# Can you quickly visualize how a quaternion would look on top of your head?

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If you see a quaternion of a certain value, can you quickly determine how this orientation would look on top of your head? If you can, how can you do that handily? Thanks Jack

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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 :)

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Quaternions.

All I can say is thank God for Quaternion.eulerAngles (With unity anyway)

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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

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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.

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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

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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

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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).

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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.

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