0
122kviews
A pentagonal prism having an edge of base 25 mm, axis height 60 mm has one of its corner in the HP. The axis is inclined at $30^0$ to the HP.
1 Answer
2
23kviews

Q. 5

Given Data

Side of prism = 20 mm

Axis length = 60 mm

Corner in the HP (One of the corner must be on right side)

Axis inclined to HP = 30°

  1. Draw XY line.

  2. A prism is resting on HP, so pentagon will be seen in TV and rectangle in FV.

  3. Draw TV as a pentagonone corner must be on right side (corner 4 is resting corner). (If the prism is resting on side of base then the initial position in TV would be different).

  4. Name the pentagon 1 2 3 4 5.

  5. Take the projections of all corners into FV.

  6. Complete the FV taking axis length 60 mm.

  7. Name the FV on both top base as well as bottom base 1’ 2’ 3’ 4’ 5’.

Stage 2

  1. As axis is inclined at 30° to XY line, so the line 1’4’ will be at 60° to XY line.

  2. So first mark point 4’ at some convenient distance on XY line. Then draw line 1’4’ at angle of 60°, so that the axis will be inclined at 30°.

  3. Using compass mark points 3’ and 5’ on the line 1’4’.

  4. Complete the FV in inclined position using compass.

  5. Take the projections of all 10 points 1’2’3’4’5’ from both the bases in TV.

  6. Take the horizontal projections of pentagon corners towards right side.

  7. 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, Intersectionof 4 and 4’ will give point 4 and Intersection of 5 and 5’ will give point 4 for both the bases.

  8. Sides of bottom base 3’ 4’ and 4’ 5’ will not be visible in TV, also vertical edge 4’4’ also not visible. Draw this lines as hidden lines.

enter image description here

Please log in to add an answer.