Looking for a math wiz!

Okay, I know I've seen a lot of engineers and technical folks on here.
I have a complex math problem relating to the classic wind triangle
that I posted on sci.math and received little response. I don't know
if they're stumped or just not interested. :-)

Here is a copy of my original post and the only useful response I
received. Anyone have a solution?

(For the controllers here, this is an enhancement we are trying to add
to the Falcon program that centers will see next year, which was
developed by a controller here at ZKC.)


Chad Speer
PP-ASEL, IA
ATCS, Kansas City ARTCC



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************************************************** *

My original post:
*****
I am helping someone with a program that estimates wind speed and
direction using radar data from aircraft. I need help finding a
formula that can determine the wind speed and direction when given the
following information for multiple aircraft:

direction of travel
speed across the ground
speed through the air

The direction of travel and the speed across the ground are taken from
the radar data. The speed through the air is taken from the pilot's
flight plan. We're air traffic controllers trying to improve our
training tools, so we get access to all the goodies.

I know that with information from just one aircraft, the possibilities
are endless for the wind speed and direction. I think it is possible
to use the same data from two or more aircraft to determine the wind
speed and direction.

I thought I could come up with a formula to solve this, but the need to
reference everything to north in order to achieve actual directions
instead of just angles took it way above my head.

Basically, you are given the lengths of two adjacent sides of many
different triangles. You also know the angle of one of those sides
(aircraft direction) with respect to a known reference (north). The
properties common to all of the triangles, which are unknown, are the
length of the third side (wind speed) and the angle of that third side
(wind direction) with respect to a known reference (north).

If I have not described this well enough, I can upload some diagrams to
a web page to simplify the problem. I would really be grateful if
someone is able to solve this!
*****



A useful response:
*****
Interesting problem.

For the single aircraft, case, let:
Wd, Ws be wind direction and speed.
Ad, As be aircraft direction and speed (relative to the air, not the
ground).
Gd, Gs be aircraft direction and speed (relative to the ground).

Representing each vector as a 2-tuple of (direction, magnitude) gives:
(Gd, Gs) = (Ad, As) + (Wd, Ws)

The unknowns are Ad, Ws, and Wd. (Note that As is known due to filed
information, i.e. a pilot will know how fast his aircraft cruises.)

Splitting into x- and y-components will result in 2 equations with 3
unknowns. I agree that there are infinitely many solutions in the
single-aircraft case.

For the two-aircraft case (and I'll just suffix with 1 and 2), we have:
(Gd1, Gs1) = (Ad1, As1) + (Wd, Ws)
(Gd2, Gs2) = (Ad2, As2) + (Wd, Ws)

where of course we assume that the wind affecting each aircraft (since
they are presumably not too many tens of miles apart) is the same.
Breaking into x- and y-components leads to 4 equations and 4 unknowns
(Ad1, Ad2, Wd, Ws). The 2-aircraft case probably has a unique solution.
We have:

Gs1 * cos(Gd1) = As1 * cos(ad1) + Ws * cos(Wd)
Gs1 * sin(Gd1) = As1 * sin(ad1) + Ws * sin(Wd)
Gs2 * cos(Gd2) = As2 * cos(ad2) + Ws * cos(Wd)
Gs2 * sin(Gd2) = As2 * sin(ad2) + Ws * sin(Wd)

One can probably square equations, add, and make use of the
relationship that sin^2(x) + cos^2(x) = 1. I'm not sure what the form
of the solution would be. I'm too lazy to work it out. It will be
messy.

I believe at first glance that the 2-aircraft case has a single unique
solution (4 equations, 4 unknowns).

However, moving on to more than 2 aircraft ...

If there are more than 2 aircraft, the system is "overspecified" (there
is a mathematical term for this, but it has been so long ...). You
probably want a way to pick a single best solution among the infinitely
many, assuming that you have some "noise" in the data.

