Topic: Machining and Machine Tool Operations
In a cutting operation the cutting speed was reduced by 20%. Assuming n=0.5 and C=350 in Taylor's equation, the increase in tool life is
- A. 46%
- B. 48%
- C. 59%
- D. 56%
Answer: D. 56%
Explanation: According to Taylor's tool life equation, $V_1 T_1^n = V_2 T_2^n$. The new cutting speed is reduced by 20%, so $V_2 = 0.8 V_1$. Given $n = 0.5$, we have $V_1 T_1^{0.5} = 0.8 V_1 T_2^{0.5}$. Solving for $T_2$, we get $(T_2 / T_1)^{0.5} = 1 / 0.8 = 1.25$. Squaring both sides yields $T_2 / T_1 = (1.25)^2 = 1.5625$. The percentage increase in tool life is $(1.5625 - 1) \times 100 = 56.25\%$, which is approximately 56%.