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How do you calculate the average of a set of circular data?

I have two angles, a=20 degrees and b=350 degrees. The average of those two angles are 185 degrees. However, if we consider that the maximum angle is 360 degrees and allows for wrap around, one could see that 5 degrees is a closer average.

I'm having troubles coming up with a good formula to handle that wrap around when calculating average. Anyone got any hints?

Or am I shooting myself in the foot here? Is this considered "bad practice" in math?

Tim Cooper
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Mizipzor
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3 Answers3

1

if you have a look at the angular circle, you will see that there are 2 opposite "angles" that corresponds to your "average".

So both 185° and 5° are correct.

But you mentionned the closer average. So in that case, you may choose the angle that is closer.

Usually, the "average" of angles concerns the counterclockwise direction. The "average" is not the same if you switch your two angles (or if you use the clockwise direction).

For example, with a=20° and b=350°, you are looking for the angle that comes after a and before b in the counterclockwise direction, 185° is the answer. If you are looking for the angle that comes before a and after b in the counterclockwise direction (or after a and before b in the counterclock wise direction), 5° is the answer.

The answer of this post is the right way to do.

So the pseudo-code for the solution is

if (a+180)mod 360 == b then
  return (a+b)/2 mod 360 and ((a+b)/2 mod 360) + 180 (they are both the solution, so you may choose one depending if you prefer counterclockwise or clockwise direction)
else
  return arctan(  (sin(a)+sin(b)) / (cos(a)+cos(b) )
Community
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ThibThib
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-2

Try this (example in C#):

    static void Main(string[] args)
    {
        Console.WriteLine(GetAngleAverage(0,0));
        Console.WriteLine(GetAngleAverage(269, 271));
        Console.WriteLine(GetAngleAverage(350, 20));
        Console.WriteLine(GetAngleAverage(361, 361));
    }

    static int GetAngleAverage(int a, int b)
    {
        a = a % 360;
        b = b % 360;

        int sum = a + b;
        if (sum > 360 && sum < 540)
        {
            sum = sum % 180;
        }
        return sum / 2;
    }

I think it works, the output is

0
270
5
1
weiqure
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  • I think you want to remove the "return" inside the if() block. GetAngleAverage(350,20) should be 5, not 10. – Adam Liss Sep 02 '09 at 02:04
  • The "sum > 360" line should be "sum >= 360". Otherwise the average angle of 350 and 10 will come out at 180, which is not what you want. – naroom Jul 27 '13 at 23:31
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    Also the average of 190 and 190 by this method would come out to 10. So... I think this method doesn't work. – naroom Jul 27 '13 at 23:33
  • I agree, this method doesn't work. Please downvote. – Tim Cooper Sep 15 '20 at 06:20
-4

Just take a normal average and then take it mod 180. In your example this gives 5 degrees, as expected.

PaulJWilliams
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  • Stecys answer states to take mod 360, which seems to work in the few attempts I tried it in. Where does 360 fails and 180 succeeds? – Mizipzor Jul 21 '09 at 12:47
  • Taking the average like (a+b)/2 and then mod 180 is exactly the same as first doing (a+b) mod 360 and then divide by 2. Both are equivalent. – Ralph M. Rickenbach Jul 21 '09 at 13:14
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    This does not work. Example: angle1 = 280, angle2 = 10, their mean is 325. Your formula gives 145. – lenooh Jan 13 '16 at 20:38
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    Not quite. In the sense that the average is the angle that bisects the two angles stated you can see with the aid of a simple diagram that the tow answers are equivalent.. – PaulJWilliams Jan 14 '16 at 11:06
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    I downvoted because the formula doesn't work. This question is deceptively difficult. – Tim Cooper Sep 16 '20 at 04:20