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: Note what changes and what stays fixed when RPM rises.
The plate thickness and feed per revolution stay the same, so a higher RPM only cuts each hole faster, not longer.

Step 2: Turn hole counts into tool life expressed in time.
Time for one hole is proportional to $1/N$, so total tool life is (number of holes)$/N$: $T_1 = 150/200 = 0.75$, $T_2 = 60/300 = 0.20$.

Step 3: Set the two Taylor equations equal to each other.
$V_1 T_1^{\,n} = V_2 T_2^{\,n}$ becomes $200(0.75)^n = 300(0.20)^n$.
Rearranged: $(0.75/0.20)^n = 300/200$, so $(3.75)^n = 1.5$.

Step 4: Take logs on both sides to isolate n.
$n = \ln(1.5)/\ln(3.75) = 0.4055/1.3218 = 0.307$.

Final Answer:
This gives an exponent of about 0.31 in Taylor's tool life equation. \[ \boxed{n \approx 0.31} \]