Development of a Truncated Hexagonal Prism
A right regular hexagonal prism of side 30 mm and axis 70 mm stands on its base on HP with one side of the base parallel to VP. A section plane perpendicular to VP and inclined at 45° to HP passes through a point on the axis 40 mm above the base and slopes upward towards the left. The prism is to be truncated, and its lateral surface is to be developed. In the front view the section plane appears as a line at 45° to XY. The six vertical edges are labelled A, B, C, D, E and F around the prism in top view, starting with the rightmost edge A. A candidate develops the lateral surface by unrolling from edge A and draws the cut line as a straight horizontal line at 40 mm above the base because ‘the section plane passes through the axis at 40 mm’. The development is 180 mm long, but the cut line does not meet the same height at the two ends of the seam.
1. Calculate the height at which the section plane cuts each vertical edge A–F. State whether the plane cuts the top base or only touches it, and identify any edge that is not cut.
Read the worked answer
Set x positive to the right and z upward from the base. The section plane is perpendicular to VP, so its front view is the line z = 40 − x (45°, rising to the left, passing through x = 0 at z = 40). For the hexagon with one side parallel to VP and the rightmost vertex A, the x-coordinates are A = +30, B = +15, C = −15, D = −30, E = −15, F = +15 mm.
Cut heights: A: 40 − 30 = 10 mm; B: 40 − 15 = 25 mm; C: 40 − (−15) = 55 mm; D: 40 − (−30) = 70 mm; E: 55 mm; F: 25 mm.
All six vertical edges are cut within the 0–70 mm height. At D the plane reaches z = 70 mm, exactly the top base, so it touches the top edge/vertex at D but does not cross the top face. No edge is uncut; the section is a six-sided polygon.
2. Develop the lateral surface starting at edge A. Give the development length, the positions of the six edges along the development, and the cut heights to be plotted. Describe the shape of the cut line.
Read the worked answer
Lateral perimeter = 6 × 30 = 180 mm. Starting at A = 0 mm, the successive edges are A 0, B 30, C 60, D 90, E 120, F 150, and A again at 180 mm.
Plot the cut heights at these stations: 10, 25, 55, 70, 55, 25, and 10 mm at 180 mm. Join adjacent stations with straight lines because each lateral face is a plane and the section plane cuts it in a straight line.
The resulting cut line is a zigzag polyline, not a horizontal line. The highest point is at D (90 mm along the development, 70 mm high); the lowest is at A (10 mm high). The development outline is a 180 mm × 70 mm rectangle, with the cut line replacing the top edge over the cut portion. On the flat sheet the cut line runs from A at 0 mm to the same edge A at 180 mm at the same height, so it closes when the surface is rolled.
3. The candidate draws the cut line as a horizontal straight line at 40 mm. Explain the error and state what the 40 mm dimension actually represents.
Read the worked answer
A horizontal cut line in the development would mean every vertical edge is cut at the same height, which happens only if the section plane is parallel to the base (perpendicular to the axis). Here the plane is inclined at 45° to HP, so its height varies linearly with x: z = 40 − x.
The 40 mm is the height where the plane crosses the axis (x = 0), not the height on the six vertical edges. Only an edge at x = 0 would be cut at 40 mm, and this hexagon has no vertical edge at x = 0; the actual cuts range from 10 mm at A to 70 mm at D. The candidate's development would also fail to close because the start and end heights at edge A would not match the actual 10 mm.