# How to find the least count of Vernier Calipers when Vernier Scale Divisions are greater than Main Scale Divisions?

When Vernier Scale Divisions are smaller than Main Scale Divisions, the least count is given by $1 \text{MSD} - 1 \text{VSD}$.

But what if say, $3 \text{VSD}=5 \text{MSD}$ ?

Do we write the least count as $1 \text{VSD} - 1 \text{MSD}$ ?I am having difficulty in visualizing the situation. Can someone please explain it with the help of diagrams?

• Practically, does the situation of a Vernier scale division being bigger than the main scale division ever arise? Commented Apr 24, 2017 at 2:45

The least count is $\frac{1}{3}$MSD.

If 3VSD = 5MSD then 1VSD =$\frac{5}{3}$MSD. The Vernier scale division is just short of 2MSD. They differ by $\frac{1}{3}$MSD. Therefore the least count is $\frac{1}{3}$MSD.

Graphically it looks like this (for zero distance):

Main scale
0  1  2  3  4  5  6  7  8  9  10
|  |  |  |  |  |  |  |  |  |  |
|    |    |
0    1/3  2/3
Vernier scale


A measurement of $5\frac{2}{3}$ would look like this:

Main scale
0  1  2  3  4  5  6  7  8  9  10
|  |  |  |  |  |  |  |  |  |  |
|    |    |
0    1/3  2/3
Vernier scale


In general you can find the least count by finding the minimum nonzero difference between any integer multiple of the VSD and any integer multiple of the MSD. $$\text{least count}=\min_{\text{non zero}}(|n\cdot MSD-m\cdot VSD|) \quad m,n\in \mathbb{N}$$ If you fill in $m=n=1$, you obtain your original equation.

• So, if 8 vsd = 5 msd then why is least count not 3/8 mm . why is it 1/8 mm? Commented Mar 5, 2019 at 6:17