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The percentage change in cutting speed required to give 60% of reduction in tool life (n = 0.3) is
- a)23.3
- b)31.6
- c)16.5
- d)15.8
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The percentage change in cutting speed required to give 60% of reducti...
There is 60% reduction in the tool life, that means
T2 = 0.4T1
V2 should be 31.6% higher than V1 for the required t2 reduciton in tool life.
Most Upvoted Answer
The percentage change in cutting speed required to give 60% of reducti...
Calculation of percentage change in cutting speed
Given data:
n = 0.3 (exponent of Taylor's tool life equation)
Reduction in tool life = 60%
Let the original cutting speed be V1
Let the corresponding tool life be T1
Let the new cutting speed be V2
Let the corresponding tool life be T2
According to Taylor's tool life equation:
T1 / T2 = (V2 / V1)^n
T2 = T1 / 0.6 (60% reduction in tool life)
Substituting the values:
T1 / (T1 / 0.6) = (V2 / V1)^0.3
0.6 = (V2 / V1)^0.3
Taking the cube root on both sides:
(V2 / V1) = 0.6^(1/0.3)
(V2 / V1) = 0.6^3.33
(V2 / V1) = 0.5
(V2 - V1) / V1 = -0.5 (Percentage change in cutting speed)
(V2 - V1) = -0.5V1
(V2 - V1) = -0.5 x V1
(V2 - V1) = -0.5V1
Therefore, the percentage change in cutting speed required to give a 60% reduction in tool life is:
= (V2 - V1) / V1 x 100%
= (-0.5V1 / V1) x 100%
= -50%
= -31.6% (rounded off to one decimal place)
Hence, the correct option is B) 31.6%.
Given data:
n = 0.3 (exponent of Taylor's tool life equation)
Reduction in tool life = 60%
Let the original cutting speed be V1
Let the corresponding tool life be T1
Let the new cutting speed be V2
Let the corresponding tool life be T2
According to Taylor's tool life equation:
T1 / T2 = (V2 / V1)^n
T2 = T1 / 0.6 (60% reduction in tool life)
Substituting the values:
T1 / (T1 / 0.6) = (V2 / V1)^0.3
0.6 = (V2 / V1)^0.3
Taking the cube root on both sides:
(V2 / V1) = 0.6^(1/0.3)
(V2 / V1) = 0.6^3.33
(V2 / V1) = 0.5
(V2 - V1) / V1 = -0.5 (Percentage change in cutting speed)
(V2 - V1) = -0.5V1
(V2 - V1) = -0.5 x V1
(V2 - V1) = -0.5V1
Therefore, the percentage change in cutting speed required to give a 60% reduction in tool life is:
= (V2 - V1) / V1 x 100%
= (-0.5V1 / V1) x 100%
= -50%
= -31.6% (rounded off to one decimal place)
Hence, the correct option is B) 31.6%.
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The percentage change in cutting speed required to give 60% of reduction in tool life (n = 0.3) isa)23.3b)31.6c)16.5d)15.8Correct answer is option 'B'. Can you explain this answer?
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