Frustum of a Square Pyramid. In Fig. 112 is shown the plan and elevation of the frustum of a square pyramid, placed with its base on the horizontal plane. If the frustum is turned through 30°, as shown in the plan of Fig. 113, the top view or plan must still be the same shape and size, and as the frustum has not been raised or lowered, the heights of all points must appear the same in elevation as in Fig. 112. The elevation is easily found by projecting points up from the plan, and projecting the height of the top horizontally across from the first elevation.

Fig. 112. Frustum of Square Pyramid

Fig. 112. Frustum of Square Pyramid.

Fig. 113. Frustum of Square Pyramid Rotated Through 30

Fig. 113. Frustum of Square Pyramid Rotated Through 30°.

Square Prism. The same principle is further illustrated in Figs. 114 and 115. The elevation of Fig. 114 shows a square prism resting on one edge, the lower side making an angle of 30° with the horizontal. The plan gives the true width or thickness, 5/8 in., but the true length of the side, 2 in., must be obtained from the elevation, for the length of the plan is greater than 2 in. and will vary with the angle at which the prism is slanted. Now if the prism be turned around through an angle of 45° with the vertical plane, the lower edge still being on //, and the inclination of 30° with H remaining same, the plan must remain the same size and shape, as shown in the right-hand view of Fig. 114. The elevation is found by projecting the corners of the prism vertically to the heights of the same points in the first elevation.

Rectangular Prism. Three positions of a rectangular prism are shown in Fig. 115. In the first view, the prism stands on its

Rectangular Prism. Three positions of a rectangular prism are shown in Fig. 115. In the first view, the prism stands on its.

Fig. 115, Rotated Views of Rectangular Prism

Fig. 115, Rotated Views of Rectangular Prism.

base, and its axis, therefore, is parallel to the vertical plane. In the second position, the axis is still parallel to V and one corner of the base is on the horizontal plane, but the prism has been turned on the line lh 1v as an axis, so that the inclination of all the faces of the prism to the vertical plane remains the same as before. That is, if in the first figure the side A B D C makes an angle of 30° with the vertical, the same side in the second position still makes 30° with the vertical plane. Hence the elevation of No. 2 is the same shape and size as in the first case. The plan is found by projecting the corners down from the elevation to meet horizontal lines projected across from the corresponding points in the first plan. The third position shows the prism with all its faces and edges making the same angles with the horizontal as in the second position, but with the plan at a different angle with the ground line. The plan then is the same shape and size as in No 2, and the elevation is found by projecting up to the same heights as shown in the preceding elevation. This principle may be applied to any solid, whether bounded by plane surfaces or curved; as far as it relates to heights, it is the same that was used for profile views.

End View a Means of Obtaining True Height. An end view is sometimes necessary before the plan or elevation of an object can be drawn. Suppose that in Fig. 116 it is desired to draw the plan and elevation of a triangular prism 3 inches long, the end of which is an equilateral triangle 1 1/2 inches on each side. The prism is lying on one of its three faces on H, and inclined toward the vertical plane at an angle of 30°. The plan may be drawn at once, because the width will be 1 1/2 inches, and the top edge will be projected half way between the other two. The length of the prism will also be shown. Before the elevation can be drawn the height of the top edge must be found by obtaining the altitude of the triangle forming the end of the prism. All tha is necessary, then, is to construct (Reproduced by Courtesy of "The Brickbuilder.") an equilateral triangle 1 1/2 inches on each side, and measure its altitude. A very convenient way to do this is shown in the figure by laying one end of the prism down on H. A similar construction is shown in Fig. 117, but with one face of the prism on V instead of on H.

Fig. 116. True Height on Horizontal Plane

Fig. 116. True Height on Horizontal Plane.

APARTMENT HOUSE IN ASHMONT, MASS.

APARTMENT HOUSE IN ASHMONT, MASS.

Edwin J. Lewis, Architect, Boston, Mass.

For Plans, See Opposite Page.

(Reproduced by Courtesy of "The Brickbuilder.")

FIRST AND SECOND FLOOR PLANS OF APARTMENT HOUSE IN ASHMONT, MASS.

FIRST AND SECOND FLOOR PLANS OF APARTMENT HOUSE IN ASHMONT, MASS.

