<p>The base formula for the direction <span class="math-container">$\theta$</span> is</p>

<p><span class="math-container">$\theta=atan2\left(\frac{\text{uwind}}{\text{vwind}}\right)$</span></p>

<p>Where <span class="math-container">$atan2$</span> is the four quadrant inverse tangent. Note that if you used the normal inverse tangent your results will be all collapsed to the first quadrant (0° to 90°). Also note that depending on the platform the result can be in radians or degrees (most often in radians, some platforms have a function atan2d that return values in degrees).</p>

<p>If you get a negative value (which is the same direction but measured counterclockwise), you have to add 360° (or <span class="math-container">$2\pi$</span> radians).</p>

<p>This will give you the direction the wind is coming from. For the direction is going to, you have to add/substract 180° (or <span class="math-container">$\pi$</span> radians).</p>

<p>For the values you provide, the answer would be 210°.</p>

CC BY-SA 4.0
2019-01-07T08:38:28.540Z
Query: maths & hence trigonometry uses a horizontal, x-direction, east-west line as it's reference & angles are measured anti-clockwise. Surveying and navigation uses the y-direction, north-south line as its reference & angles are measured clockwise. By simply using the *atan2* function, would this give the answer using the mathematical reference line & if so would it then need to be adjusted for surveying/navigation reference?
Fred
CC BY-SA 4.0
2019-01-07T16:14:07.003Z
@Fred You are right about the differences in the reference frame, but this formula does not know where does point the reference line the angle is measured from. In the math world uwind would be X, and vwind would be Y and you would consider the resulting angle as measured clockwise from the direction the positive X axis points to. But in the geographical world you can just use the same output angle and simply consider it as measured clockwise from the north arrow.
Camilo Rada
CC BY-SA 4.0
2019-01-06T16:14:30.207Z
2019-01-06T16:22:59.057Z
2019-01-06T16:22:59.057Z
Camilo Rada
<p>How to calculate wind direction from uwind and vwind?<br>
 if uwind and vwind are -1.82 , -3.18 respectively</p>

<meteorology><oceanography>
How to calculate Wind direction from uwind and vwind?
Asish K Sasidharan
# How to calculate Wind direction from uwind and vwind?

(Asked by: username_0 on 2019-01-06)

How to calculate wind direction from uwind and vwind?  
if uwind and vwind are -1.82 , -3.18 respectively

**Tags:** meteorology, oceanography


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## 1 answers

### Answer by username_1 on 2019-01-06. Score: 5

The base formula for the direction $\theta$ is

$\theta=atan2\left(\frac{\text{uwind}}{\text{vwind}}\right)$

Where $atan2$ is the four quadrant inverse tangent. Note that if you used the normal inverse tangent your results will be all collapsed to the first quadrant (0° to 90°). Also note that depending on the platform the result can be in radians or degrees (most often in radians, some platforms have a function atan2d that return values in degrees).

If you get a negative value (which is the same direction but measured counterclockwise), you have to add 360° (or $2\pi$ radians).

This will give you the direction the wind is coming from. For the direction is going to, you have to add/substract 180° (or $\pi$ radians).

For the values you provide, the answer would be 210°.

> **Comments:**
> **username_2**: Query: maths & hence trigonometry uses a horizontal, x-direction, east-west line as it's reference & angles are measured anti-clockwise. Surveying and navigation uses the y-direction, north-south line as its reference & angles are measured clockwise. By simply using the *atan2* function, would this give the answer using the mathematical reference line & if so would it then need to be adjusted for surveying/navigation reference?  
> **username_1**: @username_2 You are right about the differences in the reference frame, but this formula does not know where does point the reference line the angle is measured from. In the math world uwind would be X, and vwind would be Y and you would consider the resulting angle as measured clockwise from the direction the positive X axis points to. But in the geographical world you can just use the same output angle and simply consider it as measured clockwise from the north arrow.  

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