I have a series of vectors (current speeds) based on an ENU (East North Up) system. I would like to estimate the current speed of the water headed in a specific direction.

In this example I am interested in isolating the water flowing in the direction of 150 degrees, and am looking at the best way to calculate flow speeds of water in this direction. Is it possible to rotate the y component of the flow 30 degrees (180-150=30) and use this to estimate the magnitude of flow at 150 degrees?


u=0.0407 m/s
v=-0.1392 m/s

Speed=0.1451 m/s  Direction = 163 degrees

Can I rotate the south v vector component like so

New_v= v*cos(30) + u*sin(30)  
New_v=-0.1392*cos(30) + 0.047*sin(30)

And say the velocity of water flowing at 150 degrees would be .10023 m/s?

If I am way off, I'd be happy to learn a better way to do this


1 Answer 1


First note that your angle is being measured in a clockwise sense with $\theta=0$ being the $y$-axis.

the thing you want to calculate is the magnitude of the component of velocity along a particular axis right? Then the way I would do it is not look at $u$ and $v$ but instead look at the Speed "$s$" and Direction "$\theta$" that you are also given. If you want to know the magnitude of the component of velocity along a direction $\phi$, compute $s \cos(\theta - \phi)$.

You could do it with the $u$'s and $v$'s. Then let $c = \cos(\phi)$ and redefine $s$ to be $\sin(\phi)$. Then the answer would be $us+vc$. The idea being that you are rotating your vector by $\phi$ so that what was in the $\phi$ direction is now pointing along the $y$ axis. Then you take the $y$ component to get what was the component along the $\phi$ direction before the rotation.

Note that what you did wrong is you have $\pi - \phi$ instead of $\phi$.

  • $\begingroup$ These are my calculations- Speed=0.1451 and Direction=150. So s cos(theta-direction) would be 0.1451*cos(0-150)= -0.1256603. Meaning the magnitude of my component velocity along 150 degrees is 0.1256 $\endgroup$
    – Vinterwoo
    Sep 17, 2013 at 20:29
  • $\begingroup$ Sorry I was not clear enough in my answer. By the direciton $\theta$, I was talking about where it says Direction = 163 degrees. So instead of 0.1451*cos(0-150), you do 0.1451*cos(163-150). You should get something close to the full speed. $\endgroup$ Sep 17, 2013 at 20:33
  • $\begingroup$ Yet if I work through the u and v- if I let c= cos(150)=-0.8660254 and s=sin(150)=0.5. Then us + vc= 0.0407*0.5 + 0.1392*-0.8660=-0.1001972. $\endgroup$
    – Vinterwoo
    Sep 17, 2013 at 20:34
  • $\begingroup$ you made a sign error when you plugged in $v$. If you fix it, you get the right answer. $\endgroup$ Sep 17, 2013 at 20:45
  • $\begingroup$ Yes, Now that I am using theta =163-150 and corrected my error- everything adds up. Thanks for your help! $\endgroup$
    – Vinterwoo
    Sep 17, 2013 at 20:50

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