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Runout Effect on Chip Load

How toolholder runout redistributes chip load between flutes — and why one flute dies first.

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Heaviest flute load (mm/tooth)
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Lightest flute load (mm/tooth)
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Overload vs programmed (%)

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

f_max ≈ fz + TIR/2 · f_min ≈ fz − TIR/2
References: ISO 20816-1 (ex 10816) — Mechanical vibration evaluation; ISO 21940-11 — Rotor balancing; ISO 3685 — Tool-life testing with single-point turning tools

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

  1. 1Enter your values — Programmed chip load, Measured runout (TIR), Flutes (sensible defaults are pre-filled).
  2. 2Read the live results: Heaviest flute load, Lightest flute load, Overload vs programmed.
  3. 3Check the "with your numbers" line to see f_max ≈ fz + TIR/2 · f_min ≈ fz − TIR/2 substituted step by step.
  4. 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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