Tool Life (Taylor Equation) and Cost Per Part Calculator
Running faster cuts cycle time but burns tools quicker, and the cheapest speed is rarely the fastest one. This calculator uses Taylor's equation to predict tool life at a new speed and then finds the speed with the lowest cost per part.
Your numbers
Speed at which you know the tool life.
Typical ranges: HSS 0.08 to 0.2, carbide 0.2 to 0.4, ceramic 0.4 to 0.6.
Insert price divided by usable edges.
Your results
Taylor's equation is an empirical fit valid near your test speeds. Do not extrapolate far outside the speed range you have data for.
Cost per part vs cutting speed
Sweeps New cutting speed from 60 to 450, holding your other inputs fixed.
Real tooling cost in your ERP
Optional: discuss tracking tool consumption and cost per part from ERP and MES data.
Optional. The tool and every export work without this. No spam.
How to use it
Enter one known point: the speed you ran and the tool life you got. Choose an exponent n for your tool material. Then slide the new speed and watch tool life, parts per edge and cost per part change. Add insert cost per edge, machine rate and tool change time to capture the real economics. The chart makes the cost-minimum speed obvious: it is the bottom of the curve.
- Measure tool life as minutes of actual cutting to a fixed wear limit, such as VB 0.3 mm.
- Use edge cost, not box cost.
- Include the time to index and re-touch off.
Formulas
Taylor's equation is V x T^n = C. Given a reference pair, C = V1 x T1^n, and life at a new speed is T2 = (C / V2)^(1/n). Cutting time per part scales inversely with speed. Cost per part is machine rate times cutting time, plus the cost of one edge and one tool change divided by the number of parts per edge. Because n is small, a modest speed increase cuts tool life sharply: with n of 0.25, 25 percent more speed cuts life by about 60 percent.
Worked example
With 200 m/min giving 15 minutes of life, n of 0.25 and a 2 minute cut, going to 250 m/min drops life to about 6.1 minutes while cutting time falls to 1.6 minutes. With a 3 dollar edge, 60 dollar per hour rate and 1.5 minute tool change, cost per part is about 2.20 dollars at 250 versus about 2.31 dollars at 200. Sweeping the chart shows the true optimum for your numbers.
Common mistakes
The biggest mistake is extrapolating far from tested speeds, where the power law breaks down. Others include ignoring that feed and depth also change tool life, mixing minutes of cut with minutes of part cycle, and treating the optimum as fixed when machine rates or insert prices change. In production, optimise for the minimum cost speed only if the machine is not a bottleneck; if it is, choose the maximum production rate speed instead.
- Re-measure n when you change grade or coating.
- Check for edge breakage, not just wear, at high speed.
- Bottleneck machines favour higher speeds than the cost optimum.
Frequently Asked Questions
What is the Taylor equation?
It is the empirical relation V x T^n = C between cutting speed V and tool life T, where n and C depend on the work material, tool grade and conditions. It lets you predict life at another speed from one test point.
What exponent n should I use?
Typical values are about 0.1 to 0.2 for HSS, 0.2 to 0.4 for carbide, and 0.4 to 0.6 for ceramics. Better still, derive n from two of your own tests at different speeds.
Why is the lowest cost speed lower than the fastest?
Increased speed shortens cutting time but disproportionately reduces tool life, so insert and tool change costs per part rise. The cost curve has a minimum, usually at a moderate speed.
Should I always run the minimum cost speed?
Not on a bottleneck machine. If capacity limits throughput, the maximum production rate speed, which is higher, may earn more. Compare the cost saved with the revenue from extra output.
If tooling usage and cost per part are tracked in an ERP, Netray can help bring actual insert consumption into your costing so these models use real numbers.
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