Calculate the rotational angles of a transformed coordinate system

Hello,

I'm struggeling with a coordinate system transformation. The original coordinate system (in black colour) is give by three vectors (XV, YV, ZV), each for every axis and the location of the origin (NP). The System of three Vectors is transformed (translation and rotation) so that the new system (in red colour) is formed (Picture below).

I need to calculate the rotational angle of the system regarding rotation aroound x, y, and z axis. I know I need the rotaional matrix but I don't have a clue to solve this problem. Maybe anyone of the community knows how get rid of the problem?

With best regards

Steffen B.

[Yazan]

Do you need to calculate the required rotation and translation? or do we need just the rotational matrices?

[Steffen Bxyz]

@Yazan I only need the rotation of the red coordinate system (formed by three vectors) around X,Y and Z axis compared to the former position (black system)

Answer 1

[Matt J]

You can use the attached file to get the rotation angles (in degrees).

Rblack=normalize( [Xb(:),Yb(:),Zb(:)]-NPb(:)  ,1,'norm'); %The black XV,YV,ZV, and NP
Rred=normalize( [Xr(:),Yr(:),Zr(:)]-NPr(:)  ,1,'norm');  %The red XV,YV,ZV, and NP

angles  = flip(rot2taitbryan(Rred*Rblack.' ,'ZYX'));

[Steffen Bxyz]

@Matt J thank you for your quick reply. When I apply the code I get an error message

[Matt J]

I can't copy/paste/run your code (because it's an image).

[Steffen Bxyz]

@Matt J sorry here is the used code:

clc,close all

Xb=[-68.218;-69.454;-69.373];
Yb=[-1.1486;-0.53105;-1.4751];
Zb=[-7.6597;-7.962;-6.9121];
NPb=[-69.136;-1.476;-6.4578];

Xr=[-68.579;-69.809;-69.732];
Yr=[-1.1047;-0.46942;-1.4178];
Zr=[-7.3962;-7.6854;-6.6392];
NPr=[-69.503;-1.4188;-7.6128]; 

Rblack=normalize( [Xb(:),Yb(:),Zb(:)]-NPb(:)  ,1,'norm'); %The black XV,YV,ZV, and NP
Rred=normalize( [Xr(:),Yr(:),Zr(:)]-NPr(:)  ,1,'norm');  %The red XV,YV,ZV, and NP
angles  = flip(rot2taitbryan(Rred*Rblack.' ,'ZYX'));

[Matt J]

Let's run it here:

clc,close all
Xb=[-68.218;-69.454;-69.373];
Yb=[-1.1486;-0.53105;-1.4751];
Zb=[-7.6597;-7.962;-6.9121];
NPb=[-69.136;-1.476;-6.4578];

Xr=[-68.579;-69.809;-69.732];
Yr=[-1.1047;-0.46942;-1.4178];
Zr=[-7.3962;-7.6854;-6.6392];
NPr=[-69.503;-1.4188;-7.6128]; 

Rblack=normalize( [Xb(:),Yb(:),Zb(:)]-NPb(:)  ,1,'norm'); %The black XV,YV,ZV, and NP
Rred=normalize( [Xr(:),Yr(:),Zr(:)]-NPr(:)  ,1,'norm');  %The red XV,YV,ZV, and NP
angles  = flip(rot2taitbryan(Rred*Rblack.' ,'ZYX')),

angles = 1×3

46.7902 -5.6695 -2.6966

[Steffen Bxyz]

@Matt J it's still not working even I copy and paste your code

I got always the same error:

Error using -

Matrix dimensions must agree.

Error in Untitled (line 10)

Rblack=normalize( [Xb(:),Yb(:),Zb(:)]-NPb(:) ,1,'norm'); %The black

XV,YV,ZV, and NP

Maybe it's due to my Matlab version? I use an old R2015a version.

[Matt J]

Yes, my code assumes R2016b and higher. Simply replace with,

Rblack=normalize( [Xb(:)-NPb(:),Yb(:)-NPb(:),Zb(:)-NPb(:)]  ,1,'norm')

and similarly for Rred.

[Steffen Bxyz]

@Matt J Thank you very much, it is working just fine. I have only one question regarding the theory behind it. When I try the different order of angles to return, the angles differ. If I use ZYX and XYZ the angles should be equal but in different order. ´The results are not equal. Is it because of the rotation and translation? Do I miss something?

[Matt J]

Thank you very much, it is working just fine.

You're welcome, but please Accept-click the answer to indicate so.

If I use ZYX and XYZ the angles should be equal but in different order.

No, that's not true, in general. Not sure why you think so.