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Metal Removal Rate and Spindle Horsepower

Guide · Updated October 28, 2026

Before you push a bigger cut, check whether the spindle has the power for it. Metal removal rate (MRR) is the volume of chips made per minute. Multiply it by the unit power of the material (horsepower needed to remove one cubic inch per minute) and divide by machine efficiency to get the motor power the cut needs. The numbers are approximate, but they tell you quickly whether a cut is reasonable or will stall the spindle.

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Step by step

  1. Get the cut geometry: Radial width of cut (WOC), axial depth of cut (DOC) and table feed in in/min for milling; depth, feed per rev and surface speed for turning.
  2. Calculate MRR: Milling: MRR = WOC × DOC × feed (in³/min). Turning: MRR = 12 × SFM × feed per rev × depth.
  3. Look up unit power: Use a published unit-power value for the material and hardness, in hp per in³/min.
  4. Find power at the cutter: hp at cutter = MRR × unit power. Correct up for a dull tool.
  5. Allow for efficiency: Motor hp = cutter hp ÷ efficiency, often about 0.75 to 0.85 for a geared or belt spindle.
  6. Compare to the machine: Stay well under the rated motor power at the spindle speed you are running, since many spindles lose power at low rpm.

The formulas

Unit power changes with hardness, tool sharpness and rake angle, so published values are given as ranges. Typical approximate values in hp per in³/min at the cutter are: aluminum alloys 0.25 to 0.35, brass and bronze 0.5 to 0.8, gray cast iron 0.6 to 1.1, low and medium carbon steel 0.8 to 1.4, stainless steel 1.0 to 1.5 and titanium alloys 1.1 to 1.4. A dull tool can raise power by 25 percent or more.

Worked example: end mill in steel

A 1/2 in end mill takes a 0.250 in radial width, 0.500 in axial depth at 20 in/min in 1018 steel. MRR = 0.250 × 0.500 × 20 = 2.50 in³/min. Using about 1.0 hp per in³/min, the cut needs about 2.5 hp at the cutter. At 80 percent efficiency the motor must supply 2.5 ÷ 0.80 ≈ 3.1 hp.

On a 2 hp knee mill this cut will bog down, so you would drop to about 0.250 in axial depth (1.25 in³/min, about 1.6 hp at the motor) or reduce feed. On a 7.5 hp machining center it is a light cut.

The same cut in 6061 aluminum at about 0.3 hp per in³/min needs only 0.75 hp at the cutter, which is why aluminum can be cut so much faster on the same machine.

Worked example: turning

Turning 4140 at 400 SFM with 0.010 in/rev feed and 0.100 in depth of cut: MRR = 12 × 400 × 0.010 × 0.100 = 4.8 in³/min. At about 1.2 hp per in³/min that is 5.8 hp at the tool and about 7.2 hp at the motor at 80 percent efficiency. A 10 hp lathe can take it; a 3 hp toolroom lathe cannot, so you would cut the depth or feed in half or both.

Remember that turning MRR uses the depth of cut, which is half the change in diameter. Taking 0.200 in off the diameter is a 0.100 in depth.

Worked example: metric milling

A face mill takes ap = 3 mm, ae = 40 mm at vf = 800 mm/min in steel with a specific cutting force kc of about 2000 N/mm². P = 3 × 40 × 800 × 2000 ÷ 60,000,000 = 3.2 kW at the cutter, or about 4.0 kW at the motor with 0.8 efficiency. Specific cutting force rises as chip thickness gets thinner, so very light feeds need more power per unit volume than this simple number suggests.

Low-rpm power and torque

Most spindle motors make rated power only above a base speed. Below that they are torque limited, so the available power falls roughly in proportion to rpm. Torque in in-lb = hp × 63,025 ÷ rpm. A 3.1 hp cut at 600 rpm needs 3.1 × 63,025 ÷ 600 ≈ 326 in-lb at the spindle. Check the spindle power curve in the machine manual before running large drills or face mills at low rpm.

Common problems

Using the change in diameter as depth in turning, forgetting efficiency, using unit power for annealed steel on hardened material, and ignoring the low-rpm power curve. Watch the spindle load meter on the first part and back off if it runs near 100 percent.

FAQ

What is MRR?

Metal removal rate, the volume of material removed per minute, usually in in³/min or cm³/min.

What is unit power?

The power needed to remove one cubic inch of material per minute, in hp per in³/min. It depends on material, hardness and tool condition.

Why does my machine stall below rated horsepower?

Many spindles cannot make full power at low rpm, and a dull tool raises the power needed.

How do I check part size after a heavy cut?

Let the part cool, then measure. Gage blocks set mics and indicators; build any stack free at gageblockcalc.com.

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