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

Solution 1:

Compute unit vectors from the angles and take the angle of their average.

Solution 2:

This question is examined in detail in the book: "Statistics On Spheres", Geoffrey S. Watson, University of Arkansas Lecture Notes in the Mathematical Sciences, 1983 John Wiley & Sons, Inc. as mentioned at http://catless.ncl.ac.uk/Risks/7.44.html#subj4 by Bruce Karsh.

A good way to estimate an average angle, A, from a set of angle measurements a[i] 0<=i

                   sum_i_from_1_to_N sin(a[i])
a = arctangent ---------------------------
                   sum_i_from_1_to_N cos(a[i])

The method given by starblue is computationally equivalent, but his reasons are clearer and probably programmatically more efficient, and also work well in the zero case, so kudos to him.

The subject is now explored in more detail on Wikipedia, and with other uses, like fractional parts.

Solution 3:

I see the problem - for example, if you have a 45' angle and a 315' angle, the "natural" average would be 180', but the value you want is actually 0'.

I think Starblue is onto something. Just calculate the (x, y) cartesian coordinates for each angle, and add those resulting vectors together. The angular offset of the final vector should be your required result.

x = y = 0
foreach angle {
    x += cos(angle)
    y += sin(angle)
}
average_angle = atan2(y, x)

I'm ignoring for now that a compass heading starts at north, and goes clockwise, whereas "normal" cartesian coordinates start with zero along the X axis, and then go anti-clockwise. The maths should work out the same way regardless.

Solution 4:

FOR THE SPECIAL CASE OF TWO ANGLES:

The answer ( (a + b) mod 360 ) / 2 is WRONG. For angles 350 and 2, the closest point is 356, not 176.

The unit vector and trig solutions may be too expensive.

What I've got from a little tinkering is:

diff = ( ( a - b + 180 + 360 ) mod 360 ) - 180
angle = (360 + b + ( diff / 2 ) ) mod 360

• 0, 180 -> 90 (two answers for this: this equation takes the clockwise answer from a)
• 180, 0 -> 270 (see above)
• 180, 1 -> 90.5
• 1, 180 -> 90.5
• 20, 350 -> 5
• 350, 20 -> 5 (all following examples reverse properly too)
• 10, 20 -> 15
• 350, 2 -> 356
• 359, 0 -> 359.5
• 180, 180 -> 180