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No2 In Fig. 1, the shadow falls on the vertical, and in Fig. 2, on the horizontal coordinate plane.
SCHOLIUM. This is the foundation of shadowing from artificial lights, as in interior night representations.
To find the shadow of a given line, the rays being parallel. 1. Let a b be the given line, terminated at a and b: then the shadows of the extremities being joined (these being found by Prop. 1.), we shall have the shadow required : the shadow falls wholly upon and within the horizontal coordinate, as in Fig. 1.; or wholly on the vertical, as in Fig. 2.
2. Should, however, the trace of the shadow line through one extremity of the line fall upon one plane and the other on the other, a different method, or rather an extended method, must be employed.
Since the ray through the given line at either extremity and the line itself are in one plane, the trace of that plane upon the coordinate planes will be the position of the shadow; and its limits have already been found. It is hence only necessary to find in addition to a, and B, the horizontal trace B, of the ray through b; which enables us to find the intersection Q of the shadow plane with the axis. The shadow will hence be composed of the two parts a, Q, Q Ba.
SCHOLIUM 1. Should the position of the given line and the ray render it more convenient, one of the traces of the given line ab will answer the purpose of finding Q; for both its traces are also in the traces of the shadow plane.
SCHOLIUM 2. If the shadow of the line ab, merely given in position, without reference to its extremities be required, the process will be somewhat simplified in practice by finding the traces of ab, and of a line parallel to the ray through one of the traces of ab.
To find the shadow of a given line, the rays emanating from a given point.
In this case let ab be the given line, and s the given origin of light: then the extremities a, b are to be considered as points through which lines are drawn from s to meet the coordinate planes : that is, the visible traces of sa, sb are the shadows of a and b.
In Fig. 1, these are both on the horizontal coordinate plane, and the shadow is a, B.; in Fig. 2, they are both on the vertical, and the shadow is a, b,; whilst in Fig. 3, one visible trace is on each plane, and, as in the preceding proposition, another trace of one of the shadow lines, as B2 is necessary for finding Q the point in which the shadow cuts the axis.
To find the shadow of a given point in a given plane, the rays being either parallel or emanating from a given point.
1. Let a be the given point, P the given plane, and R' the ray: find the traces of the ray through a parallel to the given ray; then any two planes Q, R through this line will cut P in lines which pass through the point of intersection, or the shadow of a. Find the projections of these lines, viewed as the intersection of P with Q and R respectively: then the intersections of their projections on the coordinate planes will be the projections of the shadow sought.
2. When the light emanates from a given point, the only difference will be that the shadow line instead of being parallel to R' will pass through the projections of this point.
CASES OF PROP. V. It will often happen that, under particular conditions of the data, the practical construction becomes much simplified; and the simplified cases which occur in usual practice require to be specially noticed.
1. Let the plane be parallel to the elevation, and P, be its trace, a, a, being the given point, and R, R, the direction of
Draw the parallels through a, and a, to the projections of the ray; and let a, au cut P, in an: then Q, is the horizontal projection of the shadow point on Pi, and drawing, as usual, we get az for the projection on the vertical.
2. Let the plane be parallel to the plan, and P, be its trace, a, a, as before. Then by merely reversing the order of the last operations we get ang &g for the horizontal and vertical projections of the point of shadow,
3. Let the plane be perpendicular to one of the coordinate planes, but not parallel to the other; as P in the two annexed figures.
Draw a, a, or az az (as the case may be) to meet PP or PP,, in a, or ex: and proceed, as in the preceding cases, to find az or aj: then Q is the point required.
4. Let the plane be perpendicular to both planes: that is profile.
Draw a, ang and ag og to meet the traces (parallel to the ray as before): then a, b, is the shadow point required.
For all these operations are only finding the intersection of the ray with the given plane, by the methods of Descriptive Geometry, as already explained and studied.
To find the projections of the shadow of a given line on a given plane.
Let P be the given plane, a, b, the given line, and R the direction of
1. Through any point c of the given line, draw a parallel to the ray R: then the plane of shadow of the given line will pass through this line and the given line; and its intersection with P will be the shadow
Find one of the traces of the line through c parallel to the ray, as Ye; then trace the plane Be Yoa, which will be the shadow plane. Find its intersection with P: then the lines Qıq, and Qe 92 will be the projections of the shadow cast upon the plane P by the line a Bz.
2. If the light emanate from a point, the difference of construction