Question

Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.

Answer 1

Answer:21.3%

Explanation:

Given

80 % reduction in tool life

According to Taylor's tool life

$VT^n$=c

where V is cutting velocity

T=tool life of tool

80 % tool life reduction i.e. New tool Life is 0.2T

Thus

$VT^{0.12}=V'\left ( 0.2T\right )^{0.12}$

$V'=\frac{V}{0.2^{0.12}}$

$V'=\frac{V}{0.824}$=1.213V

Thus a change of 21.3 %(increment) is required to reduce tool life by 80%