How to calculate Wind direction from uwind and vwind?
$begingroup$
How to calculate wind direction from uwind and vwind?
if uwind and vwind are -1.82 , -3.18 respectively
meteorology oceanography
asked Jan 6 at 15:57
ThehalfarisedPheonixThehalfarisedPheonix
111
$endgroup$
add a comment |
$begingroup$
How to calculate wind direction from uwind and vwind?
if uwind and vwind are -1.82 , -3.18 respectively
meteorology oceanography
asked Jan 6 at 15:57
ThehalfarisedPheonixThehalfarisedPheonix
111
$endgroup$
$begingroup$
colaweb.gmu.edu/dev/clim301/lectures/wind/wind-uv if you want more detail for the formula provided in the answer
$endgroup$
– gansub
Jan 7 at 3:56
add a comment |
$begingroup$
How to calculate wind direction from uwind and vwind?
if uwind and vwind are -1.82 , -3.18 respectively
meteorology oceanography
asked Jan 6 at 15:57
ThehalfarisedPheonixThehalfarisedPheonix
111
$endgroup$
How to calculate wind direction from uwind and vwind?
if uwind and vwind are -1.82 , -3.18 respectively
meteorology oceanography
meteorology oceanography
asked Jan 6 at 15:57
ThehalfarisedPheonixThehalfarisedPheonix
111
asked Jan 6 at 15:57
ThehalfarisedPheonixThehalfarisedPheonix
111
asked Jan 6 at 15:57
ThehalfarisedPheonixThehalfarisedPheonix
111
asked Jan 6 at 15:57
ThehalfarisedPheonixThehalfarisedPheonix
111
asked Jan 6 at 15:57
ThehalfarisedPheonixThehalfarisedPheonix
111
111
$begingroup$
colaweb.gmu.edu/dev/clim301/lectures/wind/wind-uv if you want more detail for the formula provided in the answer
$endgroup$
– gansub
Jan 7 at 3:56
add a comment |
$begingroup$
colaweb.gmu.edu/dev/clim301/lectures/wind/wind-uv if you want more detail for the formula provided in the answer
$endgroup$
– gansub
Jan 7 at 3:56
$begingroup$
colaweb.gmu.edu/dev/clim301/lectures/wind/wind-uv if you want more detail for the formula provided in the answer
$endgroup$
– gansub
Jan 7 at 3:56
$begingroup$
colaweb.gmu.edu/dev/clim301/lectures/wind/wind-uv if you want more detail for the formula provided in the answer
$endgroup$
– gansub
Jan 7 at 3:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The base formula for the direction $theta$ is
$theta=atan2left(frac{text{uwind}}{text{vwind}}right)$
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 $2pi$ 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°.
answered Jan 6 at 16:14
Camilo RadaCamilo Rada
8,54012966
$endgroup$
$begingroup$
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?
$endgroup$
– Fred
Jan 7 at 8:38
$begingroup$
@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.
$endgroup$
– Camilo Rada
Jan 7 at 16:14
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The base formula for the direction $theta$ is
$theta=atan2left(frac{text{uwind}}{text{vwind}}right)$
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 $2pi$ 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°.
answered Jan 6 at 16:14
Camilo RadaCamilo Rada
8,54012966
$endgroup$
$begingroup$
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?
$endgroup$
– Fred
Jan 7 at 8:38
$begingroup$
@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.
$endgroup$
– Camilo Rada
Jan 7 at 16:14
add a comment |
$begingroup$
The base formula for the direction $theta$ is
$theta=atan2left(frac{text{uwind}}{text{vwind}}right)$
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 $2pi$ 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°.
answered Jan 6 at 16:14
Camilo RadaCamilo Rada
8,54012966
$endgroup$
$begingroup$
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?
$endgroup$
– Fred
Jan 7 at 8:38
$begingroup$
@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.
$endgroup$
– Camilo Rada
Jan 7 at 16:14
add a comment |
$begingroup$
The base formula for the direction $theta$ is
$theta=atan2left(frac{text{uwind}}{text{vwind}}right)$
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 $2pi$ 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°.
answered Jan 6 at 16:14
Camilo RadaCamilo Rada
8,54012966
$endgroup$
The base formula for the direction $theta$ is
$theta=atan2left(frac{text{uwind}}{text{vwind}}right)$
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 $2pi$ 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°.
answered Jan 6 at 16:14
Camilo RadaCamilo Rada
8,54012966
edited Jan 6 at 16:22
answered Jan 6 at 16:14
Camilo RadaCamilo Rada
8,54012966
answered Jan 6 at 16:14
Camilo RadaCamilo Rada
8,54012966
answered Jan 6 at 16:14
Camilo RadaCamilo Rada
8,54012966
8,54012966
$begingroup$
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?
$endgroup$
– Fred
Jan 7 at 8:38
$begingroup$
@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.
$endgroup$
– Camilo Rada
Jan 7 at 16:14
add a comment |
$begingroup$
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?
$endgroup$
– Fred
Jan 7 at 8:38
$begingroup$
@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.
$endgroup$
– Camilo Rada
Jan 7 at 16:14
$begingroup$
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?
$endgroup$
– Fred
Jan 7 at 8:38
$begingroup$
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?
$endgroup$
– Fred
Jan 7 at 8:38
$begingroup$
@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.
$endgroup$
– Camilo Rada
Jan 7 at 16:14
$begingroup$
@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.
$endgroup$
– Camilo Rada
Jan 7 at 16:14
add a comment |
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$begingroup$
colaweb.gmu.edu/dev/clim301/lectures/wind/wind-uv if you want more detail for the formula provided in the answer
$endgroup$
– gansub
Jan 7 at 3:56