I have a series of vector directions and I need to find the "average" direction. I am not looking for the overall direction which would be the sum of the directions and this can't be used in cases where the vectors all cancel out.

I know of spherical linear interpolation but its not really practical when I have to do this for a large set of vectors.

EDIT: I misinterpreted my problem and so here is the updated version, sorry for any inconvenience.

What I have is a set of directions and one of the directions is selected as "leading" (shown as the red line) direction and then I need to find the average direction produced by the remaining vectors (the black ones) and it seems by drawing it out I have figured out one way of how I can do it.

However, Rahul's comment about means of circular quantities does seem to be a much better method which I will be looking into.

The dashed red vector is meant to be the opposite of the red vector.

  • 3
    $\begingroup$ By average direction, do you mean an average of the directions without considering their magnitude? In which case you could normalise all the vectors (so that they have the same direction but magnitude 1) and then add them $\endgroup$
    – Mathmo123
    Jul 7 '14 at 23:46
  • $\begingroup$ How would that work if the resulting sum is a vector with zero magnitude? And, yes the magnitude is irrelevant in my case. $\endgroup$ Jul 7 '14 at 23:49
  • $\begingroup$ I think the point then is that the average direction is not defined - for example if I pull in one direction and you pull in the opposite direction, then there is no total pull, so you can't really say that the average direction is in any particular direction. $\endgroup$
    – Mathmo123
    Jul 7 '14 at 23:50
  • $\begingroup$ Would it not be perpendicular to the two vectors in the case? $\endgroup$ Jul 7 '14 at 23:56
  • $\begingroup$ Perpendicular in which direction? If we factor in 3 dimensions, that gives infinitely many options (e.g. up, down, at a 30 degree angle to the ground etc) And I'm not convinced perpendicular is the answer anyway. $\endgroup$
    – Mathmo123
    Jul 7 '14 at 23:58

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