A Line Inclined to Both Reference Planes
A line AB is 70 mm long. End A is 15 mm above HP and 20 mm in front of VP; end B is 55 mm above HP and 60 mm in front of VP. A candidate draws the front view 70 mm long at 45° to XY and the top view also 70 mm long at 45°, arguing that “the line is 70 mm, so both views must show 70 mm.” The sheet shows no construction arcs, no loci and no distinction between true angles and apparent angles. The examiner asks for a corrected projection and a numerical check. The candidate has a calculator but no drawing of the auxiliary plane. The line lies wholly in the first quadrant, so no trace is expected on the extensions within the sheet.
1. Calculate the true inclinations with HP and VP, and the lengths of the front and top views. Show the formulas.
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Take the coordinate differences: Δz = 55 − 15 = 40 mm (height above HP), Δy = 60 − 20 = 40 mm (distance in front of VP), and Δx is the horizontal separation along XY. Since the true length is 70 mm, Δx² + Δy² + Δz² = 70², so Δx² = 4900 − 1600 − 1600 = 1700, giving Δx = 41.23 mm. The true inclination with HP satisfies sin α = Δz/TL = 40/70 = 0.5714, so α = 34.85°. The true inclination with VP satisfies sin β = Δy/TL = 40/70, so β = 34.85°. The front-view length is √(Δx² + Δz²) = √(1700 + 1600) = 57.45 mm. The top-view length is √(Δx² + Δy²) = 57.45 mm. The apparent angle of the front view to XY is tan⁻¹(Δz/Δx) = tan⁻¹(40/41.23) = 44.1°, and the top view has the same apparent angle. Neither view is 70 mm long, and neither apparent angle is the true inclination.
2. Explain the projection construction you would use, including loci and rotation, and identify the candidate’s error.
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Draw XY and the projector through A. Mark a′ 15 mm above XY and a 20 mm below XY because the top view lies below the reference line in first-angle projection. Draw the locus of b′ at 55 mm above XY and the locus of b at 60 mm below XY. To obtain true length, rotate the line about A until it is parallel to VP; the front view then shows the true length and the true inclination with HP. Project the rotated end down to the top-view locus to obtain the corresponding top-view length. Rotate that length about a in the top view until it meets the locus of b; this fixes b, and the projector fixes b′. The angle through which the front view was rotated is the true inclination with HP, not the angle of the final front view. The candidate’s error is assuming that a line oblique to both HP and VP can show true length in both views. In fact both projections are foreshortened to about 57.45 mm, and the apparent angles are about 44.1°, not the true 34.85°.
3. Is the line parallel to the profile plane? Justify with the coordinate differences and state the consequence for the traces.
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A line parallel to the profile plane would have no extension along the XY direction, so Δx would be zero. Here Δx = 41.23 mm, so the line is not parallel to the profile plane; it is oblique to HP, VP and the profile plane. Both endpoints are above HP and in front of VP, so the whole segment lies in the first quadrant. Its extensions would meet HP and VP somewhere outside the segment, but those traces are not on the drawn line and are not needed for this projection. The practical consequence is that neither the front view nor the top view can be treated as a true-length view, and the traces cannot be located by simply extending the given views without first producing the line. The candidate should use the rotation construction, not a trace shortcut.