# What is a "correction factor" wrt this problem below?

A surveyor has a steel measuring tape that is calibrated to be 100.000 m long (i.e., accurate to ±1 mm ) at 20 ∘C.

A) If she measures the distance between two stakes to be 75.175 m on a 3 ∘C day, does she need to add or subtract a correction factor to get the true distance?

So, I realize steel will shrink when it's colder, so she'll have to subtract. But, for part B, I don't even know what a correction factor is, so, what?

B) How large, in mm, is the correction factor?

### 2 Answers

- DixonLv 78 months agoFavorite Answer
You need to investigate how much steel contracts between 20 and 3 Celcius. It will be probably given as some factor (ie ratio) per degree C over a reasonable working range of temperature, ie it will be linearised even if it isn't truly linear.

- Anonymous8 months ago
If it's colder than the tape contracts so she had to add a correction factor.