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I'm stuck on the following riddle found on the geocaching web site:

A helicopter pilot is flying at an altitude of 1981 meters. He is heading inbound an airfield 335 meters high. There are no particular obstacles (such as mountains, ...) between the helicopter and the airfield. What maximal distance (in nautical miles) can the pilot expect in order to be able to communicate with the airfield control tower?

Unfortunately I don't know anything about air control rules and multiple Google searches came to no avail.

Please could you give me a clue (not the answer as it would spoil the fun) how to answer this riddle?

Edit

For those who are interested here is the link to the geo-cache

DeltaLima
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Johann
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    I would say it's about the curvature of the earth and assumes the signal is blocked by the horizon – ratchet freak Sep 13 '14 at 19:20
  • google for altitude / height / elevation and then VHF line of sight range calculations ... – Radu094 Sep 13 '14 at 19:25
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    I guess the "correct" answer is the theoretical maximum considering the curvature of the earth, but I doubt in reality the radio signal can be transmitted that far with adequate quality. – kevin Sep 13 '14 at 19:49
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    I wouldn't automatically assume that radio power is more limiting than curvature-of-the-earth. It certainly could be, but plenty of aviation radios have adequate power to reach out a couple of hundred miles, and since we're talking pretty low altitudes here, those radios would have ample power for far greater ranges than what's being discussed in the answers. – Ralph J Jul 25 '15 at 17:58
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    Propagation for aviation bands was studied by ITU and factors quantified in REC P528 with several graphs and tables A zip file contains the related code and there is a graphical implementation on GitHub. As one can see it's not a just a matter of line of sight. From a height of 50m, radio horizon (including refraction effect) is at a mere 30 km, but VHF range may be several times this distance. Example. – mins May 22 '21 at 16:34

6 Answers6

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The formula for VHF range taught in flight schools1 is

$$ d = 1.25 * \sqrt{h} $$

where d is the range in nautical miles and h is the height in feet of the receiver above the transmitter.

So in your case it would be (spoiler: mouse-over to see hidden text):

$$ 1.25 * \sqrt{(1981 - 335) * 3} \approx 87.84 $$

However, keep in mind that in some countries the range of Tower VHF radio tranmitters is limited so they only achieve a certain range. Uncontrolled airfields in Germany usually have a range of 30nm, which is achieved by limiting their power.

1: "Motorflug kompakt" - Winfried Kassera / 2012 - p. 296

ChrisW
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SentryRaven
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    taught in flight school - I think I need a partial refund. Anyway, Reminds me: Flt. engineer asks capt. crusty what's our wing span? Pilot: "I don't know". FE: "Don't you think you should know? What if you wanted to put the plane in the hanger?" Pilot: "Well, I don't know how wide the hangar is either." – radarbob Sep 14 '14 at 01:41
  • @radarbob what's a 27 MHz CB radio ground station got to do with an aircraft mounted (as in: probably lower powered) 108-137 MHz radio? It's a good rule of thumb. – falstro Sep 14 '14 at 09:59
  • I'm afraid your formula is not correct. If you have an aircraft flying at 10000ft and a radio station on top of a mountain at 10000ft the range is 0? – Emil Sep 14 '14 at 10:08
  • @emilTthe formula is rather a rule of thumb, you cannot use it for extreme values like the one from your example. – SentryRaven Sep 14 '14 at 10:14
  • @SentryRaven: The correct formula is 1.25*(sqrt(ht)+sqrt(hr)), because the height of the transmitter matters too. The higher the transmitter, the longer the range. The same goes for the reciever. That's because the range (ignoring transmitter power and obstacles) is limited by curvature of the earth. The higher you place any of the antennas, the longer it will take until the signal is blocked by the earth. – Emil Sep 14 '14 at 11:31
  • @SentryRaven I'm afraid neither your or Emil's formula validate on the geocaching site (which doesn't mean they're wrong, but looks like it's not the answer the geo-cache owner expected). Thanks anyway! – Johann Sep 15 '14 at 09:22
  • @Johann The question then is, which formula or algorithm did the geocaching site use? I see where Emils formula comes into play and I have found both formula equally on google for quick reference. The one I have in my answer is from my PPL exam preparation book, but I am looking forward to anybody who can shed a light on which one would be better. Maybe invite the guys from Physics.SE over? – SentryRaven Sep 15 '14 at 09:25
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For the first step try this line of sight calculator. Line of sight is an imaginary line that exists between two objects. Radio transmissions require a clear path between antennas known as radio line of sight. Formula:

Line of Sight = √ (2*height1) + √ (2*height2) in Miles

Nathan Tuggy
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Carnatic
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    It would be more helpful to include some sort of explanation for how the calculator works, like the other current answers do, in case the link is broken or otherwise does not work for someone. – fooot May 26 '15 at 15:41
  • Here you go:Line of Sight Tutorial Definition:

