--- title: "Taylor tool life equation numerical" subject: "Manufacturing and Production Processes" chapter: 7 key: "semester_4/manufacturing_and_production_processes:taylor-tool-life-equation-numerical" source: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/taylor-tool-life-equation-numerical site: BCT Engineering --- # Taylor tool life equation numerical Manufacturing and Production Processes · Chapter 7: Material Removal Processes: “ Chip-forming” - Asked in: year not given, 5 marks - Video: [Taylor Tool Life Equation | Numerical Problem Solved Step-by-Step (Machining)\GATE Numerical](https://www.youtube.com/watch?v=TQBNpK7Ltpo) — ProdTech Edu - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/manufacturing_and_production_processes/taylor-tool-life-equation-numerical - Whole chapter: https://www.bctengineer.com/md/semester_4/manufacturing_and_production_processes/chapter-7 ## Question State Taylor's tool life equation. In a turning test with a HSS tool, a tool life of 60 minutes was obtained at a cutting speed of 100 m/min and 150 minutes at 80 m/min. Determine the constants n and C, the cutting speed for a tool life of 90 minutes, and the tool life if the speed is increased by 20 percent over 100 m/min. Comment on the economics. ## Short answer Taylor's equation: V T^n = C. Constants n = ln(1.25)/ln(2.5) = 0.2435, C = 271 m/min. For T = 90 min, V = 90.6 m/min. At 120 m/min, T = 28.4 min. A 20% speed rise more than halves tool life; an economic cutting speed minimizes total cost per piece. ## Answer **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the constant C in V T^n = C mean? A. The tool life at 1 m/min B. The cutting speed for a tool life of 1 minute C. The tool life at maximum speed D. The exponent depending on tool material Answer: B. When T = 1 min, V = C, so C is the speed giving 1 min life. (Key point: Taylor equation) 2. How is n found from the two tests? A. n = ln(T2/T1)/ln(V1/V2) = 4.1 B. n = ln(V1/V2)/ln(T1/T2), which is negative C. n = ln(V1/V2)/ln(T2/T1) = ln 1.25/ln 2.5 = 0.2435 D. n = V1/V2 = 1.25 Answer: C. From V1T1^n = V2T2^n, (V1/V2) = (T2/T1)^n, so n = 0.2435. (Key point: Finding n) 3. What is the tool life at 120 m/min? A. About 72 min B. About 50 min C. About 12 min D. About 28 min, using T = (C/V)^(1/n) Answer: D. T = (271/120)^4.107 = 28.4 min, less than half of 60 min. (Key point: Tool life at higher speed) 4. What is the economic effect of raising the cutting speed? A. Machining time falls but tool-change time and tool cost per piece rise; an economic speed minimises cost B. Both machining time and tool cost per piece fall C. Tool life increases so total cost always falls D. Speed has no effect on cost per piece Answer: A. Total cost has a minimum at the economic speed, below the speed of maximum production rate. 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