Transformations

[thomthom]

If you always calculate from one fixed state then there is no need to undo the current transformation, you just set a new transformation overriding the old one.

[zitoun]

@thomthom said:

If you always calculate from one fixed state then there is no need to undo the current transformation, you just set a new transformation overriding the old one.

Yes indeed, but for some reason I thought that the method entity.transformation= was working differently (but I may have been really tired at the time I tested it).
It completely make sense that it works the way you describe it.

[zitoun]

Ahem.

Just remember:

@object.transform! @object.transformation.inverse
@object.transform! @transformationAt[toPage]

@object.transform! @transformationAt[toPage] * @object.transformation.inverse

Note the order of the transformations... I had forgotten this well known pb !

[thomthom]

@zitoun said:

@object.transform! @object.transformation.inverse

This just resets the transformation - same as @object.transformation = Geom::Transformation.new

@zitoun said:

@object.transform! @transformationAt[toPage] * @object.transformation.inverse

Isn't this just the same as: @object.transformation = @transformationAt[toPage] ?

[AdamB]

@thomthom said:

If you always calculate from one fixed state then there is no need to undo the current transformation, you just set a new transformation overriding the old one.

And you will end up with many problems if you build your code around inverting the CTM and multiplying by a new transform.

So if you're rotating an object, a low quality approach is to repeatedly apply a (say) 5 degree rotation rather than as Thomthom suggest, calculate the rotation at a time T and then calc what rotation you need.

The problem is that floating point does have finite precision, so if you repeatedly perform incremental transform as it seems you're doing, you'll end up with a non-orthonormal matrix. (Thats a bad thing).

[zitoun]

@thomthom said:

@zitoun said:

@object.transform! @object.transformation.inverse

This just resets the transformation - same as @object.transformation = Geom::Transformation.new

@zitoun said:

@object.transform! @transformationAt[toPage] * @object.transformation.inverse

Isn't this just the same as: @object.transformation = @transformationAt[toPage] ?

Yes, the inverse is not useful in my case.
I use @object.transformation= @transformationAt[toPage], as you suggest, and it works (well I still have some awkward pbs but I expect to solve them soon).

The code was just for my tests, I wanted to underline here that if you wish to do

entity.transform! entity.transformation.inverse
entity.transform! transfA
entity.transform! transfB

you could as well do

entity.transform! transfB * transfA * entity.transformation.inverse

or even better, as you say Thomthom

entity.transformation = transf[b]B[/b] * transf[b]A[/b]

[zitoun]

@adamb said:

And you will end up with many problems if you build your code around inverting the CTM and multiplying by a new transform.

Not really: it's more like reseting the transformation and set the one you like. No inverse is needed, and this solution is actually the only solution I can see so far.

But you're right, ignoring each object's current orientation is a real problem.

[thomthom]

@zitoun said:

But you're right, ignoring each object's current orientation is a real problem.

But you know that from the previous transformation.

[zitoun]

Another thing: angle_between only gives absolute angles.
If you need oriented angles, you might need something like

def angBtw(v1, v2)
	u1 = v1.normalize
	u2 = v2.normalize
	a = u1.angle_between u2
	if ( u1.cross u2 ).dot( Z_AXIS ) > 0
		return a
	else
		return -a
	end
end

Note 1: normalization is used only because there seems to be some trouble with angle_between for small values...
Note 2: I took Z_AXIS as my scene reference vector for now, but the good way to do would be to take the up vector of your object.

[Cleverbeans]

@zitoun said:

I have no clue what these are yet, but I guess I somehow can retrieve my rotation matrix from such component...

Note that a transformation has the .to_a method as well which gives you exactly the information you're looking for. The array it returns is 16 floats for the four by four matrix which defines the transformation from the components defining coordinate system to the current coordinate system. Since you mentioned quaternions I'll assume you're familiar with with the essentials of matrices and linear transformations. First, we can interpret the scaling and rotation of a component as the action of a matrix on the basis vectors of the component definition. Say for example you define your component relative to the standard basis of x=[1,0,0], y=[0,1,0], z=[0,0,1]. Rather than keeping these vectors separately, you can just encode them as the 3x3 identity matrix, and then define all the entities in the component relative to this matrix. From here, any sort of rotation/scaling can be interpreted as a change of basis. The trouble of course is that any linear transformation preserves the zero vector, so you can't model a translation in this way. This means to fully describe the relative coordinates you require a an equation of the form Ax + b where A is an invertible 3x3 change of basis matrix and b is the translation vector with x being the "defining" vector of some entity internal to the component.

This is where the clever bit comes in. You can model an affine transformation of the form Ax + b as a single matrix transformation by embedding it into a space with one higher dimension. This is why the transformation.to_a method returns a 16 entry array, it's a 4x4 matrix with the first four entries being the first column, the second four entries being the second column and so on. It's easy enough to extract the original 3x3 matrix and the translation vector as well, the "upper-left" 3x3 block matrix is exactly the change of basis matrix A, and the final column vector is of the form [b, 1] where b is the translation vector. From here extracting the rotation matrix is a simple matter of matrix algebra since any change of basis matrix A can be decomposed as A = QS where Q is a rotation matrix and S is a diagonal matrix representing the scaling of each axis.

It's worth getting to know the 4x4 representation if you haven't before since it's also the way the OpenGL standard models the objects. The ability to model vector addition as matrix multiplication is only one of the reason they do it, the other being that when you're rendering perspective views rather than simply orthographic views the value of the bottom right entry plays an important role in correcting the scale of the transformation after applying the perspective matrix.