Using the Taylor's equation, $V T^{n} = C$ , calculate the percentage increase in tool life when the cutting speed is reduced by 50%.
(Assume n = 0.5 and C = 400)

300%

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400%

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100%

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50%

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Solution

The correct option is A 300%
$V_{1} T_{1}^{n} = V_{2} T_{2}^{n}$

$V_{1} T_{1}^{0.5} = V_{2} T_{2}^{0.5}$

$T_{2} = 2^{2} T_{1}$

$T_{2} = 4 T_{1}$

The Percentage Increase in the Tool Life = $\frac{T_{2} - T_{1}}{T_{1}} \times 100 %$

The Percentage Increase in the Tool Life = $\frac{4 T_{1} - T_{1}}{T_{1}} \times 100 % = 300 %$

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