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



On Dec 16, 4:13 pm, 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.

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

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

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

"Stefan" wrote in message
...
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.)

Of course you are correct, I meant to include Track in the knowns, (
afterall if you dont know the track whether you are the pilot or the ATC you
are really in trouble.)
terry

"Stefan" wrote in message
...
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.

While that statement is true, it doesn't cover all the possibilities.
We're talking about the case of a single aircraft? We know its HDG,
GS, TAS, but nothing else? That's only three of the six wind triangle
variables. We must know at least four to get an exact answer. With
only three knowns, there are a whole range of possible answers, since
we don't know what the wind correction angle is without knowing a CRS.

For example, for the following HDG, TAS and GS:

HDG 0
TAS 120
GS 100

All the following CRS, WD an WS are valid solutions, plus many more
in-between:

CRS WD WS
=== === ===
310 054 095
330 056 060
350 039 028
000 000 020 CRS=HDG
010 321 028
030 304 060
050 306 095

I think you were visualizing a triangle, and thought of the two obvious
solutions. But there are a lot more. Again, use the E6B method and
you'll see that any drift angle along the TAS arc contains a valid WS
and WD answer.

Regards, Kev

Chad,

I think that some clarification of the use might help. I can imagine two
scenarios:

1) enroute - in this case, you know the track and ground speed from radar.
You have the filed TAS. You have no idea of the crab angle or heading. This
sounds like the information you presented in your question. However,
enroute, who cares what the wind correction is. I am having trouble
imagining how this would help a controller - in training or not.

2) terminal - in this case, you still know the track and ground speed from
radar. However, now the filed TAS is probably no longer valid since it is
probably at a lower altitude than filed and used to compute the TAS. The
plane may also be under speed constraints. On the other hand, the controller
may be vectoring so now you have an assigned heading and maybe even an
assigned speed. You also have the pressure for the area and the TAS can be
computed from that and radar data. If this were an automated computation, it
could be done on every plane that was given an assigned speed (or even IAS
queried from the pilot) and an assigned heading (or even queried from the
pilot). Now maybe not every plane can provide all of the data but certainly
some could. Enough to update the calculated wind periodically.

Scenario 2 seems to me the one that would make knowledge of the wind most
desirable. It is also the scenario that could provide additional data and
certainly provides different data than what was posed in your question.

What say you?

--
-------------------------------
Travis
Lake N3094P
PWK

"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 ?