The Answer Sheet That Was Wrong by Two
You are marking the Workshop Calculation and Science paper of 30 trainees. Question 3 reads: 'A mild steel job is 50 mm at the large end and 38 mm at the small end and is to be taper turned over its full 120 mm length. (a) Find the compound rest swivel angle. (b) Find the tailstock set-over.' Trainee R has written: swivel 5°43′ (5.71°), set-over 12 mm. Question 4 asks for the spindle speed to drill a 20 mm hole in mild steel with an HSS drill at a cutting speed of 25 m/min; R has written 3,980 rpm. Later, in the inspection bay, a 200 mm aluminium alloy component (α = 23 × 10⁻⁶ per °C) with a drawing tolerance of 200.00 ± 0.05 mm is measured straight off the machine at 35 °C on a steel micrometer calibrated at 20 °C, and reads 200.09 mm. The trainee wants it booked as rejected. The key gives full method marks and 1 mark per correct numerical answer.
1. Diagnose R's two errors as a pattern rather than as two mistakes, and decide the mark allocation.
Read the worked answer
Pattern. Both answers are wrong by an exact power of the taper ratio or of ten - R has a systematic conversion fault, not a calculational one. That matters for marking and for what you teach.
Question 3. R used the taper ratio directly: tan θ = (D − d)/L = 12/120 = 0.10, so θ = 5.71°. The correct relation uses the half angle because the taper is measured across the diameter, not the radius: tan θ = (D − d)/2L = 0.05, θ = 2.86°. The set-over 12 mm is (D − d) instead of (D − d)/2 - the same missing factor of two. So R knows the formulae, knows which quantities to take, and loses the radius-versus-diameter conversion in both parts.
Question 4. 3,980 rpm is 1000 × 25 / (π × 2). R put the diameter in as 2 instead of 20 - centimetres instead of millimetres. Again the formula is right; the unit handling is not.
Mark decision. Full method marks in both cases, because the method is genuinely there and is correctly stated. Award 0 of 1 for each numerical answer. Write 'D/2, not D' against Question 3 and 'D in mm - you used cm' against Question 4, and separately note 'unit check missing' on the front sheet.
Why not partial marks. A numerical answer that is 100% or 200% out is not a rounding slip; it is a wrong answer. Partial credit for it teaches the batch that unit discipline is optional. The method marks already protect R's grade, and the written diagnosis is the instructional record.
2. Show the correct working for Question 3, in the form you would put on the board for the batch.
Read the worked answer
Given. D = 50 mm, d = 38 mm, taper length = 120 mm (full length, so set-over is computed over the full length).
Step 1 - taper ratio. Taper = (D − d) / L = (50 − 38) / 120 = 12 / 120 = 1:10
Step 2 - half taper angle. The taper is a diametral change, so the compound rest swivels to the half angle: tan θ = (D − d) / (2L) = 12 / 240 = 0.05 θ = tan⁻¹(0.05) = 2.862° ≈ 2°52′
Step 3 - tailstock set-over. Set-over = (D − d) / 2 = 12 / 2 = 6 mm (If the taper covered only part of the job, set-over would be (D − d)/2 × L_total / L_taper. Here they are equal, so no correction is needed - but the batch must know the correction exists.)
The two traps written on the board.
- Diameter versus radius: every taper formula halves because you work on one side of the axis.
- Swivel angle is always the half angle. Setting 5.71° on the compound rest produces a taper of about 1:5, twice as steep as the drawing, and the part is scrap.
Check habit to install. After computing the swivel angle, sanity-check it against the ratio: 1:10 should give a small angle, roughly 3°, not 6°. A 1:10 taper is a gentle self-holding taper, not a steep one.
3. For the 200 mm aluminium component, decide whether the part is accepted or rejected, and justify the decision with the numbers.
Read the worked answer
Decision: accept the part, after correcting for temperature. Do not book it against the raw reading.
Correction. ΔL = L × α × ΔT = 200 × 23 × 10⁻⁶ × (35 − 20) = 200 × 23 × 10⁻⁶ × 15 = 0.069 mm The part is 0.069 mm larger at 35 °C than it will be at 20 °C, so: True size at 20 °C = 200.09 − 0.069 = 200.021 mm
Compare with the drawing. Limits are 199.95 to 200.05 mm. 200.021 mm lies inside them. The part conforms.
Why the raw reading must not be used. 200.09 mm looks like a rejection, and a trainee or operator who acts on it will either scrap a good part or, worse, take another 0.04 mm off and produce 199.98 mm of true size... which would still just pass here, but the same reasoning on a tighter tolerance will not. Rejecting on an uncorrected reading is a decision made on a measurement artefact.
The micrometer's own expansion is second-order: the frame grows by roughly 25 × 11.5 × 10⁻⁶ × 15 ≈ 0.004 mm in the anvil gap, and it reads smaller, partly cancelling. Mention it, then set it aside - it is around 6% of the workpiece correction.
The real lesson. The correction is 0.069 mm against a total tolerance band of 0.10 mm. Measurement at 35 °C consumes nearly 70% of the tolerance. So the shop rule is not 'always correct the reading', it is 'let the part soak to 20 °C before final inspection'. Correction is the fallback; temperature control is the control measure.