, also of equal size, are drawn within these larger circles in the construction of the present problem; more may be necessary in some, none at all in others.
It will be seen that the portions of the diagonals and diameters of squares which are cut off between the circles represent radiating planes, occupying the position of the spokes of a wheel.
Now let the line AEB, Fig.69., be the profile of the vase or cup to be drawn.
Inclose it in the rectangle CD, and if any portion of it is not curved, as AE, cut off the curved portion by the vertical line EF, so as to include it in the smaller rectangle FD.
[p90]
Draw the rectangle ACBD in position, and upon it construct two squares, as they are constructed on the rectangle ACD in Fig.68.; and complete the construction of Fig.68., making the radius of its large outer circles equal to AD, and of its small inner circles equal to AE.
The planes which occupy the position of the wheel spokes will then each represent a rectangle of the size of FD. The construction is shown by the dotted lines in Fig.69.; c being the center of the uppermost circle.
[Geometric diagram]
Fig.69.
Within each of the smaller rectangles between the circles, draw the curve EB in perspective, as in Fig.69.
Draw the curve xy, touching and inclosing the curves in the rectangles, and meeting the upper circle at y.[Footnote 32]
Then xy is the contour of the surface of the cup, and the upper circle is its lip.
If the line xy is long, it may be necessary to draw other rectangles between the eight principal ones; and, if the curve of profile AB is complex or retorted, there may be several lines corresponding to XY, inclosing the successive waves of the profile; and the outer curve will then be an undulating or broken one.
[p91]
[Geometric diagram][Geometric diagram]
Fig.70.
III. All branched ornamentation, forms of flowers, capitals of columns, machicolations of round towers, and other such arrangements of radiating curve, are resolvable by this problem, using more or fewer interior circles according to the conditions of the curves. Fig.70. is an example of the construction of a circular group of eight trefoils with curved stems. One outer or limiting circle is drawn within the square EDCF, and the extremities of the trefoils touch it at the extremities of its diagonals and diameters. A [p92] smaller circle is at the vertical distance BC below the larger, and A is the angle of the square within which the smaller circle is drawn; but the square is not given, to avoid confusion. The stems of the trefoils form drooping curves, arranged on the diagonals and diameters of the smaller circle, which are dotted. But no perspective laws will do work of this intricate kind so well as the hand and eye of a painter.
IV. There is one common construction, however, in which, singularly, the hand and eye of the painter almost always fail, and that is the fillet of any ordinary capital or base of a circular pillar (or any similar form). It is rarely necessary in practice to draw such minor details in perspective; yet the perspective laws which regulate them should be understood, else the eye does not see their contours rightly until it is very highly cultivated.
[Geometric diagram]
Fig.71.
Fig.71. will show the law with sufficient clearness; it represents the perspective construction of a fillet whose profile is a semicircle, such as FH in Fig.60., seen above the eye. Only half the pillar with half the fillet is drawn, to avoid confusion.
[p93]
Q is the center of the shaft.
PQ the thickness of the fillet, sight-magnitude at the shaft’s center.
Round P a horizontal semicircle is drawn on the diameter of the shaft ab.
Round Q another horizontal semicircle is drawn on diameter cd.
These two semicircles are the upper and lower edges of the fillet.
Then diagonals and diameters are drawn as in Fig.68., and, at their extremities, semicircles in perspective, as in Fig.69.
The letters A, B, C, D, and E, indicate the upper and exterior angles of the rectangles in which these semicircles are to be drawn; but the inner vertical line is not dotted in the rectangle at C, as it would have confused itself with other lines.
Then the visible contour of the fillet is the line which incloses and touches[Footnote 33] all the semicircles. It disappears behind the shaft at the point H, but I have drawn it through to the opposite extremity of the diameter at d.
Turned upside down the figure shows the construction of a basic fillet.
The capital of a Greek Doric pillar should be drawn frequently for exercise on this fourteenth problem, the curve of its echinus being exquisitely subtle, while the general contour is simple.
