Q.A line PF 65 mm long has its end P. 15 mm above the HP and 15 mm infront of the VP. It is inclined at 55 to the HP 35 to the VP. Draw its projections of the line.
SOLUTION:
Steps:
1. Since the sum of inclinations (6+4) of the line with the principle planes is 90°, the projections of the line lie in the same projector.
2. The projections of Pare marked. The lengths of the top view and front view are determined in usual manner.
3. The line is first assumed to be parallel to the VP and inclined at 55° to the HP. Draw p’f; of length 65 mm inclined at 55° to xy. Project fi to fi on a line drawn through p parallel to xy. pf1 is the length of the top view. (Fig. ).
4. Next, find the length of the front view assuming the line to be parallel to the HP and inclired at 35° to the VP.
5. For this, draw pfe of length 65 mm inclined at 35º to xy. Project 12 to f½ on the line draw through p’ parallel to xy.
6. With pas centre and pf, as radius, draw an arc to cut the path through f2 at f. With p’ as centre and p’; as radius, draw an arc to cut the path through fi at f’.
7. Draw lines pf and p’f’ to get the required projections. It is found that pf and p’f’ lie in the same projector perpendicular to xy.