Edwin J. Lewis, Architect, Boston, Mass.

For Exterior, See Opposite Page.

In all the work thus far the plan has been drawn below and the elevation above. This order is sometimes inverted and the plan put above the elevation, but the plan still remains a top view no matter where placed, so that after some practice it makes but little difference to the draughtsman which method is employed.

Illustrative Examples

1. Construct plan and elevation of a regular hexagonal pyramid. It is evident that two distinct geometrical views are necessary to convey a complete idea of the form of the object; an elevation to represent the sides of the body, and to express its height; and a plan of the upper surface to express the form horizontally.

It is to be observed that this body has an imaginary axis or center-line, about which the same parts are equally distant; this is an essential characteristic of all symmetrical figures.

Draw a horizontal straight line L T through the center of the sheet, Fig. 118; this line will represent the ground line. Then draw a perpendicular Z Z' to the ground line. For the sake of preserving the symmetry of the drawing, the center of the lower range of figures are all in the same straight line M N, drawn parallel to the ground line.

In delineating the pyramid, it is necessary, in the first place, to construct the plan. The point S', where the line Z Z' intersects the line M N, is to be taken as the center of the figure, and from this point, with a radius equal to the side of the hexagon which forms the base of the pyramid, describe a circle, cutting M N in A' and D'. From these points with the same radius, draw four arcs of circles, cutting the primary circle in four points. These six points being joined by straight lines, will form the figure. A' B' C' D' E' F', which is the base of the pyramid; and the lines A' D', B' E', and C F', will represent the projections of its edges foreshortened as they would appear in the plan. If this operation has been correctly performed, the opposite sides of the hexagon should be parallel to each other and to one of the diagonals; this should be tested by the application of the square or other instrument proper for the purpose.

Fig. 117. True Height on Vertical Plane

Fig. 117. True Height on Vertical Plane.

By the help of the plan obtained as above described, the vertical projection of the pyramid may be easily constructed. Since its base rests upon the horizontal plane, it must be projected vertically upon the ground line; therefore, from each of the points A', B", C", and D', raise perpendiculars to that line. The points of intersection, A, B,C, and D, are the true positions of all the angles of the base; and it only remains to determine the height of the pyramid, which is to be set off from the point G to S, and to draw S A, SB, SC, and S D, which are the only edges of the pyramid visible in the elevation. Of it is to be remarked that S A and S D alone, being parallel to the vertical plane, are seen in their true length; and moreover, that from the assumed position of the solid under examination, the points F" and F" being situated in the lines B B' and C C", the lines S B and are each the projections of two edges of the pyramid.

Fig. 118. Construction of Regular Hexagonal Pyramid

Fig. 118. Construction of Regular Hexagonal Pyramid.

2. Construct the projections of the pyramid, Example 1, having its base set in an inclined position, but with its edges S A and S D still parallel to the vertical plane, Fig. 118.

It is evident, that with the exception of the inclination, the vertical projection of this solid is precisely the same as in the preceding example, and it is only necessary to show the same view of the prism in its new position. For this purpose, after having fixed the position of the point D upon the ground line, draw through this point a straight line D A, making with L T an angle equal to the desired inclination of the base of the pyramid. Then set off the distance D A, equal to that used in Example 1; erect a perpendicular on the center, and set off G S equal to the height of the pyramid. Transfer also from the first example the distance B G and C G to the corresponding points, and complete the figure by drawing the straight lines AS, B S, C S, and D S.

In constructing the plan of the pyramid in this position, it is to be remarked, that since the edges S A and S D are still parallel to the vertical plane, and the point D remains unaltered, the projection of the points A, D, and S, will still be in the line M N. The position of A' is determined by the intersection of the perpendicular A A' with M N. The remaining points, B', C', etc., in the projection of the base, are found in a similar manner, by the intersections let fall from the corresponding points in the elevation, with lines drawn parallel to M N at a distance (set off at o, p,) equal to the width of the base. By joining all the contiguous points, the figure A' B' C' D' E' F' is obtained representing the horizontal projection of the base, two of its sides, however, being dotted, as they must be supposed to be concealed by the body of the pyramid. The vertex S having been similarly projected to S', and joined by straight lines to the several angles of the base, the projection of the solid is completed.