In the paper "Applications of Quaternions to Computation with Rotations" by Eugene Salamin, 1979 (click here), they get 126.5 degrees as the mean value of the rotation angle of a random rotation (by integrating quaternions over the 3-sphere).

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How can I make sense of this result? If rotation angle around a given axis can be 0..360°, should not the mean be 180? or 0 if it can be −180..180°? or 90° if you take absolute values?

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    $\begingroup$ I think the angles they're talking about must be in the range from $0^\circ$ to $180^\circ$. If you rotate a circle $181^\circ$ clockwise, that's the same as $179^\circ$ counterclockwise, and I think the latter is the angle they're talking about. But the rotations above are rotations in $3$-dimensional Euclidean space, so just plus or minus signs do not suffice to express the direction in which the sphere was rotated. $\endgroup$ Aug 10, 2013 at 16:50
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    $\begingroup$ The key phrase is "Random rotations are to be chosen uniformly distributed over the 3-sphere" $\endgroup$ Aug 10, 2013 at 16:54

1 Answer 1


First, SO(3) of course has its unique invariant probabilistic measure. Hence, “random rotation” is a well-defined SO(3)-valued random variable. Each rotation (an element of SO(3)) has an uniquely defined rotation angle θ, from 0 to 180° (π) because of axis–angle representation. (Note that axis is undefined for θ = 0 and has two possible values for θ = 180°, but θ itself has no ambiguity.) Hence, “angle of a random rotation” is a well-defined random angle.

Why is its average closer to one end (180°) than to another (0)? In short, because there are many 180° rotations, whereas rotation by zero angle is unique (identity map).

Note that I ignore Spin(3) → SO(3) covering that is important in quaternionic discourse, but it won’t change the result: Haar measure on Spin(3) projected onto SO(3) gives the same Haar measure on SO(3), hence there is no difference whether do we make computations on S3 of unit quaternions (the same as Spin(3)) or directly on SO(3).


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