Magnitude and direction from north and east components

I have displacement data components along north and east how to find its magnitude and direction with reference to north(0-360) in matlab. The components may have same sign or opposite both possible.

[Deepak Gupta]

Hi Mithun,

Can you give sample data? And expected result.

[TAPAS]

Let north=-0.34 east=0.54

Answer 1

[Deepak Gupta]

Hi Mithun,

You can think of North and East as your X and Y. As you have taken north as reference so use below formulas to calculate magnitude and angle.

Magnitude = sqrt(North^2+East^2);
Theta = atan((-East)/North);

I am using -East because East is 90 degree closewise to North and atan calculates angles in counter clockwise directions from reference.

Thanks,

Deepak

[TAPAS]

Is the direction from North and is it in the range[0, 360]

[Deepak Gupta]

To caluculate direction directly between [0, 360], you can use this:

atan2d(-East,North) + 360*(-East<0)

Yes, it will give you direction in reference North direction. Try it yourself.

[TAPAS]

Are you sure we need to do -East because we are already doing atan2d(East, north)

[Deepak Gupta]

Read comments added already and think before asking further questions. Test the code with a known values.

[TAPAS]

Thank you

Answer 2

[KSSV]

If (x,y) is a componenet

m = sqrt(x^2+y^2)  ;   % magnitude 
theta = atan(y/x) ;   % direction 

[TAPAS]

What should be my y??

[TAPAS]

And shall I get the direction from North reference?