Preserving up as much as possible while generating a new orientation

nullsquared · 2007-11-04T08:36:04

Hi! What I'm trying to do today [smile], is generate an orientation from a normal vector and a desired "up" vector. I want forward (local z axis) to be a normal I have generated off of some world geometry. So far so good. Now, I want up (local y axis) to be as close to possible to some vector, desiredUp. However, the important vector here is the forward vector, so I need the up vector to be "perfectly" perpendicular to the forward vector, just as close to the desiredUp up as possible. (I generate the local x axis via a cross product of the local z and local y.) I'm using quaternions, and need a quaternion for the final orientation. Though if matrices would work better or are needed, I can freely convert between the two. Ideas? So far I've tried just using desiredUp and the normal directly, and that works great when the 2 are perfectly perpendicular. But when things start sloping, it fails (the ground fails, too, for that matter, since desiredUp and the normal would be the same or exactly the opposite): Notice how [1] is aligned well to the wall (which is perpendicular to "up"), but [2] is sort of "upwards"-ish (I tried to outline the quads themselves, they were a bit vague). Thanks a bunch in advance! Hopefully this doesn't sound too stupid [embarrass]

Math and Physics Programming

Started by nullsquared November 01, 2007 03:37 PM

14 comments, last by nullsquared 16 years, 5 months ago

nullsquared

128

Author

November 01, 2007 05:30 PM

Quote:Original post by Sneftel
Not sure what you mean by that. What values are you passing in for za and desiredUp, what values are you getting out, and what do you expect to get out? In this situation, numbers are far more reliable than pictures.

Hm. Actually, I seem to be getting zero vectors for the x and y axises. For z I've tried [0, -1, 0], [0, 1, 0], to demonstrate the problem most clearly.

www.nullular.com

Zipster

2,420

November 01, 2007 08:10 PM

There are only two ways the math works out. You either need to do:

right = desiredUp x forward
up = forward x right

or

right = forward x desiredUp
up = right x forward

Note how the forward vectors never "line up" in a column. And if you're getting zero vectors for strange reasons, you should definitely check to make sure the cross product is correct in the first place :)

Sneftel

1,788

November 01, 2007 09:19 PM

Quote:Original post by agi_shi
For z I've tried [0, -1, 0], [0, 1, 0], to demonstrate the problem most clearly.

If your up-vector and your forward-vector are collinear, then the problem is underconstrained (has an infinite number of solutions).

nullsquared

128

Author

November 02, 2007 05:34 AM

Quote:Original post by Sneftel
Quote:Original post by agi_shi
For z I've tried [0, -1, 0], [0, 1, 0], to demonstrate the problem most clearly.

If your up-vector and your forward-vector are collinear, then the problem is underconstrained (has an infinite number of solutions).

Yeah, I sort of realized that... so how do I fix it?

@ Zipster: Thanks, I get it now :D

www.nullular.com

Sneftel

1,788

November 02, 2007 08:25 AM

Quote:Original post by agi_shi
Yeah, I sort of realized that... so how do I fix it?

You'd hardcode an alternate desiredUp (or two), if you actually expected the situation to ever arise. Of course, in the vast majority of cases this won't be necessary.

nullsquared

128

Author

November 04, 2007 08:36 AM

Quote:Original post by Sneftel
Quote:Original post by agi_shi
Yeah, I sort of realized that... so how do I fix it?

You'd hardcode an alternate desiredUp (or two), if you actually expected the situation to ever arise. Of course, in the vast majority of cases this won't be necessary.

Ah, I see... I hard-coded the forward vector as the "alternate" desired up vector, and it works wonders!

I would like to really thank you two! [smile]

www.nullular.com

Preserving up as much as possible while generating a new orientation

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