Representing a CFrame with the minimum amount of numbers possible

[IdiomicLanguage]

@1waffle1

Using spherical coordinates with no radius, x = cosθsinφ, y = sinθsinφ, z = cosφ converts 2 angles into a direction.

Hmm, interesting. While two angles can define any 3D direction, they can’t directly represent any rotation applied to any vector (for the reasons above). Given a start vector orthogonal to the first axis of rotation, any rotation can be represented though. So if we assume that a rotation about two predefined axes start with a predefined orthogonal vector to the first axis of rotation, then we can convert the rotation applied to another format which can be applied regardless of the starting vector. Does that count as defining a 3D rotation? xD

[acreol (arto)]

What’s a way to get 3 components to represent the orientation of a cframe that avoids trig functions?

[suremark]

@Acreol

are the equations:

r00^2+r01^2+r02^2=1
r10^2+r11^2+r12^2=1
r20^2+r21^2+r22^2=1
r00*r10+r01*r11+r02*r12=0
r00*r20+r01*r21+r02*r22=0
r10*r20+r11*r21+r12*r22=0

is there any easy way to see that they are independent?

If you know vector calculus, you could look at the Jacobian matrix of the system of equations. Where the Jacobian has full rank, the equations are independent.

I didn’t (still don’t) understand what your explanation has to do with Euler angles, I thought you were talking about matrices

My bad, I didn’t realize Idiomic was talking about Euler angles.

I guess bijection is not the right word, showing an injection from orthogonal rotation matrix to Euler angles would suffice since that means there exists an Euler angle for every orthogonal rotational matrix

I thought gimbal lock occurs because you can’t lerp the components independently, what does not being bijective have to do with gimbal lock?

I think you mean to say that Euler angles are surjective, not injective.
I’m not sure what you mean by lerping the components independently, but the fact that Euler angles aren’t bijective implies that the derivative is degenerate at certain points (to be more clear, they are not injective), which gives you gimbal lock.

That works too if you assume A is orthogonal (like I was assuming B is orthogonal)

You can assume B is orthogonal, but that doesn’t prove B^t is orthogonal, which would be necessary for A^t * A = I to apply for A=B^t.

[acreol (arto)]

@suremark

If you know vector calculus

nope xd

I think you mean to say that Euler angles are surjective, not injective.

i’m trying to say that for every 3x3 orthogonal matrix with the property that right:Cross(up)==back, there needs to be at least one 3 component euler angle. If that is true, then using using your proof that there exists a matrix to transform one 3x3 orthogonal matrix to any arbritrary 3x3 orthogonal matrix, it is safe to say that there 3 component euler angles can represent all 3d rotations

don’t we want the injectivity and not care about the surjectivity as illustrated in the 2x2 image matrix here

I’m not sure what you mean by lerping the components independently

local function lerp(r1,r2,t)
return v3(r1.X*(1-t)+r2.X*t,r1.Y*(1-t)+r2.Y*t,r1.Z*(1-t)+r2.Z*t)
end

The fact that Euler angles aren’t bijective implies that the derivative is degenerate at certain points (to be more clear, they are not injective), which gives you gimbal lock.

I don’t understand, even if it were bijective (or injective) it could be a crazy mapping that is discontinuous at many points (this is not saying I am convinced that it is not bijective/injective though)

[suremark]

@Acreol

i’m trying to say that for every 3x3 orthogonal matrix with the property that right:Cross(up)==back , there needs to be at least one 3 component euler angle. If that is true, then using using your proof that there exists a matrix to transform one 3x3 orthogonal matrix to any arbritrary 3x3 orthogonal matrix, it is safe to say that there 3 component euler angles can represent all 3d rotations

We are in agreement. I’m saying that for Euler angles to be injective, the Euler angle representation has to be unique when it exists. Surjectivity means that there is always an Euler angle representation (possibly several).

@Acreol

I don’t understand, even if it were bijective (or injective) it could be a crazy mapping that is discontinuous at many points (this is not saying I am convinced that it is not bijective/injective though)

Yeah, but the Euler angles representation is continuous and differentiable. Not sure what you’re getting at here.

[acreol (arto)]

@suremark

We are in agreement. I’m saying that for Euler angles to be injective , the Euler angle representation has to be unique when it exists. Surjectivity means that there is always an Euler angle representation (possibly several).

ok, my bad that makes sense

@suremark

Yeah, but the Euler angles representation is continuous and differentiable. Not sure what you’re getting at here.

I just don’t understand the relation between gimbal lock and bijections

(i now understand why rotation matrix → euler angles isnt a bijection though)

[suremark]

@Acreol

I just don’t understand the relation between gimbal lock and bijections

Erm, my bad. Actually, gimbal lock isn’t really related to bijections, although it is related to the notion of a local bijection (more precisely, a local homeomorphism) which is a more elusive concept.

[EgoMoose]

So, I didn’t take the time to read all the posts in here so some of this may have been answered already, but a lot of the answers to your questions can be found here. Specifically, the quaternions stuff, the Rodrigues rotation stuff, and the CFrame stuff.