5. Having found the traces of the plane containing the given triangle PQR (fig. 217), determine the true shape of the triangle, (i) by rotating it about the h.t. of the plane into the H.P.; (ii) by rotating it about the v.t. of the plane into the V.P.; (iii) by finding the true length of each side separately.

Hint. (i) Rabat about the h.t. as in Prob. 145 or by Prob. 147. (ii) rabat each point about the v.t., as in the case of point P, Prob.148. (iii) by Prob. 107.

6. The given square abcd of 2" side represents the plan of a trap-door opened through an angle of 35°, about the hinge-line at de, which is flush with the ground. Draw the traces of the plane

x

6

containing the trap-door, as referred to the given xy line, show the representation of the elevation of the door, and, from the corner a, draw the plan of a line scribed on the trap-door inclined at 30° to the ground.

Hint.-Draw a new xy at right angles to cd so as to find the height of AB as in fig. 188. When AB is known, find the v.t. of the plane on ry by Prob. 137, the hinge-line at cd being the h.t., and project up the outline of the door. Complete by Prob. 143, either case I. or II.

7. Determine the planes inclined at 45° to the V.P. and containing the given line PQ (fig. 218); also determine a third plane containing the same line PQ, so that its inclination to the H.P. is the same as that of the line.

Hint.-Refer to the observation to Prob. 146. The last plane will have its h.t. perpendicular to pq. All the traces must pass through the respective traces of the given line.

8. Draw the projections of an equilateral triangle of 13′′ side lying in the given oblique plane RST of Ex. 2 (a), when one of its corners is in the h.t., a second corner is in the v.t., and the side joining those two corners makes 70° with the h.t.

Hint. Having rabatted the plane about one of its traces, proceed by the observations to Probs. 147-8. There are two possible solutions.

9. The v.t. of an inclined plane makes 40° with xy. From the point in xy where the traces meet, draw a 2′′ line lying in the plane, but inclined to the H.P. at 30°.

Hint.-Draw a line ac, a'c' in the plane as in fig. 218, Case I., but 2" long. From o' draw the required plan parallel to ac. The elevation of the line will remain a'c'. 10. Draw a rough perspective view of the solution to Ex. 9.

N.B.-Enlarge the figures four times the given sizes.

11. Draw the projections of a horizontal line lying in the given plane and 1.25" above the horizontal plane of projection. (1888) 12. The traces of a plane are given. Determine the plan and elevation of a point which is in the given plane, 1" in front of the vertical plane, and the plan of which is 14′′ from the horizontal trace of the given plane. (1890)

Hint. The plan of the point is the intersection of two lines-one parallel to and "below xy, and the other parallel to and 1" from the h.t. of the plane. Determine the elevation by Prob. 138.

13. ov is the vertical trace of a plane, and pp' are the projections of a point lying in it. Draw the horizontal trace. (April 1899)

14. The traces of a plane both make 40° with xy. Draw the projections of two lines lying in this plane, one parallel to the horizontal, and the other parallel to the vertical plane of projection. (1892)

15. HT is the horizontal trace of a plane. AB is a line in the plane. Complete the elevation of AB. Draw the vertical trace of the plane. Find and measure the inclination of the plane to the horizontal plane. (April 1902)

Hint.-See Probs. 137 and 139 or 129 Inclination = 40°.

16. a' is the elevation of a point in the given plane lom; b the plan of a point in the given plane no,p. Determine the real length of the line joining these points.

PR. GEO.

(1887)

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17. Draw a plane to contain the given line ab, a'b', and the horizontal trace of which makes an angle of 45° with the ground line. (1885)

18. ab, a'b' are the projections of a given line; cd is the plan, and p'a point on the elevation of a second line intersecting the first. Determine the traces of the plane containing the two lines.

(1886) Hint.-Let the plans ab, cd intersect in i'; project up i to ' in a'b' ; join p'ï', and determine the plane containing AB and PI by Prob. 142, noting that PI is a segment of CD.

19. oh is the horizontal trace of a plane, and ab the plan of a line lying in the plane. The real length of AB being 2", draw the vertical trace of the plane, and the elevation of the line.

(1896)

Hint. Find the h.t. of the line in oh by producing ab to meet it. By Prob. 85 determine the heights of A and B, aided by Probs. 11 and 12, so as to get the elevation a'b'. Join o and the v.t. of AB.

