HF radio question . . .

[phonetic]

Any help with this would be appreciated, it's probably easy but it's got the better of me . . .

An aircraft transmits on HF from an altitude of 16000ft to an ATC which has its receiver aerial situated on top of the control tower at a height of 120ft. At what range could the aircraft captain expect to make radio contact?

[HEATHROW DIRECTOR]

It all depends what HF frequency he is using - under good solar conditions he could make the contact over many thousands of miles. Under the conditions we are currently experiencing, with no sunspots, it will be somewhat less.

Now, the big question is..... did you really mean HF?

[Matt.V]

Hi there,

I am not a pilot and don't know how you are supposed to calculate this sort of thing, but from what you have written, assuming you meant VHF and with careful application of common sense and general knowledge it should be:

Lets work in meters:

altitude=16000ft=4876.8m
tower height=120ft=36.5m



I attached a picture which should help to comprehend my thinking. My drawing skills are a bit sh*t, but if it was done properly you could see right angle near the tower. That means we get a triangle with a right angle and can use Pythagoras theorem which says c^2 = a^2 + b^2 where c is hypotenuse. So in our case:

(R + altitude)^2 = (R + tower height)^2 + (distance)^2

R here is the radius of our lovely blue planet and it is equals 6378100m
(NOTE: for calculation purpose every number has to be the same unit)

If you rearrange this equation then:
distance = square root(( R + altitude)^2 - (R + tower height)^2)

and insert the numbers you get:
distance = 248548m

from here you can convert it to any units you like
248548m = 248.548km = 154.4m = 134.2nm

This is very rough calculation. We assumed that distance was a straight line, which in reality is not. Electromagnetic waves are bent by gravity. Also, earth radius is not the same everywhere. Moreover, wave propagation depends on weather conditions and obstacles along the way. I heard, that engineers, who deal with vhf, uhf and shf communications use coefficient of 4/3 to get more realistic results. In our case it would equate to 178nm. How much of it is true I don't know.

Sorry for my broken English
Bottle of wine is empty, I guess it's bed time
Goodnight

[phonetic]

Hi!

Thanks for the quick responses. Brilliant.

I pondered the HF vs VHF typo myself, but using the good old 1.23xsqr(alt AGL) didn't seem to match.

The question is from a piece of JAA ATPL training software. The whole question (including answers) reads . . .

-------------
46. An aircraft transmits on HF from an altitude of 16000ft to an ATC which has its receiver aerial situated on top of the control tower at a height of 120ft. At what range could the aircraft captain expect to make radio contact?


A) 137,44 NM.

B) 171,8 NM.

C) 158,71 NM.

D) 1.044.500 ft.

(The answer is, apparently, B)

-----------

Thanks for the responses. It's got me a tad confused. I'm convinced I'm missing something obvious though.

Cheers!

[The Old Fat One]

Phonetic

The radar horizon for an aircraft at 16000 ft is about 170 nms. So for line of sight comms (VHF, UHF) answer B would appear to be the best fit.

If the question really means HF, the question and all the answers are wrong, as BelArgUSA has already explained.

From an old HF operator.

[delaneyslad]

vhf ME THINKS?

Good morning phonetic
If you're heading towards the ATPL's I wouldn't have thought you'd have the time in the exam to do Matt.V's calculations (plausible as they are and very commendable)
Your original assumption (dangerous word) is correct. i banged in 1.25*sq rt of the Tx + sq rt of Rx and got answer b.
Or am I missing something
Good luck with the studies

[rodthesod]

Matt.V,

Sorry but that is not a right angle (considerably less in fact), even if it was drawn accurately.

Rubbish question, as others have said.

[Fratemate]

Strange question and one that I can't work out either.

If the question meant to say VHF, then you'd use the formula:

Line of sight dist (NM) = 1.06*Sq Rt of Transmitter Ht (Ft) + 1.06 * Sq Rt Of Receiver Ht (Ft)
= 134.08 + 11.61......145.69nm.

I've no idea how they got 171.8nm.

As for HF......who knows. Sky waves + Ionosphere + Tx power + .......... Too many variables I think.

[davey1082]

hi guys, just thought id add a reply...

well we all understand that hf and vhf are line of sight. i always understood the calculation to be

1.25*sq root transmitter height(ft) + 1.25*sq root a/c height=dist nm.

so 1.25*sq rt 16000ft + 1.25* sq rt 120ft= 171.80694.

171.80nm.

hope this helps guys and please correct me if im wrong.

cheers davey

[Matt.V]

Sorry but that is not a right angle (considerably less in fact), even if it was drawn accurately.

Well, it's no good saying that it would be "considerably less in fact" if you don't know what it is exactly. But you are right, it is not a right angle. If you want accuracy, you should divide that triangle into two. Apply Pythagoras theorem twice and add the two answers. I bet you, you'd get 171.8nm.

Btw, if drawn properly that angle would be 89.8 degrees.

[paco]

davey - it's officially 1.23 not 1.25

Phil

[The Old Fat One]

well we all understand that hf and vhf are line of sight.

please correct me if im wrong.

Since you ask, you are wrong.

HF is not line of sight. HF does have a ground wave component which can be considered line of sight, but the function of HF is medium to longer range (2-3 thousand miles typically) and for that it uses the skip wave. In certain ionospheric conditions HF comms can extend to very long range.

So...

The question is crap, and meant VHF. The answer is B.

Are we done?