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

Concept: Taylor's tool life equation describes the empirical relationship between cutting velocity and tool wear life: \[ V T^n = C \] Where:
• \( V \) = Cutting speed.
• \( T \) = Tool life.
• \( n \) = Tool life exponent (dependent on tool material).
• \( C \) = Constant. For two distinct operating conditions with the same tool-workpiece combination, the relation can be written as: \[ V_1 T_1^n = V_2 T_2^n \]

Step 1: State the given problem parameters. Let the initial cutting speed be \( V_1 \) and the initial tool life be \( T_1 \). The problem states that the tool life exponent is: \[ n = 0.25 = \frac{1}{4} \] The final cutting speed \( V_2 \) is reduced to half of the initial speed: \[ V_2 = \frac{V_1}{2} \]

Step 2: Set up the ratio using the tool life equation. Substitute \( V_2 \) into the relation: \[ V_1 T_1^n = \left(\frac{V_1}{2}\right) T_2^n \] Dividing both sides by the initial velocity \( V_1 \) simplifies the equation to: \[ T_1^n = \frac{1}{2} T_2^n \quad \Rightarrow \quad \frac{T_2^n}{T_1^n} = 2 \] Using the properties of exponents: \[ \left(\frac{T_2}{T_1}\right)^n = 2 \]

Step 3: Solve for the final tool life ratio. Substitute \( n = 0.25 = \frac{1}{4} \) into the expression: \[ \left(\frac{T_2}{T_1}\right)^{\frac{1}{4}} = 2 \] Raise both sides of the equation to the power of 4 to isolate the ratio: \[ \left[\left(\frac{T_2}{T_1}\right)^{\frac{1}{4}}\right]^4 = 2^4 \] \[ \frac{T_2}{T_1} = 16 \quad \Rightarrow \quad T_2 = 16 T_1 \] Therefore, cutting the speed in half causes the tool life to increase by a factor of 16.