into nine equal squares, which are the outer upright surfaces of the nine cubes in the side AD of the square ABCD.

From the points where the lines, which are parallel to AD in this square, meet the side CD thereof, draw short lines to LN, all parallel to DL, and they will divide that side into the outer upright surfaces of the nine cubes which compose it: and then the outsides of all the cubes that can be visible to an observer, placed at a proper distance from the corner D of the square, will be finished.

As taught in Prob. 17. place the pyramid AE upright on its square base A t va, making it as high as you please; and the pyramid DH on its square base hu w D, of equal height with AE.

Draw EH from the top of one of these pyramids to the top of the other; and EH will be parallel to AD.

Draw ES and HS to the point of sight S, and HP to the point of distance P, intersecting ES in F.

From the point F, draw FG parallel to EH; then draw EG, and you will have a perspective square EFGH (parallel to ABCD) with its two diagonals EG and FH, intersecting one another in the centre of the square at I. The four corners of this square, E, F, G, H, give the perspective heights of the four pyramids AE, BF, CG, and DH; and the intersection I of the diagonals gives the height of the pyramid MI, the centre of whose base is the centre of the perspective square ABCD.

Lastly, place the three pyramids BF, CG, MI, upright on their respective bases at B, C, and M ; and the required perspective representation will be finished as in the figure.

PROB. 23. To put upright pyramids in perspective, on the sides of an oblong square or parallelogram; so that their distances from one another shall be equal to the breadth of the parallelogram.

In most of the foregoing operations we have considered the observer to be so placed, as to have an ob lique view of the perspective objects in this, we shall suppose him to have a direct view of fig. 26. that is, standing right against the middle of the end AD which is nearest to his eye, and viewing AD under an angle of 60 degrees.

Having cut AD in the middle, by the perpendicular line Ss, take S therein at pleasure for the point of sight, and draw ES for the horizon, parallel to AD.-Here Ss must be supposed to be produced downward, below the limits of the plate, to the place of the observer; and SE to be produced towards the left hand beyond E, far enough to take a proper point of distance therein, according to the foregoing rules.

Take A dat pleasure, and D g equal to A d, for the breadths of the square bases of the two pyramids AE and DF next the eye: then draw AS and d'S, and likewise DS and g S, to the point of sight S; and DG on to the point of distance, intersecting AS in G: then, from G draw GI parallel to AD, you will have the first perspective square AGID of the parallelogram ABCD.

From I draw. IH to (or toward) the point of distance, intersecting AS in H: then, from H draw VOL. XVI. Part I.

+

HK parallel to AD, and you will have the second perspective square GHKI of the parallelogram.—Go on in this manner till you have drawn as many perspective squares up toward S as you please.

Through the point e, where DG intersect g S, draw bf parallel to AD; and you will have formed the two perspective square bases A b c d and e f D g of the two pyramids at A and D.

From the point f (the upper outward corner of ef D g) draw fh toward the point of distance, till it meets AS in h; then, from this point of meeting, draw h m parallel to GI, and you will have formed the two perspective squares Ghik and Im In, for the square bases of the two pyramids at G and I.

Proceed in the same manuer to find the bases of all the other pyramids, at the corners of the rest of the perspective squares in the parallelogram ABCD, as shown by the figure.-Then,

Having placed the first two pyramids at A and D upright on their square bases, as shown in Prob. 9. and made them of any equal heights at pleasure, draw ES and FS from the tops of these pyramids to the point of sight S place all the rest of the pyramids upright on their respective bases, making their tops touch the straight lines ES and FS; and all the work, except the shading part, will be finished.

PROB. 24. To put a square pyramid of equal sized cubes in perspective.

Fig. 27. represents a pyramid of this kind; consist- Fig. 27. ing as it were of square tables of cubes, one table above another; 81 in the lowest, 49 in the next, 25 in the third, 9 in the fourth, and 1 in the fifth or uppermost. These are the square numbers of 9, 7, 5, 3, and I.

If the artist is already master of all the preceding operations, he will find less difficulty in this than in attending to the following description of it: for it cannot be described in a few words, but may be executed in a very short time.

In fig. 28. having drawn PS for the horizon, and ta- Fig. 28. ken S for the point of sight therein (the observer being at O) draw AD parallel to PS for the side (next the eye) of the first or lowermost table of cubes. Draw AS and DS to the point of sight S, and DP to the point of distance P, intersecting AS in the point B. Then, from B, draw BC parallel to AD, and you will have the surface ABCD of the first table.

