Cutting Force, Power and Torque
Specific cutting force, how it rises as the chip gets thin, spindle power and torque checks, and whether the machine can take the cut.
Specific cutting force
The specific cutting force kc is the cutting force per unit area of chip (N/mm²). It is a property of the material, and it rises as the chip gets thinner. A simple, widely used model is kc = kc1 × h−mc, where h is the chip thickness in mm, kc1 is the value at 1 mm and mc is an exponent of about 0.2 to 0.3.
| Material | kc1 (N/mm² at h = 1 mm) | mc | kc at h = 0.1 mm | Unit power at h = 0.1 mm |
|---|---|---|---|---|
| Aluminum (6061) | 700 | 0.25 | 1 245 | 0.021 kW per cm³/min (0.46 hp per in³/min) |
| Brass (free-cutting) | 750 | 0.25 | 1 334 | 0.022 kW per cm³/min (0.49 hp per in³/min) |
| Mild steel (1018) | 1 800 | 0.25 | 3 201 | 0.053 kW per cm³/min (1.17 hp per in³/min) |
| Alloy steel (4140) | 2 100 | 0.25 | 3 734 | 0.062 kW per cm³/min (1.37 hp per in³/min) |
| Stainless (304) | 2 400 | 0.21 | 3 892 | 0.065 kW per cm³/min (1.43 hp per in³/min) |
| Titanium (Ti-6Al-4V) | 1 400 | 0.23 | 2 378 | 0.040 kW per cm³/min (0.87 hp per in³/min) |
| Gray cast iron | 1 100 | 0.28 | 2 096 | 0.035 kW per cm³/min (0.77 hp per in³/min) |
Values are typical (they vary with hardness, the tool and the maker); machining data from the tool maker or a handbook is better for a specific material.
Cutting power from the removal rate
Pc = Q × kc ÷ 60 000
Worked example: slotting aluminum with a 12 mm tool
- The cut: 6 mm deep, 12 mm wide, at 7 958 rpm and 0.08 mm per tooth (Vf = 1 910 mm/min). The removal rate is 6 × 12 × 1 910 ÷ 1 000 = 138 cm³/min.
- The average chip thickness for a full slot is fz × (1 − cos φ) ÷ φ with φ = π, which is 0.051 mm. So kc = 700 × 0.051−0.25 = 1 474 N/mm².
- The cutting power is Pc = 138 × 1 474 ÷ 60 000 = 3.4 kW. With a drive efficiency of 0.8 the spindle motor must supply 4.2 kW.
- The cutting force along the rim is Pc ÷ Vc = 3 377 W ÷ 5 m/s = 675 N (about 69 kg-force) on the tool.
Spindle power, torque and the speed range
T = 9 549 × P ÷ n
A spindle makes full power only above a certain speed; below it the torque is the limit, so a large cutter at a low RPM (a face mill, a big drill) is often torque-limited, and a small cutter at a high RPM is power-limited.
Worked example: power and torque limits
- The cut above needs 3.4 kW at 7 958 rpm, which is a torque of 9 549 × 3.4 ÷ 7 958 = 4.1 N·m.
- On a 2.2 kW spindle (efficiency 0.8) the available cutting power is 1.8 kW: the most that can be removed in this material is 1.8 × 60 000 ÷ 1 474 = 72 cm³/min, only 52 % of the 138 cm³/min asked. Reduce the depth or the feed.
- A spindle with a torque limit of 15 N·m gives 1.6 kW at 1 000 rpm but only 0.8 kW at 500 rpm: large tools at low speed are torque-limited.
- Speed range matters: a gearbox or a belt step multiplies torque at low speeds.
Why thin chips cost more
Worked example: mild steel at two chip thicknesses
- At h = 0.02 mm, kc = 1 800 × 0.02−0.25 = 4 786 N/mm²; at h = 0.10 mm it is 3 201 N/mm². The thin chip needs 50 % more energy per cubic millimeter.
- A thin chip is less efficient because the edge’s radius is a larger share of it: the tool pushes and rubs as well as cuts.
- So the smallest possible chip is not the gentlest: a chip load that is too small raises the force per unit of metal removed, the heat and the wear (see Chip Load, Chip Thinning and Engagement).
