In a machining operation, doubling the cutting speed reduces the tool life to \(\frac{1}{8}\) of the original value. The exponent ‘n’ in Taylor’s tool life equation \(VT^n = C\) is:
Show Hint
Always reduce Taylor’s equation by taking ratios—this eliminates constant \(C\) and simplifies calculations significantly.
Concept:
Taylor’s tool life equation is one of the most fundamental empirical relationships in machining science, expressing the relation between cutting speed and tool life:
\[
VT^n = C
\]
where:
• \(V\) = cutting speed
• \(T\) = tool life
• \(n\) = tool life exponent (depends on tool-work material combination)
• \(C\) = constant for a given system
This equation implies that if cutting speed increases, tool life decreases nonlinearly.
Step 1: Write initial condition clearly.
\[
V_1 T_1^n = C
\]
Step 2: Write changed condition based on problem.
Given:
• Cutting speed is doubled → \(V_2 = 2V_1\)
• Tool life becomes \(\frac{1}{8}\) of original → \(T_2 = \frac{T_1}{8}\)
Thus:
\[
V_2 T_2^n = C
\]
Step 3: Substitute values into second equation.
\[
2V_1 \left(\frac{T_1}{8}\right)^n = C
\]
Step 4: Expand the expression very carefully.
\[
2V_1 \cdot \frac{T_1^n}{8^n} = C
\]
Step 5: Compare both equations.
From Step 1:
\[
V_1 T_1^n = C
\]
Divide second equation by first:
\[
\frac{2V_1 \cdot \frac{T_1^n}{8^n}}{V_1 T_1^n} = 1
\]
Cancel common terms \(V_1\) and \(T_1^n\):
\[
\frac{2}{8^n} = 1
\]
Step 6: Solve the exponential equation.
\[
8^n = 2
\]
Express 8 as power of 2:
\[
(2^3)^n = 2^1
\]
\[
2^{3n} = 2^1
\]
\[
3n = 1
\]
\[
n = \frac{1}{3}
\]
Step 7: Interpret the result physically.
This indicates that tool life is highly sensitive to cutting speed. A small increase in speed drastically reduces tool life.
Final Answer:
\[
\boxed{\frac{1}{3}}
\]