The style of solution you want is probably about the same as a
"least-squares" solution to a system of linear equations, i.e.

http://www.mathresource.iitb.ac.in/l...hapter8.5.html

I would need to do some thinking about how to phrase this problem as a
least-squares problem (the sines and cosines above put doubts in my
head), but there is probably a way to do it. So, out of a group of
data for at least 2 aircraft, you should be able to grind out a
solution that is unique according to some constraints and assumptions.

To summarize my thoughts:

a)1 aircraft -- system not solvable.
b)2 aircraft -- system has one solution, but I'm too lazy to do the
algebra.
c)3 or more aircraft -- system is overspecified, and some least
squares approach should give a solution.

One more thought: I've spent a fair amount of time in little Cessnas.
It has been my experience that wind direction and speed won't vary too
much over distance, but may vary EXTREMELY with altitude. I've flown
on days when the winds at 3,000 feet were 15 knots and the winds at
6,000 feet were 50 knots (or at least this is my memory). I accept the
assumption that winds affecting aircraft at the same altitude that are
within maybe 20NM of each other are about the same. But I do not
accept the assumption that winds at different altitudes are similar --
my experience says otherwise. If you agree, this adds yet another
dimension to the problem.

(BTW, near the end of my student training, I used to like to fly in
heavy crosswinds to practice technique. We have an airport about 20NM
North of our local airport with a roughly perpendicular runway, and I
discovered that if the wind was blowing straight down the runway here I
could get a perfect crosswind to practice with just by going N and
using the other airport. The surface winds were always about the same
at both airports. That is why I'm comfortable with the assumption that
winds don't vary much over relative short distances.)

Good problem.

I am not a mathematician. I hope others can add more insight.
*****

"Chad Speer" wrote in message
ups.com...
Okay, I know I've seen a lot of engineers and technical folks on here.
I have a complex math problem relating to the classic wind triangle
that I posted on sci.math and received little response. I don't know
if they're stumped or just not interested. :-)

Here is a copy of my original post and the only useful response I
received. Anyone have a solution?

(For the controllers here, this is an enhancement we are trying to add
to the Falcon program that centers will see next year, which was
developed by a controller here at ZKC.)


Chad Speer
PP-ASEL, IA
ATCS, Kansas City ARTCC



************************************************** *
************************************************** *

My original post:
*****
I am helping someone with a program that estimates wind speed and
direction using radar data from aircraft. I need help finding a
formula that can determine the wind speed and direction when given the
following information for multiple aircraft:

direction of travel
speed across the ground
speed through the air

The direction of travel and the speed across the ground are taken from
the radar data. The speed through the air is taken from the pilot's
flight plan. We're air traffic controllers trying to improve our
training tools, so we get access to all the goodies.

I know that with information from just one aircraft, the possibilities
are endless for the wind speed and direction. I think it is possible
to use the same data from two or more aircraft to determine the wind
speed and direction.

I thought I could come up with a formula to solve this, but the need to
reference everything to north in order to achieve actual directions
instead of just angles took it way above my head.

snip

Chad
if you know HDG ( ie where you are pointing), GS and TAS then there is only
1 possibility for the wind speed and direction. these can be calculated
from the cosine rule. If you know the cosine rule and the sine rule for
triangles you can calculate a lot of things.
to apply both of these rules draw yourself a little triangle and mark the
sides small a,b and c.
then mark the angles capital A,B and C where angle A is opposite side a
and angle B is opposite side b.
cosine rule a^2 = b^2 +c^2 - 2bc cos( A)
sine rule a/sin A = b/sin B = c /sin C

in the case of the NAV triangle

WS= SQRT( GS^2+TAS^2 =2*GS*TAS*cos(HDG-TR))
where WS= windspeed
GS = ground speed
TAS = airspeed
HDG = heading(where you are pointing)
TR = track ( where you are going)

If you want I can email you an excel spreadsheet that has this already
coded. you just enter your TAS, GS and HDG and it will give you the WS and
Wind direction.

Terry
PPL

d&tm schrieb:

if you know HDG ( ie where you are pointing), GS and TAS then there is only
1 possibility for the wind speed and direction.

Actually, there are two.

But the question was a different one. It has already been answered
pretty well, all what remains is to do the dirty work and shuffling some
formulas.

