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a'b'd'd' of the base, by turning the square ABCD through the angle 0, thence find its plan abcd.

The elevations of the edges perpendicular to the plane mnp will be perpendicular to np, and equal in length to the edge of the cube ; draw d'a", 6'", d'I", and c'c" at right angles to np', and equal to BC; then join a"c", the figure a"c' will be the elevation ; the plans of the

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points a'b'd"c“ will be the points in which perpendiculars to xy, drawn from these points, cut the parallels to xy drawn from A, B, D, and C.

Problem 35.

To find the projection of a cube when one of its diagonals is perpendicular to the plane of projection.

Construct a square A'B'C'D', and let it represent a face of a cube. Draw ay perpendicular to B'D', one of the diagonals of this face. Then taking A'B'C'D as an elevation of the cube, its plan will be ADCEFG, as explained in Pr. 9. Join EC, one of the diagonals of the cube, and draw a'y' at right angles to it. Then the projection of the solid will be obtained as before. For example, the projection of G will be in the projector drawn from G at right angles to my, and its height above ay is equal to cc'. Therefore make gG% equal to cc'. Also, the projection of Fis F found by setting off

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H'F' equal to D'B' along the projector drawn from F.
The points E, D', &c., are found in a similar manner.

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Problem 36. To draw the plan of a square when its surface is inclined 42°, and one of its sides is horizontal.

Here as the surface of the square is to be inclined 42°, we commence by assuming the traces of a plane inclined at that angle, and rotate the figure from a horizontal position into this plane.

Then cd, the horizontal trace, is perpendicular to xy, and the vertical trace makes an angle of 42° with it. The square ABCD must be drawn with AD, one of its sides parallel to cd. Through C and D projectors must be determined meeting xy in m' and n'.

Then with cas centre, describe the arcs m'C" and n'D', intersecting the vertical trace în C and D'.

These arcs will thus represent the journey of the points C and D whilst being rotated into the plane e'cd. C'D' is thus the elevation

of the whole square, because AD and BC, the sides of the square, being horizontal and perpendicular to the vertical plane, their elevations are points.

The intersections of projectors through D and C', with lines

parallel to xy through A, B, C, and D, are the plans of the corners of the square.

Problem 37.

To draw the plan and elevation of a hexagonal prism, which has its axis inclined 40° to the paper and one face parallel to the vertical plane.

Draw the hexagon with one side parallel to xy, which is the plan of the solid when standing with its base upon the paper. By arranging the figure in this way, it will be seen that one face of the object is parallel to the vertical plane. The elevation must be determined from this plan, according to the methods described in the foregoing problems.

Upon the elevation draw G G" to represent the axis of the solid, and produce it. Let us now assume a new 'y', making an angle of 40° with the line GʻG” produced, the elevation will then be that of the solid, with its axis inclined as regards the new horizontal plane. The student may easily see this by folding his paper upon the new ir'y, so as to show both a horizontal plane and a vertical. The first

elevation will then be seen under quite a different aspect, viz., that of a solid tilted over.

Now in order to determine the plan, projectors must be drawn through every point of the elevation, at right angles with the assumed a'y', and lengths must be measured along each of these projectors, equal to the distances of the points in the first plan, from the first base line sy. For example, take the point A, a projector al passes through a at right angles to the new a'y'. The distance A A" is transferred along this projector to the point a beyond the ground line ; that is, AA" is equal to la. In the same manner, all the other points of the base are projected, and thus a new plan of it is obtained.

The plan of the other end of the solid is determined in like manner by projectors through its points in elevation, And as the

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first plan is that of both ends, the distances to be measured along the projectors first drawn will be precisely the same as before. To illustrate this, take point D. The distance measured upon the projector beyond 'y' is equal to 2D.

The whole plan is completed by joining the similar points in each base, as shown in the figure. We may notice, that lines which are parallel in the solid are still parallel, however they may be projected ; e.g., 3 4 is parallel to 5 6.

A little consideration of the position of the solid will show that that part of the base in which A is situated is hidden, and that the opposite end is wholly seen in the plan. The edges being dotted indicate this.

Problem 38.

To draw the plan and elevation of a solid hexagonal column, the height of which is 30', and length of one side 10' on a scale of 20 to the inch; the front face being parallel to the vertical plane.

Here the plan will be a regular hexagon. Draw a line AB parallel to ry, in length half an inch; on it construct a regular hexagon, ABC, &c. This will be the required plan.

In order to find the elevation, we draw the projectors Aa', Bb', &c. From a'b'c'f' raise perpendiculars a'A', U'B', c'C', and f'F', each i}

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inch in height. Then through F', A', B', C', draw a line parallel to wy, and the elevation, consisting of a series of rectangles, will be complete.

It may be remarked that the projectors from A and B, which are the angles at the base of the front face, are coincident with those that are drawn from E and D, and are therefore invisible. The rectangle A'a'b'B' is an elevation of the front face which rises up from AB, and also of the face which rises up from ED. Also, the rectangle F'f'a'A' is the elevation of the faces which rise up both from AF and FE; as also the rectangle B'b'd'Cis the elevation of faces which rise up both from BC and DC.

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