    Line of Sight is an imaginary line that exists between two objects. Radio transmissions require a clear path between antennas known as radio Line of Sight. Formula : Line of Sight = √ (2height1) + √ (2height2) in Miles

     – Carnatic May 27 '15 at 07:17
  • Height in the formula above is in feet... – Carnatic May 27 '15 at 07:23
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    @Carnatic Are those miles in your answer (and in the link) statute or nautical miles? – Ralph J Jul 25 '15 at 18:01
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    Radio transmissions in general do not require a clear line-of-sight between antennas. Most radio communications do not have line-of-sight between antennas. However, the signal will generally be significantly stronger within line-of-sight. When you actually have the Earth itself between the transmitters, that will usually cause enough attenuation that the communication won't work, which is why this formula works for aircraft radio transmissions. It's worth noting, though, that even this rule doesn't apply to all radio communications. Some curve with the ground or reflect off the ionosphere. – reirab Jul 25 '15 at 18:19
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    Just don't forget, that both are heights above ground. Substituting 1981 and 335 would not give realistic results. – Jan Hudec Jan 12 '16 at 14:13
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    1/ For radio LoS, Earth is 30% more flat that in reality (diffraction). 2/ If LoS is clear but 90% of 1st Fresnel ellipsoid is obstructed, a loss of 20 dB occurs. Its radius for 110 MHz and a distance of 50 km is 180 m, that's a lot when stations are on the ground (radio links towers in flat regions are tall for this reason). 3/ Overhorizon communication is still possible in VHF/UHF, with a loss of 0.5 dB/NM. 4/ There is no fixed range, it depends on the link budget (antennas, power used, losses, dry/wet air, receiver sensitivity, ...) – mins May 23 '21 at 18:01
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The range is calculated by this formula:

$Range = 1.25 (√Ht + √Hr)$

'Range' - in nautical miles
Ht - height of transmitter in feet
Hr - height of receiver in feet

Farhan
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MK Aviation
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  • I tried this: 1.25*(80.618+33.152) which gives 142 nautical miles but unfortunately it doesn't seem to be the expected answer. Thanks anyway! – Johann Feb 25 '15 at 19:13
  • @Johann, you are substituting the wrong numbers. They are both heights (which in aviation means above ground), not altitudes. – Jan Hudec Jan 12 '16 at 14:15
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    You can't really apply this simple model in a situation where there's a substantially varying difference between height and altitude in the region of interest, because it assumes a spherical earth. If one radio is atop an isolated mountain, that altitude is as good as height above the curvature of the earth. On the other hand if both are over low terrain and there is a mountain in between, that may block a signal that would otherwise have line of sight across more typical curvature of the earth. – Chris Stratton Jul 08 '18 at 20:14
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Line of sight calculation is based on earth radius and height above ground of transmitter (and receiver). R = 21000000 (21 million feet) approximately. Based on plane trigonometry, d squared = h * (2 * R + h). This is distance to horizon for an observer at height h above surface of radius R. If R >> h then a good approximation is: d squared = 2 * R * h; or d = square root (2 * R) * square root (h)

and since R is essentially fixed, we can calculate it and are left with

d = 6480 * square root (h) = 6480 h^1/2

which gives d in feet.

Using another approximation 1nm (nautical mile) = 6076 feet

d = 1.07 h^1/2 (h in feet above surface, d in nautical miles) or d = 1.23 h^1/2 (h in feet above surface, d in statute miles)

So the well known formula: d = 1.25 h^1/2 is an approximation for the horizons range in statute miles when h is in feet.

The problem stated above is incomplete. You are left with some assumptions when attempting its solution. 1) are we to assume adequate TX power and RX sensitivity to reach? 2) are we limited only be line of sight (i.e direct line between TX and RX that does not penetrate the earths surface?) 3)Should we assume the airfield height to be the height of the surface under the aircraft and at essentially all points between the two?

If all these assumptions can be made (maybe there are more?) then the height of the aircraft above the ground is 1646 m = 5400 ft and d = 81 nm.

Note that if you assume the airfield to be 335 m above ground level you will get a drastically different answer to the line of sight range. With the aircraft at 1981 m (6500 ft) and the field at 335 m (1100 ft) then: d1 (aircraft) = 89 ft and d2 (field) = 36 nm so d = d1 + d2 = 125nm.

ramjet
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I was taught similarly. √ftALT×1.26 = Distance in NM. So 1981 m = 6500 ft, 335 m=1100ft, then difference =5400 ft. √5400*1.26=73.4*1.26=92.5 nm.

Joe
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Convert first from meters to feet :

1981/3.28 = 603.96 =604 feet
335 / 3.28 = 102.13 feet

use formula $Range=1.25(√Ht+√Hr)$ = 1.25 ( 24.55 + 10.10 )= 44.56 nm

Jamiec
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