Make XRRay.matrix unique, add steps for obtaining it #655
Conversation
index.bs
Outdated
|
|
||
| <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. |
There was a problem hiding this comment.
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?
There was a problem hiding this comment.
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.
|
Thanks Manish! LGTM too. |
4 participants
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