Problem 39. To draw the plan of a pentagonal pyramid when one edge of its base is inclined at an angle of 45°. Take a straight line A'B', and let it be inclined to my at an angle of 45°. Then upon A'B' construct a regular pentagon A'B'C'DE', and join each of the points A’, B', C', &c., to F' the centre, which is the vertex of the pyramid. To find the plan of the pyramid ; from F', the vertex of the pyramid, draw FF, and set off from F, FG equal to the height of the pyramid. The line FG represents the axis of the pyramid. Through point G, draw a line parallel to xy, and from the points A', B', C', &c., drop projectors, cutting this line in the points A, E, C, &c. Join each of these points to F, and we have the required plan of the pyramid, having an edge of its base inclined at 450. Problem 40. To determine the plan of a hexagonal pyramid when lying on one of its faces on the horizontal plane. Describe the hexagon ABCDEF, find its centre 0, and join it to each of the angular points A, B, C, &c. This will complete the plan of the pyramid, when its base is horizontal, O being its vertex. Now, the projection of the solid required is that which will result after the pyramid has revolved upon one of the edges of its base, as AB, until the face OAB rests in the horizontal plane. To obtain this projection, then, we must first determine the angle which the faces of the pyramid make with its base. The face OAB represents a plane whose horizontal trace is AB, and we must construct a right-angled triangle to find its inclination, having os, drawn at right angles to AB, for its base, and the height of the pyramid, that is, the height of O above A and B, for its perpendicular. Produce AB and draw xy at right angles to it. The vertical projection of BA is B'A' in xy. Draw the projector 00', making O'M' equal to the height of the pyramid, and join O' to B'A'. Then, the angle O'B'M' expresses the inclination of the face of the pyramid to its base, and consequently to the horizontal plane. Make A'O” equal to A'O, and from A' draw A'E', making the angle O'A'E' equal to O'B'M'. Next produce CF, DE, meeting cy in M' and N', and from centre A', with radii A'M', A'N', describe arcs cutting A'E' in F', C', and Join O'E', O"F", which completes the elevation of the pyramid when resting on one of its faces. This will be understood by joining O'N', when О'A'N would be the elevation of the pyramid resting with its base upon the horizontal plane, and O' A' Ê is ('A'N' after the latter has revolved until A'O' assumes the position of A'0". The plan is thus determined--the plan of C is C, being the point of intersection let fall from C" with a line drawn from C parallel to xy. The plan of the other points may be similarly obtained. Problem 41. To project a hexagonal pyramid whose axis is inclined at an angle of 50° to the horizontal plane, but parallel to the vertical. The projections of the given pyramid having its axis parallel to the vertical plane, and at right angles to the horizontal plane, are shown in the following figure No. 1, its plan being a hexagon, and its elevation an isosceles triangle. Place elevation No. 1 at the given inclination, for elevation No. 2. The projection of the various points for the plan of the pyramid are found by letting fall perpendiculars from the various points in the elevation of No. 2 to intersect the parallels to wy, drawn from the corresponding points in the plan of No. 1. Hence the vertical projectors from B, 6, C', A', falling on the parallel DE produced, determine the position of these points on the plan. The projector from a', which meets FG and HK produced (No. 1), determines the position of the angles F and H on the plan No. 2. And so the projector froin c', which meets FG and HK produced (No. 1), determines the positions of angles G and K in plan No. 2. Join P, the plan of the apex of the pyramid, with D, H, K, E, G, F, the contiguous angular points in the base, and the plan of the whole pyramid will be complete. The edges of the base GE, EK, are dotted, because they are hidden by the body of the pyramid. It will be seen that the axis BP and the diameter DE are in the same line, which is parallel to the vertical plane. NOTE.—Pour faces are seen in the plan : e.g., PFG is the plan of the face of which FG is the plan in No. 1; PFD is the plan of the face of which FD is the plan in No.1 ; PDH is the plan of the face of which DH is the plan in No. 1; and PHK is the plan of the face of which HK is the plan in No. 1. It will be observed that two faces are hidden in the inclined plan, viz., those of which PGE and PKE are the plans. Problem 42. The projections of any solid being given, to determine other projections from them. Let the figure AC be the plan of a square prism, of which A'B'C'D' is the end elevation. It is required to determine a new elevation upon a vertical plane, making an angle, 0, with the long edges of the solid. Assume x'y making the required angle with either of the plans of the sides, as AA. Projectors through A, B, C, and D in the plan, at right angles to X'y', will contain the required elevations of the points A'B'C'D. Now, as the heights of these points above the horizontal plane are shown in the given end elevation, it is only necessary to transfer them from one elevation to the other. For example, the distance AA' is the same as A'l. Similarly, we obtain the elevation of the end A', B', C', D'; and since the edges of the solid are parallel to the horizontal plane, the heights of the corners are the same as those of A, B, C, D. Problem 43. To project a cone whose axis is inclined at an angle of 45° to the horizontal plane, but parallel to the vertical. The projections of a cone having its axis parallel to the vertical plane, and at right angles to the horizontal plane, are shown in the following figure No. 1, its plan being a circle, and its elevation an isosceles triangle. Now, in order to project a cone whose axis is inclined at an angle of 45° to the horizontal plane, we draw B'C', the elevation of the diameter of the cone at an angle of 45°; because when the axis is inclined at 45°, the diameter of the base is also inclined at the same angle. Transfer the elevation B'A'C' of No. 1 to B'A'C' of No. 2. Draw a diameter BC on No. 1 parallel to xy, and another DE at right angles to it. Then the length of the diameter BC on plan No. 2 is determined by projectors dropped from B and to meet the diameter BC, No. 1 produced. |