face of the cylinder; i. e. it is the projection of the shadow of the original of the point G in the circumference of the base, on the inward surface of the cylinder.

b being the seat of the point D on the surface of the water, the reflection d of the point D is found by producing the perpendicular ᎠᏏ until ld is equal to bD.

This is evident, because the known law of reflections is, that the reflection of all objects appear to be as much on one side of the reflecting plane, as the real objects are on the other side of it. In AS, make As equal to AS, any point q may be found in the shadow on the surface of the inward cylinder, in the reflection is found, in the same manner as Q in the real figure, using the point s instead of S.

The shadow of the cylinder on the surface of the cone, is found by just such another expedient as the shadow of the line BD on the surface of the cylinder.

Ex. 5. (Fig. 23.) In this scheme every thing else being easy to comprehend, by the explanations already given, it only remains to shew the manner o finding the reflection in the looking-glass of the picture on the eazle.

A is the centre of the picture, and AB the vanishing line of the ground; the distance of the picture being equal to AB. AC is the vanishing line of the picture on the eazle, and CD the vanishing line of the looking-glass.

Through a, where the edge ba of the leg of the table meets the surface of it, draw ae, and through & draw bd, both parallel to AB, bd meeting the intersection cd of the surface of the picture on the eazle with the ground in d, and then draw de parallel to AC, meeting ae in e, and drawing Ae the projection, Ae is obtained of the common intersection of the surface of the table and of the picture on the eazle. For ae being parallel to AB is the projection of a line in the surface of the table parallel to the picture, and, for the same reason, bd is the projection of a line on the ground, and de is the projection of a line in the plane of the picture on the eazle, both of them parallel to the picture. Therefore, abde is the projection of a trapezium, parallel to the picture, whose angle e is in the common intersection, the surface on the table, and of the picture on the eazle; but A being the common intersection of the vanishing lines of those two planes, is the vanishing point of their common intersection, and, therefore, eA is the projection of that intersection (Cor. 2, Theo. 10). For the same reason, o being the projection of the point where the surface of the glass touches the table, and E being the common intersection of the vanishing lines AB and CD, oE is the projection of the common intersection of the surface of the table and the surface of the glass. Therefore ƒ, where oE and eA meet, is the projection of the point where the three planes meet of the surface of the table, the glass, and the picture on the eazle. Therefore, drawing fC is the projection of the common intersection of the picture on the eazle and the looking-glass.

Having found the vanishing point P of lines perpendicular to the plane of the looking-glass, whose vanishing line is CD (by Prob. 17) draw PA through the vanishing point A of the line GH, and meeting CD in D, D is the vanishing point of the seat of GH on the plane of the glass. Therefore, GH cutting Cf in i, Di is the projection of that seat. Then draw GP cutting Di in k, k is the seat of the point G on the glass. Wherefore, in GP make kg to represent a line equal to that represented by Gk (by Prob. 3), g is the projection of the reflection of G, and gi is the reflection of Gi, and drawing PH cutting gi in h, gh is the reflection of GH; and, in the same manner, may any other lines in the reflection be found.

The reflection of the picture on the eazle may also be described by its vanishing une, in the same manner as the projection of the picture itself was described; for, in PAD making aD to represent a line equal to that represented by AD, a is the vanishing point of the reflected line gh, and Ca is the vanishing line of the reflected picture on

the eazle.

PERSPECTIVE.
PART II.

THE Inverse method of Perspective, or the manner of finding the Original Figures from their given Projections.

Prob. 21. Given the intersecting and vanishing lines of a plane, and the centre and distance of the vanishing line, to find the original c a given projection.

Fig. 38. Let it be required to find the original of the object klmnp. Having continued the projections pk, kl, lm, mn, until they meet the intersecting line in the points Q, R, T, U, and the vanishing line in the points V, G, H, I: draw OV, OG, OH, OI, and QP, RK, TM, UN, parallel to OV, OG, OH. OI, meeting respectively in K, L, M. Draw Op and On, and produce them to meet QP and UN in P and N; finally, drawing NP, then PKLMN is the original figure required.

The truth of this construction is evident from Prob. 10, being the reverse operation to this.

Prob. 22. Having the vanishing and intersecting lines of a plane, the centre and distance of the vanishing line, and the projection of a line, to find the length of the original of the projection given.

Fig. 39. Let ab be the projection given: produce ab till it meets the vanishing line in its vanishing point V; join OV, and make V3 equal to VO, and draw 3a and 36 cutting the intersection in 1 and 2, then the distance between 1 and 2 will be the length required.

Let W be the intersection of ab, V3 being equal to VO, the distance of the vanishing point V, and W2 being parallel to V3, the point 3 may be considered as the point of sight, and W1, W2, as the original line, and 3, 1 and 3, 9 as visual rays generating the projection ab.

Prob. 23. Having the projection of a line divided, and the vanishing point, to find the proportion of the parts of the origina'.