Divide AD into nine equal parts, as A a, a b, b cd, &c. then make AK and DL equal to A a, and perpendicular to AD. Draw KL parallel to AD, and from the points of equal division at a, b, c, &c. draw lines to KL, all parallel to AK. Then draw hS to the point of sight S, and from the division points a, b, c, &c. draw lines with a black lead pencil, all tending towards the point of sight, till they meet the diagonal BD of the square.

From these points of meeting draw black lead lines to DC, all parallel to AD; then draw the parts of these lines with black ink which are marked 1, 2, 3, 4, &c. between h E and DC.

Having drawn the first of these lines q with black ink, draw the parts a i, b k, cl, &c. (of the former lines which met the diagonal BD) with black ink also; and rub out the rest of the black lead lines, which Ꮓ would

would otherwise confuse the following part of the work. Then, draw LF towards the point of sight S; and, from the points where the lines 1, 2, 3, 4, &c. meet the line DC, draw lines down to LF, all parallel to DL; and all the visible lines between the cubes in the first table will be finished.

Make i G equal and perpendicular to 6 i, and q M equal and parallel to i G; then draw GM, which will be equal and parallel to i q. From the points k, l, m, n, &c. draw k, n, lo, m p, &c. all parallel to i G, and the outsides of the seven cubes in the side G q of the second table will be finished.

Draw GS and MS to the point of sight S, and MP to the point of distance P, intersecting GS in H; then, from the point of intersection H, draw HI parallel to AD; and you will have the surface GHIM of the second table of cubes.

From the points n, o, p, q, &c. draw black lead lines toward the point of sight S, till they meet the diagonal MH of the perspective square surface GHIM; and draw SM, with black ink, toward the point of sight.

From those points where the lines drawn from 22, 0, P, q, &c. meet the diagonal MH, draw black lead lines to MI, all parallel to AD; only draw the whole first line y I with black ink, and the parts 2, 3, 4, &c. and ut, ou, pv, &c. of the other lines between y N and MI, and GM and y 1, with the same; and rub out all the rest of the black lead lines, to avoid further confusion. Then, from the points where the short lines 1, 2, 3, &c. meet the line MI, draw lines down to q E, all parallel to M q, and the outer surfaces of the seven cubes in the side ME will be finished; and all these last lines will meet the former parallels 2, 3, 4, &c. in the line q E. q

Make O equal and perpendicular to y t, and y P equal and parallel to t O; then draw OP, which will be equal and parallel to t y.-This done, draw OS and PS to the point of sight S, and PP to the point of distance P in the horizon. Lastly, from the point Q, where PP intersects OS, draw QR parallel to OP; and you will have the outlines OQRP of the surface of the third perspective table of cubes.

From the points u, v, w, x, draw upright lines to OP, all parallel to t O, and you will have the outer surfaces of the five cubes in the side O y of this third table.

From the points where these upright lines meet OP, draw lines toward the point of sight S, till they meet the diagonal PQ; and from these points of meeting draw lines to PR, all parallel to OP, making the parts 2, 3, 4, 5, of these lines with black ink which lie between ZY and PR. Then, from the points where these lines meet PR, draw lines down to y N; which will bound the outer surfaces of the five cubes in the side PN of the third table.

I

Draw the line with black ink; and, at a fourth part of its length between and Z, draw au upright line to S, equal in length to that fourth part, and another equal and parallel thereto from Z to V: then draw SV parallel to Z, and draw the two upright and equidistant lines between ♪ Z and SV, and you will have the outer surfaces of the three cubes in the side SZ of the fourth table.

Draw SS and VS to the point of sight S in the ho

rizon, and VP to the point of distance therein, intersecting SS in T; then draw TU parallel to SV, and you have STUV, the surface of the fourth table, which being reticulated or divided into 9 perspective small squares, and the uppermost cube W placed on the middlemost of the squares, all the outlines will be finished; and when the whole is properly shaded, as in fig. 27. the work will be done.

PROB. 25. To represent a double cross in perspective.

In fig. 29. let ABCD and EFGH be the two per- Fig. 29. spective squares, equal and parallel to one another, the uppermost directly above the lowermost, drawn by the rules already laid down, and as far asunder as is equal to the given height of the upright part of the cross; S being the point of sight, and P the point of distance, in the horizon PS taken parallel to AD.

