Make XRRay.matrix unique, add steps for obtaining it#655
Make XRRay.matrix unique, add steps for obtaining it#655toji merged 2 commits intoimmersive-web:masterfrom
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| <section class="unstable"> | ||
| The <dfn attribute for="XRRay">matrix</dfn> attribute is a [=matrix=] which represents the transform from a ray originating at <code>[0, 0, 0]</code> and extending down the negative Z axis to the ray described by the {{XRRay}}'s {{XRRay/origin}} and {{XRRay/direction}}. | ||
| The <dfn attribute for="XRRay">matrix</dfn> attribute is a [=matrix=] which represents the transform from a ray originating at <code>[0, 0, 0]</code> and extending down the negative Z axis to the ray described by the {{XRRay}}'s {{XRRay/origin}} and {{XRRay/direction}}. Such a matrix MUST be one that has a rotation component which leaves any vector perpendicular to {{XRRay/direction}} and the <code>Z</code> axis unchanged. This attribute MUST be computed by [=XRRay/obtain the matrix|obtaining the matrix=] for the {{XRRay}}. This attribute SHOULD be lazily evaluated. |
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Such a matrix MUST be one that has a rotation component which leaves any vector perpendicular to {{XRRay/direction}} and the
Zaxis unchanged.
I'm pretty sure I get what you're trying to say here, but this was a bit confusing to me. Perhaps just because 3D math isn't my bread-and-butter. Would it be simpler to refer to the cross product?
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The cross product has no direction when the vectors are parallel, so I'd have to have an extra bit of text saying that the rotation is zero in such a case. This phrasing conveniently avoids this problem since when dir and Z are parallel, there are infinite perpendicular vectors, and keeping them all unchanged implies rotation=0.
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Thanks Manish! LGTM too. |
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I'm not 100% sure if I like the existence of this attribute, but I took a stab at making it conceptually unique. While there are many 3D rotations between two vectors, there is only one 2D rotation (and also, it is the easiest to compute -- most algorithms for "rotation between two vectors" will give you this). Such a rotation will basically keep the perpendicular to
directionand-zunchanged.I've added text specifying that the matrix must be of this type, as well as a partial algorithm for computation (I don't elaborate on how the axis-angle rotation matrix is to be obtained)
r? @NellWaliczek
fixes #582
cc @thetuvix @bialpio