Least Count, Vernier Callipers & Screw Gauge

The smallest value an instrument can resolve is its least count. For mechanical instruments like the vernier callipers and screw gauge, the least count determines measurement precision.

Concept

Least Count

The smallest value that can be measured by an instrument:

$\text{Least count} = \frac{\text{Value of one main-scale division}}{\text{Number of vernier-scale divisions}}$

For a screw gauge:

$\text{Least count} = \frac{\text{Pitch}}{\text{Number of head-scale divisions}}$

where the pitch is the distance the spindle advances per full turn.

Vernier Callipers

A standard vernier has:

• Main scale (MS): graduated in mm.
• Vernier scale (VS): 10 divisions that match 9 mm of the main scale.

Each VS division = $9/10 = 0.9$ mm. So: $\text{LC} = 1\,\text{mm} - 0.9\,\text{mm} = 0.1\,\text{mm} = 0.01\,\text{cm}.$

Reading procedure:

1. Read the main scale just before the 0 of the vernier scale → $M$ mm.
2. Find the vernier division that exactly coincides with a main-scale line → $V$.
3. Reading $= M + V \times \text{LC}$.

Screw Gauge

• Pitch typically $0.5$ mm or $1$ mm.
• Circular (head) scale: typically $50$ or $100$ divisions.
• LC = pitch / head divisions = $0.5/50 = 0.01$ mm = $10\,\mu\text{m}$.

Reading procedure:

1. Main (linear) scale reading $L$ in mm.
2. Circular scale reading: division coinciding with the reference line $\times$ LC.
3. Reading $= L + (\text{circular division}) \times \text{LC}$.

Zero Error

If the jaws are closed/spindle touches the anvil and the zero of the vernier (or circular scale) does NOT coincide with the zero of the main scale, there is a zero error.

• Positive zero error: Vernier zero is to the right of MS zero. Subtract from each reading.
• Negative zero error: Vernier zero is to the left of MS zero. Add to each reading.

Backlash Error (Screw Gauge)

A screw gauge has slight play in the threads. Always turn the screw in one direction when taking measurements.

Worked Example

Q: A vernier callipers has LC = 0.01 cm. While measuring a rod, MS reading = 2.4 cm, and the 6th vernier division coincides with an MS line. Zero error = +0.02 cm (positive). Find the corrected length.

Solution: Apparent reading $= 2.4 + 6 \times 0.01 = 2.46$ cm.

Corrected length $= 2.46 - 0.02 = 2.44$ cm.

Q (screw gauge): Pitch = 1 mm, 100 divisions on the circular scale. LC = ?

$\text{LC} = 1/100 = 0.01\,\text{mm}$.

If MS reads $5$ mm and CS reading is 47, the diameter $= 5 + 47 \times 0.01 = 5.47$ mm. If zero error = $-0.03$ mm (negative), corrected = $5.47 - (-0.03) = 5.50$ mm.

Common Confusions

• "Pitch" is the distance moved per complete turn, not per division.
• Always subtract a positive zero error and add a negative zero error to the observed reading.
• Backlash matters only when reversing direction; just don't reverse.
• LC is the resolution, not the accuracy of the instrument.

Key Takeaways

• Vernier LC = (1 MSD) / (number of VSDs). Standard vernier: 0.01 cm.
• Screw gauge LC = pitch / head divisions. Standard screw gauge: 0.01 mm.
• Always check for zero error before measuring.
• Turn the screw gauge in one direction to avoid backlash.
• Resolution (LC) ≠ accuracy; calibration still matters.