Open Mike1425217072 opened 3 years ago
Hi, as far as I'm concerned, the calculation of Omega0 is right. in line 63, we have
t0dot = Omega0*P0*u0+P0*u0dot;
which is
$\dot{t}_0 = (\Omega_0 P_0 )u_0+ P_0 \dot{u}_0$.
so Omega0*P0
should be $\frac{\mathrm{d}}{\mathrm{d}t}P_0$,
since P0
is
[R, zeros(3);
zeros(3), eye(3)];
it's derivative will be the derivative of the top left part of P0, i.e,
R_dot = SkewSym(t0(1:3))*R
In function [t0dot,tLdot]=Accelerations(t0,tL,P0,pm,Bi0,Bij,u0,um,u0dot,umdot,robot)
line 49, %Base-link Omega0=[SkewSym(t0(1:3)), zeros(3,3); zeros(3,3), zeros(3,3)];
This should be:
Omega0=[SkewSym(t0(1:3)), zeros(3,3); zeros(3,3), SkewSym(t0(1:3))];
right?