Looking for a math wiz!

Chad Speer wrote:
This is interesting. I hadn't considered breaking them up into
headwind and crosswind components. I see the simplicity in that, but
I'll have to run that through the brain a few times to see how it fits.
I certainly can't dismiss it offhand!

No crosswind calcs needed. No trigonometry needed. Just addition and
subtraction.

In your case, aircraft always follow a specified course. Headings
don't matter. Crosswinds don't matter. The only thing that changes is
their speed, and that's simply TAS +/- wind speed for a particular
direction.

So if you calculate at startup the head/tail wind (TAS-GS) for two
aircraft flying basically North or South, and East or West, then you
have enough rough information to extrapolate the speed of the trainee's
airliner, as it turns to face different directions. It's not precise,
but it's good enough to make things more realistic by at least having
the speed change appropriately for each direction.

Kev

"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