Angle Needed to Point Directional Antennas

[Happyflyer]

This is for the math wizzards out there or not. When it comes to the directional antennas on your controller, the correct elevation to point them is interesting when calculated. Instead of using math I tried with graphpaper. It looks like it may "add up" but maybe way off. When having the quad 10,000 feet away and up at 500 feet, according to the graph design, the elevation of the controller is very little above level. Curve of the earth was not considered because of the short distance involved.
Even up at 2,000 feet (too high) the elevation of the controller was still not that great. So what do the math experts think?

 

[N017RW]

The Earth's curvature is about 8 inches per mile if recall but regardless it is not a factor due to the elevation of the aircraft.

Well I'm NO wizard but here's my pass at it:

Solving your problem using a 'right triangle':
At 10K ft away and 500 ft altitude the angle of the antenna would be 2.86 or about 3 degrees. This puts the a/c at about 10012 ft away from Tx.

At 2000 ft altitude it becomes a bit over 11 degrees (11.3). This puts the a/c at ~10.2K ft away from Tx.

 

[Happyflyer]

Well, I guess my graph worked. I showed not much of elevation at 10,000 feet away and 500 feet up.
Thanks

 

[MadMitch88]

I fly long-distance with my P2V+ sometimes (up to 2.5 miles) and I need some kind of app that can calculate the minimum angle I need to point my half-sphere antenna and minimum flying height I need to maintain LOS with my bird at that max. distance when I know there are trees that are approx. X feet in height and Y amt. of horizontal distance away from where I am standing.

Obviously I can fly my bird at 800 feet. above-ground and be confident that I will maintain LOS at long distance even with tall trees all around me --- but gut feeling tells me the FAA won't be pleased to hear about it. Even without silly FAA rules, I still want to keep the bird low as possible in order to get maximum detail of landscape in the video footage.

This all comes down to basic trigonometry but sadly, I havent taken trig since junior year of high school (30 yrs. ago), even though I scored an "A" in it when I took it. Any of you math wizzes out there know how to make this kind of app, because you would be making a pretty penny if you can upload it to the iTunes and Google Play store and charge $5.00 for it. You would be making a LOT of Phantom owners like me "shut up and give you our money".

:mrgreen: :mrgreen: :mrgreen:

 

[N017RW]

Mitch, I HEAR YA'! (for me about 40 yrs.)

Don't do apps. but in the mean while...

Go here:
http://www.pagetutor.com/trigcalc/trig.html

I'm sure there's others but I found this a while back to help me quickly check my kids homework.
(it's what I used earlier )

Make two calcs., one for the quad the other for the obstructions.

For a quick go/no-go, make sure the angle for the quad (ground antenna) is GREATER than the obstructions.
Then you know your 'shooting' above them.



Cheers!

 

[MadMitch88]

Mitch, I HEAR YA'! (for me about 40 yrs.)

Don't do apps. but in the mean while...

Go here:
http://www.pagetutor.com/trigcalc/trig.html

I'm sure there's others but I found this a while back to help me quickly check my kids homework.
(it's what I used earlier )

Make two calcs., one for the quad the other for the obstructions.

For a quick go/no-go, make sure the angle for the quad (ground antenna) is GREATER than the obstructions.
Then you know your 'shooting' above them.

Cheers!

N017RW,

Thanks, that will work! Although I wish someone could make a simple Android app that would also allow plugging in values for Y and Z as shown in this diagram -- so dummies like me can plug in all values at once and avoid multiple calculations in the field --- and better yet, get a dynamic result on A as B fluctuates (ie, I decide to fly farther than my original calculation) so I know how to adjust height above ground accordingly:

Y = height of trees
Z = horizontal distance from me to trees

 

[Happyflyer]

Ah, nice little "app." Reminds me of the old 3, 4, 5 used to square something off another object. The formula the scarecrow recited, after he got his brain, in the Wizzard of Oz that still makes mathematicians chringe when they hear his erroneous formula. :lol: