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Self GMD of 4-strand diamond conductor

Short answer

For the diamond four-strand arrangement, self-GMD for inductance (using r′ = 0.7788r) is 1.692 r, and for capacitance (using r) is 1.801 r.

Find the self GMD for inductance and capacitance calculations for the following conductor configuration, given that the radius of each sub-conductor is r. [Figure: four identical touching sub-conductors of radius r arranged in a diamond (one top, two side by side in the middle, one bottom).]

The self-GMD of a stranded conductor of nn strands is the n2n^2-th root of the product of all distances between strands, where each strand's distance to itself is r′=0.7788rr' = 0.7788r (inductance) or rr (capacitance).

          (1)
       (2)   (3)
          (4)
  2-3 = 2r, 1-2 = 1-3 = 2-4 = 3-4 = 2r
  1-4 = 2(sqrt3)r

Distances

Strands 2 and 3 touch: D23=2rD_{23} = 2r. Strands 1 and 4 each touch both 2 and 3, so 1, 2, 3 (and 2, 3, 4) form equilateral triangles of side 2r2r. Then D14=2×3r=23rD_{14} = 2 \times \sqrt{3}r = 2\sqrt{3}r.

StrandDistances to othersProduct
12r,2r,23r2r, 2r, 2\sqrt3 r83 r38\sqrt3\,r^3
22r,2r,2r2r, 2r, 2r8r38r^3
32r,2r,2r2r, 2r, 2r8r38r^3
42r,2r,23r2r, 2r, 2\sqrt3 r83 r38\sqrt3\,r^3

Product of mutual distances =(2r)10(23r)2=12288 r12= (2r)^{10}(2\sqrt{3}r)^2 = 12288\,r^{12}.

For inductance

DsL=[(0.7788r)4×12288 r12]1/16=1.6922 r\begin{aligned} D_{sL} &= \left[(0.7788r)^4 \times 12288\,r^{12}\right]^{1/16} \\ &= 1.6922\,r \end{aligned}

For capacitance

DsC=[r4×12288 r12]1/16=1.8013 rD_{sC} = \left[r^4 \times 12288\,r^{12}\right]^{1/16} = 1.8013\,r

Answer: self-GMD for inductance =1.692r= 1.692r; for capacitance =1.801r= 1.801r.

✓ Answer checked by Claude Opus 5.5 (high effort)

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