Taylor tool life equation numerical

Taylor's equation: the cutting speed $V$ and tool life $T$ (minutes to reach a given flank wear, usually 0.3 mm) are related by

$$V\,T^{\,n} = C$$

n depends mainly on the tool material (about 0.1-0.15 for HSS, 0.2-0.4 for carbide, 0.5-0.7 for ceramics), and C is the speed for a tool life of 1 minute.

Constants

Two tests give $V_1T_1^n = V_2T_2^n$:

$$\frac{V_1}{V_2} = \left(\frac{T_2}{T_1}\right)^{n}\;\Rightarrow\; \frac{100}{80} = \left(\frac{150}{60}\right)^{n}$$ $$n = \frac{\ln 1.25}{\ln 2.5} = \frac{0.2231}{0.9163} = 0.2435$$ $$C = V_1T_1^{n} = 100\times(60)^{0.2435} = 100\times2.710 = 271.0\ \text{m/min}$$

(Check: $80\times150^{0.2435} = 80\times3.388 = 271.0$.)

Speed for $T = 90$ min

$$V = \frac{C}{T^{n}} = \frac{271.0}{(90)^{0.2435}} = \frac{271.0}{2.992} = 90.6\ \text{m/min}$$

Tool life at 120 m/min

$$T = \left(\frac{C}{V}\right)^{1/n} = \left(\frac{271.0}{120}\right)^{4.107} = (2.258)^{4.107} = 28.4\ \text{min}$$

Comment on economics

A 20 percent rise in speed cuts the tool life from 60 to 28.4 min, i.e. by more than half. Machining time per piece falls, but the tool-change time and tool cost per piece rise. There is an economic cutting speed (minimum cost per piece) which lies below the speed of maximum production rate, and a speed of maximum profit between them. It is found by differentiating total cost (machining + tool-change + tool cost) with respect to speed.

Answer: $n = 0.2435$, $C = 271$ m/min, $V = 90.6$ m/min for 90 min life, $T = 28.4$ min at 120 m/min.