Q.A line EF, 85 mm long has its end E, 25 mm above the HP and 20 mm in front of the VP. The top and front views of the line have lengths of 55 mm and 70 mm respectively. Draw the projections of the line and find its true inclinations with the VP and the HP. (Fig.
Solution:
Steps:
1. Locate e, the plan of E, 20 mm below xy and e’ its front view, 25 mm above xy .
2. Assume the line to be parallel to the VP first. Its top view will be parallel to xy and its front view will have true length.
3. Hence, draw e*f_{1} of length 55 mm (length of the top view) parallel to xy.
4. Draw a projector through f_{1} With e’ as centre and 85 mm (true length) as radius, draw an arc to cut the projector through f_{1} at f_{1} ^ *
5. Then, the inclination & of the line e’ * f_{1}’ represents the true inclination of the line with the HP.
6. Draw ab, the path of F in front view, parallel to xy through f_{1}’
7. Repeat the construction with the front view. Draw e’ * f_{2}’ parallel to xy and of length 70 mm (given).
8. Draw a projector down through f_{2}’ . With e as centre and radius 85 mm, draw an arc to intersect the projector through f_{2}’ at f_{2}
9. The inclination Phi of e*f_{2} with xy shows the true inclination of the line with the VP.
10. Draw cd, the path of F in the top view.
11. With e as centre and e*f_{1} as radius, drawn an arc to cut ce at f. With e’ as centre and radius e’ * f_{2} draw arc to meet ab at f’
12. Draw lines ef and e’ * f’ as the required projections. Here θ = 49° and Φ= 35°