Question

The cutting speed is reduced by 75%. Using the Taylor equation $(V{T}^n=C)$ for tool life, the percentage increase in tool life is (Take n = 0.5 and C = 400)

• A. 400%
• B. 800%
• C. 1500%
• D. 1600%

Answer 1

The correct option is C 1500%

$V_2=V_1-0.75V_1=0.25V_1$

From Taylor equation,

$V_1{T}_1^n=V_2{T}_2^n$

$\Rightarrow {\left(\frac{T_2}{T_1}\right)}^n=\frac{V_1}{0.25V_1}=4$

$\Rightarrow \frac{T_2}{T_1}={\left(4\right)}^2=16$

$\therefore$ % increase in tool life $=(\frac{T_2}{T_1}-1)\times 100=15\times 100=1500$%