Draw AE, DH, and CG; then AEHD and DHGC shall be the two visible sides of the upright part of the cross; of which, the length AE is here made equal to three times the breadth EH.

Through these points of division, at I and K, draw MO Divide DH into three equal parts, HI, IK, and KD. and PR parallel to AD; and make the parts MN, IO, PQ, KR, each equal to HI: then draw MP and OR parallel to DH.

From M and O, draw MS and OS to the point of sight S; and from the point of distance P draw PN cutting MS in T: from T draw TU parallel to MO, and meeting OS in U; and you will have the uppermost surface MTUO of one of the cross pieces of the figure.From R, draw RS to the point of sight S; and from U draw UV parallel to OR; and OUVR shall be the perspective square end next the eye of that cross part.

Draw PMX (as long as you please) from the point of distance P, through the corner M; lay a ruler to N and S, and draw XN from the line PX-then lay the ruler to I and S, and draw YZS.-Draw XY parallel to MO, and make XW and YB equal and perpendicular to XY: then draw WB parallel to XY, and WXYB shall be the square visible end of the other cross part of the figure.

Draw BK toward the point of sight S; and from U draw UP to the point of distance P, intersecting YS in Z: then, from the intersection Z, draw Z a parallel to MO, and Z b parallel to HD, and the whole delineation will be finished.

will have a true perspective representation of a double This done, shade the whole, as in fig. 30. and you Fig. 30.

cross.

PROB. 26. To put three rows of upright square objects in perspective, equal in size, and at equal distances from each other, on an oblong square plane, the breadth of which shall be of any assigned proportion to the length thereof.

Fig. 31. is a perspective representation of an oblong Fig. 31. square plane, three times as long as it is broad, having a row of nine upright square objects on each side, and one of the same number in the middle; all equally high, and at equal distances from one another, both long-wise and cross-wise, on the same plane.

In fig. 32. PS is the horizon, S. the point of sight, P Fig. 32

the

the point of distance, and AD (parallel to PS) the breadth of the plane.

Draw AS, NS, and DS, to the point of sight S; the point N being in the middle of the line AD: and draw DP to the point of distance P, intersecting AS in the point B then, from B draw BC parallel to AD, and you have the perspective square ABCD.

Through the point i, where DB intersects NS, draw a e parallel to AD; and you will have subdivided the perspective square ABCD into four lesser squares, as Aai N, Nie D, a B k i, and i k Ce.

From the point C (at the top of the perspective square ABCD) draw CP to the point of distance P, intersecting AS in E; then from the point E draw EF parallel to AD; and you will have the second perspective square BEFC.

Through the point 7, where CE intersects NS, draw bf parallel to AD; and you will have subdivided the square BEFC into the four squares B blk, klf C, b Fm l, and Im Ff.

From the point F (at the top of the perspective square BEFC) draw FP to the point of distance P, intersecting AS in I; then from the point I draw IK rallel to AD; and you will have the third perspective square EIKF.

pa

Through the point n, where FI intersects NS, draw cg parallel to AD; and you will have subdivided the square EIKF into four lesser squares, Ec nm, m ng F, c Ion, and no Kg.

From the point K (at the top of the third perspective square EIKF) draw KP to the point of distance P, intersecting AS in L; then from the point L draw LM parallel to AD; and you will have the fourth perspective square ILMK.

Through the point p, where KL intersects NS, draw dh parallel to AD; and you will have subdivided the square ILMK into the four lesser squares Id po, oph K, dLqp, and p q Mh.

Thus we have formed an oblong square ALMD, whose perspective length is equal to four times its breadth, and it contains 16 equal perspective squares.—If greater length was still wanted, we might proceed further on toward S.

Take A 3, equal to the intended breadth of the side of the upright square object AQ (all the other sides being of the same breadth), and AO for the intended height. Draw O 18 parallel to AD, and make D 8 and 47 equal to A 3; then draw 3 S, 4 S, 7 S, and 8 S to the point of sight S; and among them we shall have the perspective square bases of all the 27 upright objects on the plane.

you

Through the point 9, where DB intersects 8 S, draw 1 10 parallel to AD, and have the three perspective square bases A 1 2 3, 4 5 6 7, 8 9 10 D, of the three upright square objects at A, N, and D. Through the point 21, where eb intersects 8 S, draw 14, 11 parallel to AD; and you will have the three perspective squares, a 14 15 16 17 18 19 20, and 21 11 e 22, for the bases of the second cross row of objects; namely, the next beyond the first three at A, N, and D.

