Mobile Cophased Dipoles

[Needle Bender]

@Needle Bender:...With this, here:

"In case the 7 feet of 75%VF or so 75Ω, times 2 = ~13.5' isn't long enough to reach the T, you can place identical lengths of 50Ω from each antenna mount to it's respective 75Ω attached to it's side of the T."

Are you saying that I should have 7' feet of RG59 running from each antenna to a T? I'm guessing that you used the "velocity factor" formula that you mentioned, but I don't quite see the math. In the way of coax, I have a Firestik Fire-Flex K-9A co-phase harness (each lead being 18') that I could easily shorten, if need be. I don't know what the velocity factor is, though, and the website doesn't say, either. The website DID mention that the coax was 72ohm, as opposed to 75ohm, if that makes a difference.

I appreciate the insight

An ELECTRICAL 1/4 wavelength of 75Ω coax acts as an impedance matching transformer so that if you give it a 50Ω load or impedance at one end, you should see about 100Ω at the other end.
If you add one to each of two 50Ω antennas, you'll end up with two 100Ω loads or, effectively, two 100Ω antennas.

If you parallel 2 identical resistances or impedances, you end up with one resistance/impedance of 1/2 the value of each.

Two 100Ω antennas in parallel = a single 50Ω load to your radio.

The velocity factor is the speed of radio wave propagation through the medium of dielectric.

Through air RF energy travels basically full speed (the speed of light) but through a solid medium (such as polyethylene foam) the RF energy is slowed (permittivity) and can only cover a smaller percentage of the normal distance in the same amount of time, hence a NINE FOOT 1/4 wave in air being only SIX FEET long inside of 66% VF coax.

PS. The 75Ω 1/4 waves should be at the T not at the antennas if you need to add a set of lengths of 50Ω coax to reach the T.

 

[rabbiporkchop]

I found it seems easier to use 270 degrees of cable as opposed to 90 degrees of cable for your phasing lines. I tried using a 90 degree phasing harness and I was unable to get my antennas to tune correctly but once I switched it out to 270 degrees per side everything just seemed to fall into place.

 

[Sarasota Slim]

An ELECTRICAL 1/4 wavelength of 75Ω coax acts as an impedance matching transformer so that if you give it a 50Ω load or impedance at one end, you should see about 100Ω at the other end.
If you add one to each of two 50Ω antennas, you'll end up with two 100Ω loads or, effectively, two 100Ω antennas.

If you parallel 2 identical resistances or impedances, you end up with one resistance/impedance of 1/2 the value of each.

Two 100Ω antennas in parallel = a single 50Ω load to your radio.

The velocity factor is the speed of radio wave propagation through the medium of dielectric.

Through air RF energy travels basically full speed (the speed of light) but through a solid medium (such as polyethylene foam) the RF energy is slowed (permittivity) and can only cover a smaller percentage of the normal distance in the same amount of time, hence a NINE FOOT 1/4 wave in air being only SIX FEET long inside of 66% VF coax.

PS. The 75Ω 1/4 waves should be at the T not at the antennas if you need to add a set of lengths of 50Ω coax to reach the T.

Awesome. Having played around with the calculator a bit, I believe I understand the math, now. Thank you for the explanation.

 

[Sarasota Slim]

I found it seems easier to use 270 degrees of cable as opposed to 90 degrees of cable for your phasing lines. I tried using a 90 degree phasing harness and I was unable to get my antennas to tune correctly but once I switched it out to 270 degrees per side everything just seemed to fall into place.

You've lost me, lol. Where are the degrees coming from? If we're still talking length of coax, I have no clue what you mean by 270 or 90 degrees. If we are not talking length, that what are we talking about? Please, excuse my ignorance on the topic.

 

[The DB]

The 1/4 wavelength or 90 degrees that is talked about actually converts the impedance to 112.5 ohms, not 100. It is close enough to work though.

To translate some of the above. When referring to degrees, 90 degrees is 1/4 wavelength. In turn, 180 degrees is 1/2 wavelength, 270 degrees is 3/4 wavelength, and 360 degrees is 1 wavelength. As you travel down a coax cable, a pattern forms. Some points, namely every 1/2 wavelength will have the same impedance. As 0 degrees, or the starting point, is 1/2 wavelength away from the 1/2 wavelength point, the 1/2 wavelength point duplicates the impedance as seen at the end of the coax. This is why many will tell you to use a 1/2 wavelength of coax when using an antenna analyzer, the analyzer sees something very close to what is at the feed point of the antenna.

Now to take this a step further, lets start with the points mentioned above for half wavelength multiples. Now take the half way points, such as 90 degrees, which is 1/4 wavelength, and 270 degrees, which is 3/4 wavelength. These points also follow the same rule as the points I mentioned above. Essentially what the coaxial transformer mentioned above is doing is going from a given starting point, and then finding the points that are the furthest away from the existing known half wavelength points. These are the points that the impedance is the most different from the impedance at the end of the coax.

NOTE: When I use the word "wavelength" above I am implying "electrical wavelength".


The DB

 

[Sarasota Slim]

@ DB: Ahh, that makes sense. Thank you.
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On a side note, I'm starting to notice that I would have done well to have given more thought and effort in advancing my math skills, back in high school. I'd likely grasp all this a lot faster. I appreciate y'alls patience with me.

 

[Needle Bender]

@ DB: Ahh, that makes sense. Thank you.
*************************************
On a side note, I'm starting to notice that I would have done well to have given more thought and effort in advancing my math skills, back in high school. I'd likely grasp all this a lot faster. I appreciate y'alls patience with me.

LOL! Me too, just in other areas.

You can try calculating it yourself, if you know the approximate impedance of a load (L) the multiply it with the 50Ω, then using the square root on a calculator ( √ ) - find the square root of those two multiplied together = coax impedance needed.
Example:

antenna load 170Ω x transmitter input 50Ω = 8500

√8500 = 92.2Ω coax needed, and guess what, 90Ω coax is available.

 

[Slowmover]

“When I look to the sky I see cherry pie”

Love the threads. Never an ending so imagining success as easy as failure.

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