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.

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 elements 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 bcde, etc. The axis is perpendicular to H and the cone is so situated that its shadow falls entirely on the V plane.

GO. It is evident that the shadow 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.

61. This has been done in Fig. 25, in which avs is the shadow of the apex a. The ellipse bvscvsdvs, etc., is the shadow of the base, found by assuming a sufficient number of points in the perimeter of the base and finding their shadows. The ellipse drawn through these points of shadow is evidently the shadow of the base. From the point avs the straight lines avssvs and avsxvs were drawn tangent to the ellipse of the shadow. This determined the shadow on V of the cone. The lines avs,svs and avsxvs are the shadows of the shade elements of the cone. It remains to determine these in the cone itself. Any point in the perimeter of the shadow of the base must have a corresponding point in the perimeter of the base of the cone, and this can be determined by drawing from the point in the shadow the projection of the ray back to the perimeter of the base. Therefore if we draw 45°-lines from the point xvS and svs back to the line cvmv in the elevation, we determine the points xv and sv. The lines drawn from these points to the apex aY are the shade elements of the cone. They can be determined in plan by projection from the elevation. The cross-hatched portion of the cone indicates its shade. It will be observed that but little of it is visible in elevation.

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62. When the plane of the base of the cone is parallel to the plane on which the shadow falls, as in Fig. 26, the work of finding its shade and shades ow is materially reduced, for the shadow of the base can then be found by finding the shadow of the center b of the base and drawing a circle of the same radius. (See Fig. 19)

63. In general, to find the shadow of any cone, find the shadow of its apex, then the shadow of its base, draw straight lines from the former, tang ent to the latter. Care should be taken, however, that both the shadow of the apex and the shadow of the base are found on the same plane.

64. It is to be noted that in a cone whose elements make an angle of 35°-15'~52" or less, that is, making the true angle of the ray of light or less, with the co-ordinate plane, the shadow of the apex will fall within the shadow of the base, and, therefore, the cone will have no shade on its conical surface.

65. Problem XI. To find the shade and shadow of a right cylinder.

In Fig. 27 is shown the plan and elevation of a right cylinder resting on H. The rays of light will evidently strike the top of the cylinder and the cylindrical surface shown in the plan between the points bh and dh At these points, the projections of the ray of light are tangent, and these points in plan determine the shade elements of the cylinder in elevation. These shade elements, avbv and evdv, are the lines of tangency of planes of light tangent to the cylinder. The shadow of these lines ab and cd. together with the shadow of the arc aefg, etc., of the top of the cylinder, form the complete shadow of the cylinder.

FIG26.

Since ab and cd are perpendicular to II, as much of their shadow as falls upon the II plane will be 45° lines drawn from bh and dh respectively. In one case the amount is the line nvbv, in the other hydv. The remainder of the lines will fall upon V and this is found by Problem II. These shadows on V will evidently be parallel to avnv and cvhv.

. The shadow of the shade line aefg, etc., on V will be found by Problem II. (§32)

66. If the cylinder had been placed so that the whole of its shadow fell upon H, the shadow of the shade line of the top would have been found by rinding the shadow of the center of the circle and drawing a circle of the same radius, since the plane of the top is parallel to H. (§ 38)

67. Problem XII. To find the shade and shadow of an oblique cylinder.

In Fig. 28 is shown the plan and elevation of an oblique cylinder whose vertical section is a circle.

68. Unlike the right cylinder we cannot apply to the plan or elevation the projections of the ray and determine at once the location of the shade elements. To find the shadow of an oblique cylinder we proceed in a manner similar to finding the shade and shadow of a cone; that is, we find first the shadow of the cylinder and from that shadow determine the shade elements.

69. In Fig. 28 the top and base of the oblique cylinder have been assumed, for convenience, parallel to one of the co-ordinate planes. The shadow of the cylinder will contain the shadows of the top and base, hence if we find their shadows and draw straight lines tangent to these shadows, we shall obtain the required shadow of the cylinder.

70. In Fig. 28, the top and bottom being circles, the shadows of their centers a and b are found at avs and bvs, and circles of the same radius are drawn. Then the lines mvsnvs and ovspvs are drawn tangent to these circles. The resulting figure is the required shadow wholly on V. Projections of the ray are then drawn back from mvs, nvs, ovs and pvs respectively to the perimeter of the top and base. Their points of intersection mv, mv, ov and pv are the ends of the shade elements in the elevation. They can be found in plan by projection. An inspection of the figure will make it evident what portions of the cylindrical surface between these shade elements will be light and what in shade.

FIG.28.

71. In this problem it will be seen that the shadow does not fall wholly upon V. The shadow leaves V at the points xvs and yvs and will evidently begin on H at points directly below, as xhs and yhs.

If projections of the ray are drawn back to the object in plan and elevation from these points, xvs, yvs, xhs, and yhs, they will determine the portion of the shade line which casts its shadow on H. It is evident that in this particular object it is that portion of the shade line of the top between the points p and y and the portion xp, of one of the shade elements. The shadows of these lines are found on H by Problem II.

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