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Inverse Distance Fields for multi-segment / multi-bay antennas?

I'm interested in learning about Inverse Distance Fields for various multi-segmented / multi-bay transmitting antenna setups.

I already know that...

a 1/4-wave monopole over a ground plane produces 305.78 mV/m
1/2-wave monopole = 381 mV/m
5/8-wave monopole = 440 mV/m
KSTP's almost-Franklin (179.2+179.2°) = 511.78 mV/m

What about a true 180+180 Franklin? Or Franklins with more segments?

Sometime ago I stumbled across a paper R Fry had posted somewhere. It was kinda low resolution, but it looked like a 2-segment Franklin was about 530 mV/m, up to a 10-segment being about 1100 mV/m or so. R Fry, if you read this, do you by any chance have a link to that?

Also same goes for FM and TV multi-bay antennas used by broadcast stations. How would I know what the equivalent inverse distance field in the horizontal plane is in mV/m @ 1 km for 1 kW for those antennas? And how do they differ from Franklins and segmented antennas?
 
I have posted a few graphics about Franklins (I don't recall where at this point), but never for more than two sections.

As for multibay FM antennas, if their centerline bay spacing along the tower is one wavelength, then the r.m.s. power gain of their elevation pattern in the horizontal plane is roughly equal to the number of bays, if linearly polarized (just horizontal or just vertical). If circularly polarized then the gain is half the above values, per polarization.

The exact, published gain value depends on the way their manufacturers rate them, and can be found on their web sites.

The (r.m.s.) inverse distance field they produce in free space can be deduced, roughly, by recalling that 1 kW of radiated power produces an E-field of about 221.7 mV/m at a distance of 1 kilometer. That field changes by the square root of the change in total radiated power.

So an FM station radiating 100 kW ERP would produce a field at 1 km = SQRT(100) x 221.7 mV/m = 2,217 mV/m.

For free space paths such fields decay at a 1/r rate.
 
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