§1Power and time
Two estimates decide whether a proposed cut is practical: the power it will demand of the machine, and the time it will take to complete. Both are quick to work out, and both are worth doing before committing to a heavy cut.
Power matters because every machine has a limited spindle motor: ask for more than it can deliver and the cut stalls, the belt slips or the tool stops. Time matters because it drives cost, scheduling and the economics of the whole job (its own econometrics page). Both estimates share a single foundation — the rate at which metal is being removed, multiplied by how hard the material is to cut. Get those two, and power and time both follow. This page builds the estimate from that foundation: the material’s specific cutting energy (§2), the power it implies (§3–4), the torque (§5) and the time (§6).
Contents§2Specific cutting energy
The key material property for power is the specific cutting energy — the energy needed to remove one cubic millimetre of the material, equal to its specific cutting force.
Cutting is work: shearing metal into a chip takes energy, and the energy per unit volume removed is a property of the work material, called the specific cutting energy (or, as a pressure, the specific cutting force). For steel it is roughly 2–3 J/mm³ (equivalently 2000–3000 N/mm²); aluminium, far softer, needs only about a third of that; hardened and high-strength materials need more. This single number links the geometry of a cut to the power it draws: however the removal rate is made up — fast and shallow, or slow and deep — the power is that rate times the specific cutting energy. It is the machining equivalent of a material’s resistance, and every power estimate starts from it.
Contents§3Power from removal rate
Cutting power is simply the metal removal rate multiplied by the specific cutting energy — the volume cut per second times the energy each unit takes.
Take the face-milling cut from the milling page, removing 57 300 mm³/min — that is 955 mm³/s. In steel at a specific cutting energy of 2.5 J/mm³, the cutting power is 2.5 × 955 = 2388 W ≈ 2.4 kW. The estimate needs nothing but the removal rate and the material, and it can be run before the cut to check the machine will cope: a 2.4 kW cut is comfortable for a mid-size mill but would overwhelm a small one. Equivalently the power is the cutting force times the cutting speed, P = Fc × V (the hero) — the two forms agree, one reckoned from the volume removed, the other from the force at the edge.
§4Motor power and efficiency
The power at the cutting edge is not the power the motor must supply — friction in the drive means the motor works harder, by the machine’s efficiency.
Between the motor and the cutting edge sit belts, gears and bearings, each losing a little to friction, so the motor must supply more than the cutting power. Dividing by a typical spindle-drive efficiency of about 0.75, the 2388 W at the edge needs 2388 ÷ 0.75 = 3183 W ≈ 3.2 kW at the motor. This is why a machine’s rated motor power always exceeds the cutting power it can actually deliver, and why the useful cut is judged from motor power times efficiency, not motor power alone. When checking whether a machine can take a cut, compare the cutting power to the motor’s rating discounted by its efficiency.
§5Cutting torque
Where power is the rate of doing work, torque is the twisting effort at the spindle — and at low spindle speeds it is torque, not power, that can run short.
For the same power, torque rises as speed falls, because power is torque times angular speed. The 2388 W cut above, taken at 300 rev/min (ω = 31.4 rad/s), needs a spindle torque of 2388 ÷ 31.4 = 76 N·m; taken faster, it would need less. This is why heavy cuts at low speed — large-diameter turning, big drills — are limited by the machine’s available torque rather than its power, and why machines have gearboxes: a low gear trades speed for the torque a big, slow cut demands. Power tells you if the motor is big enough; torque tells you if it can be delivered at the speed the cut runs.
Contents§6Machining time
The time a cut takes is the distance the tool must travel divided by how fast it feeds — a direct calculation once feed and speed are set.
Turning a 100 mm length at a feed of 0.2 mm/rev and 300 rev/min, the feed rate is f × N = 60 mm/min, so the pass takes 100 ÷ 60 = 1.67 min. Multiply by the number of passes, add tool changes and handling, and the job time follows — the basis of the cost-per-part economics on the econometrics page. Raising feed or speed cuts the time proportionally, which is the productivity gain that must always be weighed against the shorter tool life and rougher finish those same increases bring. Time, power and tool life pull against one another, and estimating all three is how a sensible cut is chosen.
§7Quick reference
The working core of the page on one card rack.
Specific energy
steel ~2–3 J/mm³
aluminium ~⅓ of that
Cutting power
P = energy × MRR
= Fc × V
Motor power
P ÷ efficiency (~0.75)
Torque
T = P/ω · rises as speed falls
Time
t = L/(f·N)