20. Assume an ry line, and draw the traces of a vertical plane s, making an angle of 38° with the vertical plane of projection. Find a point P lying in the plane s, 11" from the vertical plane of projection, and 2" above the H.P. Draw the projections of a line, lying in the plane s inclined at 60° to the H.P. and passing through P.

(June 1899)

Hint. .-Draw a plane as in fig. 187 c, and a line 13" below xy to cut the former in p. p' is 2" above xy. Set off a line from p' at 60° to xy, and complete by the cone method of Prob. 143.

21. A horizontal line 13" above the horizontal plane makes an angle of 60° with the vertical plane of projection. Draw the traces of a plane containing the line and making an angle of 70° with the vertical plane.

(June 1900)

Hint.-Draw the line like fig. 113 o; take E as the vertex of a cone with the base in the V.P. The v.t. of the plane required will touch the base circle and pass through the v.t., f', of the line. The h.t. will be parallel to ef. Two possible solutions.

22. The horizontal trace of a vertical plane makes 35° with xy. Obtain the elevation of a line lying in this plane, inclined at 45°, and passing through xy. (1891)

Hint.-Draw the plane as in fig. 187 c. From point o, where the traces meet, set off a line above xy and inclined at 45° to it. This line is rabatted into the v.P., and its plan is in xy. Now rotate its plan through 35°, and project up for the elevation required. Another method is based on Prob. 143, and another by alteration of ground line.

CHAPTER XIX.

PLANES, LINES, AND POLYGONS.

IT has already been stated in Rule 25, page 176, that when a line is perpendicular to a plane, the plan of the line is perpendicular to the h.t., and the elevation is perpendicular to the v.t. of the plane.

Fig. 227 is a photograph of a model of an oblique plane v'o'h, and a line AB, placed actually perpendicular to it, when in their natural position. The plan ab is shown in perspective when it has been drawn perpendicular to o'h, and the elevation a'b' perpendicular to 'o'.

The following additional Rules on Planes and Lines should be understood:

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28. The magnitude of the dihedral angle between two given intersecting planes the angle between the two lines of intersection of the given planes made by a third plane cutting them perpendicular to their intersection.

Thus, when we say the dihedral angle between a floor and wall, or the co-ordinate planes = a right angle, we mean that the magnitude of the angle between the two edges of the surfaces made by a third plane cutting them at right angles to their intersection ay = 90°.

29. If a given plane contain a line which is perpendicular to a second given plane, the dihedral angle between the given planes = a right angle.

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30. The angle of inclination of a given line to a given plane the complement of the angle between the given line and a second line drawn from a point in the given line perpendicular to the given plane.

PROBLEM 153.

To determine the projections of the point of intersection of a given line pq, p'q' and plane v'o'h.

1st Method.-Assume the given plan, pq, to be also the plan of a second line contained by the given plane. Find the elevation pq of such a line. In the case under consideration, r (a point in pq) being in the h.t. of the plane, its elevation is in xy; and, pq not conveniently meeting xy on the paper, a horizontal in the plane is drawn passing through q (Prob. 138). By projecting up from q, the corresponding elevation q2 is then readily obtained. The two elevations p'q', pq, intersect in ; hence, project down to pq in i. i, i̇ are the projections of the intersection required.

Of course, instead of the above assumption, p'q' may have been taken as the elevation of a second line contained by the plane, when the corresponding plan of such a line could be obtained; the intersection of the two plans would then be the plan of the point required. Pq, p2q2 may here be regarded as the intersection of the given plane with an auxiliary vertical plane whose h.t. is pq produced, on the supposition that it contains the given line; and, since the two lines are uniplanar, and one of them lies in the given plane, it follows that their intersection is the common point of the given line and plane.

Fig. 223.

2nd Method.-Draw an xy, at right angles to o'h', and convert the given plane into an inclined plane v"m"h (Prob. 139). Obtain the new elevation p"q" of the given line on xy, cutting v"m" in ". Now, since all points in the plane are to be seen in the v.t. of the inclined plane, i" is an elevation of the intersection. Project it down to i in pq for the plan, thence project up i to i in p'q' for the elevation of the required intersection.