dot_product = dot4(qa, qb); dot_product_squared = dot_product * dot_product; denominator = dot4(qa, qa) * dot4(qb, qb); return 2.0f * acos(sqrt(abs(dot_product_squared / denominator)));I haven't tried it, so there could be mistakes. The idea is to not rely on the inputs being normalized, and write the expression in such a way that you are guaranteed to get exactly 0 if you qa==qb.
Quaternion angle with itself > 0.001
20 replies to this topic
Crossbones+ - Reputation: 19098
Posted 13 August 2014 - 11:30 AM
You can also compute the angle between quaternions using