Inputs
Cutting speed vc, cutter diameter D, number of teeth z, feed per tooth fz, depth of cut ap and width of cut ae; for the power part additionally the material constants kc1.1 and mc and the machine efficiency.
Manufacturing · Machining
Milling cutting data and power demand
Calculate spindle speed, table feed and material removal rate from cutting speed, cutter diameter, number of teeth and feed per tooth. The calculator additionally gives the power demand via the Kienzle relation: the specific cutting force is evaluated at the actual mean chip thickness of the arc of engagement, not at the feed per tooth.
Mean-value model. The material constants kc1.1 and mc come from your own machining tables or manufacturer data; no material table is embedded here. Tool wear, peak loads within one tooth engagement, tool deflection and vibration are not assessed.
Pckc · Q
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Cutting data and power demand
Set the tool and the cut.
Inputs and method
Milling cutting data and power demand
Calculate spindle speed, feed and material removal rate of a milling cut, and the power it demands from the machine.
Calculation
The kinematics follow from n = vc·1000/(π·D), vf = fz·z·n and Q = ap·ae·vf. For the power, the Kienzle specific cutting force kc = kc1.1·hm^−mc is evaluated at the mean chip thickness of the arc of engagement: from cos φs = 1 − 2·ae/D follows hm = fz·(2·ae/D)/φs. Because kc is a specific energy, Pc = kc·Q and Pa = Pc/η.
Example
vc = 200 m/min on a 20 mm cutter with 4 teeth gives n ≈ 3,183 min⁻¹; with fz = 0.1 mm this is vf ≈ 1,273 mm/min. At ap = 10 mm and ae = 5 mm that is Q ≈ 63.7 cm³/min, hm ≈ 0.048 mm and, with kc1.1 = 1,500 N/mm² and mc = 0.25, about 3.4 kW cutting and 4.3 kW drive power.
Sources and limits: Machining quantities per DIN 6580/6584; specific cutting force per the Kienzle relation kc = kc1.1·h^−mc. The material constants kc1.1 and mc are table values from machining literature or manufacturer data and are deliberately not embedded here. Mean-value model without wear, peak loads, tool deflection or vibration.
Milling cutting data and power demand in detail
Calculate spindle speed, feed and material removal rate of a milling cut, and the power it demands from the machine.
What is cutting data?
Cutting data describe how fast tool and workpiece move relative to each other. The cutting speed vc belongs to the material-tool pairing; the spindle speed only follows from it via the diameter. The feed per tooth sets how thick each individual chip becomes, and together with depth and width of cut it determines how much material is removed per minute – and how much power that costs.
Formula and variables
Pc = kc1.1 · hm^−mc · Q
n = vc · 1000 / (π · D)vf = fz · z · nQ = ap · ae · vfcos φs = 1 − 2·ae/D, hm = fz · (2·ae/D) / φsPa = Pc / η
| Symbol / input | Meaning |
|---|---|
| vc, n, vf | Cutting speed and the spindle speed and table feed that follow from it. |
| fz, z, D | Feed per tooth, number of teeth and cutter diameter. |
| ap, ae | Axial depth and radial width of cut; ae = D is full slotting. |
| hm, φs | Mean chip thickness and the arc of engagement it follows from. |
| kc1.1, mc | Material constants of the Kienzle relation, from machining tables. |
Choose the inputs correctly
Cutting speed vc, cutter diameter D, number of teeth z, feed per tooth fz, depth of cut ap and width of cut ae; for the power part additionally the material constants kc1.1 and mc and the machine efficiency.
How to use the calculator
Take the cutting speed and feed per tooth from the tool manufacturer's recommendation for your material, enter cutter diameter and number of teeth, and define the cut through depth and width of cut. Spindle speed and table feed can be set directly on the machine. For the power part add kc1.1 and mc from your machining table and compare the drive power against the machine's rating plate.
Worked example
vc = 200 m/min on a 20 mm cutter with 4 teeth gives n ≈ 3,183 min⁻¹; with fz = 0.1 mm this is vf ≈ 1,273 mm/min. At ap = 10 mm and ae = 5 mm that is Q ≈ 63.7 cm³/min, hm ≈ 0.048 mm and, with kc1.1 = 1,500 N/mm² and mc = 0.25, about 3.4 kW cutting and 4.3 kW drive power.
Which value decides what?
Spindle speed and table feed are the two values set on the machine. The removal rate says how productive the cut is, the drive power whether the machine can deliver it at all. The mean chip thickness is the value that governs tool life: if it is too low the edge rubs instead of cutting and wears faster, even though the calculated load looks small.
Cutting speed in m/min, all lengths in millimetres, spindle speed in min⁻¹, table feed in mm/min, removal rate in cm³/min, specific cutting force in N/mm², power in kW.
Typical applications
Setting up milling programs and manual machines, checking whether an existing machine can deliver a planned cut, comparing roughing strategies via the removal rate, and estimating the required clamping force from the mean cutting force.
Assumptions, limits and common mistakes
Machining quantities per DIN 6580/6584; specific cutting force per the Kienzle relation kc = kc1.1·h^−mc. The material constants kc1.1 and mc are table values from machining literature or manufacturer data and are deliberately not embedded here. Mean-value model without wear, peak loads, tool deflection or vibration.
Common mistake: Do not confuse cutting speed with spindle speed: vc is a material recommendation, n only follows from it via the diameter. Do not confuse feed per tooth with feed per revolution – the factor between them is the number of teeth. And do not estimate power taking the mean chip thickness as equal to the feed per tooth: at small widths of cut hm is considerably smaller than fz, which raises the specific cutting force.
Frequently asked questions
Why is the mean chip thickness smaller than the feed per tooth?
Because in milling the edge enters the material along an arc. It takes the full thickness fz only in the middle of the arc of engagement; at the start and end the chip is thinner. Averaged over the arc, hm = fz·(2·ae/D)/φs. For full slotting this gives the familiar 2·fz/π ≈ 0.64·fz, and considerably less at small widths of cut.
Why does the specific cutting force rise for thin chips?
Because besides the actual shearing, friction and deformation at the cutting edge also cost work, and their share relative to the volume removed grows as chips get thinner. That is exactly what the exponent mc describes: kc = kc1.1·hm^−mc. Very small feeds per tooth are therefore energetically unfavourable and additionally load the edge by rubbing.
Where do I get kc1.1 and mc?
From machining tables in the technical literature or from the tool manufacturer's data, matched to the specific material. They are deliberately not embedded here as a table, because published values scatter between sources and material conditions and are not reproduced without a verified source.
Is the calculated drive power the maximum load?
No, it is a mean value. Within one tooth engagement the load varies considerably, and the fewer teeth are engaged at once, the more pronounced that variation is. At low speeds it is also often the available spindle torque rather than the power that is the real limit – which is why it is reported here as well.
What is the best way to raise the removal rate?
All three factors ap, ae and vf enter linearly, but they act differently on the tool. More depth of cut at a small width of cut spreads the wear over the length of the edge and keeps hm small; more width of cut raises force and deflection more sharply. The power calculation shows where the machine reaches its limit first.