ROTMAT will also do this. But I am not sure that this is what Phil wants? The superposition transformation includes a translation, so there is no locus of points that are left unchanged. You can generate an axis of rotation for polar angles from the R which will be quite different from t. You should be able to translate the rotation axis (change the origin of the coordinate system) to get a new transformation x' = R'x + t' which might be better for visualisation. If Phil's last sentence is right, you should be able to arrange it so that t' is parallel to the rotation axis of R'
No, I don't know how to do this off the top of my head, but it sounds very similar to the transformations done in the Schomaker and Trueblood TLS paper. Or maybe this is over-complicating things ;-) Martyn -----Original Message----- From: CCP4 bulletin board on behalf of [EMAIL PROTECTED] Sent: Tue 7/29/2008 1:32 PM To: [email protected] Subject: Re: [ccp4bb] Rotation axis Dear Phil, Because question keep popping up in the bullitin board about conversion from a rotation matrix into rotation angles, I decided to take the relevant subroutines from an old program from Groningen and make a jiffy to do these conversions. It is a small fortran program and does not need any additional libraries or subroutines. The program will take a rotation matrix and translation vector and print all kind of rotation angles and also the component of the translation vector parallel to the rotation axis, which is the number you want. All other components of the translation vector can be made zero by choosing the right position of the rotation axis. Best regards, Herman Schreuder -----Original Message----- From: CCP4 bulletin board [mailto:[EMAIL PROTECTED] On Behalf Of Phil Evans Sent: Tuesday, July 29, 2008 10:11 AM To: [email protected] Subject: [ccp4bb] Rotation axis If I've go a superposition transformation (x' = Rx + t), as it happens from a superposition in ccp4mg, how do I get the position & direction of the rotation axis (to draw in a picture)? I know that any (orthonormal) transformation can be represented as a rotation about an axis + a screw translation along that axis I'm sure I've done this before ... thanks Phil
