convert wind direction in true North to Math convention

Hi all,

I am very confused about how to convert wind direction in true North to math convention (Trigonometric style). Could anyone offer some help?

Thanks

[Geoff Hayes]

tanq - do you have any examples or can you provide some context? If the wind is coming from the north (0 degrees), is the "math convention" just

 0 + 180 = 180 degrees

and then you convert that into radians? I'm assuming that the compass is like

              N
              ^
              |
         W < - - > E
              |
              S

And so a wind coming from the north has a direction towards the south i.e. 180 degrees along the y-axis. Likewise, a wind from the west is blowing to the east and so its direction would be

 270 + 180 = 450 = 90 degrees

I suppose that given this angle you may want to determine the x- and y-components. Since the angle is relative to true north, then for wind coming from the north with speed ws

 theta = 180 (degrees)
 x = ws * sind(theta)
 y = ws * cosd(theta)

Or the above is wrong and isn't what you are looking for...

Answer 1

[Jim Riggs]

Working with wind direction vectors can be confusing, because there are two different conventions which are both widely used.

Meteorologists tend to describe the wind direction as the direction "from" which the wind is coming. i.e. the direction you would be facing with the wind hitting you directly in the face.

But many engineers prefer to use a wind vector defined by the direction "toward" which the wind is going. This is the direction you would be facing with the wind hitting you in the back of your head.

As you can see, these two "directions" are opposite even though they are describing the same vector. So the first step is to determine whether you are using the "from" convention or the "toward" convention.

(If you have data from a meteorological source, the wind direction is probably "direction from") The rest is simple - just draw a picture of the problem.

[wave_buoys]

Ok. thanks all.

So let's say I have a list of direction provided by Meteorologists (coming from):

55 130 220 290 (degree)

how could I convert this data into the engineering style (going to) using Matlab? Sorry for asking this simple question.

Thanks

[Image Analyst]

Not sure what you mean. Attach your data file. Or else simply do

windSpeeds = [55, 130, 220, 290];

Not sure what you want to do after that with the speeds though.

[Jim Riggs]

 windir_from = [55, 130, 220, 290];
 windir_toward = windir_from + 180;

or

 windir_toward = winddir_from - 180;

Either of these will work. Now, if you want the component in the North and East directions then you first must convert directions to radians. Then its

 Vwind_North = windspeed .* cos(windir_toward);
 Vwind_East =  windspeed .* sin(windir_toward);

These are the vector components of the wind (toward) which is how they would be used in a typical engineering application.

[wave_buoys]

thanks Jim Riggs, but if I just simply add 180 deg into the wind direction windir_from: windir_from = [55, 130, 220, 290];

I get: windir_toward = [235 310 400 470]

Is this correct? because the wind direction in both cases ranges from 0 to 360 deg.

Please help & thanks

[Jim Riggs]

The trigonometry functions will work with these numbers just fine. Look at the values for Vwind_North and Vwind_East, do they look right? Don't forget that the Matlab functions (sin, cos) require units of radians. If you prefer, you can apply a 0 to 360 degree limiting function;

 if x < 0
     x=x+360;
  elseif x > 360
    x=x-360;
  end

[Image Analyst]

Or you could use mod() or rem(). Or use sind() and cosd() if you want to do operations on angles in degrees.

However I still don't know what, given an angle in degrees, "convert wind direction in true North to math convention" means. What is "math convention"??? You've got degrees, so what do you want? I still have no idea.

[wave_buoys]

great thanks all for your contributions. I sorted out

Answer 2

[Mayra]

I have been struggling with the same problem and I found an easy way to deal with this.

First you should convert your original data (direction with true north, i.e. clockwise and 0° at the top and wind intensity, I will call Direction and Wind) to cartesian coordinates.

[u,v] = pol2cart(deg2rad(Direction),Wind);

Then you convert it againd to polar coordinates but swaping u and v

[theta,rho]=cart2pol(v,u);

Now theta is the wind direction in polar coordinates, i.e. counterclockwise and 0° at right.

Answer 3

[wave_buoys]

Sorry. Here is my data from meteorological source. I want to convert it into engineering style. How could I do in Matlab?

Thanks

[Image Analyst]

OK, what about it? What do you want to do with it? I have no idea.