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 tho in. ward surface of the cylinder. 6 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 Db until bd is equal to 6D. 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 la of the leg of the table meets the surface of it, draw ae, and through v draw bd, both parallel 10 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, alde 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, ó 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, E is the projection of the common intersection of the surface of the table and the surface of the glass. Therefore f, 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 vanishirg 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), & is the projection of the relection 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 a given projection. Fig. 38. Let it be required to find the original of the object klmnp. Having continued the projections pk, kl, im, 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, heing 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, 13 being equal to 10, 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 9 as visual rays generating the projection ab. |