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ArticlePublished 11 Jul 2026Updated 14 Jul 20265 min readBy Kevin Jogin
KEVOS® Knowledge Library · Engineering → Mechanical Engineering

Engineering / Mechanical Engineering

Machining Econometrics

The fastest cut is rarely the cheapest. Push the speed up and machining time falls but tools wear out faster; slow down and tools last but the machine sits longer on each part. Somewhere between lies the speed that costs least — and it can be found exactly.

  • Reading time · 5 min
  • 7 sections
  • Cost-per-part curve, computed
  • Optimum tool life worked
min costmachiningtoolingtotalcutting speed Vcost per part
Doc №KL-ENG-MECH-098
SectionEngineering → Mechanical Engineering
Sheet1 of 1
DrawnKEVOS®
Date2026-07-11

§1Speed is not free

Cutting faster always shortens machining time, but it never comes free: through Taylor’s law it shortens tool life steeply, adding tooling and downtime cost. Economics is about weighing the two.

The earlier pages gave the physics — Taylor’s V Tⁿ = C relating speed to tool life, and the time and power a cut demands. Economics adds the money. Every part carries costs that move in opposite directions as speed rises: the cost of the machine’s time, which falls as faster cutting finishes the part sooner, and the cost of tooling and tool-changing, which climbs as faster cutting wears edges out. Because one falls and the other rises, their sum has a minimum — a cutting speed, and a matching tool life, at which the part costs least (§3–4). Finding that point, rather than simply running fast, is what machining econometrics is for.

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§2The three costs per part

The cost of machining one part breaks into three pieces, and the whole analysis rests on knowing how each responds to cutting speed.

What a machined part costs, and how speed moves it
CostWhat it isAs speed rises
Machiningmachine + labour rate × cutting timefalls (shorter time)
Toolcost per cutting edge, shared over the parts it makesrises (fewer parts per edge)
Tool changemachine rate × time to change a worn tool, per partrises (more changes)
The machining cost falls with speed because the part is finished sooner; the tool and tool-change costs rise because a faster cut wears edges out quicker, so each edge makes fewer parts and is changed more often. A fixed set-up cost per part sits on top but does not depend on speed, so it shifts the whole curve up without moving its lowest point.
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§3The cost curve

Add the falling machining cost to the rising tooling cost and the total cost per part traces a U — high at both extremes, lowest at one particular speed.

At low speed the machine dwells on each part, so machining cost dominates and the part is expensive despite long tool life. At high speed the machine is quick but tools are consumed and changed so often that tooling cost dominates and the part is expensive again. Between the two, the total dips to a minimum: the minimum-cost cutting speed (the hero curve). The curve is usually shallow around its base — a broad, forgiving valley — so speeds a little either side of the optimum cost almost the same, which matters in practice because it leaves room to trade a touch of cost for more output (§6). The shape, not just the single point, is what guides the choice.

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§4Minimum-cost tool life

The optimum need not be searched for by trial — the tool life that minimises cost follows in closed form from Taylor’s exponent and the cost figures.

Tcost = 1 − nn (tc + Ct / Cm)  — n Taylor exponent, tc tool-change time, Ct tool cost/edge, Cm machine rate
Example 1 — the tool life to aim for

With a high-speed-steel exponent n = 0.25, a tool-change time of 5 min, a tool cost of $6 per edge and a machine-plus-labour rate of $1/min, the minimum-cost tool life is (1 − 0.25)/0.25 × (5 + 6/1) = 3 × 11 = 33 min. So the cutting speed should be set — through Taylor’s law — to give about a 33-minute tool life, not the far shorter life a flat-out speed would give. The result is intuitive: cheap tools and quick changes push the optimum toward faster cutting (shorter life), while expensive tools or slow changes push it toward slower cutting (longer life). The formula turns the cost figures straight into the tool life, and hence the speed, to run.

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§5Maximum-production tool life

A different aim — the most parts per hour, regardless of cost — gives a different, shorter optimum tool life, found the same way but without the tool-cost term.

Example 2 — running for output, not cost

When the goal is maximum production rather than minimum cost — a bottleneck machine, an urgent order — only time matters, not the price of edges, so the tool-cost term drops out and the optimum tool life is simply (1 − n)/n × tc = 3 × 5 = 15 min. This is shorter than the 33-minute minimum-cost life, meaning a higher cutting speed: to make the most parts per hour you run faster and change tools more often, accepting the extra tooling cost for the extra output. The two optima bracket the useful range — run no faster than the maximum-production speed (15-minute life) and no slower than the minimum-cost speed (33-minute life). Between them lies every sensible choice.

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§6The high-efficiency range

The minimum-cost and maximum-production speeds are not rivals but the two ends of a band — the high-efficiency range — within which any speed is a reasonable compromise.

Because the cost curve is shallow near its base, running anywhere between the minimum-cost speed and the (faster) maximum-production speed costs only a little more than the true minimum while giving more output — so the sensible operating window is the whole band between them, not a single point. Where in that band to sit depends on the situation: toward the minimum-cost end when the machine has spare capacity and cost rules, toward the maximum-production end when the machine is a bottleneck and throughput rules. This is why shops speak of a speed range rather than one magic number, and why the two formulae above matter more as a pair of limits than as exact set-points. Choose the end of the range that fits the constraint — cost or capacity — and cut there.

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§7Quick reference

The working core of the page on one card rack.

Two forces

machining cost ↓ with speed

tooling cost ↑ with speed

Cost curve

U-shaped · shallow base

min = cheapest speed

Min cost

T = (1−n)/n · (t_c + C_t/C_m)

→ 33 min (example)

Max production

T = (1−n)/n · t_c

→ 15 min (faster)

Range

run between the two

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