Taylor's tool-life law application In single-point turning with a cemented carbide tool on steel (Taylor exponent n = 0.25), if cutting speed is halved, by what factor does the tool life change according to V * T^n = C?

Introduction / Context:
Taylor’s tool-life equation relates cutting speed V and tool life T via V * T^n = C for given tool-work combinations. It allows quick what-if predictions when speeds or materials change.

Given Data / Assumptions:

• n = 0.25 (typical for carbide on steel).
• Initial speed V1, life T1; new speed V2 = V1 / 2.
• Same feed, depth, environment so constant C applies.

Concept / Approach:
Using V1 * T1^n = C and V2 * T2^n = C, equate and solve for T2/T1 when V2 = V1/2.

Step-by-Step Solution:
V1 * T1^n = (V1/2) * T2^nT1^n = (1/2) * T2^nT2^n = 2 * T1^n → T2 = T1 * 2^(1/n)With n = 0.25, 1/n = 4, so T2 = T1 * 2^4 = 16 * T1

Verification / Alternative check:
Trend matches intuition: lower speed increases life, and a small n magnifies the effect strongly.

Why Other Options Are Wrong:

• Half or two times contradict the power-law sensitivity with n = 0.25.
• Eight times corresponds to n = 1/3, not 1/4.
• “Unchanged” ignores Taylor’s empirical relation.

Common Pitfalls:
Mishandling exponents; using 2^n instead of 2^(1/n); changing multiple variables simultaneously.

Final Answer:
Tool life becomes sixteen times the original