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VOR Range

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Old 24th April 2006 | 19:54
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VOR Range

No doubt, the effective range of a VOR beacon will depend on your altitude, but is there a rule of thumb, or indeed, somewhere that officially has the info.

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C23
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Old 24th April 2006 | 20:06
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From: East Midlands
See AIP: http://www.ais.org.uk/aes/pubs/aip/pdf/enr/20401.PDF

Andy
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Old 24th April 2006 | 20:34
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From: S Warwickshire
DOC only works if you have line of sight with the station.

Rule of thumb for line of sight (in NM) = sqrt(height in feet)
Assuming no big hills in the way etc.
e.g 2500' - 50NM
5000' - 70NM
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Old 24th April 2006 | 21:04
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Thanks Guys.

C23
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Old 24th April 2006 | 23:13
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Rule of thumb for line of sight (in NM) = sqrt(height in feet)
I believe the RoT is 1.25 * sqrt(height in feet).
 
Old 25th April 2006 | 06:20
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From: S Warwickshire
The exact solution for length of tangent (assuming station at GL) is:
sqrt(2rh+h^2)
where r is the radius of the earth
and h is the height of the observer.

To a first order the h^2 term is negligible.
So length of tangent = sqrt(2r) * sqrt(height)
In nautical miles the coefficient works out at 1.0634
In statute miles it is 1.2245

So 1 is an easy conservative value for a rule of thumb.
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Old 25th April 2006 | 06:39
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From: warwickshire
I'm sure for the ATPLs it was something like height of Tx + height of Rx (feet), square rooted, x 1.25. Anyone more current?
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Old 25th April 2006 | 07:21
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From: UK
Originally Posted by Mark 1
The exact solution for length of tangent (assuming station at GL)...
In nautical miles the coefficient works out at 1.0634
Your exact solution assumes that the ray travels in a straight line. In fact it is curved by refraction in the atmosphere. The magnitude of the effect depends on the frequency: for light it's about 10% greater distance to the horizon than the geometrical solution, for VHF it could be rather more.
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Old 25th April 2006 | 07:56
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The exact solution for length of tangent (assuming station at GL) is:
sqrt(2rh+h^2)
I admit it, I am mathematically challenged, but I don't get that. If I'm at 4000' then my range is 4000nm plus a shed load more? Which units?
I'm sure for the ATPLs it was something like height of Tx + height of Rx (feet), square rooted, x 1.25. Anyone more current?
It would have to be 1.25 * sqrt(H of Tx) + 1.25 * sqrt(H of Rx).

Bookworm,
The ATPL assumes line of sight.
 
Old 25th April 2006 | 08:18
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From: S Warwickshire
Back to the original question....

For a rule of thumb, I prefer 1 as its easy to remember and slightly conservative.

I remember seeing 1.25 in one of the CPL text books, but this may be for a distance in stat. miles.

If the VOR is positioned significantly higher than the surrounding land then you can add a bit for this, as bookworm says.

Given the variability with local geography, I'd go for the conservative factor.

This is only for planning anyway. If you have a flag and ident, and are less than DOC distance, then it's OK to use it.
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Old 25th April 2006 | 10:15
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From: UK,Twighlight Zone
Straight Line Propagation Paths - Direct or Space Wave

D (NM)=1.25(sqrt HTTX Station + 1.2 sqrt HTRX Station)

Example.

VOR 400ftamsl
Aircraft 5000ft amsl

D= 1.25 x (sqrt400 + sqrt5000) = 113.38NM
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Old 25th April 2006 | 10:23
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None of this ATPL ground school stuff is very relevant in practice, because if you potter about the UK at low levels you won't be receiving anything a lot of the time (except GPS) and if flying UK/European airways then ATC will send you all over the place and quite happily DCT some VOR which is way outside the DOC - they simply assume you have BRNAV which means a GPS.

It's the damned exams one has to get through
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Old 25th April 2006 | 10:27
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From: UK,Twighlight Zone
Yes and didnt I learn something damned clever and utterly useless!!!
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Old 25th April 2006 | 10:44
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From: Hampshire, UK
Devil

And to help complicate things a bit more:-
We ancient master mariners were taught:-
Distance of the horizon = (1.15 x sqrt H) + (1.15 x sqrt h)
Were H = Your height of eye and
h = Height of object (normally a ruddy lighthouse)
Although they did make a consession that a constant of 1.17 a better value for average conditions.
Just thought you ought to know.
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Old 25th April 2006 | 10:45
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Given the variability with local geography, I'd go for the conservative factor...This is only for planning anyway. If you have a flag and ident, and are less than DOC distance, then it's OK to use it.
Can't argue with that. Nowt wrong with a pprune style sometimes heated, often accademic and probably pointless debate though
 

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