Averaging angles… Again

Question

I want to calculate the average of a set of angles, which represents source bearing (0 to 360 deg) - (similar to wind-direction)

I know it has been

Answer 1

Thank you all for helping me see my problem more clearly.

I found what I was looking for. It is called Mitsuta method.

The inputs and output are in the range [0..360).

This method is good for averaging data that was sampled using constant sampling intervals.

The method assumes that the difference between successive samples is less than 180 degrees (which means that if we won't sample fast enough, a 330 degrees change in the sampled signal would be incorrectly detected as a 30 degrees change in the other direction and will insert an error into the calculation). Nyquist–Shannon sampling theorem anybody ?

Here is a c++ code:

double AngAvrg(const vector<double>& Ang)
{
    vector<double>::const_iterator iter= Ang.begin();

    double fD   = *iter;
    double fSigD= *iter;

    while (++iter != Ang.end())
    {
        double fDelta= *iter - fD;

             if (fDelta < -180.) fD+= fDelta + 360.;
        else if (fDelta >  180.) fD+= fDelta - 360.;
        else                     fD+= fDelta       ;

        fSigD+= fD;
    }

    double fAvrg= fSigD / Ang.size();

    if (fAvrg >= 360.) return fAvrg -360.;
    if (fAvrg <  0.  ) return fAvrg +360.;
                       return fAvrg      ;
}

It is explained on page 51 at http://www.epa.gov/scram001/guidance/met/mmgrma.pdf

Thank you MaR for sending the link as a comment.

If the sampled data is constant, but our sampling device has an inaccuracy with a Von Mises distribution, a unit-vectors calculation will be appropriate.

Answer 2

You could do this: Say you have a set of angles in an array angle, then to compute the array first do: angle[i] = angle[i] mod 360, now perform a simple average over the array. So when you have 360, 10, 20, you are averaging 0, 10 and 20 - the results are intuitive.

Answer 3

What does it even mean to average source bearings? Start by answering that question, and you'll get closer to being to define what you mean by the average of angles.

In my mind, an angle with tangent equal to 1/2 is the right answer. If I have a unit force pushing me in the direction of the vector (1, 0), another force pushing me in the direction of the vector (1, 0) and third force pushing me in the direction of the vector (0, 1), then the resulting force (the sum of these forces) is the force pushing me in the direction of (1, 2). These the the vectors representing the bearings 0 degrees, 0 degrees and 90 degrees. The angle represented by the vector (1, 2) has tangent equal to 1/2.

Responding to your second edit:

Let's say that we are measuring wind direction. Our 3 measurements were 0, 0, and 90 degrees. Since all measurements are equivalently reliable, why shouldn't our best estimate of the wind direction be 30 degrees? setting it to 25.56 degrees is a bias toward 0...

Okay, here's an issue. The unit vector with angle 0 doesn't have the same mathematical properties that the real number 0 has. Using the notation 0v to represent the vector with angle 0, note that

0v + 0v = 0v

is false but

0 + 0 = 0

is true for real numbers. So if 0v represents wind with unit speed and angle 0, then 0v + 0v is wind with double unit speed and angle 0. And then if we have a third wind vector (which I'll representing using the notation 90v) which has angle 90 and unit speed, then the wind that results from the sum of these vectors does have a bias because it's traveling at twice unit speed in the horizontal direction but only unit speed in the vertical direction.

Answer 4

What is wrong with taking the set of angles as real values and just computing the arithmetic average of those numbers? Then you would get the intuitive (0+0+90)/3 = 30 deg.

Edit: Thanks for useful comments and pointing out that angles may exceed 360. I believe the answer could be the normal arithmetic average reduced "modulo" 360: we sum all the values, divide by the number of angles and then subtract/add a multiple of 360 so that the result lies in the interval [0..360).

Answer 5

Here you go! The reference is https://www.wxforum.net/index.php?topic=8660.0

def avgWind(directions):
    sinSum = 0
    cosSum = 0
    d2r = math.pi/180 #degree to radian
    r2d = 180/math.pi
       
    for i in range(len(directions)):
        sinSum += math.sin(directions[i]*d2r)
        cosSum += math.cos(directions[i]*d2r)
      
    return ((r2d*(math.atan2(sinSum, cosSum)) + 360) % 360)
a= np.random.randint(low=0, high=360, size=6)
print(a)
avgWind(a)

Answer 6

In my opinion, this is about angles, not vectors. For that reason the average of 360 and 0 is truly 180. The average of one turn and no turns should be half a turn.