converting atan2 output to 360 deg
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Hi, I'm using the function atan2, however my output is from -180 to 180 degrees (I converted from radians) How do I modify it such that it outputs a value from 0-360 degrees?
winddir = (atan2(Vi,Ui))*(180/pi);
1 件のコメント
Image Analyst
2015 年 4 月 13 日
There is an atan2d() function you know. All the functions have "d" versions that work in degrees instead of radians. Though it goes from -180 to +180.
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John D'Errico
2015 年 4 月 14 日
編集済み: John D'Errico
2015 年 4 月 14 日
Since it will be periodic, just add 360 if the value is less than 0. This will suffice to correct the negative angles.
winddir = winddir + (winddir < 0)*360;
You can use atan2d if you prefer to work in degrees, but it will map to the same range, [-180,180], so you will still need to correct for the negative angles.
If you wanted a simple expression that works in one line of code, this should do:
winddir = atan2d(Vi,Ui) + 360*(Vi<0);
その他の回答 (2 件)
Stephen23
2022 年 5 月 30 日
編集済み: Stephen23
2022 年 5 月 30 日
The simple, robust, and efficient solution is to use MOD:
mod(atan2d(Y,X),360)
2 件のコメント
Sam Ruegsegger
2024 年 4 月 12 日
Stephen23,
I am writing to you to thank you for your robust and simple solution. In the midst of a mehatronics lab, trial after trial went by, and my team still could not find a working solution. Despair was creeping in. And then we met you. Your solution has given us a bulwark against the demons of MATLAB. You may have written this 2 years ago, but your legacy lives on in the MathWorks chat room.
Godspeed, brother.
Stephen23
2024 年 4 月 13 日
@Sam Ruegsegger: you can also vote for answers that helped you. It is an easy way to show your appreciation.
theodore panagos
2018 年 11 月 7 日
You can use the formula:
atan(x,y)=180/pi()*(pi()-p()/2*(1+sign(x))*(1-sign(y^2))-pi()/4*(2+sign(x))*sign(y)
-sign(x*y)*atan((abs(x)-abs(y))/(abs(x)+abs(y))))
That formula give the angle from 0 to 360 degrees for any value of x and y.
x=x2-x1 and y=y2-y1
For x=y=0 the result is undefined.
2 件のコメント
Nicolas soto vallejos
2022 年 5 月 30 日
Have to say this formula is awsome if you only need to know what quarternal direction the vector is traveling in you could just use the first part. sure only returns in 45 deg but it makes it easier to work with and ofcourse you could allways use the entire thing if you need exact angles but i'm amazed at the brillians of this formula 10/10!
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