# How to encode homogeneous transformations on the root node of a gltf

Let us say we have a gltf asset. Let us say we want to apply an arbitrary linear transformation to the root node.

We will say that the root node has a rotation, translation and scaling components. For simplicity let us say the transformation is swapping the y and z components but the final goal is to encode any arbitrary linear transformation T.

We could try doing T(translation), T(rotation), T(scale).

In the example of swapping, let us say the translation of the root node was 0, the rotation the identity and the scaling 1 so we have:

trans = (0, 0, 0); rot=(1, 0, 0, 0); scale=(1, 1, 1);

Well if we apply our transformation to these components, we literally get the exact same components back, so we did nothing. But we wanted to encode the transformation that swaps the z and y axes. So what do I do? How do I modify the triplet at the root node so that when it is applied it executes the swap?

An exchange of two axes is a mirror operation. So you'll need to choose an axis on which to reflect the object using its scale triplet, then change the rotation to compensate.

Let's say we mirror the object on the x axis: newScale = (-1, 1, 1) * scale. Now a unit vector that had been pointing toward +x points toward -x, and one that had been pointing toward +z or +y still point to those directions.

We next want to rotate this object so the formerly +x, now -x vector points to +x, the +z vector points to +x, and the +y vector points to +z. (to get the expected outcome of our x/z swap). That's a 180 degree rotation about the line y = z, or as a quaternion: newRot = (0, 1.0f/sqrt(2), 1.0f/sqrt(2), 0) * rot.

(Here using XYZW component order according to the glTF spec)

But we could just as easily choose to treat this as a mirror in the y axis instead. Then we'd get: newScale = (1, -1, 1) * scale, and unit vectors in the +x and +z directions would stay put, while a unit vector in the +y direction gets mapped to -y. To turn these around, we'd want a +90 degree rotation around the -x axis: newRot = (-1.0f/sqrt(2), 0, 0, 1.0f/sqrt(2)) * rot.

Or, to round things out, we could treat it as a mirror in the z axis: newScale = (1, 1, -1) * scale. +x and +y are preserved, and +z maps to -z. So then we'd want to rotate +90 degrees around the +x axis to fix this up: newRot = (1.0f/sqrt(2), 0, 0, 1.0f/sqrt(2)) * rot.

Of course, the trouble here is that mirroring your object flips the winding of all your triangles. So you may need to detect if your new transformation includes an odd number of sign flips or axis exchanges (negative determinant when expressed as a matrix), and if so, swizzle your indices as you load them to get back to the desired winding.

In general, if you can decompose your target transformation into a translation t, rotation r, and scale s, then you can apply...

newScale = scale * (inverse(rot) * s)

newRot = r * rot;

newTrans = trans + t;


But as noted in the comments, not every affine transformation can be represented this way. Those that include non-uniform scaling along axes diagonal to the root node's coordinate axes, or shearing transformations, can't be exactly represented as a sequence of axis -aligned scale, rotation, and translation (in that order).

• Note that this solves the swapping axes example but not the general question. The full question is we have a linear transformation T represented as a 4x4 matrix. And we want to transform the components of the TRS vectors of the node such that they are equivalent to having applied that transformation to them. i.e. we want to change the root node such that the full gtlf expereinces the transformation induced by T, not just reflections or axes swapping. Nov 30, 2022 at 23:38
• Unfortunately, no algorithm solves the question in its full generality. A translation, rotation, scale triplet can only represent a subset of all affine transformations representable by a matrix - those with no diagonal scaling or shear. So there are some matrices T for which no change to the translation, rotation, and scale properties of a node will achieve the same transformation. Nov 30, 2022 at 23:56
• I see, that's unfortunate, thank you. Nov 30, 2022 at 23:57