[hiddenghost]

I want to derive the matrix that does arbitrary rotation of a point around a point.

There are a few versions of the already derived rotation matrix on the net.

I did the derivation and found mine doesn't look like what some one else has done.
It would be good to see what I'm doing wrong.

Please use this form:
T1 = transform 1 = subtraction of A
T2 = transform 2 = rotation
T3 = transform 3 = addition of A
A = original

Premultiply:
T = [T1] x [T2] x [ T3] x [A]


These attachements show the difference of my messed up derivation and another persons derivation.
My version has no -1 in the first two of the right column. One sin and cos are switched.

His version has a -1 in the first two of the right column. The sin and cos functions line up vertically in the right column.

This post has been edited by hiddenghost: 15 December 2009 - 11:40 PM

[hiddenghost]

Ok, now I see that the Px(sin) + Py(cos) are commutative so the reason for the switch of the two is apparent.

I still can't quite see how the -1 got next to the two cosines in the right column for his.

This post has been edited by hiddenghost: 15 December 2009 - 11:48 PM

[LaFayette]

edit: Ok, now I read it more carefully.


Thinking about it without any calculations, all matrices have a 1 in the bottem right corner which lead to terms looking something like Px*cos (theta) - Px = Px(cos (theta) - 1). You are probably doing some simple mistake..

What does your translation matrices look like??

I did it like this on the computer

R = {{Cos[x], - Sin[x], 0}, {Sin[x], Cos[x], 0}, {0, 0, 1} }
T1 = {{1, 0, Px}, {0, 1, Py}, {0, 0, 1}}

Inverse[T1]*R*T1

and got exactly what the other guy got.

This post has been edited by LaFayette: 17 December 2009 - 06:01 AM

[hiddenghost]

LaFayette, on 17 Dec, 2009 - 03:30 AM, said:

What does your translation matrices look like??

The translation to the origin

The rotation

The translation back to the arbitrary point


Everything is correct for me accept the -1 so yeah, it probably is a simple error.

[LaFayette]

Opposite to what I wrote yesterday, given a point p to rotate about an arbitrary point q you should first apply the translation to origin, the rotation, and then the translation back.

So
R = {{Cos[x], - Sin[x], 0}, {Sin[x], Cos[x], 0}, {0, 0, 1} } //Rotation
T1 = {{1, 0, -Px}, {0, 1, -Py}, {0, 0, 1}} //Translation to origin
T2 = {{1, 0, Px}, {0, 1, Py}, {0, 0, 1}} //Translation back

And T = T2 x R x T1 which in steps are:

A = R x T1 = { {Cos[x], -Sin[x], -Px Cos[x] + Py Sin[x]}, {Sin[x],
Cos[x], -Py Cos[x] - Px Sin[x]}, {0, 0, 1} }

T = T2 x R x T1 = T2 x A = { {Cos[x], -Sin[x], Px - Px Cos[x] + Py Sin[x]}, {Sin[x], Cos[x],
Py - Py Cos[x] - Px Sin[x]}, {0, 0, 1} }

(Hope you get the notation I'm using)
Changing the order gives a slightly different result (other signs) but given that we are in a normal coordinate system I think this should be correct. (I actually tried writing a program in Mathematica using that matrix and it worked)

This post has been edited by LaFayette: 18 December 2009 - 02:21 AM

[hiddenghost]

I think I get your notation.
Another way to write it would be:
R = {{Cos[x], - Sin[x], 0},
------{Sin[x], Cos[x], 0},
------ {0,-------0,----1} } //Rotation
T1 = {{1, 0, -Px},
------- {0, 1, -Py},
------- {0, 0, 1}} //Translation to origin
T2 = {{1, 0, Px},
------- {0, 1, Py},
------- {0, 0, 1}} //Translation back

Thank you LaFayette.

When I get time I will try to calculate it again to see if I missed something.

This post has been edited by hiddenghost: 18 December 2009 - 12:42 PM