Let AB, fig. 8, be the given projection, divided in C and V its vanishing point: draw at pleasure VO, and ab parallel to it, and, from any point O in the line OV, draw OA, OB, OC, cutting ab in a, b, and c; then will the original of AC be to the original of CB, as ac is to cb.

Cor. ac cb:: ACX BV: BC × AV.

:

Prob. 24 Having the projection of a line divided into two parts, ane the proportion of the original, to find the vanishing point.

Let AB, fig. 40, be the projection given divided in C, through C draw at pleasure aCb, and in it make aC to Cb as the original of AC is to the original of CB, and draw aA and bB meeting in O. Parallel to ab draw OV meeting AB in V, which will be the vanishing point sought.

Cor. BV BA :: Cax CB: Cbx AC- Ca x CB.

These two last problems, with their corollaries, are easily deducible from Prob. 3 and its Corollary.

Prob. 25. Having the projection of a triangle, with its vanishing line, its centre, and distance, to find the species of the original triangle.

Let abc, fig. 41, be the given projection, HG its vanishing line, and S its centre, and SO perpendicular to HG, and equal to its distance. Having continued the sides of the given projection till they meet the vanishing line in their vanishing points G, H, I, draw GO, GH, and GI, and the originals of the angles bac, abH, acb, will be equal to GOI, GOH, IOH, respectively (by Prob. 14). Whence the species of the original triangle is given.

Prob. 26. Having the projection given, of a triangle of a given species and its vanishing line, to find the centre and distance of that vanishing line.

Let ABC, fig. 42, be the given projection, and FD its vanishing line. Produce the sides of the projection till they meet the vanishing line in their vanishing points D, E, F. Bisect DE and EF in G and H, and draw GI and HK perpendicular to FD, making GI to GE as the radius is to the tangent of the angle represented by BAC, and KH to EH as the radius is to the tangent of the angle represented by BCA; so that EIG and FKH may be equal to those angles. With the centres I and K, and the radii IE and KE, describe two circles cutting each other in O, and draw OS cutting FD at right angles in S; then will S be the centre, and SO the distance sought.

Suppose S the centre, and SO the distance of the vanishing line FD, the originals of the angles BAC, and BGA will be equal to DOF, and EOF (by Prob. 14); but, by the nature of the circle, DOE and EOE are equal to GIE and HKE, which, by the construction, are equal to the angles that ought to be represented by BAC and BCA, Therefore, S is the centre, and SO the distance sought.

Prob. 27. Having the projection of a trapezium of a given species, to find its vanishing line, centre, and distance.

Let abcd, fig. 42, be the given projection: draw the diagonals ae,

bd, meeting in e, and, by the proportion of the originals of ae, ec, and be, ed, find the vanishing points E and F of the-lines ae and bd (by Prob. 22). Draw FE, which will be the vanishing line sought. Then, by the given species of the original of the triangle abe, find the centre S, and distance SO (by Prob. 24).

Prob. 28. Having the projection of a right-angled parallelopiped, to find the centre and distance of the picture, and the species of the original figure.

Let ABCDEFG, fig. 43, be the projection given. Produce the projections of the parallel sides until they meet in their vanishing points H, I, K, and draw III, HK, and IK, which will be the vanishing lines of the several faces of the figure sought, containing a solid rightangle. Draw KL perpendicular to HI, and HM perpendicular to KI, meeting in S; which will be the centre of the picture (by Cor. 2, Prob. 19).

Then, on the diameter LK, describe a circle, and draw SO perpendicular to LK cutting in O, and OS will be the distance of the picture (by Prob. 17), LOK being a right-angle because of the circle. Finally, find the distance of the vanishing lines KI and IH (by Cor. 3, Prob. 19), M and L being their centres (by Theo. 1), and then find the species of the originals of the faces DAFE and DABC (by Prob. 23).

Cor. When the vanishing line of one of the faces (suppose IH) passes through the centre of the picture, the vanishing point K of the sides perpendicular to it, will be at an infinite distance; by which means the situation of LK is indeterminate.

So that the species of the face ABCD may be taken at pleasure, and then the centre and distance of the picture may be found (by Prob. 23). And, in this case, if it were only required that the projection proposed should represent a right-angled parallelopiped in general, the place of the point of sight might be any where in the circumference of a circle described on the diameter HI, and in a plane perpendicular to the picture.

As very great inconveniences are generally experienced by the Draughtsman in making Perspective Drawings, on account of the extreme remoteness of the vanishing points, which occasion a necessity for a ruler of such a length as to render it inconvenient for use, nay, altogether impossible, as it would sometimes require the whole extent of the largest room, and, indeed, a greater space may in some cases be required, than any room can furnish. In order to obviate this difficulty, the Delineator is either obliged to shorten the distance of the point of view, which will produce a very distorted representation of the object, or he is under the necessity of submitting to the drudgery of making his drawing upon a small scale, and of enlarging it afterwards.

These difficulties have occasioned other methods to be resorted to; but all the methods that have hitherto been invented, have been accompanied with such excessive labour, and, at the same time, have been liable to such extreme inaccuracy, and, to add to the difficulty