Question

Specifications call for a hole in a machined part to be 2.314 in. in diameter. If the hole is measured to be 2.317 in., what is the machinist's percent error?

Answer 1

The percent error is defined as the percent difference between the theoretical value and the measurement:

`E_(\%)=(v_m-v_t)/(v_t)*100\%`

Replace 2.314 in for the theoretical value and 2.317 in for the measured value to find the percent error:

`E_(\%)=\frac{2.317\text{ in}-2.314\text{ in}}{2.314\text{ in}}*100\%=0.1296...\%\approx0.13\%`

Therefore, the machinist's percent error was approximately 0.13%.