| Fig. 112. |
This is simply the reverse of the previous problem. Let AB be the given line. From distance D through A draw DC, and from S, point of sight, through A draw SO. Drop OP at right angles to base, making it equal to OC. Join PB, and line PB is the actual length of AB.
This problem is useful in finding the position of any given line or point on the perspective plane.
[ LIV]
To Find these Points when the Distance-Point is Inaccessible
If the distance-point is a long way out of the picture, then the same result can be obtained by using the half distance and half base, as already shown.
Fig. 113.