wind direction

[Ipswichweather.co.uk]

Hello,

I am not to sure weather i have posted this in the right section of this forum.

I need help caculating wind direction from the U and V components of wind.

I can caculate the angle and degree, but having trouble detrerming the quadrent.

After spending many weeks on this i think i have come up with a function that returns the inverse tangent of (y/x) in the range -π ≤ return_val ≤ π, using the signs of both arguments to determine the quadrant of the return value

The formula:

r2d = 45.0/atan(1.0) (radians to degrees) or 57.2957795

dir = atan2(u, v) * r2d + 180

If some one here is able to check this formula, or knows another way please post.

I would be most greatful.

Ian Gooch

[johnholmes]

you've lost me mate.

Why do you need u and v components?

can't really help beyond that very simple idea, certainly not at this time of night!

Edited May 9, 2007 by johnholmes

[Ipswichweather.co.uk]

Hello John,

Why do you need u and v components?

I am procducing my own forecasts based on the GFS 0.05deg NWP model. Unlike others who use this data I use a program that converts GRIB data into its raw state.

and does not do the caculations for me.

The reason for this is: I can then add my own caculations based on the topography of the area i am forecasting. Plus I can change the data, if some of the values are missing - quite comman with data set as large as the GFS.

This should allow me to produce a more realistic forecast - Well thats the plan away

Now wind is packed in a GRIB message as U m/s and V m/s at the required height

Example: A wind @10 meters would return as 10 HGTL V=2.3 m/s U=2.6 m/s

Now to caculate the direction you need for find the angle

V = North to South if V<0

V= South to North if V>0

U= West to East if U>0

U= East to West if U<0

Next you need to find the Quadrent, and this is where my trouble lies.

Now

N 360

|

|

W 270-----------90 E

|

|

S 180

Now based on the example above U=2.6 and V=2.3, this to me would put it in the quadrent SW (between S and W)

As U>0 and V>0.

I caculated the angle and ask if a met office forecaster would check the anwser and my formula, before i placed it in to my math model.

The relpy was that it should be in the SE quadrant, but i can't see how.

I contacted NOAA who gave me the formula in my earlier message, with should caculate the Quadrent for me, as well as the angle.

I just need some further advice weather this formula is correct, so it can be placed in my model, and carry on with my work.

Ian

[johnholmes]

Are you sure this is the correct way

V = North to South if V<0

V= South to North if V>0

U= West to East if U>0

U= East to West if U<0

It seems logical to me but I'm no expert. Did the answer from NOAA and the Met O agree or not?

[Thundery wintry showers]

I did something very similar in my MRes project at Leeds.

My understanding is that U (+ve) = west-east, and V (+ve) = south-north, corresponding to the x and y axis respectively.

Thus, I'd have thought that if U and V were positive the wind would be coming from the SW quadrant, i.e. from west-east and from south-north.

[VillagePlank]

Here you go . . .

'a=(u,v) 'b=(0,1)

Dim ma As Single ' |a|

Dim m As Single

Dim d As Single

' Get wind speed (magnitude) and convert it to mph

ma = Sqr(u * u + v * v)

m = ma * 2.24

' Get angle(d) difference between a, and b

' because cos(d) = a.b/|a||b|, therefore d = arccos(a.b/|a||b|)

' simplified d = arccos(v/|a|)

d = 360 - (ArcCos(v / ma) * 57.295780490443)

d will give you the angle (from 0..360) so you should be able to determine what you need from that.

Edited May 10, 2007 by VillagePlank

[johnholmes]

I wish my brain still worked!

[Ipswichweather.co.uk]

I did something very similar in my MRes project at Leeds.

My understanding is that U (+ve) = west-east, and V (+ve) = south-north, corresponding to the x and y axis respectively.

Thus, I'd have thought that if U and V were positive the wind would be coming from the SW quadrant, i.e. from west-east and from south-north.

My thoughts exactly.

John - The replys I get from NOAA are very hard to understand, they do not make things clear to me. I think that because they are all hold Phd's they are unable to explain in simple terms to the likes of which i can understand.

While the Met Office forecast did make to clear, but i believe they got it wrong. I reply i got from the UKMO was infact relayed though their customer enquires, so perhaps thats where the mistake was made.

Mind you they did get the 1987 storm wrong - perhaps if they new which way the wind was blowing, they would of got it right LOL.

Ian

[VillagePlank]

Did my post give you what you needed? Do you want me to explain it in further detail? (might take a while to construct the post - so I didn't bother)

Edited May 10, 2007 by VillagePlank

[Ipswichweather.co.uk]

Here you go . . .

'a=(u,v) 'b=(0,1)

Dim ma As Single ' |a|

Dim m As Single

Dim d As Single

' Get wind speed (magnitude) and convert it to mph

ma = Sqr(u * u + v * v)

m = ma * 2.24

' Get angle(d) difference between a, and b

' because cos(d) = a.b/|a||b|, therefore d = arccos(a.b/|a||b|)

' simplified d = arccos(v/|a|)

d = 360 - (ArcCos(v / ma) * 57.295780490443)

d will give you the angle (from 0..360) so you should be able to determine what you need from that.

That is the same as my first formula 57.2957*ATAN2(u,v)+180

That, as well as yours, should determing the quadrant in the sum.

Ian