Ball-nose scallop & effective diameter
A ball of radius R stepped over by ae leaves scallops h = R − √(R² − (ae/2)²) high, about ae²/(8R). A Ø10 mm ball at a 0.5 mm stepover leaves 6.3 µm. And at a shallow depth the ball cuts on a much smaller diameter than its own: a Ø6 mm ball 0.2 mm deep cuts on Ø2.15 mm, so 250 m/min there needs about 36 900 rpm, not the 13 300 the full diameter suggests.
Work out the scallop from a stepover or the stepover for a scallop, the effective diameter at your depth (and tilt, for a ball nose), and the spindle speed for your cutting speed, for ball-nose and bull-nose cutters.
- Scallop height h
- Scallop height, µm
- Effective diameter De
How it’s calculated
Scallop h = R − √(R² − (ae/2)²), so ae = 2·√(h·(2R − h)); a bull nose on a flat floor leaves none while ae ≤ D − 2r, then the same on its corner radius r. Effective diameter De = 2·√(ap·(D − ap)), tilted D·sin(β + arccos((D − 2ap)/D)); bull nose D − 2r + 2·√(r² − (r − ap)²). n = 1000·vc ÷ (π·De) in mm, 12·vc ÷ (π·De) in inches.
Stepover and scallop
| ae/D | ae (mm) | h (mm) | h (µm) |
|---|---|---|---|
| 2 % | 0.200 | 0.0010 | 1.0 |
| 5 % | 0.500 | 0.0063 | 6.3 |
| 10 % | 1.000 | 0.0251 | 25.1 |
| 15 % | 1.500 | 0.0566 | 56.6 |
| 20 % | 2.000 | 0.1010 | 101.0 |
| 30 % | 3.000 | 0.2303 | 230.3 |
Effective diameter by depth
| ap/D | ap (mm) | De (mm) |
|---|---|---|
| 1 % | 0.100 | 1.990 |
| 2 % | 0.200 | 2.800 |
| 5 % | 0.500 | 4.359 |
| 10 % | 1.000 | 6.000 |
| 20 % | 2.000 | 8.000 |
| 30 % | 3.000 | 9.165 |
| 50 % | 5.000 | 10.000 |
How is scallop height calculated?
Seen across the passes, each pass of a ball nose is an arc of radius R, and neighbouring arcs ae apart cross half-way between them. The cusp left standing is the height of that crossing above the bottom of the arc: h = R − √(R² − (ae/2)²). For a stepover small against the ball, that is close to ae²/(8R), so halving the stepover quarters the cusp and doubles the passes. Turned the other way, the stepover for a cusp h is ae = 2·√(h·(2R − h)).
A bull nose finishing a flat floor cuts flat with its bottom, D − 2r across, so it leaves no cusp until the stepover is wider than that; past it, the corner radii leave the same circular cusp on the gap. On walls and slopes, where only the corner cuts, the corner radius r acts as a ball of radius r.
What is the effective cutting diameter?
At a depth ap a ball nose only cuts on the part of the ball below the surface, whose widest diameter is De = 2·√(ap·(D − ap)). The spindle speed for a cutting speed has to use that diameter: n = 1000·vc ÷ (π·De). Tilting the tool by β moves the cut off the tip, where the cutting speed is zero, and out to D·sin(β + arccos((D − 2ap)/D)), up to the full diameter.