53. An inspection of the figure will make it evident that the "treads" of the steps, A, B, C, D and the "risers," N, 0, P are all in light. Of the hand rail it will be evident also that the left hand top, and the front face of the post are in light: The remaining faces are in shade. This is true of both rails; there fore, in one case we must find the shadow of a broken line, abode/ on the vertical wall and on the steps, and then find the shadow of the broken line mnopqr on the vertical wall and on the ground.
56. Beginning with the shad9w of the left-hand rail, the shadow of the point a on the wall is evidently ce, since a lies in the plane of the vertical wall (§ 39) The line ab is perpendicular to V hence its shadow will be a 45° line, the point being found by Problem I. The shadow of be, the sloping part of the rail, will fall partly on the vertical wall and partly on the treads and risers. We have already found the shadow of the end b on V in the point Lys. The shadow of c on V, found by Problem I will be cvs. The portion bv.gvs is the part of the shadow of this line be that actually falls on the wall, the steps preenting the rest of the line from falling on V.
The line of shadow now leaves the vertical wall at the point directly below gvs. The ground line for finding the shadow on this-upper tread will evidently be the line Av, since that line is the projection V this tread which is a line. The horizontal projection of the tread is a plane between the lines ahin,h and bhnh. We have now determined our GL and we also have one point, 9.s in the required shadow on the upper tread A. It re mains to find the shadow of the end c on A. Draw the projection of the ray through ev until it meets the line Av, drop a perpen dicular until it intersects the projection of the ray drawn through ch at the point cap. The point cas lies on the plane A extended. DraW the line ga8cas. The portion pashas is the part actually falling on the tread A. • From this point It, the shadow leaves tread A and falls on the upper riser M. The shadow will now show in elevation and begin at the point hms directly above the point has.
We .now determine a new ground line and it will be that pro jection of the upper riser which is a line. . The vertical pro jection of M is a plane surface between the lines Av and Bv. The
II projection of the riser M is the line Mh, therefore this is our GL, and we find the shadow on M in a manner similar to the find ing of ,the shadow on A, just explained. Bear in mind that we have one point hms, already found in this required shadow on M.
In a like manner the shadow of the remainder of the shade line is found until the point f is reached, which is its own shadow on the ground. (39) 57. It is to be noted that, since the plane of the vertical wall and the planes of the risers are all parallel, the shadows on these surfaces of the same line are all parallel. For a similar reason, the shadows of the same line on the treads and ground will be paral lel. This fact serves as a check as to the correctness of the shadow.
Also note in the plan that the shadow of the vertical edge of of the post is a continuous on the ground, the lower tread D, and on the next tread C, above. While this line of shadow on the object is of course in reality a broken line, it appears in horizontal projection on plan as a continuous line.
The shadow of the shade line of the right-hand rail is simply the shadow of a broken line on the co-ordinate planes, and requires no detailed explanation.
58. Problem X. To find the shade and shadow of a cone.
The finding of the shadow of a cone is, in general, similar to finding the shadow of the polyhedron none of whose planes are perpendicular or parallel to the co-ordinate planes.
It is impossible to determine at the beginning, the shade ele ments of the cone whose shadows give the shadow of the cone, and we first find the shadow-of the cone itself and from that determine its_shade elements: that is to say we reverse the usual process in determining the shadow of an object.
59. Fig. 25 shows, in elevation and plan, a cone whose apex is a and whose base is bode, etc. The axis is perpendicular to II and the cone is so situated that its shadow falls entirely on the V plane.
60. " It is evident that the shad ow of the cone must contain the shadow of its base and also the shadow of its apex. Therefore, if we find the shadow of its apex by Problem I, and then find the shadow of its base by Problem III and draw straight lines from the shadow of the apex tangent to the shadow of the base, the resulting figure will be the required shadow of the cone.