Question

According to the Taylor's tool life equation, if the cutting speed is halved with index \( n = 0.25 \), then the tool life:

• A. Increases by 8 times
• B. Decreases by 8 times
• C. Decreases by 16 times
• D. Increases by 16 times

Hint:

A useful shortcut derived from Taylor's equation is: \[ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\frac{1}{n}} \] Substituting the values gives: \((2)^{\frac{1}{0.25}} = 2^4 = 16\).

Answer 1

The Correct Option is D

Step 1: Write down what changes and what stays fixed. Taylor's equation is \( VT^n = C \), so for the same tool and job, \( V_1 T_1^{\,n} = V_2 T_2^{\,n} \). Here the speed is cut to half, so \( V_2 = 0.5V_1 \), and \( n = 0.25 \).

Step 2: Rearrange to get tool life directly from speed. From Taylor's equation, \( T = (C/V)^{1/n} \), so tool life depends inversely on speed raised to the power \( 1/n \). This gives \[ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{1/n} \]

Step 3: Put in the numbers. Since \( V_1/V_2 = 2 \) and \( 1/n = 1/0.25 = 4 \), \[ \frac{T_2}{T_1} = 2^4 = 16 \] So halving the cutting speed makes the tool last 16 times longer. \[ \boxed{T_2 = 16T_1} \]