Taylor's tool life equation and the effect of cutting speed

VTⁿ = C, so T₂/T₁ = (V₁/V₂)^(1/n).

Worked answer.

Taylor's equation. F. W. Taylor found that, for a given tool-work pair, cutting speed V and tool life T are related by
V Tⁿ = C,
where n is an exponent that depends mainly on the tool material, and C is the cutting speed that gives a tool life of 1 minute. Typical textbook ranges of n are about 0.1–0.15 for HSS, 0.2–0.5 for carbides and higher for ceramics. A larger n means tool life is less sensitive to speed.

Step 1: Find C.
C = V₁ T₁ⁿ = 100 × 60^0.25.
60^0.25 = √(√60) = √7.746 ≈ 2.783.
C ≈ 278.3 m/min.

Step 2: New tool life.
V₁T₁ⁿ = V₂T₂ⁿ, so T₂ = T₁ × (V₁/V₂)^(1/n) = 60 × (100/200)^(1/0.25) = 60 × (0.5)⁴ = 60 / 16 = 3.75 minutes.

Check. V₂T₂ⁿ = 200 × 3.75^0.25 = 200 × 1.392 ≈ 278.3, which equals C. The answer is consistent.

Interpretation. Doubling the speed cut tool life by a factor of 16. That is why speed is the most important factor in tool wear. Higher speed means higher temperature at the tool-chip interface, which accelerates wear mechanisms such as diffusion and crater wear.

Practical use. Choosing speed is an economic trade-off. Higher speed cuts machining time per part but increases tool changes and tool cost. Textbook optimisation gives the tool life for minimum cost as T = (1/n − 1)(C_t/C_m + t_c), where C_t is tool cost per edge, C_m is machine cost per minute and t_c is tool-change time.

Common mistake. Using T₂ = T₁ × (V₁/V₂)ⁿ, which gives 60 × 0.5^0.25 ≈ 50 minutes. The exponent must be 1/n because n is attached to T, not V.

Likely follow-up. "The extended Taylor equation?" V Tⁿ f^a d^b = C, which also includes feed f and depth of cut d. Speed has the strongest effect, followed by feed, then depth of cut.