Runout Effect on Chip Load
How toolholder runout redistributes chip load between flutes — and why one flute dies first.
At 0.05 mm chip load, 0.02 mm TIR overloads one flute by 20% while another rubs — tool life falls to that worst flute. The smaller the cutter and chip load, the more runout dominates: micro-tools want TIR under 5 µm, not the 'good enough' 0.02 mm.
Formula
Note: Condition-monitoring guidance, not a substitute for the machine maker's limits or a qualified vibration analyst on safety-critical equipment.
How toolholder runout redistributes chip load between flutes — and why one flute dies first. A free cnc machining: speeds, feeds & tool wear tool — no sign-up, no upload, instant results in your browser.
About Runout Effect on Chip Load
Runout Effect on Chip Load computes the governing relationship f_max ≈ fz + TIR/2 · f_min ≈ fz − TIR/2 live as you type. At 0.05 mm chip load, 0.02 mm TIR overloads one flute by 20% while another rubs — tool life falls to that worst flute. The smaller the cutter and chip load, the more runout dominates: micro-tools want TIR under 5 µm, not the 'good enough' 0.02 mm. Defaults are pre-filled with realistic values for this exact scenario, and the worked example substitutes your numbers step by step so the math is never a black box.
How to use Runout Effect on Chip Load
- 1Enter your values — Programmed chip load, Measured runout (TIR), Flutes (sensible defaults are pre-filled).
- 2Read the live results: Heaviest flute load, Lightest flute load, Overload vs programmed.
- 3Check the "with your numbers" line to see f_max ≈ fz + TIR/2 · f_min ≈ fz − TIR/2 substituted step by step.
- 4Adjust inputs until the scenario matches yours, then copy or share the result.
Why use Runout Effect on Chip Load?
- ✓Instant, free and private — every calculation runs client-side in your browser; nothing is uploaded
- ✓Built on the stated formula f_max ≈ fz + TIR/2 · f_min ≈ fz − TIR/2 with authoritative sources cited on the page (ISO 20816-1 (ex 10816) — Mechanical vibration evaluation; ISO 21940-11 — Rotor balancing; ISO 3685 — Tool-life testing with single-point turning tools)
- ✓At 0.05 mm chip load, 0.02 mm TIR overloads one flute by 20% while another rubs — tool life falls to that worst flute.
- ✓SI ⇄ Imperial toggle converts your inputs in place, so you can work in the units your drawings use
Frequently asked questions
What formula does the runout effect on chip load use?+
It evaluates f_max ≈ fz + TIR/2 · f_min ≈ fz − TIR/2, exactly as published. Sources: ISO 20816-1 (ex 10816) — Mechanical vibration evaluation; ISO 21940-11 — Rotor balancing; ISO 3685 — Tool-life testing with single-point turning tools. The substituted worked example on the page lets you verify every step against the textbook.
How should I read the result — and how far can I trust it?+
At 0.05 mm chip load, 0.02 mm TIR overloads one flute by 20% while another rubs — tool life falls to that worst flute. Condition-monitoring guidance, not a substitute for the machine maker's limits or a qualified vibration analyst on safety-critical equipment.
When is this calculator the right tool for the job?+
How toolholder runout redistributes chip load between flutes. The smaller the cutter and chip load, the more runout dominates: micro-tools want TIR under 5 µm, not the 'good enough' 0.02 mm. For neighbouring scenarios, the related tools below cover the same engine with different presets.
Does it support both metric and imperial units?+
Yes — the SI ⇄ Imperial toggle converts the values already in the fields, preserving the physical quantity, so you can flip mid-calculation without re-entering anything.
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