Stefan

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 Speer
PP-ASEL, IA
ATCS, Kansas City ARTCC

Stefan wrote:
d&tm schrieb:

if you know HDG ( ie where you are pointing), GS and TAS then there is only
1 possibility for the wind speed and direction.

Actually, there are two.

Eh? Not to sidetrack the thread too much, but how could there be two
wind answers?

For example on the E-6B, to solve this, you'd set TRACK up, grommet
over GS, and then look for where your TAS arc meets your drift
correction angle (HDG-TRACK). The vector back to the grommet is the
single direction and speed for the wind.

Thanks, Kev

"Stefan" wrote in message
...
d&tm schrieb:

if you know HDG ( ie where you are pointing), GS and TAS then there is
only
1 possibility for the wind speed and direction.

Actually, there are two.

I give up, can you please explain how there can be 2 ?

"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.

Now we have 2 equations with 2 unknowns ( WS and WD) which should be
solvable but I am stuggling to get it out. However I did a little exercise
which suggests the end result may not be very useful. because small errors
in the ground speeds will cause very large errors in the calculated wind
speed.

By subtracting the 2 equations above you get
WS=(GS2^2-GS1^2-TAS2^2+TAS1^2)/(2(GS2COS(180-ABS(WD-TR2)-GS1COS(180-ABS(WD-T
R1))
What I then did was calculate the ground speeds for 2 aircraft for a known
wind of 20 kts from 220 M

aircraft 1 aircraft 2
TAS 120 100
GS 111.7 91.4
TR 290 150

If I then use my above equation for WS after setting wind direction to 220,
I get 20 kts as expected. However if I round off the GS to 112 and 91 kts
the Wind speed changes from 20 to 9.5kts which suggests the ground speeds
have to be super accurate to get anywhere near the right wind speed.. Given
that also you are taking the TAS from a flight plan, which will vary with
density altitude and RPM setting etc. suggests your objective is going to be
rather difficult to achieve in practice.

Hope this helps and good luck. If you come up with the solution I would love
to see it.
terry

"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.

Now we have 2 equations with 2 unknowns ( WS and WD) which should be
solvable but I am stuggling to get it out. However I did a little
exercise
which suggests the end result may not be very useful. because small errors
in the ground speeds will cause very large errors in the calculated wind
speed.

By subtracting the 2 equations above you get

WS=(GS2^2-GS1^2-TAS2^2+TAS1^2)/(2(GS2COS(180-ABS(WD-TR2)-GS1COS(180-ABS(WD-T
R1))
What I then did was calculate the ground speeds for 2 aircraft for a
known
wind of 20 kts from 220 M

aircraft 1 aircraft 2
TAS 120 100
GS 111.7 91.4
TR 290 150

If I then use my above equation for WS after setting wind direction to
220,
I get 20 kts as expected. However if I round off the GS to 112 and 91 kts
the Wind speed changes from 20 to 9.5kts which suggests the ground speeds
have to be super accurate to get anywhere near the right wind speed..
Given
that also you are taking the TAS from a flight plan, which will vary with
density altitude and RPM setting etc. suggests your objective is going to
be
rather difficult to achieve in practice.

Hope this helps and good luck. If you come up with the solution I would
love
to see it.
terry

should have mentioned
in 180- ABS( WD-TR)
ABS is absolute , ignore negatives.
terry

Lots of fun to play with the formulas... But, the fly in the ointment
here is that the airspeed filed by the pilot is a canned number and has
nothing to do with reality... I fly IFR, I file the same airspeed - 130
kt - for all flights regardless of load, altitude, and what the power
setting winds up to be... If the actual airspeed in your problem is
within 15% of the filed number you will be lucky...

denny

Kev schrieb:

Eh? Not to sidetrack the thread too much, but how could there be two
wind answers?

Mathematically: There are always two square roots which solve the
equation: A positive and a negative.

Physically: If you only know GS, TAS and HDG, then you don't know
whether the wind blows from the let or from the right. (If you also know
the track, then of course there's only one solution.)

Stefan