I took this image from old thread as did not want to hijack it. I wonder if someone could provide simple formula similar to formula in Sketch 1 for Sketch 2. Basically I’m trying to answer question how “visual” range changes if we put obstruction between antenna 1 and antenna 2.
h1 - antenna 1 height
h2 - antenna 2 height
H - obstruction height between antenna 1 and 2
L - distance between antenna 1 and obstruction
Here’s my guess … range with obstruction = cos(angle to clear obstruction) ^5 X unobstructed range
Examples …
No obstructions = 1 X 100 miles = 100 miles
30-degree obstruction = 0.866^5 X 100 = 49 miles
45-degree obstruction = 0.707^5 X 100 = 18 miles
60-degree obstruction = 0.5^5 X 100 = 3 miles
Again, just a quick guess at a simple formula to get you in the ballpark.
@kenf3 Thank you. Why ^5?
@geckoVN I’ve seen this but it answer different question - simple LoS without obstructions.
Just trial and error with the examples (started with ^2). The results with ^5 seemed more plausible.
Here is a (not so simple) formula based on the hidden height approximation:
http://walter.bislins.ch/bloge/index.asp?page=Advanced+Earth+Curvature+Calculator
The hidden height is the part of the aircraft altitude that is "below the observer’s horizon).
h1, h2, H and L are as per your post. (all heights are above assumed spherical earth surface)
dmax = maximum range case-2 of your second diagram (sorry about lack of subscript formatting here)
dHL = maximum range case-1 of your diagram.
Note that it is best to do all the calculations in one system of measurement - I use metric, and convert to or from aeronautical units as required.
I’ve got an Excel spreadsheet that does this calculation - let me know if it would be of use.
@JRG1956 Thank you. I’ll try to do calc based on these formulas, but spreadsheet would be really useful.
What does your Excel formula tell you for 30 degrees, 45, and 60 degrees obstacles? I’m curious how much the range is reduced versus my ^5 formula guess.
If you know where the obstruction is and how how tall it is, you could consider the problem as being similar to planning a point to point link, with the Tx site being on the top of a very tall mast, say 10,000 ft and the Rx being at whatever altitude your Rx antenna is.
It will be a bit of a faff to do, but try using this:-
Cambium LINKPlanner
It is free.
The table below just uses the hidden height formula. I’ve worked backwards from your 100 mile unobstructed range (assumed nautical miles) to come up with an implied altitude of 8100 feet. Also have used a 5m receiver height.
You can see that a 15 degree obstruction reduces the line of site range at that altitude from 100NM to 5 NM.
With a 5m receiver the horizon is at 4.3NM, so the formula could break down for the distances required to get angles of 30 degrees or more. (Note that these calculations are without refraction - so max ranges are shorter than those typical shown using the “hey what’s that” approach recommended in this forum.
Wow, range disappears quickly!
Possibly a result of radiation pattern of the antenna. For simple vertical dipole, the pattern is a doughnut with strongest response in the horizon and zero in the zenith.
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