Least count: vernier callipers and screw gauge

Finer than a millimetre

A school ruler has marks every millimetre. Its least count, the smallest step it can show, is $1\ \text{mm}$. That is fine for a notebook, but not for the thickness of a coin or a hair.

Two clever instruments read much finer: the vernier callipers (to $0.1\ \text{mm}$) and the screw gauge (to $0.01\ \text{mm}$).

The vernier callipers

The trick is a second, sliding scale: the vernier. Its $10$ divisions fill only $9\ \text{mm}$, so each vernier division is $0.9\ \text{mm}$, $0.1\ \text{mm}$ shorter than a main division.

In general, if $n$ vernier divisions fill $n - 1$ main divisions, the least count is $\dfrac{1\ \text{MSD}}{n}$.

How to read it.

1. Main scale reading: the last millimetre mark before the vernier zero.
2. Find the one vernier line that lines up exactly with a main scale line.
3. Reading $=$ main scale reading $+$ (that line's number $\times$ least count).

Try it. Slide the vernier slowly and watch which line coincides.

You should find that as the vernier zero moves one tenth of a millimetre, the coinciding line moves to the next vernier line. That is the whole secret.

In this picture the vernier zero is past $24\ \text{mm}$, and line $6$ coincides: $24\ \text{mm} + 6 \times 0.1\ \text{mm} = 24.6\ \text{mm}$.

The screw gauge

A screw moves forward by one pitch for each full turn. Its thimble carries a circular scale of, say, $50$ divisions. One division of turn moves the screw $\dfrac{1}{50}$ of a pitch.

Reading $=$ linear scale reading (whole and half millimetres showing on the sleeve) $+$ (circular division on the datum line $\times$ least count). Here: $5.5\ \text{mm} + 27 \times 0.01\ \text{mm} = 5.77\ \text{mm}$.

Watch out. On many gauges the half-millimetre marks are below the datum line. Miss one and your answer is $0.5\ \text{mm}$ out.

Zero error

Close the jaws, or let the screw's faces touch. The reading should be $0\ \text{mm}$. If it is not, the instrument has a zero error, a systematic error in every reading.

• Positive zero error: the instrument reads a little too much. For a vernier, the vernier zero is to the right of the main zero. For a screw gauge, the circular scale's zero is below the datum line.
• Negative zero error: it reads too little, and the zero marks are the other way.

In every case:

A worked example

A vernier callipers (least count $0.1\ \text{mm}$) closed shows its vernier zero left of the main zero, with line $8$ coinciding. Then a pebble reads $18\ \text{mm}$ plus line $3$.

1. Zero error. Left means negative. Line $8$ coincides, so the vernier is short by $(10 - 8) \times 0.1\ \text{mm} = 0.2\ \text{mm}$: zero error $= -0.2\ \text{mm}$.
2. Observed. $18\ \text{mm} + 3 \times 0.1\ \text{mm} = 18.3\ \text{mm}$.
3. Correct. $18.3\ \text{mm} - (-0.2\ \text{mm}) = 18.5\ \text{mm}$.

Think first. A negative zero error means the callipers read too little, so the true size must be bigger than the reading. $18.5 > 18.3$. Good.

Practise with new numbers

Try it in Python

The program reads the vernier of the worked example, with and without its zero error.

Remember these

• Least count: the smallest step an instrument can show.
• Vernier: least count $= \dfrac{1\ \text{MSD}}{n}$; reading $=$ MSR $+$ line $\times$ LC.
• Screw gauge: least count $= \dfrac{\text{pitch}}{\text{divisions}}$; reading $=$ linear $+$ circular $\times$ LC.
• Correct reading $=$ observed $-$ zero error (a negative zero error is added back).

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A vernier has $25$ divisions that fill $24\ \text{mm}$. What is its least count?

2. A screw gauge has a pitch of $1\ \text{mm}$ and $100$ circular divisions. The sleeve shows $6\ \text{mm}$ and division $72$ is on the datum line. What is the reading?

3. A screw gauge reads $+0.02\ \text{mm}$ with its faces touching. It then reads $1.47\ \text{mm}$ on a wire. What is the wire's diameter?

4. Why can a screw gauge measure finer than a vernier callipers?