Re: [amsat-bb] Aircraft Doppler?

["Dave G." <dmg@xxxxxxxxxx>]

John.

Beautiful exposition --- 
I appreciate it, even tho' I did not ask the question :-\

Dave KK7SS


On 19 Nov 2005 at 14:49, jmfranke wrote:

> The maximum Doppler shift would be when the aircraft was
> moving direct towards or away from the tranceiver.  When the
> aircraft is approaching, the shift would be to a higher
> frequency, when moving away, the shift would be to a lower
> frequency.  One cycle of Doppler occurrs for every half
> wavelength of change of distance.  This is because the
> transmitter to receiver path changes one wavelength when the
> aircraft moves a half wavelength.  For a direct approach or
> departure, the Doppler shift is independent of distance.  For
> other angles, just multiply the direct approach valus by the
> Cosine of the approach or departure angle.
> 
> The Doppler shift "D" is found by starting with calculating
> the wavelength L:
> 
> L(in feet) = 984,000,000/F(in Hz)
> 
> D = V/(L/2)
> 
> But V is in mph and it must be converted to feet/sec.  An easy
> convesion to remember is 60mph = 88 ft/sec
> 
> Or, D(in Hz) = F(in MHz) times V(in mph) times 88 divided by
> ((984/2) times 60)
> 
> Running the number for a police traffic radar using 10.525 GHz
> and a speed of 1mph:
> 
> D = (10525*88)/((984/2)*60) = 926200/29520 = 31.375 Hz/mph
> which is correct
> 
> Now, returning to your problem;
> F = 144, let V = 1mph
> 
> D = (144*88)/((984/2)*60) = 12672/29520 = 0.429 Hz/mph
> 
> So:
> for 200mph, the maximum shift is 200* 0.429 = 85.8 Hz
> for 300 mph, the maximum shift is 300* 0.429 = 128.7 Hz
> for 500 mph, the maximum shift is 500* 0.429 = 214.5 Hz
> 
> One resuly of the calculation is that it shows that aircraft
> talking to each other do not need to worry about the Doppler
> shift moving the signals more than the channel bandwidth.
----
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