Problem 48. To draw the plan and elevation of a dodecahedron, when one edge of its base is inclined at an angle of 30° to the ground line. Take a straight line ab, and let it be inclined to the ground line at an angle of 30°, and upon it describe the regular pentagon abcde. Describe a circle about abcde. Then bisect each side of the pentagon, and draw straight lines through the points of bisection and the opposite angles. These lines will cut the circumference of the circle in A, B, C, D, E. Join these points, and make m5 equal to mo, and from centre o, with radius ob, describe a circle. By means of the lines already drawn, divide this circle into ten equal parts in the points 1, 2, 3, ....., 10; and join these points. To find the elevation of the dodecahedron, project the points a, b, c, d, e, on the base line. We next find the height of any of the points 2, 4, 6, 8, 10, above the points a, b, &c. Thus take the point 10; the line d10 is the projection of a line whose real length is equal to ab, &c. We then find the height of 10 above d, as previously shown, and set it off on the perpendicular from f to g. We now find the height of any of the points 1, 3, 5, 7, 9, above the points 2, 4, 6, 8, 10. For instance, from 9 to 10 is the projection of à line whose real length is ab, &c. Thus the height of 9 above 10 is found as before, and is gh. Lastly, make hk equal to fg, and through g, h, k, draw lines parallel to xy. The points A, B, C, D, E, will be projected on the line drawn from k, the points 2, 4, 6, 8, 10, on the line drawn from g, and points 1, 3, 5, 7, 9, will be on the line from h. Then let the points in elevation be joined in the same order as in the plan. Section VI.-SECTIONS. In the preceding sections, we have treated on the plans and elevations of the various solids as wholes. Now it is often necessary to furnish certain essential details respecting a solid figure, which cannot be obtained by either plans or elevations. To this end, certain portions of it are shown, termed sections; the consideration of which forms the subject of the present section. . Problem 49. To draw the section of a cone when cụt by a plane parallel to its base. Let the triangle B' A'C' represent the elevation of the given cone, F and the circle DEF its plan, also let G'H' be the elevation of a section plane cutting the cone parallel to its base B'C'. Draw DE, the diameter of the plan, parallel to ty, and from G' and H' draw projectors cutting DE in G and H. Then the line GH is the diameter of the plan of the section. On GH construct the circle GFII, which will be the plan of the section of the cone cut off by the plane GH. Then let the section be shaded by parallel lines drawn at an angle of 45°, as is usual when sectional drawings are indicated. Note 1.-If the cone be cut by a plane passing through its axis, the section will be a triangle. NOTE 2.—There are other sections of a cone, such as the ellipse, the hyperbola, and the parabola. Problem 50. To draw the gection of a hollow pipe. Let a'b'd represent the elevation of the end of a pipe of which DEFG is the plan, and let a'd be the elevation of the section plane. Drop projectors at right angles to xy from the points where the section plane cuts the figure, as a'd'é'c. In consequence of the section GF plane dividing the figure exactly in half, the projections of the lines represented by the points a'c will exactly coincide with the lines in the plan DG, EF. Also, the plan of the semicircle a'c' will coincide with the line GF. DE represents the plan of the other semicircular end of the pipe. It only remains, therefore, to show the thickness, which is done by means of the projectors drawn from d'é' parallel to DG and EF. Problem 51. Let mn be the section plane, then mn represents a horizontal plane, i.e., a plane at right angles to the vertical plane. It will thus be seen that the cube is cut from the anterior face (from the face turned towards the eye) to the posterior face (the face turned from the eye). The section produced will be an oblong, having its breadth equal to h'k', and length equal to the edge of the cube. In order to represent the section, find first the plan of the whole cube, as in Pr. 9. From h' and k' drop perpendiculars, and we obtain efkh, the plan of the required section. Problem 52. To draw the section of a cylinder lying on the ground with its ends parallel to the vertical plane, and at right angles to the horizontal plane ; the plane of section being parallel to the axis of the cylinder. Let A'B'C' represent the elevation of a cylinder lying on the ground, and DEFG its plan. Also let A'B' represent the elevation of the section plane cutting the cylinder parallel to its axiş. |