5.Development of a truncated cone:

A cone of base diameter 40 mm and slant height 60 mm is kept on the ground on its base. An AIP inclined at 45° to the HP cuts the cone through the midpoint of the axis. Draw the development.

Solution:
The development of the truncated cone is shown in figure 5.

Figure 5. Development of the truncated cone (problem 5).

Draw the Front view and top view of the cone.  Dive the circumference of the circle (Top View) in to 12 equal parts  1, 2, 3, 4, …., 12. Project these points on the Front view to obtain the points 1’, 2’, 3’, …., 12’. Draw a line inclined at 45 ° to the horizontal and passing through the mid pint of the axis of the cone to represent the AIP. The locate the  intersection points of the AIP with the generators O’-1’,  O’-2’, …. O’-12’ as P1’, p2’, p3’, …. P12’.  Draw the projection  (figure shown on the right of the Front view)   by drawing  the line O1 parallel to O’ 7’. Obtain the included angle of the sector. θ = (20/60)* 360 = 120° (following the procedure shown in problem 4). Then draw sector O–1–1– O with O as a centre and included angle 120°.  Divide the sector into 12 equal parts (i.e., 10° each). Draw lines O–2, O–3, O–4, …, O–12. Draw horizontal projectors from P1’, P2’, …., P12’ such that it meets the line O1 at p1, p2, p3, …, p12. With O as centre and radius O’P1’,  mark point P1 on line O1. With O as centre and radius equal to Op2, draw an arc to intersect the radial line O2 at point P2. Similarly obtain points P3, P4, …, P12, and P1. Join points P1, P2, P3, …., P12 and P1 to obtain the development of the truncated cone.