14.12 Economics of Machining & Tool Life Optimisation

The Trade-Off

Raising cutting speed cuts the machining time for each component, which sounds like an unambiguous good. But Taylor's tool life equation

$$V\,T^{n} = C$$

shows that tool life falls steeply as speed rises, so tools must be changed more often. Each change costs machine downtime and a fresh cutting edge.

Machining economics finds the speed that balances these opposing effects. Note the exponent: since $n$ is typically 0.1 to 0.25 for high speed steel and 0.2 to 0.5 for carbide, tool life is extremely sensitive to speed. With $n = 0.2$, a 20% increase in cutting speed cuts tool life by roughly two thirds.

Building the Cost Model

Let:

The number of tool changes per component is $t_m/T$, since a tool lasting $T$ minutes covers $T/t_m$ components.

$$\text{Cost per component} = \underbrace{C_m t_l}_{\text{non-productive}} + \underbrace{C_m t_m}_{\text{machining}} + \underbrace{C_m t_c \frac{t_m}{T}}_{\text{tool changing}} + \underbrace{C_t \frac{t_m}{T}}_{\text{tool cost}}$$

The non-productive term is independent of cutting speed, so it does not affect the optimum — an important point, because it means the optimum speed is unchanged by improvements in loading and unloading, even though the total cost falls.

Expressing Everything in Terms of Speed

For turning a bar of diameter $D$ and length $L$ at feed $f$ and speed $V$:

$$t_m = \frac{\pi D L}{1000\,f\,V}$$

so machining time is inversely proportional to $V$. Writing $t_m = k/V$ and, from Taylor, $T = (C/V)^{1/n}$:

$$\frac{t_m}{T} = \frac{k}{V}\cdot\frac{V^{1/n}}{C^{1/n}} = \frac{k\,V^{(1/n)-1}}{C^{1/n}}$$

Substituting into the cost equation and differentiating with respect to $V$ gives the classical results.

Optimum Tool Life for Minimum Cost

$$\boxed{T_{\text{opt,cost}} = \left(\frac{1}{n} - 1\right)\left(t_c + \frac{C_t}{C_m}\right)}$$

and the corresponding economic cutting speed follows from Taylor's equation:

$$V_{\text{opt}} = \frac{C}{T_{\text{opt}}^{\,n}}$$

Notice what this result contains and what it does not. It depends on the tool change time, the ratio of tool cost to machine rate, and the Taylor exponent. It does not depend on the length of the cut, the diameter, or the non-productive time.

The grouped term $\left(t_c + C_t/C_m\right)$ is the total cost of a tool change expressed in minutes of machine time. A cheap insert on an expensive machine gives a small value and therefore a short optimum tool life — run fast and change often. An expensive tool on a cheap machine reverses the conclusion.

Optimum Tool Life for Maximum Production

If the objective is minimum time per component rather than minimum cost, the tool cost term drops out entirely:

$$\boxed{T_{\text{opt,production}} = \left(\frac{1}{n} - 1\right)t_c}$$

This is the tool life for maximum production rate, and the corresponding speed $V_{\max}$ is higher than the economic speed.

The Relationship Between the Two Optima

Since $C_t/C_m$ is always positive,

$$T_{\text{opt,cost}} > T_{\text{opt,production}} \quad\Longrightarrow\quad V_{\text{opt,cost}} < V_{\text{opt,production}}$$

The speed for maximum production is always higher than the speed for minimum cost. This is intuitive once stated: pushing for output ignores what the tools cost, so it justifies running faster and wearing tools out sooner.

The band between these two speeds is called the high-efficiency machining range or the Gilbert range. Operating below $V_{\text{cost}}$ is wasteful on both counts — it costs more and produces less, which is unambiguously wrong. Operating above $V_{\text{production}}$ is likewise wrong on both counts. Any speed inside the band is a defensible choice depending on whether the shop is cost-constrained or capacity-constrained.

This gives a practical decision rule: when the order book is full and the machine is the bottleneck, work towards the production-rate end of the range; when work is scarce and machine time is not the constraint, work towards the minimum-cost end.

Worked Example

A turning operation uses a carbide insert with Taylor constants $n = 0.25$ and $C = 150$. Tool changing takes 3 minutes, the insert edge costs 60 rupees and the machine plus operator rate is 12 rupees per minute.

Minimum cost tool life:

$$T = \left(\frac{1}{0.25}-1\right)\left(3 + \frac{60}{12}\right) = 3\times(3+5) = 3\times8 = 24 \text{ minutes}$$

Economic cutting speed:

$$V = \frac{150}{24^{0.25}} = \frac{150}{2.213} = 67.8 \text{ m/min}$$

Maximum production tool life:

$$T = 3\times3 = 9 \text{ minutes}$$

Maximum production speed:

$$V = \frac{150}{9^{0.25}} = \frac{150}{1.732} = 86.6 \text{ m/min}$$

So the high-efficiency range runs from 67.8 to 86.6 m/min. Running at 50 m/min or at 100 m/min would both be indefensible.

Effect of Feed and Depth of Cut

Taylor's equation in its extended form recognises that feed and depth also affect tool life:

$$V\,T^{n}\,f^{a}\,d^{b} = C'$$

with exponents satisfying $n < a < b$ in the usual ordering of influence. Because cutting speed affects tool life far more strongly than feed, which in turn affects it more than depth of cut, the standard rule for maximising removal rate follows:

Take the greatest depth of cut the machine and setup allow, then the greatest feed the surface finish and tool strength allow, and only then optimise the cutting speed.

This ordering is one of the most useful practical results in production engineering and is a common examination point.

Constraints on the Optimum

The calculated optimum is frequently unattainable, and a competent answer recognises the constraints:

When a constraint binds, the operating point moves to the constraint boundary rather than to the unconstrained optimum.

Interpretation for a Maintenance Workshop

In a coal subsidiary's central workshop, the parameters differ sharply from a mass-production shop. Batch sizes are small, setup and non-productive time dominates the cycle, and a machine standing idle is common. The cost model responds accordingly: the non-productive term is large but does not shift the optimum speed, while the ratio $C_t/C_m$ tends to be high because tooling is bought in small quantities at retail prices and machine rates are low. That pushes the economic tool life up and the economic cutting speed down — conservative speeds with long tool life are genuinely correct in that setting, not merely traditional.