In a similar manner the shadow of any point or line in the chimney can be found on the roof.

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Before completing the shadow of the chimney upon the roof let us consider the shadow of the flat band on the main part of the chimney. This band projects the same amount on all sides. On the left-hand and front faces it will cast a shadow on the chimney proper. Only the shadow on the front face will be visible in elevation. To find this, draw the profile projection of the ray through the point qp until it intersects the line apvP, the profile projection of the front face. From this point q1P draw a horizontal line across until it meets the vertical projection of the ray drawn through qv. From qv, the shadow of qvwv on the front face will be parallel to qvwv for that line is parallel to that face; therefore draw q,vzv.

Now that the visible shadows on the chimney itself have been determined, its shadow on the roof can be found as explained in the first part of this problem. A portion of the shade line of the flat band, zvwv, wvnv, etc., falls beyond the chimney on the roof, as shown by the line zsws, wsns, etc.

52. It is to be noted in the shadow on the roof that:

(a) The shadows of the vertical edges of the chimney make angles with a horizontal line equal to the angle of the slope of the roof (in this case 60 ).

(b) The horizontal edges which are parallel to V cast shadows which are parallel to these same edges in the chimney.

(c) The horizontal edges which are perpendicular to V cast shadows which make angles of 45° with a horizontal line.

53. The above method would also be used in finding shad-ows on sloping surfaces when the objects are given in elevation and side elevation, as, for example, a dormer window.

54. Problem IX. To find the shades and shadows of a hand rail on a flight of steps and on the ground.

Fig. 24 shows the plan and elevation of a flight of four steps situated in front of a vertical wall, with a solid hand rail on either side, the hand rails being terminated by rectangular posts. At a smaller scale is shown a section through the steps and the slope of the hand rail.

This problem amounts to finding the shadow of a broken line, that is to say, the shade line, on a series of planes. Each of the planes requires its own ground line, which in the case of each plane will be that projection of the plane which is a line. Since the planes of the steps and rails, with one exception, are all parallel or perpendicular to the co-ordinate planes we can determine at once what planes are in light and what in shadow and thus determine the shade line.

55. An inspection of the figure will make it evident that the "treads" of the steps, A, B, C, D and the "risers," M, N, O, P are all in light. Of the hand rail it will be evident also that the left-hand face, the top, and the front face of the post are in light. The remaining faces are in shade. This is true of both rails; therefore, in one case we must find the shadow of a broken line, abcdef 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.

5G. Beginning with the shadow of the left-hand rail, the 6hadow of the point a on the wall is evidently av, 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 bvs being found by Problem I. The shadow of bc, 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 bvs. The shadow of c on V, found by Problem I will be cvs. The portion bvsgvs is the part of the shadow of this line bc that actually falls on the wall, the steps preventing the rest of the line from falling on V.

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The line of shadow now leaves the vertical wall at the point gas, 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 of this tread which is a line. The horizontal projection of the tread is a plane between the lines ahmh and bhnh. We have now determined our GL and we also have one point, gas 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 cv until it meets the line Av, drop a perpendicular until it intersects the projection of the ray drawn through ch at the point cas. The point cas lies on the plane A extended. Draw the line gascas. The portion gashas is the part actually falling on the tread A.

From this point h, 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 M, which is a line. The vertical projection of M is a plane surface between the lines Av and Bv. The H 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 pointy 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 parallel. 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 ef of the post is a continuous 45° line 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.