This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.
42. The methods for finding the shadows of solids vary with the nature of the given solid. The shadows of solids which are bounded by plane surfaces, none of which are parallel, or perpendicular, to the co-ordinate planes, can in general, be found only by finding the shadows of all the bounding planes. These will form an enclosed polygon, the sides of which are the shadows of the shade lines of the object, and the shade lines of the solid are determined in this way. The following is an illustration of this class of solids.
43. Problem IV. To find the shade and shadow of a polyhedron, none of whose faces are parallel or perpendicular to the co=or= dinate planes.
Fig 20 shows a polyhedron in such a position and of such a shape that none of its faces are perpendicular or parallel to the co-ordinate planes. It is impossible, therefore, to apply to this figure the projections of the rays of light and determine what faces are in light and what in shade. Consequently we cannot determine the shade line whose shadows would form the shadow of the object.
Under these circumstances we must cast the shadows of all the boundary edges of the object. Some of these lines, of shadow will form a polygon, the others will fall inside this polygon. The edges of the object whose shadows form the bounding lines of the polygon of shadow are the shade lines of the given object. Knowing the shade lines, the light and shaded portions of the object can now be determined, since these are separated by the shade lines.
In a problem of this hind care should be taken to letter or number the edges of the given object.
44. The edges of the polyhedron shown in Fig. 20 are ab, be, cd, da, ac and bd.
Cast the shadows of each of these straight lines by the method shown in Problem II.
We thus obtain a polygon bounded by the lines bvscvs, cvsavs, avsbvs, and this polygon is the shadow of the given solid.
The lines which cast these lines of shadow, bvscvs, cvsavs, and avsbvs are therefore the shade lines of the object, and, therefore, the face abc is in light and the faces abd, bed and acd are in shade.
The shadows of the edges bd, dc, and ad falling within the polygon, indicate that they are not shade lines of the given object, and, therefore, they separate two faces in shade or two faces in light. In this example bd and cd separate two dark faces.
In architectural drawings the object usually has a sufficient number of its planes perpendicular or parallel to the co-ordinate planes, to permit its shadow being found by a simpler and more direct method than the one just explained.
45. Problem V. To find the shade and shadow of a prism on the co-ordinate planes, the faces of the prism being perpen= dicular or parallel to the V and H planes.
In Fig. 21 such a prism is shown in plan and elevation. The elevation shows it to be resting on II, and the plan shows it to be situated in front of V, its sides making angles with V. Since its top and bottom faces are parallel to II and its side faces perpendicular to that plane, we can apply the projections of the rays of light to the plan and determine at once which of the side faces are in light and which in shade. The projections of the rays R1 and R2 show that the faces abgf and adif receive the light directly, and that the two other side faces do not receive the rays of light and are. therefore, in shade. The edges bg and di are two of the shade lines. R3 and R4 are the projections of the rays which are tangent to the prism along these shade lines.
Applying the projection R5 in the elevation makes it evident that the top face of the prism is in light and the bottom face is in shade since the prism rests on H. This determines the light and shade of all the faces of the prism, and the other shade lines would therefore be he and ed.
Casting the shadow of each of these shade lines, we obtain the required shadow on V and H.
It is evident that the shadows of the edges bg and di on H will be 45° lines since these edges are perpendicular to II (§ 31) Also, their shadows on V will be parallel to the lines themselves since these shade lines are parallel to Y. (§ 30)
46. In general, to find the shadow of an object whose planes are parallel or perpendicular to H or V:
(1) Apply to the object the projections of the ray of light to determine the lighted and shaded faces.
(2) These determine the shade lines.
(3) Cast the shadows of these shade lines by the method followed in Problem II.
47. Problem VI. To find the shade and shadow of one object on another.
In Fig. 22 is shown in plan and elevation a prism B, resting on II and against Y. Upon this prism rests a plinth A: To find the shadow of the plinth on the prism and the shadow of both or. the co-ordinate planes. Since these objects have their faces either perpendicular to, or parallel to, the co-ordinate planes, we can determine immediately the light and shade faces and from them the shade lines.
48. Considering first the plinth. A, it is evident that its top, left-hand and front faces will receive the light, that the lower and right-hand faces will he in shade. The back face resting against the V plane will be its own shadow on V. (§ 39) The shade lines of A will be, therefore, ef,fg,gc and cd.
Cast the shadows of these lines. A rests against V and part of its shadow will fall on V; also, since it rests on B the remainder will fall on B. Begin with the point e, one end of the shade line; this point, lying in V, is its own shadow on V. (§ 39).
The line ef being perpendicular to V, its shadow, or as much of its shadow as falls on V, will be, therefore, a 45° line drawn from ev. The point th, in plan, shows the amount of the line cf which falls on V, the rest tf falls on the side face of the prism, and this shadow is not visible in elevation or plan.