Machining Power Calculator (MRR & Spindle HP)
Work out metal removal rate, spindle horsepower, torque and cutting force for milling, turning or drilling, with specific cutting force that varies with chip thickness.
How to use this calculator
- 1Enter the cut you actually intend to take, not the machine's capacity - this page tells you what that cut costs.
- 2Get the feed right: it drives both the removal rate and the chip thickness, and the chip thickness is what sets the specific force.
- 3Put in your machine's spindle power to get a straight answer on whether the cut fits.
- 4If it does not fit, reduce the depth of cut before the feed. Cutting the feed thins the chip and pushes the specific force up, so it saves less power than it looks like it should.
- 5On a low-speed cut, check the machine's torque curve rather than its horsepower rating.
How the calculation works
MRR (milling) = axial depth x radial depth x feed rate
MRR (turning) = depth of cut x feed per rev x rpm
Kc = Kc(1.1) x hm^-mc x (1 - rake/100)
Power (hp) = MRR x Kc / 2730.3
Torque (in-lb) = hp x 63025 / rpm- Kc(1.1)
- Specific cutting force at a 1 mm chip, N/mm². A property of the material - 1350 for mild steel, 750 for aluminium, 3300 for nickel alloys
- mc
- The Kienzle exponent, typically 0.2 to 0.28. It is what makes thin chips expensive: force per unit volume rises as the chip thins
- hm
- Mean chip thickness in millimetres, which depends on the feed, the engagement and the lead angle - not just the feed
- 2730.3
- Unit conversion. A cubic inch is 16387 mm³, so 1 in³/min at 1 N/mm² is 0.2731 W, and a horsepower is 745.7 W
Power scales linearly with removal rate at a fixed chip thickness, so doubling the depth of cut doubles the power. Doubling the feed does not quite double it, because the thicker chip is cheaper per cubic inch.
Torque and horsepower are not interchangeable. Below its base speed a spindle usually holds constant torque, so available horsepower falls with rpm - which is why a low-speed cut can stall a machine whose nameplate looks ample.
The rake correction of about 1% per degree is the standard allowance. Positive rake reduces the force and negative rake increases it.
Worked example
A half inch end mill roughing mild steel
- 1.Removal rate = 0.25 x 0.25 x 45 = 2.81 cubic inches a minute.
- 2.Feed per tooth is 45 / (3800 x 4) = 0.00296 in; at half diameter engagement the mean chip is 2/pi of that in the slot sense - here 0.0019 in, or 0.048 mm.
- 3.Kienzle for 1018 at that chip: 1350 x 0.048^-0.21 = 2556 N/mm².
- 4.Power = 2.81 x 2556 / 2730.3 = 2.63 hp at the cut, so 3.10 hp at the motor at 85% efficiency.
Result: 3.10 hp at the motor for 2.81 in³/min
The same metal, a much heavier cut
- 1.A 2 in face mill, 1.5 in wide and 0.15 in deep at 30 in/min: 0.15 x 1.5 x 30 = 6.75 in³/min.
- 2.Feed per tooth = 30 / (950 x 5) = 0.0063 in, and with 75% radial engagement the mean chip is 0.115 mm - much thicker than the end mill case.
- 3.A thicker chip is cheaper: Kc comes out at 2127 N/mm² instead of 2556.
- 4.Power = 6.75 x 2127 / 2730.3 = 5.26 hp at the cut, 6.19 at the motor. Two and a half times the metal for twice the power - 0.78 horsepower per cubic inch against 0.94.
Result: 6.19 hp for 6.75 in³/min - thick chips are efficient
Turning stainless, checked against a 5 hp lathe
- 1.Removal = 0.125 x 0.012 x 600 = 0.90 cubic inches a minute.
- 2.The chip is the feed times the sine of the lead angle: 0.012 in at 90 degrees, which is 0.305 mm.
- 3.Austenitic stainless at that chip: 2150 x 0.305^-0.20 = 2727 N/mm².
- 4.Power = 0.90 x 2727 / 2730.3 = 0.90 hp at the cut, 1.06 at the motor - about 21% of a 5 hp lathe, so comfortable.
Result: 1.06 hp - 21% of a 5 hp lathe
A big drill at low speed, where torque bites
- 1.A 1.5 in drill removes a full circle: (pi/4) x 1.5² x 0.012 x 150 = 3.18 cubic inches a minute.
- 2.Each of the two lips takes half the feed, so the chip is 0.131 mm and Kc is 2070 N/mm².
- 3.Power = 3.18 x 2070 / 2730.3 = 2.41 hp at the cut, 2.84 at the motor - 95% of a 3 hp machine, so inside it on paper.
- 4.But at 150 rpm that is 1013 in-lb, over 84 ft-lb of torque. Very few 3 hp machines deliver rated power at 150 rpm; most hold constant torque below base speed and would stall. This is the case for a pilot hole.
Result: 2.84 hp on paper, but 84 ft-lb at 150 rpm is the real constraint
Removal rate is the whole story, until it is not
Cutting power is remarkably simple at first glance: it takes a certain amount of energy to turn a cubic inch of a given metal into chips, so multiply that by the cubic inches per minute and you have the power. Metal removal rate is trivially calculated - depth times width times feed for milling, depth times feed times rpm for turning - and the multiplication is arithmetic.
The complication is that the energy per cubic inch is not fixed. It depends strongly on how thick the chip is, and only weakly on everything else. That is not a small correction: across the working range of chip thicknesses the specific energy for a given material varies by a factor of two or more.
Handbook tables of "unit power" hide this by quoting a single figure per material, implicitly at a typical roughing chip. Used for roughing, they are about right. Used for finishing, they underestimate substantially - and finishing cuts at low spindle speed are exactly where machines stall.
