Max Material Removal Rate from Spindle Power
The power-limited max removal rate a spindle can drive.
Example
You enter
- Available motor horsepower (hp) 5
- Unit power (hp per in3/min) 1
- Spindle drive efficiency (%) 80
You get
- Max material removal rate 4 in3/min
- Cutting hp at the limit 4
No published worked example is on file for this calculation. It is checked against a hand-worked derivation of its method and an independent reference test.
Details, formula, and sources
Scopemax MRR = motor hp x efficiency / unit power. A 5 hp spindle at 80% efficiency removes up to 4.0 in3/min of carbon steel (unit power 1.0), or 12.1 in3/min of aluminum (0.33). The stall limit only -- depth/feed, tool strength, and rigidity are separate. The tool and machine govern the real cut.
Formula or tablemax MRR (in3/min) = available motor hp x (efficiency / 100) / unit power (hp per in3/min); the specific-cutting-energy relation solved for the removal rate.
Source and editionCutting power solved for the removal rate - first-principles specific-cutting-energy relation with Machinery's Handbook (Industrial Press) unit-power values, by name.
Free accessThe specific-cutting-energy arithmetic is public; the motor horsepower, unit power, and efficiency are user-supplied.
What governsEstimate. AHJ and licensed professional govern.
API field namesavailable_motor_hp, unit_power_hp, efficiency_pct, max_mrr_in3_min, cutting_hp
- Power limit only the motor-stall ceiling; depth/feed, tool strength, and rigidity are separate limitsmachining practice
- Unit power specific cutting energy, hp per in3/min: ~1.0 carbon steel, ~0.33 aluminum, ~1.5 stainless/titanium; default 1.0Machinery's Handbook
- Drive efficiency spindle drive efficiency; default 80%machine data