How to use this calculator
- Set the diameters. Bar OD to 0 for cutoff; OD to groove root for grooving.
- Program speed and feed. Constant surface speed and radial feed per rev from the insert maker’s parting data.
- Enter the real clamp. The G50/G92 limit your chuck and stick-out actually allow — often well under the spindle maximum.
- Read the phases. CSS time, the clamp diameter, the clamped tail, and what the same cut costs at constant RPM.
How it works
Under G96 the spindle holds surface speed by spinning faster as the tool feeds toward center, so the plunge time is the integral of dr / (f·n(r)) — quadratic in radius until the RPM clamp, linear after:
t = π(R² − r₁²)/(f·V) + (r₁ − r_end)/(f·n_max) · r₁ = V/(2π·n_max)
The per-revolution arithmetic underneath comes from the SFM to RPM converter; longitudinal-turning time and cost roll-ups live in the machining time calculator and machining cost calculator; and the threading cousin of this pass planning is the G76 infeed calculator.
Worked example
Verified against the live calculator
Parting a 2 in steel bar to center at 400 SFM,
0.002 in/rev, spindle clamped at 3,000 RPM:
clamp at Ø0.51 in · t = 18.4 s (CSS) + 2.5 s (clamped) ≈ 21 s
The spindle starts at 764 RPM on the OD and winds up to 3,000 by Ø0.51 in; the last half-inch runs clamped with falling surface speed. Fixed at 764 RPM the whole way, the same cut takes 39 seconds — G96 gives back 18 of them, every part, which across a 500-piece bar job is two and a half hours of spindle time.
Frequently asked questions
How do you calculate parting (cutoff) time on a lathe?
At constant surface speed the plunge time is π(R² − r²)/(f·V), plus a constant-RPM tail once the spindle hits its clamp at diameter V/(π·n_max). A 2 in steel bar at 400 SFM and 0.002 in/rev with a 3,000 RPM clamp parts in about 21 seconds of feed-on time.
Why does parting a bigger bar take so much longer?
The R² term: material removed grows with the square of radius, and at constant surface speed the time does too. Doubling the bar from 2 to 4 inches quadruples the plunge time — the linear-looking plunge hides a quadratic.
How much time does G96 constant surface speed save when parting?
The same 2 in cutoff at a fixed OD-safe 764 RPM takes about 39 seconds; under G96 with a 3,000 RPM clamp it takes 21. The spindle buys the difference by accelerating as the radius shrinks, keeping the insert at its rated speed instead of slowing below it.
What happens at the center of a cutoff?
RPM would need to go infinite to hold surface speed, so the spindle pegs at its G50/G92 clamp and surface speed falls toward zero at the very center — which is why the last bit of a part-off sounds different and why many programmers drop the feed near center.
Method & assumptions
- Exact closed-form integral of the constant-surface-speed plunge with an RPM clamp; no approximation beyond instant spindle response.
- Feed-on time for one plunge: approach, retracts, pecking and the stepped passes of wide grooves are excluded and additive.
- Constant feed per rev throughout — shops often reduce feed near center where clamped surface speed gets slow, adding to this figure.
- The comparison cut uses the OD-safe constant RPM (or the clamp, if lower) — the honest like-for-like baseline G96 replaces.