The shadow of the point f evidently falls on the edge of the prism at fs, see plan. This point f is one end of the shade line fg, therefore f is one point in the shadow of fg on the front face of the prism B. The line fg being parallel to this front face, its shadow will be parallel to the line, therefore from the point fs we draw the horizontal line f,s.r1s

If from the point r1s we draw the projection of a ray of light back to the shade line fvgv we determine the amount of the line casting a shadow on the front face of B, that is to say, the distance fvrv. The shadow of the remainder, rvgv, falls beyond the prism on the V plane, and is evidently the line rvsgvs. Thus the shadow of the shade line from e to g has been determined.

The next portion of the shade line, gc is a vertical line and we have already obtained the shadow of the end g. Since it is a vertical line its shadow on V will be vertical and equal in length. Therefore draw gvscvs.

There remains now only the edge, cdt of which to cast the shadow. The end d being in the V plane must be its own shadow on that plane. (§ 39) We have already found the shadow of the other end c, at cvs. Therefore dvcvs is the shadow of dc and completes the outline of the shadow of the plinth.

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It will be noted that dvavs is a 45° line, which would be expected since the edge do is perpendicular to V.

49. Considering next the prism B, we find by applying the projections of the rays of light to the plan, that the front and left-hand faces are in light, and that the right-hand face is in shade. Therefore the only shade line in this case is the edge mn. The upper part of this, myr1s is in the shade of the plinth and therefore cannot cast any shadow.

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It is to be noted that the ray of light from the point rv in the plinth A passes through the point r1s in the shade line of the prism B. In finding the shadow of this point at rvs we therefore have found the shadow also of one end of the shade line nr. Since nr is vertical, its shadow will be vertical on Y. Therefore draw rvswvs. This line completes the shadow of the two objects upon the V plane.



Wood, Donn & Deming, Architects, Washington, D. C. Walls of Stucco on Wire Lath. A Three-Quarter Engaged Ionic Column Used on Entrance Feature. For Plans, See Opposite Page.



Wood, Donn & Deining, Architects, Washington, D. C.

From the point of shadow wvs draw the V projection of the ray back to the line nvr1s. This shows how much of the total line nr falls upon V, and how much upon II.

The shadow on II of the portion nw will be a 45° line since nw is perpendicular to II. The point n being in the H plane is its own shadow on that plane.

It is to be noted that the point whs is on the perpendicular directly below wvs.

50. Problem VII. To find the shade and shadow of a pedestal. Fig. 23 shows the plan and elevation of a pedestal resting on the ground and against a vertical wall. This is an application of the preceding problem in finding the shades and shadows of one object upon another. The profile of the cornice moulding on the left, at A, can be used as a profile projection in finding the shadows of those mouldings on themselves and upon the front face B, of the pedestal. By drawing the profile projections of the rays tangent to this profile of mouldings, it will be seen what edges are shade lines and where their shadows will fall on the surface of B. The line avbv can be assumed to be the profile projection of the front face of B, and being a line is used as the ground line for finding the shadow on B. As this collection of mouldings is parallel to the V plane their shades and shadows will be parallel in the elevation. Otherwise the shadows of this pedestal are found in a manner similar to the preceding problem.

51. Problem VIII. To find the shadow of a chimney on a sloping roof.

Fig. 23a shows in elevation and side elevation the chimney and roof. The chimney itself being made up of prisms with their planes parallel or perpendicular to the V plane, its light and shade faces can be determined at once, as in Problem V. It will be evident from the figure that the top, front, and left-hand faces of the chimney in elevation will be in light. The remaining faces will be in shade, and the shade lines will be therefore, yd, on the back, dc, cb, and bx. Not all of bx and yd will cast shadows for the shadow of the flat band, running around the upper part of the chimney, will cause a portion of these two lines yd and bx to be in shadow and such portions cannot cast any shadows. (See Problem VI-the shadow of one object upon another.)

It is evident that, to find the shadow of the shade line of the chimney upon the sloping roof,we must have for a ground line a projection of the roof which is a line. The roof in elevation is projected as a plane, but the side elevation (or in other words the profile projection) shows the roof projected as a line in the line hPgp. This line will be then the ground line for finding the shadow of any point in the chimney on the roof. For example, take the point b. If we draw the profile projection of the ray through the point bP until it intersects the ground line ApgP, and draw from this point of intersection a horizontal line across until it intersects the vertical projection of the ray drawn through bv, this last point of intersection bs, will be the shadow of b upon the roof.