Through the point w, where CE intersects 8 S, draw a line parallel to BC; and you will have three perspective squares, at B, k, and Ĉ, for the base of the third row of objects; one of which is set up at B.

Through the point x, where fc intersects 8 S, draw a line parallel to bf; and you will have three perspective squares, at b, l, and æ, for the bases of the fourth cross row of objects.

Go on in this manner, as you see in the figure, to find the rest of the square bases, up to LM; and you will have 27 upon the whole oblong square plane, on which you are to place the like number of objects, as in fig. 31.

Having assumed AO for the perspective height of the three objects at A, N, and D (fig. 32.) next the observer's eye, and drawn O 18 parallel to AD, in order to make the objects at N and D of the same height as that at O; and having drawn the upright lines 4 15, 7 W, 8 X, and D 22, for the heights at N and D; draw OS and RS, 15 S and WS, XS and 22 S, all to the point of sight S: and these lines will determine the perspectively equal heights of all the rest of the upright objects, as shown by the two placed at a and B.

To draw the square tops of these objects, equal and parallel to their bases, we only need give one example, which will serve for all.

Draw 3 Rand 2 Q parallel to AO, and up to the line RS; then draw PQ parallel to OR, and OPQR shall be the top of the object at A, equal and parallel to its square base A 1 2 3.-In the same easy way the tops of all the other objects are formed.

When all the rest of the objects are delineated, shade them properly, and the whole perspective scheme will have the appearance of fig. 31.

PROB. 27. To put a square box in perspective, containing a given number of lesser square boxes of a depth equal to their width.

Let the given number of little square boxes or cells Fig. 33. be 16, then 4 of them make the length of each side of the four outer sides a b, b c, c d, da, as in fig. 33. and the depth af is equal to the width a e. Whoever can draw the reticulated square, by the rules laid down towards the beginning of this article, will be at no loss about putting this perspective scheme in practice. PROB. 28. To put stairs with equal and parallel steps in perspective.

In fig. 34. let a b be the given breadth of each step, Fig 34. and ai the height thereof. Make bc, cd, de, &c. each equal to ab; and draw all the upright lines ai, bl, cn, dp, &c. perpendicular to a h (to which the horizons S is parallel); and from the points i, 1, n, p, r, &c. draw the equidistant lines i B, C, n D, &c. parallel to a h; these distances being equal to that of i B from a h.

Draw a i touching all the corner-points 1, n, p, r, t, v; and draw 2 16 parallel to r i, as far from it as you want the length of the steps to be.

Toward the point of sight S draw the lines a 1, i 2, k 3, 14, &c. and draw 16 15, 14 13, 12 11, 10 9, 87, 6 5,4 3, and 2 1, all parallel to A h, and meeting the lines w 15, u 13, s 11, &c. in the points 15, 13, 11, 9, 7, 5, 3, and 1: then from these points draw 15 14, 13 12, 11 10,9 3, 7 6, 5 4, and 3 2, all parallel to ha; and the outlines of the steps will be finished. From the point 16 draw 16 A parallel to h a, and A x 16 will be part of the flat at the top of the uppermost step. Z 2

This

This done, shade the work as in fig. 35. and the whole will be finished.

PROB. 29. To put stairs with flats and opening in perspective, standing on a horizontal pavement of

squares.

In fig. 36. having made S the point of sight, and drawn a recticulated pavement AB with black lead lines, which may be rubbed out again; at any distance from the side AB of the pavement which is nearest to the eye, and at any point where you choose to begin the stair at that distance, as a, draw Ga parallel to BA, and take a b at pleasure for the height of each step.

Take a b in your compasses, and set that extent as many times upward from F to E as is equal to the first required number of steps O, N, M, L, K; and from these points of division in EF draw 1 b, 2d, 3f, 4 h, and Ek, all equidistant from one another, and parallel to Fa: then draw the equidistant upright lines ab, td, uf, vh, wk, and Im, all perpendicular to Fa: then draw m b, touching the outer corners of these steps

at m, k, h, f, d, and b; and draw n s parallel to mb, as far from it as you want the length of the steps K, L, M, N, O to be.

