Distance to the Horizon
 

I very often fly with a passenger in the right hand seat. Without fail, they always ask - "How far can I see from up here?".

I am forced to answer that I don't know exactly (or even inexactly), so my question is, does anybody out there know a rule of thumb for working out the (theoretical) distance to the Earth's horizon given your height in 1000's of feet?

Many Thanks in advance...

I'm not sure about it. Remember something about 200 NM at lower altitudes.

regards

Is it 1.25 x Square Root of Altitude expressed in thousands of feet (the theoretical range of VHF)? Unless of course, when over Europe when it's about 30-40 nm, if you are high enough, due polution!

I've tried timing how long I'm able to track the odd passing contrail from one horizon to another, from the ground and working the distance out by using about 7nm a minute, some days are more impressive than others:sad:

Theoretically
 

Assuming:
The earth is perfectly round and flat, no high buildings, mountains, etc

Given:
r = radius of earth = 20925000 feet
l = seeing distance (in feet)
h = altitude (in feet)
1 nm = 6080 feet

We can calculate
r^2+l^2 = (r+h)^2
r^2+l^2 = r^2 + 2rh + h^2
l = sqrt(2rh + h^2)
l = sqrt(h*(2r+h) )
l = sqrt(2rh) -- drop h, because it is insignificant compared to 2r
l = sqrt(2*20925000*h)
l = 6500*sqrt(h)
l (converted to nm) = 6500*sqrt(h) / 6080 = 1.06*sqrt(h)

We find
Alt (ft) - Dist (nm)
01000 - 34
05000 - 75
10000 - 106
15000 - 130
20000 - 150
25000 - 168
30000 - 184
35000 - 198
40000 - 212

Does that sound/look right?
I drew a picture, but I don't have an area to upload.

Regards,
PieterPan

Ahh maths...
 

Well, someone ad to do it...

Imagine a triangle between you, the earths centre and the point on the horizon you wish to find the distance of. You have a right angle triangle. (your line of sight is a tangent to the circle of the earth, therefore forming a right angle with its radius)

Assuming:
1. the earth has a uniform radius of 3440.227 nautical miles
2. the earth is perfectly round (and no mountains)
3. you can actually see farther than your radome

Here's my 'guess' of how far it is to the horizon:

1000ft = 33nm
2000ft = 48nm
5000ft = 75nm
10,000ft = 106nm
20,000ft = 150nm
30,000ft = 184nm
40,000ft = 213nm


and to confirm CATCHUP's comment, my calculations say...

ISA Tropopause @ 36,000ft = 202nm

There you go, it's time for bed now!! :8 :zzz:



DAMN YOU PIETER PAN!!

Oops
 

Sorry PigDog: Beat you by a few minutes... Glad to see we both came up with roughly the same numbers.

Would like to see someone try this out for real at 40000. Find 2 points along the route with a distance of around 200 nm, fly over one, and try to see the other.

I suspect the air might never be clear enough to see all the way, let alone having 40000/212 eyesight...

Yes, good idea, off to bed:D.

PieterPan

I always thought it was 1.17*sqrt(alt_in_feet). It's a bit less than the HF formula.

so at 30 000' = 202nm.

Check this website, it states the theoretical distance:

http://www.auf.asn.au/comms/vhfradio.html

Could have saved me and PigDog a bit of time if I searched for this first... :)

I suppose when you're not at the equator, and you factor in that trees, mountains and buildings exist, the seeing distance will increase.

Regards
PieterPan

(edited, typo)

Pieter Pan's and PIGDOG's calculations are correct geometrically. A complicating factor is that neither light nor radio waves travel in perfectly straight lines in the atmosphere. The rays bend around the earth, increasing the distance to the visible or radio horizons by something like 10% more than the geometrical result.

Thus the rule of thumb tends to be

distance/nm = 1.2 * sqrt (height/ft)

(rather than the geometrical factor which is about 1.07)

@bookworm
 

It's much more complicated, cause it also depends on your speed. If you are traveling close to the speed of light you can't use that formula anymore.
;)

regards

catchup
 

catchup, you win!

Don't make bookworm pull out his book on String Theory :8

Thx for the explanation, makes sense...

Regards,
PieterPan

Spoilsport! I can't get enough of the chapter on wormholes. ;)

Bit geeky I know, but am currently doing this on a course.

Radar Horizon
Single Object Above Earth
d = sqrt(2kRh)
k = ~4/3 or -3/2 for super-refraction
R (earth) = 6370,000m
h = metres

or roughly 1.23sqrt(height(ft)) = nm as previously posted

Hilltop to Hilltop or Aircraft to Large Antenna
d = sqrt((2kRh1) + (2kRh2))
h1 / h2 = object heights respectively

Hope this helps