Wire bundle diameter calculator
The approximate diameter of a wire bundle from the count and outside diameter of each wire, for choosing a backshell entry, sleeving or heat shrink.
Bundle diameter
10.2mm
- Wire bundle alone
- 10.2 mm
- Average wire diameter
- 1.697 mm
- Total wires
- 24
- Glenair multiplication factor
- 6
- Wires as share of the bundle cross-section
- 72.5 %
Wires in the bundle
| Group | Count | OD (mm) | OD (in) | Count × OD (mm) |
|---|---|---|---|---|
| 1 | 20 | 1.473 | 0.05799 | 29.46 |
| 2 | 4 | 2.819 | 0.111 | 11.28 |
Method
How this is calculated
d_avg = Σ (n_i × d_i) / Σ n_i D = d_avg × F(n) F from the Glenair factor table, interpolated D_out = D + 0.025 in (braid) + 2 × jacket wall
- Factors
- Glenair multiplication factors for 1 to 300 wires: for example 2.0 for 2 wires, 4.0 for 10, 5.6 for 20, 12.2 for 100. Counts between listed values are interpolated linearly (our choice; the Glenair table lists the steps only). Beyond 300 the last step is extended and flagged.
- Mixed sizes
- Different diameters are reduced to a count-weighted average diameter before applying the factor. This is the published method; it is approximate when one or two large wires sit among many small ones.
- Inputs
- Use the outside diameter over the insulation from the wire datasheet. Presets use NEC Chapter 9 Table 5 diameters for THHN: 0.111 in (14 AWG) and 0.130 in (12 AWG).
FAQ
Questions
Multiply the wire outside diameter by a bundle factor for the count. Glenair's table gives 5.6 for 20 wires, so 20 wires of 0.058 in come to 0.058 × 5.6 = 0.32 in (8.2 mm).
Average the diameters weighted by count, then apply the factor for the total count. Glenair's example: 20 wires at 0.058 in and 4 at 0.111 in average 0.067 in, and 24 wires take a factor of 6.0, giving 0.40 in.
Glenair adds 0.025 in (0.64 mm) for an overall braid shield, plus twice the jacket wall thickness. A 0.5 mm jacket adds 1 mm to the diameter.
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