Talk:Rotation matrix

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Don't see why the 2x2 matrix given in the introduction applies to the x-y plane specifically. It applies to any 2D subspace with a defined origin/axis surely.

The rotation matrices about the x, y, and z axis do not seem to be equate to the "generator" written to their right (the rotation matrices with the matrix exponentials). It seems that one is the transpose of the other. Can someone please explain?

Thanks- James

Looking at the General 3D rotation case, I can’t seem to generate the same final transformation matrix as indicated. Can you show more of the details? In particular the “ij" = 12, 13, 22, & 23 matrix elements, containing many trig factors and terms, is a problem. For a R to L multiplication, you first mult the pitch matrix x the yaw matrix, right? Can you show that intermediate matrix? Then you perform roll matrix x the intermediate matrix to get the final transformation, right?. Found my error. Also applied the associative property to perform: ( roll * pitch ) * yaw.

But also, a bit confused why the angles for extrinsic are α, β, γ, about axes x, y, z ( respectively ?) while for intrinsic, they are α, β, γ, about axes z, y, x, respectively. I understand the multiplication order is reversed between intrinsic and extrinsic but the angles associated with each axis should be the same, right? Thanks. 2601:647:6480:1D20:16C:3AED:F2BB:97EF (talk) 15:51, 2 October 2024 (UTC)Reply

 ;)

see http://nghiaho.com/?page_id=846 95.91.233.242 (talk) 19:52, 2 June 2025 (UTC)Reply

There's a bit of discussion at § Euler angles. Perhaps there could be something a bit more explicit here; you can find more detail about this at Euler angles § Rotation matrix. Feel free to make some concrete suggestions. –jacobolus (t) 19:59, 2 June 2025 (UTC)Reply

According to the right-hand rule, it's clear that R_z(\theta) has the proper rotation direction, but R_y(\theta) is clearly backwards, and R_x(\theta) is ambiguous due to the arrow placement.

This should be fixed. DlittlemanDungeon (talk) 18:13, 13 June 2025 (UTC)Reply

Good catch. I agree the circular arrows should all point in the direction of conventional positive angles. Does anyone feel like fixing this or making an alternate image? –jacobolus (t) 08:15, 14 June 2025 (UTC)Reply

It says "avoid mixing <math>...</math> and {{math}} in the same expression", but I only found one example of that (and promptly fixed it). Anyone know any more about this or should the {{MOS}} be removed? Davie53K (talk) 09:04, 16 March 2026 (UTC)Reply

I did a quick skim of the article and couldn't find any expression that mixed the two of them. I think the issue is resolved now, so I'll go boldy remove the the {{MOS}} template from the article as its resolved. If an editor finds more issues related to the expressions, they can put back the template and we can hopefully address that. Gramix13 (talk) 05:09, 2 April 2026 (UTC)Reply

In this equation, $${\displaystyle {\begin{aligned}\\R=R_{x}(\gamma )\,R_{y}(\beta )\,R_{z}(\alpha )&={\overset {\text{roll}}{\begin{bmatrix}\cos \gamma &-\sin \gamma &0\\\sin \gamma &\cos \gamma &0\\0&0&1\\\end{bmatrix}}}{\overset {\text{pitch}}{\begin{bmatrix}\cos \beta &0&\sin \beta \\0&1&0\\-\sin \beta &0&\cos \beta \\\end{bmatrix}}}{\overset {\text{yaw}}{\begin{bmatrix}1&0&0\\0&\cos \alpha &-\sin \alpha \\0&\sin \alpha &\cos \alpha \\\end{bmatrix}}}\\&={\begin{bmatrix}\cos \beta \cos \gamma &\sin \alpha \sin \beta \cos \gamma -\cos \alpha \sin \gamma &\cos \alpha \sin \beta \cos \gamma +\sin \alpha \sin \gamma \\\cos \beta \sin \gamma &\sin \alpha \sin \beta \sin \gamma +\cos \alpha \cos \gamma &\cos \alpha \sin \beta \sin \gamma -\sin \alpha \cos \gamma \\-\sin \beta &\sin \alpha \cos \beta &\cos \alpha \cos \beta \\\end{bmatrix}},\end{aligned}}}$$ the expression ${\textstyle R_{x}(\gamma )}$ is expanded to $${\displaystyle {\begin{bmatrix}\cos \gamma &-\sin \gamma &0\\\sin \gamma &\cos \gamma &0\\0&0&1\\\end{bmatrix}}.}$$ However, this is the matrix for ${\textstyle R_{z}(\gamma )}$.

Similarly, $${\displaystyle {\begin{bmatrix}1&0&0\\0&\cos \alpha &-\sin \alpha \\0&\sin \alpha &\cos \alpha \\\end{bmatrix}}}$$ is the matrix for ${\textstyle R_{x}(\alpha )}$ instead of ${\textstyle R_{z}(\alpha )}$. I don't know whether the mistake is in the wrong index or in the wrong expansion and would appreciate someone knowledgeable to check it. Georg Stillfried (talk) 11:44, 16 April 2026 (UTC)Reply