Line AB: Longer Front View but Not Parallel to VP
During a check of a line-projection exercise, a candidate submits the following data for line AB: true length 85 mm; end A is 15 mm above HP and 20 mm in front of VP; end B is 55 mm above HP and 45 mm in front of VP. The candidate’s front view measures 70 mm and the top view measures 60 mm. They conclude that AB is parallel to VP because the front view is longer than the top view, and they draw the front view horizontal. The instructor points out that both endpoints have different heights and different distances from VP, so the line cannot be parallel to either principal plane. The candidate is asked to recompute the projections from the coordinates, to explain why a longer front view does not imply parallelism to VP, and to correct the orientation of both views.
1. Using the coordinates, compute the horizontal component, the inclinations to HP and VP, and the correct front-view and top-view lengths. Compare with the candidate's 70 mm and 60 mm.
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Take x along XY, y as distance in front of VP (depth), and z as height above HP. Differences: Δy = 45 − 20 = 25 mm; Δz = 55 − 15 = 40 mm. True length L = 85 mm, so Δx = √(85² − 25² − 40²) = √(7225 − 625 − 1600) = √5000 = 70.71 mm.
Inclination to HP: sin θ = Δz/L = 40/85 = 0.4706 → θ = 28.1°. Inclination to VP: sin φ = Δy/L = 25/85 = 0.2941 → φ = 17.1°.
Front view length = √(Δx² + Δz²) = √(5000 + 1600) = √6600 = 81.2 mm. Top view length = √(Δx² + Δy²) = √(5000 + 625) = √5625 = 75.0 mm.
The candidate's 70 mm and 60 mm are both too short; they are not the correct projections. The front view is longer than the top view because φ < θ, not because the line is parallel to VP.
2. The candidate drew the front view horizontal. Explain the contradiction and give the correct angles the front and top views make with XY.
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A horizontal front view means the two endpoints have the same height above HP in the front view, i.e. Δz = 0. But the given data have Δz = 40 mm, so the front view cannot be horizontal.
The front view is inclined to XY at tan⁻¹(Δz/Δx) = tan⁻¹(40/70.71) = 29.5°. The top view is inclined to XY at tan⁻¹(Δy/Δx) = tan⁻¹(25/70.71) = 19.5°.
If the candidate forced the front view horizontal, they probably projected the endpoints onto a line parallel to XY or measured the true length along the wrong direction. Both views must be reconstructed from the same Δx, Δy and Δz; the front view keeps Δx and Δz, the top view keeps Δx and Δy.
3. Is there any principal-plane condition under which the front view would equal the true length? State it and explain why it does not apply here.
Read the worked answer
The front view equals the true length only when the line is parallel to VP, i.e. φ = 0°, so Δy = 0. Here Δy = 25 mm, so the line is inclined to VP and the front view is foreshortened to L cos φ = 85 cos 17.1° = 81.2 mm.
Similarly the top view would equal true length only if θ = 0° (Δz = 0), but Δz = 40 mm. A view can show true length only when the line is parallel to the plane of that view; otherwise an auxiliary plane parallel to the line is needed.