Can you quickly visualize how a quaternion would look on top of your head?
I recognise a few specific quaternion values (90 degree rotations in various axis, etc), but I couldn't tell you what direction an arbitrary-value quaternion is facing.
I'm sure some people can, but it's beyond me :)
If you take a general quaternion like 0.6 + 0.3*i + 0.2*j - 0.71*k you may first look at the real part and observe this should be equal to cos(alpha/2) for some angle alpha. 0.6 is greater than 0.5 which is the cosine of 60 degrees, but less then sqrt(2)/2 ~= 0.71 which is the cosine of 45 degrees. It should then be something in between, let's say 52.5 which is in the middle (the correct angle is 53.13 degrees) and thus we have a rotation of approximately 105 degrees around the axis defined by its imaginary part. This is basically how you can get a rough idea of what a quaternion does.
around the axis defined by its imaginary part
Which is equally cumbersome to get, since you have to divide first by the sin of half the angle you got... I seriously recommend having a little mobile app to convert from quaternions to euler or something... for example: https://play.google.com/store/apps/details?id=quaternion.angel
around the axis defined by its imaginary part
Which is equally cumbersome to get, since you have to divide first by the sin of half the angle you got... I seriously recommend having a little mobile app to convert from quaternions to euler or something... for example: https://play.google.com/store/apps/details?id=quaternion.angel
Unless you brain can only work with unit-length vectors, you don't need to divide by anything.
As you explained, we can determine the angles of rotation around an
imaginary axis by looking at the real part of the quaternion,
can we also easily work out the axis in the imagainary part?
Thanks
Jack
Ha ha, I don't understand why humans, like me, always like to run circles around things.
Thanks everybody for participation in this discussion.
Sometimes I really want to work out quaternions quickly without relying on calculators,
this makes things way quicker
Thanks
Jack
An axis can be defined by any non-zero vector, not just the unitary ones. You do not need to convert the imaginary part at all.. I can clearly take that imaginary part and normalize it if I need/want, but it isn't necessary to do so. If I look at my original example I can say that the axis is in the X>0,Y>0,Z<0 octant and that the greater component is along the z-axis. This component is indeed more than twice the x-component and more than three times the y-component. I would probably draw this axis using the vector (3,2,-7.1).
around the axis defined by its imaginary part
Which is equally cumbersome to get, since you have to divide first by the sin of half the angle you got... I seriously recommend having a little mobile app to convert from quaternions to euler or something... for example: https://play.google.com/store/apps/details?id=quaternion.angel
Unless you brain can only work with unit-length vectors, you don't need to divide by anything.
That's pretty much my case, but I see your point :P My mistake.