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SHADOWS OF LINES.

26. Problem II. To find the shadow of a given line on a given plane.

A straight line is made up of a series of points. Rays of light passing through all of these points would form a plane of light. The intersection of this plane of light with either of the co-ordinate planes would be the shadow of the given line on that plane. This shadow would be a straight line because two planes always intersect in a straight line. This fact, and the fact that a straight line is determined by two points, enables us to cast the shadow of a given line by simply casting the shadows of any two points in the line and drawing a straight line between these points of shadow.

In Fig. 10, avbv and ahbh are the elevation and plan respectively of a given line ab space. Casting the shadow of the ends of the line a and b by the method illustrated in Problem 1 and drawing the line aysbvs, we obtain the shadow of the given line ab on 27. Fig. 11 shows the construction for finding the shadow of the line ab when the shadow falls upon H.

28. Fig. 12 shows the construction for finding the shadow of a line so situated that part of the shadow falls upon IT and the remainder on H. To obtain the shadow in such a case, it must be found wholly on either one of the co-ordinate planes. In Fig. 12, it has been found wholly on V, ays being the actual shadow of that end of the line, and bvs being the imaginary shadow of the end b on V. Of the line aysbvs we use only the part a"cvs, that being the shadow which actually falls upon V.

The point where the shadow leaves V and the point where it begins on II are identical, so that the beginning of the shadow on II will be on the lower ground line directly below the point cvs; ehs will then be one. point in the shadow of the line ou and casting the shadow of the end b we obtain Lb.. The line chsbhs, drawn between these points, is evidently the required shadow on II.

29. Another method of casting the shad ow of such a line as ab is to determine the entire shadow on each plane independently. This will cause the two shadows to cross the ground lines at the same point c, and of these two lines of shadows we take only the actual shadows as the required result. This method involves unnecessary construction,lut should be understood..

30. Fig. 13 shows the construction of the

shadow of a given line on a plane to which it is parallel. It should be noted that the shadow in-this ease is parallel and equal in length to the given line.

31. Fig. 14 shows the construction of the shadow of a given line on a plane to which the given line is perpendicular. It is to • be noted that the shadow coincides in direction with the projection • of the Ivy of light on that plane, and is equal in length to the diagonal of a spare of which the given line is one side.

32. Fig. 15 shows the construction for finding the shadow of a curved line' on a given plane. Under these conditions we find. by Problem 1, the shadows of a number of points in the Tine -the greater the number of points taken the more accurate the re sulting shadow. The curve drawn through these points of shadow is the required shadow.

33. In Fig. 16 the given line ab is in space and the prob. lem to find its shadow on two rectangular planes mnop and nrso, both perpendicular to H.

Consider first the shadow of ab on the plane mnop. The edge no is the limit of this plane on the right. Therefore from the point mh draw back to the given line the projection of a ray of light. This 45° line in tersects the given line at It is evident that of the given line ab, the part ac falls on the plane mnop and the remainder, cb, on the plane nrso.

To find the shadow of ac on the left-hand plane we must first determine our ground ling. The ground line will be that pro. jection of the plane receiving the shadow which is a line. In this example the vertical projection of the plane mnop is the rectangle 9onvowpv. This projection cannot, therefore, be used as a GU. The plan, or II projection, of this plane is, hoWever, a line This line, therefore, will be used as the ground line for finding the shadow of ac on mnop.

We find the shadow of a to be at as and the shadow of a at es, Problem I. The line oes is, therefore, a part of the required shadow. The remaining part, esP is found in a similar manner.

34. The above illustrates the method of determining the GT, when the shadow falls upon some plane other than a co-ordinate plane. In case neither projection of the given plane is a Erie, the shadow must be deter mined by methods which will be ex plained later.