Conductor Bundles: Reduce Impedance Per Meter

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http://www.kabculus.com/capacitance-and-inductance-matrices/node7.html
this short page describes conductor bundles, which are power transmission lines hung parallel to each other. i think the page is trying to explain why, when they are hung that way that they reduce impedance per meter, but i don't get it. i understand all the variables they are using except what it is they are calculating.
r is an approximation for the radius of the conductors' cross section (i think)
R is how far away the cables are hung from the center of the bundle (i think)
e0 is the permeability of free space

potential? that means voltage right? why are they using the greek letter for angle? why isn't it 120v rms?
and what is a2 and a4? equivalent radius?
 
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You mention 120V AC, but bundled conductor power lines are used at 345 KV or higher. Shift your horizon a whole lot.

If we used a single but thicker conductor, the volts/meter gradient would be higher. That increases corona losses.

You also should think of the skin effect. AC current tends to run on the outside of the conductor. i.e. the skin. So the ideal profile would be a hollow cylinder, not a solid wire. But 3 conductors in a triangle shape, approximate a hollow cylinder. Even more conductors, the better the approximation.
 
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chopficaro said:
i think the page is trying to explain why, when they are hung that way that they reduce impedance per meter, but i don't get it.
The same technique is used for radio antennas where they are called cage lines, or cage dipoles.

The electrostatic radius is greater for a cage than for a single wire. That reduces corona discharge, but it also increases the capacitance per unit length to other bundles and to ground.

Because the currents are shared by the conductors of the cage, those currents are further apart than they would be on a single conductor, so there is less magnetic coupling of the currents. That reduces the self inductance per unit length of the cage.

The characteristic impedance of a transmission line, Zo = √ ( L / C ). When using a cage, L is reduced, and C is increased, so the impedance of a line, or a dipole over ground, will be lower than that of a single wire.

Another advantage of a cage for dipole elements is that the radius can be reduced in the element as it approaches a conductive tower or supporting structure. That reduces impedance changes along the element, so it reduces losses and raises the Q of the dipole.
 
According to Electrostatics Theory the surface integral around an electric charge Q of the electric field flux density D it is equal to this charge [Gauss' law].D=ε.E where E it is the electric field [intensity].
So ε.E*2.п.r.length=Q or E(r)=Q/(2.п.ε)/r/length
The electric potential[aka voltage] V=ʃE.dr=Q/(2.п.ε).ln(r2/r1) per unit length.
The Green function it seems to be the potential per unit length and unit charge.
For bundled parallel wires see the link-for instance:
http://users.ece.utexas.edu/~baldick/classes/369/Lecture_6.ppt