is vertical, being at right angles to the plan of one of the lateral edges of the pyramid. First, construct the square ABCD; join the diagonals AC, BD, and we have the plan of the pyramid ; the point E, where the diagonals cut each other, being its vertex. We now draw mnp, the trace of the section plane, at right angles to any of the lateral edges of the pyramid, as ED. Draw xy parallel to mnp. We now find the elevation of the pyramid, which is E'AC". The points m and p, being two points in the base, will be projected in m' and p', and there remains only to find the elevation of the point n. The vertical projection of ED is E'D' at right angles to xy; and to determine n in E'D' we want a separate construction. Now, what we require is to find the height of n above the horizontal plane, which height must be set off from D' along D'E'. This can be done by finding the elevation of ED' when viewed at right angles to a vertical plane conceived to pass through it, that is, in the direction of AC. Now we have such an elevation of EC in EC', the vertical projection of EC. Hence, transfer the point n to s, by describing an arc, with centre E, and radius En to cut EC' in s. Problem 58. To draw the horizontal projection of the section of a square pyramid, when cut by a plane parallel to its base, the plane of the base of the pyramid being inclined at 40°. First construct the square ABCD; join the diagonals AC, BD, and we have the plan of the pyramid ; the point E, where the diagonals cut each other, being its vertex. Draw FG, inclined to xy, at an angle of 40°. Then draw HDK parallel to sy, and set off upon FG, AD, and DC equal respectively to HD and DK. From D, erect DE perpendicular to FG. Join EA, EC, and we have an elevation of the pyramid upon the plane FG. Its plan will be found as previously shown. We have now to project the section upon the plan. Draw adc parallel to FG, to represent the section plane. If the projection of one point in the section be understood, it will be easily seen how to find the others. For example, the point c is a point in the lateral edge EC. Now, the plan of EC is E'C', hence the plan of c must be in E'C', viz. d'. For a similar reason, the plan of d must be in E'D', viz. d'. The point d represents one diagonal of the section, as D represents a diagonal of the base of the pyramid. It is clear that the other extremity of this diagonal must be in E'D"; and since it must also be in the projector let fall from d, it will be in d". Now, the plan of a is a' on E'C'; because the plan of EA will be on E'C'. Join the points a'd'c'd", and we have the horizontal projection of the required section. NOTE.-The lines E'd", E'a', and E'd' are dotted in the figure, because this part of the pyramid is supposed to be removed. Problem 59. To draw the sectional elevation of a right cone, when cut by a vertical plane. Let ABC and A'B'C' be the plan and elevation of the given cone, DE being the trace of the cutting plane. The plane DE cuts the base of the cone in points a, b, whose elevations are a', b' in B', C'. We have now found two points in the required section. From centre A, describe a circle cutting DE in points c, d. This circle represents the base of another cone, the elevation of which is A'é'f'. Now c and d are two points in the base of the second cone, in the same manner as a and 6 are two points in the base of the given cone, and their elevations will be upon e'f', just as the elevations of a, b are upon B'C'. We thus get c'd', two more points in the required section. Further, from centre A describe a circle tangential to DE, and touching it at m. This circle represents the base of a third cone, whose elevation is A'g'h'. The elevation of m is m', which gives the height of the section. Through points a', d', m', d', b', describe a curve as shown in the diagram, and thus obtain the section required. NOTE.—The curve which shows the outline of the section is a hyperbola. Problem 60. To find the projection of a cone standing on its base, when the section plane is perpendicular to the vertical plane, and making an angle with the horizontal plane ; also a projection of the cone, showing the true form of the section. I First, through the vertex of the cone draw a line V'E to any point within the base A'B'; this line is to be considered as the vertical projection of a generatrix of the cone, and the point e' where it cuts the line mn, is the projection of that point on the surface of the solid, where the cutting plane actually passes through the generatrix EV. The point é' may be projected upon the plan by letting fall a perpendicular from E, cutting the circumference of the base in E, and joining EV; then another perpendicular let fall from é will intersect EV in a point e, which will be the horizontal projection of a point in the required curve. By drawing another line, e.g., V'D', and projecting its point of intersection d' with the cutting plane, to d, a second point in the curve is obtained ; and so on for any number of points required. The exterior generatrices A'V' and B'V', being both projected upon the line AB, the extreme limits of the curve sought will be at the points a and b, on that line, which are the projections of the points of intersection a' and b', of the cutting plane with the outlines of the cone. And, as the line ab will clearly divide the curve symmetrically into two equal parts, the points f, g, h, &c., will be readily obtained by setting off above that line, and on their respective perpendiculars, the distances, dd, ee, &c. A sufficient number of points having thus been determined, the curve drawn through them (which will be found to be an ellipse) will be the outline of the required section. This curve may be obtained by another, and perhaps simpler method, depending on the principle that all sections of a cone by planes parallel to the base are circles. Thus, let line F'G' represent a cutting plane; the section which it makes with the cone will be denoted, on the horizontal projection, by a circle drawn from the centre V with a radius equal to half the line F'G; and by projecting the point of intersection H', of the horizontal and oblique planes, by a perpendicular H'H, and noting where this line cuts the circle above referred to, we obtain the points H and K in the curve required. Similarly, any number of additional points may be found. Secondly, let the cutting plane mn be conceived to turn upon the point b', so as to coincide with the vertical line b'k', and let b'k' be transferred to a'b', which will represent as before the extreme limits of the required curve. Now, taking any point, such as d', it is clear that in this new position of the cutting plane, it will be represented by d", and that if we make the further supposition that the cutting plane were turned upon a'b', as an axis, till it should be parallel to the vertical plane, the point which had been projected at d" would then have described round a'b' an arc of a circle whose radius is the distance dd, No. 2. This distance, therefore, being set off at d' and f', on each side of a'b', gives two points in the required curve. By a similar mode of operation, any number of points may be obtained, through which, if we draw a curve, it will be an ellipse, of the true form and dimensions of the section. Or, having found the axes, major and minor, the ellipse may be constructed by any of the methods referred to in Plane Geometry. R |