Length — Question 4

Question 4

Reading a vernier from its geometry

A direct vernier caliper has main-scale spacing $1.0\ \mathrm{mm}$. Ten equal vernier intervals span $9.0\ \mathrm{mm}$. To read it, find the main-scale mark immediately left of the vernier zero, then determine which vernier mark aligns exactly with a main-scale mark.

When the jaws are correctly closed, this caliper indicates $+0.20\ \mathrm{mm}$. Assume that this is a constant additive zero error, with no other error for this exercise.

1. Determine the length of one vernier interval and the difference between one main interval and one vernier interval. Explain how this difference permits a fraction of a millimetre to be read.

2. Read the displayed separation of the open jaws. Identify the main-scale reading and the aligned vernier mark, rather than estimating the gap by eye alone.

3. Correct for the closed-jaw indication. Explain the correction’s sign.

4. A student adds the full physical distance from the vernier zero to the aligned mark to the main reading. Explain why the added fraction instead uses the difference between the two scale spacings.

Question 4 - Solution

Reading a vernier from its geometry

(a) The spacing difference supplies the readout increment.

$$\mathrm{LC}=1.00\ \mathrm{mm}-\frac{9.00\ \mathrm{mm}}{10}=0.10\ \mathrm{mm}.$$

One vernier interval is $(9.0/10)\ \mathrm{mm}=0.90\ \mathrm{mm}$. The spacing difference is $(1.00-0.90)\ \mathrm{mm}=\boxed{0.10\ \mathrm{mm}}$. After each successive vernier mark, its position falls another $0.10\ \mathrm{mm}$ behind the corresponding main-scale step.

(b) Use the observed coincidence.

The vernier zero lies between 12 and 13 mm, and vernier mark 3 aligns with the main mark at 15 mm. If $\delta$ is the zero’s distance beyond 12 mm, then $$12\ \mathrm{mm}+\delta+3(0.90\ \mathrm{mm})=15\ \mathrm{mm}.$$ Hence $\delta=0.30\ \mathrm{mm}$ and the indicated separation is $\boxed{12.30\ \mathrm{mm}}$.

(c) Subtract the positive zero error.

The caliper reads $0.20\ \mathrm{mm}$ too high at true zero. Under the stated additive-error model, $$L=L_{\mathrm{indicated}}-0.20\ \mathrm{mm} =\boxed{12.10\ \mathrm{mm}}.$$ Adding the offset would increase an already high reading.

(d) The coincidence compares two scales.

The aligned mark is $2.70\ \mathrm{mm}$ from the vernier zero, but the matching main mark is $3.00\ \mathrm{mm}$ beyond the main reading. Their difference, $0.30\ \mathrm{mm}$, locates the zero. The alignment encodes the accumulated spacing mismatch, not an extra $2.70\ \mathrm{mm}$ of jaw opening.