Tool deflection
A cutter is a cantilever, so it bends under the cutting force. The tip deflection is δ = F L³ ÷ (3 E I), with I = π d⁴ ÷ 64 (the actual value for a fluted tool is about 70 to 80 % of that of a solid shank). Carbide has E ≈ 600 GPa, steel 210 GPa. The deflection grows with the cube of the stick-out and falls with the fourth power of the diameter.
Worked example: a 6 mm carbide cutter
- With 100 N of side force and 25 mm of stick-out the tip deflects 0.0136 mm. At 50 mm (twice the length) it deflects 0.1092 mm, 8 times as much.
- A 12 mm cutter (twice the diameter) at 25 mm deflects only 0.0009 mm: 16 times less.
- To hold a 0.01 mm tolerance on a wall with this tool, the side force must stay below 73 N: use a finishing pass with a small stepover. A tool also bends away from the wall and leaves a taper.
- Reduce the stick-out, use the largest tool that fits, and take a spring pass (repeat the finishing pass) to remove the deflection error.
Can the machine take the cut?
Worked example: mild steel on a 750 W hobby spindle
- A 0.75 kW spindle with an efficiency of 0.7 delivers about 0.53 kW at the cutter. For mild steel at h = 0.02 mm (kc = 4 786 N/mm²), the largest removal rate is 0.53 × 60 000 ÷ 4 786 = 6.6 cm³/min.
- A 6 mm four-flute tool at 60 m/min (3 183 rpm) and 0.03 mm per tooth feeds at 382 mm/min; with 3 mm depth and 3 mm stepover it removes 3.4 cm³/min and needs 0.28 kW: within the limit.
- Drilling is a power check too: a 12 mm drill at 25 m/min (663 rpm) and 0.15 mm/rev removes 11 cm³/min and needs about 0.7 kW (see Drilling, Reaming, Tapping and Boring).
- If the calculated power exceeds the machine’s, the cut stalls the spindle, the machine chatters, or the controller reports a following error: reduce the depth first (it lowers the power in proportion and does not change the chip thickness).
Troubleshooting
| Symptom | Likely cause | Try |
|---|---|---|
| Spindle slows or stalls in the cut | Cut exceeds the available power or torque | Reduce the depth of cut or the feed; check the power at the actual speed |
| Tool deflects and leaves a taper | Long stick-out or a small tool | Shorten the tool, take a spring pass, use a stiffer tool |
| Machine skips steps or shows following errors | Force exceeds the drive’s capability | Reduce the feed rate or depth; check the acceleration |
| Light cut takes surprising effort | Thin chip: high specific force | Raise the chip load (apply chip thinning) |
| Spindle motor gets hot | Long periods near its limit | Lower the load to 70 % of the rating for continuous work |
| Tool breaks on entry | Overload at the first engagement | Ramp or arc into the cut and reduce the entry feed |
Related
- Speeds and Feeds: the overview of this guide.
- Chip Load, Chip Thinning and Engagement: chip thickness, which sets the force.
- Tool Life, Wear and Cutting Economics: how power and speed shorten tool life.
- Beam Bending and Deflection (the cantilever formula) and Sizing a Motion Axis.
FAQ
How do I calculate the power needed for a cut?
P in kW = Q (cm³/min) × k_c (N/mm²) ÷ 60 000, with the specific cutting force at the chip thickness of the cut; divide by about 0.8 for the motor.
Why is the spindle power rating not the whole story?
The power is available only above a base speed; below it the torque limit applies, so a large tool at low RPM is torque-limited.
Why does a light cut need so much force?
The specific cutting force rises as the chip gets thinner, and the edge radius becomes a larger share of the chip.
How do I reduce tool deflection?
Shorter stick-out (deflection ∝ L³), a larger diameter (∝ 1/d⁴), lighter radial cuts and a spring pass.
References
- Sandvik Coromant, Modern Metal Cutting (specific cutting force k_c1 and m_c, power calculation).
- Oberg, E., et al., Machinery’s Handbook (power consumption in cutting, unit power).
- Kronenberg, M., Machining Science and Application, Pergamon (specific cutting energy).
Last reviewed: October 2026. Specific cutting forces vary with material hardness and tool: use the maker’s data for exact values.