By interpenetration is meant the intersection of two bodies of similar or different form, resulting in a regular or irregular figure, as the case may be.

Fig. 86.

Take a simple case, namely, a cylinder penetrating a rectangular prism at an angle of 60 degrees.

Let abcd (Fig. 86) be the side elevation of a prism; ef the axis of a right cylinder, and ghij the outline of the same.

It will be seen that the line of intersection gj is on a plane oblique to the axis of the cylinder ef; and gj will therefore be of greater length than hi, the diameter. Thus at the line of junction of the two bodies the contact line will be an ellipse, with major axis gj and minor axis = hi.

This is a simple form of interpenetration, and one frequently met with in masonry work.

## Problem

To determine the interpenetration of a cone and a cylinder.

Let abc (Fig. 87) be the side elevation of a cone, and defg a cylinder with its axis at right angles to that of the cone.

Fig. 87.

In this case the two bodies in question have curved surfaces. In front elevation the face of the cylinder shows as a circle. To find the curves of interpenetra-tion in plan and side elevation, divide the semi-circumference of the circle on side of x into any convenient number of equal parts, say 16, and draw lines through the points thus determined at right angles to the axis of the cone, prolonging them to the plan and side elevation. Where these lines cut the slant side of the cone, project vertical lines to the base; and with the axis as centre and the points marked off by these lines in the base as the extremities of the radii draw concentric arcs on plan from the centre of the plan of the cone, cutting the lines previously drawn. Through the points of contact draw a curve; which will be the curve of interpenetration on plan. The curve of contact on the side elevation is found in the following manner: Make hi, kl, and k1m on side elevation equal to hi, kl, and km on plan respectively. A curve drawn through l, i, and m, and points similarly found, on the horizontal dotted lines of Fig. 87, will give the curve of penetration on side elevation.

## Problem

On the interpenetration of a horizontal cylinder with a vertical prism.

Let Abdc (Fig. 88) be the plan of a cylinder with its axis forming an angle of 18 degrees with the L. of I.; and Efgh the plan of a vertical prism penetrated by the cylinder.

Project lines of construction from A, B, C, and D of the cylinder, and from the ends of the axis as shown by NO, above the L. of I. Project similar lines from the points E, F, G, and H of the prism. Let the cylinder be above the L. of I. Make O1P and O1Q equal to OD and OC on plan. The lines PQ and D1C1 will form the axes of an ellipse, which can be drawn by any of the usual methods. Proceed with the other end of the cylinder in the same manner, as at Rb1sa1; that portion shown by Ra1s being dotted as not visible. Draw in the outline of the prism.

To find the curves formed by the contact of the two bodies.

Fig. 88.

With N as centre and NA as radius describe a semicircle, and divide the semi-circumference into any convenient number of equal parts. In this case a quadrant only is shown, divided into eight parts.

Project these points, by lines parallel to the axis of the cylinder, on to the side FH of the prism. From the points of contact thus found project vertical lines at a convenient distance above the L. of I., as shown.

From the divisions of the quadrant draw lines parallel to the diameter BA, cutting NT in 1, 2, 3, 4, etc., and from N1R in the elevation cut off parts Njl1,, N121, N131, etc., equal to N1, N2, N3, etc., on the plan, and through 11, 21, 31, 41, etc., draw lines parallel to B1C1 cutting the lines projected from the intersections of the side of the prism FH. The crossings of these two sets of lines mark points on the curve of penetration, and through them the required curve is drawn.

Fig.89.

The curves of penetration of the cylinder with the other three faces of the prism are also shown, but the construction lines have been left out to prevent confusion of the diagram.

Fig. 89 shows the method of finding the curve of penetration of a semi-cylinder intersecting another semi-cylinder, as is the case of vaults. Abcd is the plan of the large cylinder, and Aed is its section. Fghk is the smaller cylinder, and Glh its section. Divide GH into any convenient number of equal parts 1, 2, 3, 4, etc., and through these points draw lines parallel to the axis of the smaller cylinder and cutting the section at 11, 21, 31, 41, etc. From ME cut off parts M12, M22, M32, M42,, etc., equal to 111, 221, 331, 441, etc. Through 12, 22, 32, 42, etc., draw lines parallel to AD, cutting the section of the large cylinder at a, b, c, d, etc., and from a, b, c, d draw lines parallel to FM and cutting the lines through 1, 2, 3, 4, etc., at a1, b1, c1, d1, etc. A curve drawn through these points a1, b1, c1, etc., gives the plan of the line of penetration.

The curve shown on the left-hand side of Fig. 89 is that of the plan of the line of intersection of a cylinder cutting another cylinder obliquely, and is similarly obtained.

Fig. 90 shows the method of finding the curves of penetration for a semi-cylinder intersecting a sphere, as is the case of a cylindrical vault interpentrating a spherical dome, the semi-cylinder standing on a prism, whose width is equal to the diameter of the semi-cylinder, and therefore springs from the line XY. Abc is the plan of the sphere, and Adec the plan of the cylinder. Draw the sections of the sphere and cylinder as B1GF and Dhe respectively. Divide the line DE into any convenient number of parts at 1, 2, 3, 4, etc., and through these points draw lines parallel to BH, cutting Dhe at 11, a1 31, 41, etc.

Set the heights 111, 221, 331, 441, etc., up XG above the springing line XY, as X12, X22, X32, X42, etc., and through 12, 22, 32, 42, etc., draw lines parallel to B,F cutting the circumference B1GF at a, b, c, d, etc., respectively. Through a, b, c, d, etc., draw lines parallel to GK, cutting BH at a1 b1 c1 d1, etc. With O as centre and Oa1, Ob1 ,Oc1 ,Od1 ,etc., as radii draw concentric arcs, cutting the lines drawn through 1, 2, 3, 4, etc., at a2, b2, c2 ,d2 ,etc. A curve drawn through the points a2 ,b2 ,c2 ,d2, etc., gives the plan of the curve of penetration required.

To find the curve of penetration along BH, draw f12E1 and mHl representing the springing line and crown line respectively of the semi-cylinder. Project the points a2 ,b2 ,c2 ,d2 ,etc., to a12, b12 ,cl2 ,dl2 ,on the line f12E1

Fig. 90.

and set up the heights b12q, c12p, dl2O,e12n, etc making them respectively equal to 111,221,331,441, etc. A line drawn through m, n, 0, p, q, and Y2 gives the curve of penetration of the semi-cylinder with the dome. The curves of penetration below the line fl2E1 is found in the following manner. Divide rF and r1C1 each into 3 equal parts at s, t and s1, t1 respectively, and erect the perpendiculars sv, tu, s1v1, and t1u1. Make s1v1 equal to sv, and t1u1 equal to tu. A line joining Y3 v1 and u1 will give the required curve of penetration on the section BH.

Fig. 91.

The curve for the penetration of the prism with this hemisphere on plan is obviously a straight line.

Fig. 91 shows an example of two semi-cylinders of the same span intersecting at right angles. The plan of the lines of intersection are obviously straight lines in this case. It will be necessary in working the stones for such a vault, to determine the curve of penetration along one of these diagonal intersection lines. This is done in the following manner: Divide the diameter of either vault up into any number of equal parts as at a, b, c, etc., and through a, b, c, etc., draw lines parallel to the axis of the vault to cut the circumference of the vault in a1, b1, c1, etc., and the diagonal XY at a2, b2 c2, etc. At a2, b2, c2, etc., draw lines at right angles to XY, making a2a3 equal to aa1 and b2b3 equal to bb1, and so on. A curve drawn through the points a3 b3, c3, etc., gives the required curve of the section along XY.