Why thin chips cost more per cubic inch
A cutting edge is not a knife edge. It has a radius - a micron or two on a fresh carbide edge, more once it has worn - and the material has to be thicker than several times that radius before it can be sheared off cleanly as a chip.
When the chip is thick relative to the edge radius, almost all the energy goes into shearing along a well-defined plane, and the process is efficient. When the chip approaches the edge radius, an increasing share goes into ploughing: pushing material down and around the edge, deforming it, generating heat without producing a chip. That energy still has to come from the spindle, and it does not remove any metal.
The Kienzle model captures this empirically with a power law, Kc = Kc(1.1) times chip thickness to the power of minus mc. Kc(1.1) is the specific force at a one millimetre chip and mc is typically 0.2 to 0.28. For mild steel, that means a 0.05 mm chip needs about 2530 N/mm² against 1350 at a millimetre - nearly double.
The practical consequence runs against intuition: when a cut is too much for the machine, reducing the feed is a poor way to reduce power. It cuts the removal rate but thins the chip and raises the specific force, so the power falls by less than the arithmetic suggests. Reducing the depth of cut lowers the rate without touching the chip thickness, and takes the power down proportionately.
Horsepower, torque, and which one stops you
These are the same quantity related by speed - horsepower is torque times rpm over 63,025 in inch units - but a machine does not deliver them the same way across its range.
Above its base speed a spindle motor delivers roughly constant power, so torque falls as speed rises. Below base speed it typically holds constant torque instead, which means available power falls in proportion to rpm. A 5 hp spindle with a 1,200 rpm base speed has about 2.5 hp available at 600 rpm and about 0.6 at 150, whatever the nameplate says.
This is why big drills and large face mills stall machines that ought to manage them. The horsepower calculation says the cut fits; the torque at the speed involved is more than the spindle can produce. A geared head machine exists precisely to solve this, trading speed for torque mechanically so the motor can work where it is efficient.
The number worth checking is therefore the torque figure against the machine's torque curve at the actual spindle speed, not the horsepower against the nameplate. Where no curve is published, a rough guide is that below base speed the available power scales with rpm.
What the force figure is for
The tangential force is the one that produces the torque, and it is useful for a different question: whether the workholding and the fixture can take the cut.
A few hundred pounds at the cutting edge is entirely normal for a moderate cut in steel, and it acts at the tool, some distance from wherever the part is clamped. That leverage is what lifts parts out of vices and moves them on magnetic chucks. It is also what deflects long tools - and deflection is not linear in overhang but cubic, so a tool held a little further out is very much worse.
For drilling there is a second force this page does not compute: the axial thrust, which for a normal point is of comparable magnitude to the tangential force. That is the load the quill, the table and the clamps resist, and on a large drill it is usually what governs rather than the torque. It is the reason for pilot holes - a pilot removes the web, which is the part of a drill point that does no cutting and all of the pushing.
What this assumes, and where it stops
Assumptions
- Kienzle constants are representative published values for the material group. Actual hardness and condition move Kc(1.1) considerably.
- Sharp tools in good condition. A worn tool takes substantially more power - a factor around 1.1 to 1.3 is a common allowance and is not applied here.
- Mean chip thickness is used for the specific force, which is the usual convention.
- Drive efficiency is a single figure covering belts, gears and bearings.
- Drilling assumes a normal two-lip point removing a full circle, with each lip taking half the feed.
Limitations
- No tool wear factor is applied. A dull tool can need 30% more power than a sharp one.
- The Kienzle relation is fitted over a normal working range and extrapolates poorly below about 0.02 mm of chip, where ploughing dominates.
- Axial thrust in drilling and radial force in turning are not computed - only the tangential component that produces torque.
- It does not model the machine's torque curve, which is often the real constraint at low spindle speed.
- Coolant, coating and edge preparation all affect the force somewhat and none are modelled.
Common questions
How do I calculate metal removal rate?
In milling it is the axial depth times the radial depth times the feed rate in inches per minute, giving cubic inches per minute. In turning it is the depth of cut times the feed per revolution times the rpm. In drilling it is the full circle the drill cuts: pi/4 times diameter squared, times feed per revolution, times rpm.
How much horsepower do I need to cut steel?
Around 1 horsepower per cubic inch per minute for mild steel at a normal roughing chip, but the figure rises sharply for thin chips - a light finishing pass can need 1.8 or more for the same volume. Aluminium is about a third of steel and nickel alloys two to three times it.
Why is unit power not a constant?
Because a cutting edge has a real radius, and a chip has to be several times thicker than that radius before it shears cleanly. Thinner chips get ploughed and deformed rather than cut, which consumes energy without removing metal. The Kienzle model expresses this as a power law: specific force goes as chip thickness to the power of about minus 0.2.
My machine has enough horsepower but it still stalls - why?
Almost certainly torque at low speed. Below its base speed a spindle usually holds constant torque, so the horsepower available falls in proportion to rpm. A 5 hp spindle with a 1200 rpm base speed has around 0.6 hp at 150 rpm. Check the torque figure against the machine's torque curve at the speed you are actually running.
If a cut is too heavy, should I reduce feed or depth?
Depth, usually. Reducing the depth of cut lowers the removal rate without changing the chip thickness, so the power falls proportionately. Reducing the feed also thins the chip, which raises the specific cutting force, so the power falls by less than you expect - and the tool is then working in a less efficient regime as well.
Sources
- Machining Doctor - machining power and specific cutting force — Machining Doctor
Formula and content last reviewed on .
Results are estimates for information only, not professional advice.
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