- #1
zokos
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Hello to all,
I am currently studying computer graphics and I have came up with the following problem. Consider that we have three coordinate systems, let's say CSA1, CSA2 and CSA3 that have the same origin and differ by a rotation. That is to CSA2 connects to CSA1 by R12 and CSA3 to CSA1 by R13. Assume that Ai represents a base for the corresponding coordinate system CSAi. That is Ai comprises from 3 unit orthogonal vectors that are the base of CSAi. Then A2 connects to A1 via:
A2=R12*A1 (eq. 1)
and moreover
A3=R13*A1 (eq. 2)
eq. 1 becomes A1=inv(R12) * A2 (eq. 3) (inv stands for inverse) and eq. 2 using eq. 3 becomes
A3=R13*inv(R12)*A2 (eq. 4)
now we also know from theory that the coordinates (P1, P2, P3) of a point P in the three coordinates systems are connected via:
P2=inv(R12)*P1 (eq. 5)
P3=inv(R13)*P1 (eq. 6)
eq. 5 becomes P1=R12*P2 and substituting it in eq. 6 we have that
P3=inv(R13)*R12*P2 (eq. 7)
moreover applying the same rule in eq. 4, we get that the coordinates of P in the second and third coordinate system are connected via:
P3=inv(R13*inv(R12))*P2 that is
P3=R12*inv(R13)*P2 (eq. 8)
equations 7 and 8 gives us that
inv(R13)*R12=R12*inv(R13)
which is obviously wrong.
can anyone help me and show me my mistake?
Many thx,
zokos
I am currently studying computer graphics and I have came up with the following problem. Consider that we have three coordinate systems, let's say CSA1, CSA2 and CSA3 that have the same origin and differ by a rotation. That is to CSA2 connects to CSA1 by R12 and CSA3 to CSA1 by R13. Assume that Ai represents a base for the corresponding coordinate system CSAi. That is Ai comprises from 3 unit orthogonal vectors that are the base of CSAi. Then A2 connects to A1 via:
A2=R12*A1 (eq. 1)
and moreover
A3=R13*A1 (eq. 2)
eq. 1 becomes A1=inv(R12) * A2 (eq. 3) (inv stands for inverse) and eq. 2 using eq. 3 becomes
A3=R13*inv(R12)*A2 (eq. 4)
now we also know from theory that the coordinates (P1, P2, P3) of a point P in the three coordinates systems are connected via:
P2=inv(R12)*P1 (eq. 5)
P3=inv(R13)*P1 (eq. 6)
eq. 5 becomes P1=R12*P2 and substituting it in eq. 6 we have that
P3=inv(R13)*R12*P2 (eq. 7)
moreover applying the same rule in eq. 4, we get that the coordinates of P in the second and third coordinate system are connected via:
P3=inv(R13*inv(R12))*P2 that is
P3=R12*inv(R13)*P2 (eq. 8)
equations 7 and 8 gives us that
inv(R13)*R12=R12*inv(R13)
which is obviously wrong.
can anyone help me and show me my mistake?
Many thx,
zokos