Question

A drill bit during its lifetime can produce \(150\) through holes in a plate at a drill speed of \(200\) RPM. If the drill speed increases to \(300\) RPM, it can produce \(60\) through holes in the same plate before the drill bit fails. Assuming all other parameters remain constant, the value of the exponent in Taylor's tool life equation is ________ (rounded off to 2 decimal places).

Hint:

Convert the hole counts into actual tool life in time, since time per hole changes with drill speed, before using Taylor's equation.

Answer 1

Correct Answer: 0.31

Step 1: Recall Taylor's tool life equation.
Taylor's relation is \(V T^n = C\), where \(V\) is cutting speed, \(T\) is tool life in minutes, and \(n\) is the exponent to find.

Step 2: Convert the given hole counts into actual tool life times.
Cutting speed scales with drill RPM for a fixed diameter, so \(V_1 : V_2 = 200 : 300\).
Time per hole scales as \(1/N\), so tool life (time) is holes divided by RPM: \(T_1 = 150/200 = 0.75\) and \(T_2 = 60/300 = 0.20\) (in the same relative time unit).

Step 3: Apply Taylor's equation between the two conditions.
\(V_1 T_1^n = V_2 T_2^n\) gives \(\left(\dfrac{T_2}{T_1}\right)^n = \dfrac{V_1}{V_2}\).
\(\left(\dfrac{0.20}{0.75}\right)^n = \dfrac{200}{300}\), so \((0.2667)^n = 0.6667\).

Step 4: Solve for n using logarithms.
\(n = \dfrac{\ln(0.6667)}{\ln(0.2667)} = \dfrac{-0.4055}{-1.3218} = 0.307\).

Final Answer:
The Taylor tool life exponent works out close to 0.31. \[ \boxed{n \approx 0.31} \]