Find the shadow of any point av,for example, by Problem On that point of the shadow construct a similar square whose side equals 1 inches.
89. Problem XXII. To construct the shadow on V of a circular plane which is parallel to H, or which lies in a profile plane.
Given avovbv, the projection of a circular plane perpendicular to V and II, 2 inches in diameter, its center being 21- inches from V, to construct the shadow on V. Fig. 42. The shadow of ov, the center of the circular plane is found by Problem XVII. About ovs as a center, construct the parallelogram ABCD made up of the two right triangles ADB and DBC, the sides adjacent to the right angles being equal in length to the diameter, 2 inches, of the circular plane. Draw the diameters and diagonals of this par allelogram. The diameter TW is equal to the diameter of the given circle and parallel twit.
With ovs as a center and OD and 013 as radii, desCribe the arcs cutting the major diameter of the parallelogram in the points E and F. Through E and F draw lines parallel to the short diam eter, cutting the diagonals in the points G, H, M and N. These last four points and the extremities of the diameters R, S, T, and IV, are eight points in the ellipse which is the shadow of the given circular plane on V. A similar construction is followed for find ing the shadow on V of a circular plane parallel to II. Fig. 43.
90. Problem XXIII. To construct the shade line of a cylin der whose axis is perpendicular or parallel to the ground line.
Given the elevation of a cylinder, its axis being AB perpen dicular to H. To construct shade lines. Fig. 44.
Let CD be any horizontal line drawn through the cylinder.
Construct the 45° isosceles tri angle .AGD on the right half of the diameter.
With the radius AG describe the semi-circular arc mGii, cutting the horizontal line CD in the points m and n.
These two points will determine the shade elements me and np.
91. Problem XXIV. To con= struct the shadow on a plane (parallel to its axis) of a circular cylinder whose axis is either perpendicular, or parallel to the ground line.
Let a=the distance, in the elevation., between the projection of the axis of the cylinder and, the projection of the visible shade element. Let L=the distance between the axis of the cylinder and the plane on which the shadow falls, to be obtained from the plan.
Then the distance, between the visible shade element and its shadow on the given plane, will be equal to a+b. • The width of the shadow on the given plane will be equal to 4a. Given the circular cylinder CDEF (Fig. 45), its axis AB pendicular to II. To construct its shadow on the V plane which is 1 inches distant from the axis AB. The shade elements ono and iip can be constructed by Problem XXIII. Draw RS the shadow of the shade element up, parallel to up, and distant from it a + 1 inches. The width of the shadow on the given plane will be 4 times the distance An.
92. Problem XXV. To construct the shadow on a right cylinder of a horizontal line.
a. It will be the arc of a circle of the same radius as that of the cyl inder.
b. The center of the circle will be on the axis of the cylinder as far be loW the as that line is in front of the axis.
Given a right circular cylinder CDEF, whose diameter is 14 inches, and a horizontal line ab, 1,1- inches in front of the axis of the cylinder. To construct the shadow. Fig. 46.
Locate the point o on the axis 1i inches below cebv. With as.a center, and radius equal to inch describe the arc ramp, the required shadow.
93. Problem XXVI. To con struct the shadow of a verti cal line on a series of mould ings which are parallel to the ground line.
The shadow reproduces the actual profile of the mouldings.
Given a vertical line ay by which casts a shadow on the moulding M, which is parallel to the ground line, and whose profile is shown in the section ABCD. The line avbv, , is 1 inch in front of the fillet AB. To construct its shadow, Fig. 47: Construct the shadow on the fillet AB, of the end of the line av, or any other convenient point in the line, by Problem XVII.