Question:
A modified Taylor tool life equation is given as follows:
\[ V T^n f^m = constant \]
where \( V \) is the cutting speed (m/s), \( T \) is the tool life in minutes and \( f \) is the feed in mm/rev, \( n = 0.25 \) and \( m = 0.5 \). Under two different cutting conditions \( (V_1, f_1) \) and \( (V_2, f_2) \) the tool life \( (T_1, T_2) \) was found to be the same.
If the ratio of the cutting speeds \( (V_1/V_2) \) used is 2/3, then the ratio of corresponding feeds \( (f_1/f_2) \) must be ______ (rounded off to two decimal places).
A modified Taylor tool life equation is given as follows:
\[ V T^n f^m = constant \]
where \( V \) is the cutting speed (m/s), \( T \) is the tool life in minutes and \( f \) is the feed in mm/rev, \( n = 0.25 \) and \( m = 0.5 \). Under two different cutting conditions \( (V_1, f_1) \) and \( (V_2, f_2) \) the tool life \( (T_1, T_2) \) was found to be the same.
If the ratio of the cutting speeds \( (V_1/V_2) \) used is 2/3, then the ratio of corresponding feeds \( (f_1/f_2) \) must be ______ (rounded off to two decimal places).
\[ V T^n f^m = constant \]
where \( V \) is the cutting speed (m/s), \( T \) is the tool life in minutes and \( f \) is the feed in mm/rev, \( n = 0.25 \) and \( m = 0.5 \). Under two different cutting conditions \( (V_1, f_1) \) and \( (V_2, f_2) \) the tool life \( (T_1, T_2) \) was found to be the same.
If the ratio of the cutting speeds \( (V_1/V_2) \) used is 2/3, then the ratio of corresponding feeds \( (f_1/f_2) \) must be ______ (rounded off to two decimal places).
Show Hint
Cancel the equal tool-life term \( T^n \) from both conditions, then solve \( (f_1/f_2)^m = V_2/V_1 \) for the feed ratio.
Updated On: Aug 3, 2026
Show Solution
Verified By Collegedunia
Correct Answer: 2.25
Solution and Explanation
Step 1: Understanding the Concept:
The modified Taylor equation links cutting speed, tool life and feed through fixed exponents \( n \) and \( m \).
We are told the tool life is the same under both cutting conditions, so the \( T^n \) term is identical on both sides and can be cancelled.
We then need to relate the feed ratio to the given speed ratio.
Step 2: Key Formula or Approach:
Writing the constant equation for both conditions and cancelling \( T^n \):
\[ V_1 f_1^m = V_2 f_2^m \]
Taking logarithms of both sides lets us solve for the feed ratio directly using the exponent \( m \).
Step 3: Detailed Explanation:
Rearranging \( V_1 f_1^m = V_2 f_2^m \) gives \( \left( \frac{f_1}{f_2} \right)^m = \frac{V_2}{V_1} \).
Taking log on both sides:
\[ m \log\left(\frac{f_1}{f_2}\right) = \log\left(\frac{V_2}{V_1}\right) \]
Since \( V_1/V_2 = 2/3 \), we get \( V_2/V_1 = 3/2 = 1.5 \), so \( \log(1.5) = 0.17609 \).
\[ \log\left(\frac{f_1}{f_2}\right) = \frac{0.17609}{0.5} = 0.35218 \]
Taking the antilog:
\[ \frac{f_1}{f_2} = 10^{0.35218} = 2.25 \]
Final Answer:
The ratio of corresponding feeds \( f_1/f_2 \) works out to 2.25. \[ \boxed{2.25} \]
The modified Taylor equation links cutting speed, tool life and feed through fixed exponents \( n \) and \( m \).
We are told the tool life is the same under both cutting conditions, so the \( T^n \) term is identical on both sides and can be cancelled.
We then need to relate the feed ratio to the given speed ratio.
Step 2: Key Formula or Approach:
Writing the constant equation for both conditions and cancelling \( T^n \):
\[ V_1 f_1^m = V_2 f_2^m \]
Taking logarithms of both sides lets us solve for the feed ratio directly using the exponent \( m \).
Step 3: Detailed Explanation:
Rearranging \( V_1 f_1^m = V_2 f_2^m \) gives \( \left( \frac{f_1}{f_2} \right)^m = \frac{V_2}{V_1} \).
Taking log on both sides:
\[ m \log\left(\frac{f_1}{f_2}\right) = \log\left(\frac{V_2}{V_1}\right) \]
Since \( V_1/V_2 = 2/3 \), we get \( V_2/V_1 = 3/2 = 1.5 \), so \( \log(1.5) = 0.17609 \).
\[ \log\left(\frac{f_1}{f_2}\right) = \frac{0.17609}{0.5} = 0.35218 \]
Taking the antilog:
\[ \frac{f_1}{f_2} = 10^{0.35218} = 2.25 \]
Final Answer:
The ratio of corresponding feeds \( f_1/f_2 \) works out to 2.25. \[ \boxed{2.25} \]
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