Towards the point of sight S draw mn, 15, ko, 16, hp, fq, dr, and bs. Then parallel to the bottomline BA) through the points n, o, p, q, r, s, draw n 8; 5, 14; 6, 15; 7, 16; 1, 17; and 2 s: which done, drawn 5 and 06 parallel to Im, and the outlines of the steps K, L, M, N, O will be finished.

At equal distances with that between the lines marked 8 and 14, draw the parallel lines above marked 9 10 11 12 and 13; and draw perpendicular lines upwards from the points n, o, p, q, r, s, as in the figure.

Make H m equal to the intended breadth of the flat above the square opening at the left hand, and draw HW toward the point of sight S, equal to the intended length of the flat: then draw WP parallel to Hm, and

the outlines of the flat will be finished.

Take the width of the opening at pleasure, as from F to C, and draw CD equal and parallel to FE. Draw GH parallel to CD, and the short lines marked 33, 34, &c. just even with the parallel lines 1, 2, &c. From the points where these short lines meet CD draw lines toward the point of sight S till they meet DE; then from the points where the lines 38, 39, 40, &c. of the pavement meet Cy, draw upright lines parallel to CD; and the lines which form the opening will be finished.

The steps P, Q, R, S, T, and the flat U above the arch V, are done in the same manner with those in fig. 34. as taught in Prob. 28. and the equidistant parallel lines marked 18, 19, &c. are directly even with those on the left-hand side of the arch V, and the upright lines on the right-hand side are equidistant with those on the left.

From the points where the lines 18, 19, 20, &c. meet the right-hand side of the arch, draw lines toward the point of sight S; and from the points where the pavement lines 29, 30, 31, 32, meet the line drawn from A towards the point of sight, draw upright lines toward the top of the arch.

Having done the top of the arch, as in the figure, and the few steps to the right hand thereof, shade the whole as in fig. 37. and the work will be finished.

PROB. 30. To put upright conical objects in perspective, as if standing on the sides of an oblong square, at distances from one another equal to the breadth of the oblong.

In fig. 38. the bases of the upright cones are per- Fig. 38. spective circles inscribed in squares of the same diameter; and the cones are set upright on their bases by the same rules as are given for pyramids, which we need not repeat here.

In most of the foregoing operations we have considered the observer's eye to be above the level of the tops of all the objects, as if he viewed them when standing on high ground. In this figure, and in fig. 41. and fig. 42. we shall suppose him to be standing on low ground, and the tops of the objects to be above the level of his eye.

oblong square ABCD; and let Aa and Dd (equal to In fig. 38. let AD be the perspective breadth of the Fig. 38. A a) be taken for the diameters of the circular bases of shall be AE and DF. the two cones next the eye, whose intended equal heights

Having made S the point of sight in the horizon patherein, draw AS and a S to contain the bases of the rallel to AD, and found the proper point of distance cones on the left-hand side, and DS and d S for those on the right.

Having made the two first cones at A and D of equal height at pleasure, draw ES and FS from their tops to the point of sight, for limiting the perspective parallelogram ABCD into as many equal perspective heights of all the rest of the cones. Then divide the squares as you please; find the bases of the cones at the corners of these squares, and make the cones thereon, as in the figure.

If you would represent a ceiling equal and parallel EF, then EFGH shall be the ceiling; and by drawto ABCD, supported on the tops of these cones, draw ing ef parallel to EF, you will have the thickness of the floor-boards and beams, which may be what you please.

This shows how any number of equidistant pillars may be drawn of equal heights to support the ceiling of a long room, and how the walls of such a room may be represented in perspective at the backs of these pillars. It also shows how a street of houses may be drawn in perspective.

PROB. 31. To put a square hollow in perspective, the depth of which shall bear any assigned proportion to its width.

Fig. 41. is the repesentation of a square hollow, Fig. 41. of which the depth AG is equal to three times its width AD; and S is the point of sight over which the observer's eye is supposed to be placed, looking perpendicularly down into it, but not directly over the middle.

Draw AS and DS to the point of sight S; make ST the horizon parallel to AD, and produce it to such a length beyond T that you may find a point of distance therein not nearer S than if AD was seen under an angle of 60 degrees.

Draw DU to the point of distance, intersecting AS in B; then from the point B draw BC parallel to AD;

and you will have the first perspective square ABCD, equal to a third part of the intended depth.

Draw CV to the point of distance, intersecting AS in E; then from the point E draw EF parallel to AD; and you will have the second perspective square BEFC, which, added to the former one, makes two thirds of the intended depth.

