The equivalent radius of an antenna conductor is defined as:[1]

$r_e = \exp\left\{ {1\over L^2} \oint_\ell \oint_\ell \ln \vert \boldsymbol{x} - \boldsymbol{y} \vert \; dx \; dy \right\}$

where $\scriptstyle\ell$ denotes the conductor's circumference, $\scriptstyle{L}$ is the length of the circumference, $\scriptstyle{\boldsymbol{x}}$ and $\scriptstyle{\boldsymbol{y}}$ are vectors locating points along the circumference, and $\scriptstyle{dx}$ and $\scriptstyle{dy}$ are differentials segments along it. The equivalent radius allows the use of analytical formulas or computational or experimental data derived for antennas constructed from small conductors with uniform, circular cross-sections to be applied in the analysis of antennas constructed from small conductors with uniform, non-circular cross-sections. Here "small" means the largest dimension of the cross-section is much less than the wavelength $\scriptstyle{\lambda}$.

## Formulas

The following table lists equivalent radii for various conductor cross-sections derived assuming 1) all dimensions are much less than $\scriptstyle{\lambda}$, 2) for cross-sections composed of multiple conductors, the distances between conductors are much greater than any single conductor dimension. The formula for the flat, infinitely thin conductor is a closed-form solution. Formulas for the square and triangular cross-sections follow from numerical evaluation of the double integral.

Two circular conductors,
$\sqrt{rs}$
Two circular conductors,
$\exp\left\{ { r_1^2 \ln r_1 + r_2^2 \ln r_2 + 2r_1r_2 \ln S \over (r_1+r_2)^2} \right\}$
Equilateral triangle,
identical circular conductors
$\left( rs^2 \right)^{1/3}$
Square,
identical circular conductors
$\left( \sqrt{2}\,rs^3 \right)^{1/4}$
Pentagon,
identical circular conductors
$\left( 2.62\,rs^4 \right)^{1/5}$
Hexagon,
identical circular conductors
$\left( 6\,rs^5 \right)^{1/6}$
Approximate cylinder,
identical circular conductors,
conductors uniformly spaced,
$N \gg 6$
$\displaystyle{ r^{1/N} R^{1-1/N} }$
Flat conductor,
infinitely thin
$\displaystyle{ We^{-3/2} \approx 0.22\, W}$
Square conductor $\displaystyle{ 0.58\, W }$
Equilateral triangle conductor $\displaystyle{ 0.41\, W }$

## Derivation

The equivalent radius is derived by equating the average potential at the surface of a charged conductor of arbitrary cross-section with the electric potential on the surface of a charged cylinder.

Because a conductor's cross-section dimensions are small compared to the wavelength, the current distribution slowly varies along the conductor's length, charge is uniformly distributed along its circumference (owing to the skin effect) and the electric field is perpendicular to the surface. Furthermore, only the charge in a neighborhood around any point on the conductor significantly contributes to the electric potential at that point. Time dependence is ignored, as it may be incorporated by multiplying the current distribution by a time-varying sinusoid. These conditions imply that an electrostatic condition exists and that the geometry is, effectively, one of an infinitely long conductor with a constant surface charge density $\scriptstyle{Q}$ (charge per area), thereby reducing a three-dimensional problem to a two-dimensional one.

First, consider the potential at a fixed point $\scriptstyle{\boldsymbol{y}}$ on the circumference of the arbitrary cross-section. With the circumference divided into differential segments $\scriptstyle{dx}$, the charge distribution may be approximated by placing a vertical line charge within each segment, each with a linear charge density of $\scriptstyle{Q\, dx}$ (charge per length). It is well known that the potential of such a line charge is $\scriptstyle{-2k_e \, Q\, dx \ln \vert \boldsymbol{x} - \boldsymbol{y} \vert}$, where $\scriptstyle{k_e}$ is Coulomb’s constant. The potential at $\scriptstyle{\boldsymbol{y}}$ is the sum of the potentials for all the strips, which is

$V(\boldsymbol{y}) = -2k_e Q \oint_\ell \ln \vert \boldsymbol{x} - \boldsymbol{y} \vert \; dx .$

The average potential is then

$\bar{V} = {1 \over L} \oint_\ell V(\boldsymbol{y}) \; dy \, = -2k_e Q {1 \over L} \oint_\ell \oint_\ell \ln \vert \boldsymbol{x} - \boldsymbol{y} \vert \; dx \; dy .$

Now consider the case of cylinder with the same linear charge density as the conductor of arbitrary cross-section. It is also well known that the potential at any point on its surface, which is also equal its average potential, is

$V_c = -2k_e QL \ln \left( r_e \right) .$

Equating $\scriptstyle{\bar{V}}$ and $\scriptstyle{V_c}$ yields

$\ln\left( r_e \right) = {1\over L^2} \oint_\ell \oint_\ell \ln \vert \boldsymbol{x} - \boldsymbol{y} \vert \; dx \; dy .$

Exponentiation of both side leads to the formula for the equivalent radius.

The formula for the equivalent radius provides consistent results. If the conductor cross-section dimensions are scaled by a factor $\scriptstyle{\alpha}$, the equivalent radius is scaled by $\scriptstyle{|\alpha |}$. Also, the equivalent radius of a cylindrical conductor is equal to the radius of the conductor.

## References

1. ^ E.A. Wolff, Antenna Analysis, Chapter 3, John Wiley & Sons, New York, NY, Second Edition, 1966.