Making a Hexbeam Air Dielectric Coaxial Centerpost Feedline from Scratch

Most everybody I know has an NA4RR or K4KIO centerpost - myself included.   I have used a K4KIO centerpost in the past and had been happy with it.  I then decided to build center-post out of PVC pipe with the coax connections inside to prevent water from making contact with the foam core insulation and ruining the dielectric insulation between center conductor and shield.   I used it for about 6 months and decided to make a change to NA4RR's version and this has been on my home station for the better part of 2 years.  I did some research and captured the following equations to construct the equivalent of a 50 ohm (or as close to it) coaxial feedline.

For round coax, make the inside diameter of the outside conductor 2.302 times larger than the diameter of the inside conductor.

If the shield is square, and the inner conductor is still round, make the inside length of one side of the shield 2.134 times larger than the diameter of the inside conductor.

The derivation is as follows:

A coaxial transmission line has some inductance per unit length L$L$, and some capacitance per unit length C$C$. The conductors also have some (very low) resistance R$R$, and the dielectric has some (very low) conductance G$G$. Though it's important to note that these properties are distributed throughout the entire length of the transmission line, lumping them all together we can model a transmission line as:

The characteristic impedance is then:

The characteristic impedance is then:

Z0=R+jωLG+jωC−−−−−−−−√$$Z_0=\sqrt{\frac{R+j\omega L}{G+j\omega C}}$$

where:

• R$R$ is the resistance per unit length, considering the two conductors to be in series,
• L$L$ is the inductance per unit length,
• G$G$ is the conductance of the dielectric per unit length,
• C$C$ is the capacitance per unit length,
• j$j$ is the imaginary unit, and
• ω$\omega$ is the angular frequency.

If we assume a good conductor and dielectric, then R$R$ and G$G$ are negligible¹. The equation then simplifies to:

Z0≈LC−−√$$Z_0\approx \sqrt{\frac{L}{C}}$$

For round coaxial transmission lines, we can calculate C$C$ and L$L$ as:

C=2πϵ0ϵrln(D/d)

L=μ0μr2πln(D/d)$$L=\frac{{\mu }_0{\mu }_r}{2\pi }\ln (D/d)$$

where:

• d$d$ is the diameter of the inner conductor,
• D$D$ is the inside diameter of the outer conductor,
• ϵ0${ϵ}_0$ is the permittivity of free space, approximately 8.854⋅10−12F⋅m−1$8.854\cdot {10}^{-12}F\cdot m^{-1}$
• μ0${\mu }_0$ is the permeability of free space, approximately 1.256⋅10−6H⋅m−1$1.256\cdot {10}^{-6}H\cdot m^{-1}$
• ϵr${ϵ}_r$ is the relative permittivity of the dielectric, and
• μr${\mu }_r$ is the relative permeability of the dielectric.

Putting these together and simplifying we get:

For an air dielectric, ϵr${ϵ}_r$ and μr${\mu }_r$ are so close to 1 than you can further simplify:

Z0≈59.96ln(D/d)$$Z_0\approx 59.96\ln (D/d)$$

You can also see that only the ratio of the diameters of the conductors are relevant, so it doesn't matter what units of length are used, as long as they are the same.

We can solve for D/d$D/d$, and set Z0=50Ω$Z_0=50\mathrm{\Omega }$:

56.40ln(D/d)ln(D/d)D/dD/dD/d≈Z0≈Z059.96≈eZ0/59.96≈e50/59.96≈2.302

So in my case - 0.625" diameter and 1.5" diameter aluminum tubing are available from DX Engineering.  If I use average wall diameter for the outside tube and the OD of the inside tube, the ratio of D/d in this case is 2.3  which is really close. 

My plan is to put fiberglass sleeves around the bolts to insulate it from the outer shield along with nylon washers on the inside and outside.  Here is a scaled picture below