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Self GMD of 5-strand conductor, 6 mm strands

Short answer

For the five-strand configuration with strand radius 6 mm, the self-GMD for inductance is 11.47 mm (approximately 1.91r).

Determine the self GMD of the following conductor. Radius of each strand is 6 mm. [Figure: five identical touching strands, two in the top row resting on three in the bottom row.]

The self-GMD (GMR) of an nn-strand conductor is the n2n^2-th root of the product of all inter-strand distances, with the self-distance taken as r′=0.7788rr' = 0.7788r for inductance (the usual meaning of self-GMD).

        (4)   (5)
     (1)   (2)   (3)
  touching strands, centre spacing 2r

Data: r=6r = 6 mm, so 2r=122r = 12 mm, r′=0.7788×6=4.673r' = 0.7788 \times 6 = 4.673 mm.

Distances

Strands 4 and 5 sit in the grooves between 1-2 and 2-3, so the centres lie on a triangular grid of side 2r2r.

PairDistance
1-2, 2-3, 4-5, 1-4, 2-4, 2-5, 3-52r=122r = 12 mm
1-34r=244r = 24 mm
1-5, 3-423r=20.782\sqrt3 r = 20.78 mm

Check: 7 + 1 + 2 = 10 pairs =(52)= \binom{5}{2}.

Self-GMD

Ds=[(r′)5×((2r)7(4r)(23r)2)2]1/25=1.9114 r=1.9114×6=11.47 mm\begin{aligned} D_s &= \left[(r')^5 \times \left((2r)^7 (4r)(2\sqrt3 r)^2\right)^2\right]^{1/25} \\ &= 1.9114\,r = 1.9114 \times 6 = 11.47\ \text{mm} \end{aligned}

For capacitance (self-distance rr instead of r′r'): Ds=2.0094r=12.06D_s = 2.0094r = 12.06 mm.

Answer: self-GMD (inductance) =11.47= 11.47 mm ≈1.147\approx 1.147 cm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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