% puzzler3

% wb1413, Multibody Dynamics B, Spring 2009.
%     Arend L. Schwab 04-Mar-2009
%     Copyright (c) 2009 by TU-Delft, the Netherlands.

% find a basis for the non-equilibrum force space
% of the double pendulum, two bars of length 1
l1=1; l2=1;
% in the upright position
s1=1; c1=0; s2=1; c2=0;
% D is the jacobian of the contraints
D=[1 0  l1/2*s1 0 0 0
   0 1 -l1/2*c1 0 0 0
  -1  0  l1/2*s1 1 0  l2/2*s2
   0 -1 -l1/2*c1 0 1 -l2/2*c2]
D =

    1.0000         0    0.5000         0         0         0
         0    1.0000         0         0         0         0
   -1.0000         0    0.5000    1.0000         0    0.5000
         0   -1.0000         0         0    1.0000         0
% f=D'*sigma+f_nonequilibrium where D' spans the equilibrium force space
% the dimension or rank is 
rank(D')
ans =
     4
% the null space of D are the forces which can span 
% the non-equilibrium force space
fne=null(D,'r')
fne =

    0.5000    0.2500
         0         0
   -1.0000   -0.5000
    1.0000         0
         0         0
         0    1.0000
% non draw these