Theater Calculators

Amplification & power

Speaker Wire Gauge (AWG) Distance Calculator

Long runs eat damping factor. A common budget is 5% of the nominal impedance for the round trip. 10 m of run is 20 m of copper.

Inputs

Units
m
Ω

Results

Round-trip length
16.0m
5% resistance budget
0.400Ω
Needed copper area
0.69mm²
Recommended AWG
18

The Result is indicative and educational. It does not constitute a design, a technical opinion or medical advice and does not replace calculations or verification by a person with the appropriate qualifications and authorisations. Before purchase, installation or use in practice – in particular in electrical installations – the Result should be verified by a qualified specialist. Details: Terms of Service, point 4.

Example products

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Calculation method

R_max = 0.05 Z. Needed copper area = ρ · 2L / R_max, then the thinnest AWG (highest number) whose area meets it.

Exact formula

R_max = 0.05·Z. A = 0.0172 · 2L / R_max (mm², m). Limits: 20 °C copper, AWG table 18–10 (thicker still reports 10), no in-wall fire rating, no 70 V distributed lines.

Why the cap is 5% of speaker impedance

The tool budgets cable resistance at 5% of the nominal load (R_max = 0.05 Z). That keeps insertion loss around 0.4 dB and damping from collapsing. A longer run or a 4 Ω speaker needs thicker copper than the same watts at 8 Ω.

10 AWG is the thickest this table reports

AWG 18 through 10 are in the table. Anything that still wants thicker copper is reported as 10 AWG — that is a cap, not a claim that 10 AWG is always enough for a 30 m 2 Ω run. Do not use this for 70 V distributed lines or in-wall fire ratings.

Frequently asked questions

Why 5%?

It keeps the loss near 0.4 dB at the nominal impedance, and the response ripple small where the speaker’s impedance swings. Tighter budgets need thicker wire.

Metric mm²?

The tool reports both area and AWG. 2.5 mm² is near 13 AWG; 4 mm² near 11 AWG.

Bi-wire?

Size each run on its own length: the bass run still carries the woofer current through one cable. Only two runs joined in parallel on the same terminals halve the resistance.

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