# Not Isometric BTN-Matrix

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I have always thought that the transformation btn-matrix into tangent space has to be affine. A guy told me that it don't as long as you normalize the vectors after transformation. Is he right? [Edited by - 51mon on February 13, 2008 6:22:43 AM]

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The transformation associated with the BTN matrix isn’t an affine transform but a simple change of basis.

Edit: As other have said they aren’t isometry because the axis doesn’t have to be orthonormal.

[Edited by - apatriarca on February 14, 2008 7:26:04 PM]

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Quote:
 Original post by apatriarcaThe transformation associated with the BTN matrix isn’t an affine transform but an isometry. It’s a simple change of basis.

Sorry, I checked up the words and you are right, I meant isometric matrix. Anyway, what he was saying was that the BTN matrix don't have to be made out of perpendicular unit vectors.

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Yeah, they don't necessarily have to be perpendicular, in fact that's very often not the case. It can be handy to introduce an approximation and make the matrix orthogonal however, so that you can throw one vector away and recreate it in the shader.

stoo

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Quote:
 Original post by 51monI have always thought that the transformation btn-matrix into tangent space has to be affine. A guy told me that it don't as long as you normalize the vectors after transformation. Is he right?

Yes he's right, but as Stuart says there are advantages to having an orthonormal basis for your TBN. Aside from being able to throw a vector away and recreate, you can also skip the normalization step.

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BTN matrices are often not orthogonal. This is actually a very important and often-overlooked point when doing bump-mapping. A common thing to do is just orthogonalize the BTN matrix and use that as a basis for a normal map. This is _not_ equivalent to proper bump mapping, and can look significantly worse in many situations. For a discussion of how to do it right, Blinn's original bump mapping paper is actually a good reference.

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