Biology, asked by Alluarjun7108, 1 year ago

A right circularcylinder of base 50 mm and height 70 mm has one of the circumference point of base in the hp such that its axis is inclined at 30 degree to the hp and axis appears to be inclined at 45 degrees to the vp in the tv

Answers

Answered by kburton
0

Procedure –

Draw XY line.

As cylinder is resting in VP, so circle will be seen in FV and rectangle in TV.

Draw FV as a circle of 50 mm diameter above XY line at some convenient distance.

Draw horizontal and vertical axis of circle.

Divide the circle into 12 equal parts (You can divide it into 8 parts also). Name all the points 1’, 2’, 3’ ,….12’.

Take the projections of all point into TV.

Complete the TV taking axis length 60 mm. Draw all the generators of circle.

Name the TV on both top base as well as bottom base 1, 2, 3 …12.

Stage 2

As axis is inclined at 30° to XY line the base 1 7 line will be at 60° inclined to XY line.

So first mark point 7 at some convenient distance on XY line. Then draw line 1 7 at angle of 60°, so that the axis will be inclined at 30°.

Using compass mark all the points from stage 1 to stage 2 on line 1 7.

Complete the TV in inclined position using compass.

Take the horizontal projections of all 12 points 1’, 2’, 3’, …12’ towards right side in FV.

Take the projections of inclined TV into FV of all the points 1, 2, 3, …12 from both bases top and bottom.

Intersection of 1 and 1’ will give point 1, Intersection of 2 and 2’ will give point 2, Intersection of 3 and 3’ will give point 3…….up to point 12.

Half circle from the bottom base will be invisible from point 4’ to 10’. Draw it as a hidden curves.

Draw a smooth curve passing through remaining points.

Similar questions