Draw FW to the point of distance, intersecting AS in G; then from the point G draw GH parallel to AD; and you will have the third perspective square EGHF, which, with the former two, makes the whole depth AGHD three times as great as the width AD, in a perspective view.

Divide AD into any number of equal parts, as suppose 8; and from the division-points a, b, c, d, &c. draw lines toward the point of sight S, and ending at GH; then through the points where the diagonals BD, EC, GF, cut these lines, draw lines parallel to AD; and you will have the parallelogram AGHD reticulated, or divided into 192 small and equal perspective squares.

Make AI and DM equal and perpendicular to AD; then draw IM, which will be equal and parallel to AD; and draw IS and MS to the point of sight S.

Divide AI, IM, and MD, into the same number of equal parts as AD is divided; and from these points of division draw lines toward the point of sight S, ending respectively at GK, KL, and LH.

From those points where the lines parallel to AD meet AG and DH draw upright lines parallel to AI and DM; and from the points where these lines meet IK and LM draw lines parallel to IM; then shade the work, as in the figure.

PROB. 32. To represent a semicircular arch in perspective as if it were standing on two upright walls, equal in height to the height of the observer's eye. After having gone through the preceding operation, this will be more easy by a bare view of fig. 42. than it could be made by any description; the method being so much like that of drawing and shading the square bollow. We need only mention, that a Tb EA and DF ctd are the upright walls on which the semicircular arch is built; that S is the point of sight in the horizon T t, taken in the centre of the arch; and din fig. 41. is the point of distance; and that the two perspective squares ABCD and BEFC make the parallelogram AEFD of a length equal to twice its breadth AD.

PROB. 33. To represent a square in perspective, as viewed by an observer standing directly even with one of

its corners.

both to right and left hand from 9, and divided all the way in the same manner.

From these points of equal division, 8, 9, 10, &c. draw lines to the point of sight S, and also to the two points of distance p and P, as in the figure.

Now it is plain, that acb9 is the perspective representation of A 9 BC, viewed by an observer even with the corner C and diagonal C 9.-But if there are other such squares lying even with this, and having the same position with respect to the line 1 17, it is evident that the observer, who stands directly even with the corner C of the first square, will not be even with the like.. corners G and K of the others; but will have an oblique view of them, over the sides FG and IK, which are nearest his eye and their perspective representations will be egƒ6 and h ki 3, drawn among the lines in the figure of which the spaces taken up by each side lie between three of the lines drawn toward the point of distance p, and three drawn to the other point of distance P.

PROB. 34.

To represent a common chair, in an oblique perspective view. The original lines to the point of sight S, and points Fig. 43of distance p and P, being drawn as in the preceding operation, choose any part of the plane, as l m n 13, on which you would have the chair L to stand.-There are just as many lines (namely two) between / and m-or 13 and n, drawn toward the point of distance p, at the left hand, as between 7 and 13, or m and n, drawn to the point of distance P on the right: so that Im, mn, n 13, and 13, form a perspective square.

From the four corners l, m, n, 13, of this square raise the four legs of the chair to the perspective perpendicular height you would have them: then make the seat of the chair a square equal and parallel to l m n 13, as taught in Prob. 18. which will make the two sides of the seat in the direction of the lines drawn toward the point of distance p, and the fore and back part of the seat in direction of the lines drawn to the other point of distance P. This done, draw the back of the chair leaning a little backward, and the cross bars therein tending toward the point of distance P. Then shade the work as in the figure; and the perspective chair will be finished.

PROB. 35. To present an oblong square tuble in an oblique perspective view.

In fig. 43. M is an oblong square table, as seen by Fig. 43an observer standing directly even with C 9 (see Prob. 33.), the side next the eye being perspectively parallel to the side a c of the square abe 9.-The forementioned lines drawn from the line 1 17 to the two points of distance p and P, form equal perspective squares on the ground plane.

Choose any part of this plane of squares for the feet of the table to stand upon; as at p, q, r, and s, in direction of the lines op and r s for the two long sides, and ts and qr for the two ends; and you will have the oblong square or parallelogram qrst for the part of the floor or ground-plane whereon the table is to stand : and the breadth of this plane is here taken in proportion to the length as 6 to 10; so that, if the length of the table be ten feet, its breadth will be six.

On the four little perspective squares at q, r, s, and t,