# Numerical Examples for Chain Surveying | Errors in Surveying

#### Numerical Examples for Errors in Chain Surveying

We will now move on with different numerical problems on the concept of Errors in Chain Surveying. Going through these numericals will actually give you an idea as to how the calculations are done inspite of errors occurring in the Chain Surveying.

#### Correction due to incorrect length of chain

This is like a formula list which is to be kept in mind while making Calculations:

True distance = L’/L*measured distance

True area = (L’/L)2*measured area

True Volume = (L’/L)3 * measured volume

Where, L’ = incorrect length of chain

L = correct length of chain

The length of a line measured with 20m chain was found to be 500m. It was subsequently found that the chain was 0.04m too long. What is the length of line?

Correct length of chain, L’ = 20 + 0.04 = 20.04m

Length, L = 20m

Measured length, = 500m

True length = (L’/L) * measured length

= (20.04/20) * 500

= 501m

We will continue with more examples on the same topic in our successive articles…

### 5 thoughts on “Numerical Examples for Chain Surveying | Errors in Surveying”

1. I need some help here please, the distance between two stations was 1200m when measured with a 20 metre chain, the same distance when measured with a 30 metre chain was found to be 1195.if the 20 m chain was 0.05 too long what was the error in the 30 metre chain?

2. This methods are very use full to the survey

3. I want more examples

4. 5. 