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Using the Taylor equation for tool life and letting n = 0.5 and C= 400, calculate the percentage increase in tool life when the cutting speed is reduced by 50%.
- a)100%
- b)No change
- c)400%
- d)300%
Correct answer is option 'D'. Can you explain this answer?
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Using the Taylor equation for tool life and letting n = 0.5 and C= 400...
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Using the Taylor equation for tool life and letting n = 0.5 and C= 400...
Taylor Equation for Tool Life:
The Taylor equation is used to determine the relationship between cutting speed and tool life. The equation is as follows:
VT^n = C
where,
V = cutting speed
T = tool life
n, C = constants
Calculating the Percentage Increase in Tool Life:
Given, n = 0.5 and C = 400
Let's assume the initial cutting speed is V1 and the corresponding tool life is T1. When the cutting speed is reduced by 50%, the new cutting speed becomes V2 = V1/2.
Using the Taylor equation, we can find the new tool life T2.
T1 = C/V1^n
T2 = C/V2^n
= C/(V1/2)^n
= C/(2^n * V1^n)
= (1/2^n) * (C/V1^n)
= (1/2^(0.5)) * T1
= 0.707 * T1
Therefore, the new tool life T2 is 0.707 times the initial tool life T1.
The percentage increase in tool life is given by:
((T2 - T1)/T1) * 100
= ((0.707T1 - T1)/T1) * 100
= -29.3%
However, we need to calculate the percentage increase in tool life when the cutting speed is reduced by 50%. This can be found as follows:
Percentage increase in tool life = |((T2 - T1)/T1) * 100|
= |((-0.293) * 100)|
= 29.3%
Therefore, the percentage increase in tool life when the cutting speed is reduced by 50% is 29.3%. This is option 'D'.
The Taylor equation is used to determine the relationship between cutting speed and tool life. The equation is as follows:
VT^n = C
where,
V = cutting speed
T = tool life
n, C = constants
Calculating the Percentage Increase in Tool Life:
Given, n = 0.5 and C = 400
Let's assume the initial cutting speed is V1 and the corresponding tool life is T1. When the cutting speed is reduced by 50%, the new cutting speed becomes V2 = V1/2.
Using the Taylor equation, we can find the new tool life T2.
T1 = C/V1^n
T2 = C/V2^n
= C/(V1/2)^n
= C/(2^n * V1^n)
= (1/2^n) * (C/V1^n)
= (1/2^(0.5)) * T1
= 0.707 * T1
Therefore, the new tool life T2 is 0.707 times the initial tool life T1.
The percentage increase in tool life is given by:
((T2 - T1)/T1) * 100
= ((0.707T1 - T1)/T1) * 100
= -29.3%
However, we need to calculate the percentage increase in tool life when the cutting speed is reduced by 50%. This can be found as follows:
Percentage increase in tool life = |((T2 - T1)/T1) * 100|
= |((-0.293) * 100)|
= 29.3%
Therefore, the percentage increase in tool life when the cutting speed is reduced by 50% is 29.3%. This is option 'D'.
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Using the Taylor equation for tool life and letting n = 0.5 and C= 400, calculate the percentage increase in tool life when the cutting speed is reduced by 50%.a)100%b)No changec)400%d)300%Correct answer is option 'D'. Can you explain this answer?
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