Solving the question of how to manage the antenna elevation knowing distance and difference in altitude (vertical displacement) was the topic of a couple of articles I wrote 17 months ago. As often happens (even IRL!), as we progress into studying and the knowledge broadens, so new solutions to old problems are found.

Although I habitually use the table I generated using trigonometry as it is an immediate and handy solution (it’s always pinned in front of my nose), this article is about a handy formula mentioned in the P-825 R05-02.

## A formula to rule them all

The angle is estimated by means of this simple formula:

ElevationAngle = Δ_{ALT}/ (SR * 100)

### Example I

_{F-14}= 25,000ft

ALT

_{TGT}= 15,000ft

Δ

_{ALT}= 10,000ft

SR = 25 nm

ElevationAngle = Δ

_{ALT}/ (SR * 100) =

*4°*

### Simplification and Example II

As the altitude is pretty much always in thousands of feet, it and the denominator can be simplified: just drop two zeros from the altitude and divide by the SR:

_{F-14}= 30,000ft

ALT

_{TGT}= 15,000ft

Δ

_{ALT}= 15,000ft

SR = 15 nm

ElevationAngle = 15,0

*10°*

## Verification

For the sake of completeness, these are the values from my model, based on trigonometry. The first example is highlighted in red, the second in green. As you can see, they are quite close.

So, which one should you use? Well, that’s entirely up to you and probably the ones the makes finding the angle faster.

10 degrees for every 10000ft at 10nm. Double if 5nm, half if 20nm.

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That’s a good observation. Another way to put it is dropping three figures at 10nm (so, ten-thousand → ten degrees), then double or half as you suggested. I usually use my table, but this relation at 5nm / 10nm / 20nm is really handy. Thanks!

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