"d&tm" wrote in message
...

"Chad Speer" wrote in message
oups.com...
Terry - thanks for the reply, but heading is not known.

Stefan - we need to be able to plug these known values into a formula
and kick out a result. Assuming the original responder was on the
right track, I still don't know what to do with his suggestion. Any
ideas on that? I'm not lazy, this just went over my head a long time
ago. :-)

Chad, sorry misread the question. But I like a challenge so I had another
go.
Firstly I dont believe the equations given by the original poster are
correct. if you apply the cosine rule to the wind triangle for 2 aircraft
you get the following 2 equations.

TAS1^2=WS^2+GS1^2-2WSGS1COS(180-ABS(WD-TR1)
TAS2^2=WS^2+GS2^2-2WSGS2COS(180-ABS(WD-TR2)

WHERE TAS1 AND TAS2 ARE TRUE AIRSPEEDS FOR AIRCRAFT 1 AND 2
WS = WIND SPEED
WD =WIND DIRECTION IN DEGREES MAG
TR1 AND TR2 ARE TRACKS MAGNETIC FOR AIRCRAFT 1 AND 2
GS1 AND GS2 ARE GROUND SPEEDS FOR AIRCRAFT 1 AND 2.
(180-ABS(WD-TR) WILL GIVE YOU THE ACTUAL ANGLE BETWEEN WIND DIRECTION AND
TRACK.

Further to this post ,using the above 2 equations , with the following
correction
cos( abs( abs(wd -tr)-180)), the problem was solved graphically by solving
the 2 quadratics , or 3 for a 3 aircraft case for wind speed, for each of
the 360 degrees of possible wind direction. this gives you a parabola for
each aircraft and the overlap of the parablolas gives you the required
answer. For 2 aircraft there can be either 1 or 2 solutions. having a
3rd aircraft will give one unique solution..
This solution is available as an Excel spreadsheet if anybody is interested.
terry