There are 2 critical distances to be calculated. The first, the vertical distance between antennas. Second, the length of the phasing harness (those 2 lengths of Co-Ax which would run from the feed point of each antenna to, in this case, a Co-Ax "Tee" adapter). It's not at all difficult once you know some of the deign parameters. The first, is the physical, vertical separation of the 2 antennas for a specific result from what will be an antenna array.
Just brainstorming here but I'm thinking the physical separation value might be based on the result being a compromise between gain and frequency "broadbandedness." The mfg might already have that value and you might reach out to them and ask. Or, perhaps someone here has modeled an array using those antennas using EzNEC. The second value necessary to calculate the length of the phasing harness is based on the value of the first. That is, what is fraction of a wavelength should the length of the phasing harness be to accommodate the physical spacing value used to achieve the goal of your array.
You will also need the velocity factor of the specific brand of (probably RG-6) co-ax used to make the phasing harness. That probably going to be something like .82, .85 or something like that.
Lets put it all together. I made an 800 Mhz, 4 vertical dipole stacked array. Each vertical dipole needed to be spaced 57" apart to achieve the design goal which was flattening the pattern toward the horizon. I used 3/4 wavelength phasing sections. To determine the electrical length of those cables, I first divided 2952 by the center frequency of the antenna in mhz or, 2952/855 then multiplied that result by the velocity factor of the cable, in my case, .85 which yields the electrical length of my 1/4 wavelength phasing line. However, I needed 3/4 wavelength lines so, I then then multiplied THAT result by 3. THAT yielded the electrical length of my phasing line. So, the formula looks something like this:
2952/855= P (P* .85) = L1 (L1 * 3) = L3
Where P= Physical cable length
L1= Electrical 1/4 wave cable length
L3= Electrical 3/4 wave cable length (desired length)
So, for 98 mHz, 2952/ 98= 30.122"
30.122 * .85= 25.603" or, right at 25 5/8" for each of the 2 harnesses.
Multiply that times 3 and you get a cut length for each leg of your phasing harness of 76 7/8". Remember that an FM Tee is about 3/4 inches long so SUBTRACT HALF that value from EACH of your 2 phasing harnesses.
So, that's what the formula basically looks like. What I am lacking to make it completely accurate to your situation is the mfg's or modeling result to provide recommended spacing between your antennas.
Th example above should not be regarded as a "build to" example in your situation but rather, an example of how do perform the calculations to determine the phasing harness length once you have:
1- The recommended physical vertical antenna spacing in wavelengths for a design frequency.
2- The velocity factor of the cable you will be using to construct the phasing harnesses.
Best
Quote:
Originally Posted by skatingrocker17
Okay, so correct me if I'm wrong.
This article gives the following formula: 467 / (lowest frequency).
So... 467 / 88Mhz = 5.3 feet. |