# unable to use solve or fsolve for linear system

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Dhanush Srikanth on 26 Jun 2022
Commented: Dhanush Srikanth on 26 Jun 2022
syms R1 R2 R3 R4
COM = [-6,0,10]
r1 = [-54.2,-27.5,-16.4]
r2 = [-54.2,27.5,-16.4]
r3 = [42.9,-27.5,-16.4]
r4 = [42.9,27.5,-16.4]
rr1 = [0,0,R1]
rr2 = [0,0,R2]
rr3 = [0,0,R3]
rr4 = [0,0,R4]
mg = [0,0,980.65*16631]
eq(1) = cross(r1-r2,rr2) + cross(r1-r3,rr3) + cross(r1-COM,mg) + cross(r1-r4,rr4) == [0,0,0]
eq(2) = cross(r2-r1,rr1) + cross(r2-r3,rr3) + cross(r2-COM,mg) + cross(r2-r4,rr4) == [0,0,0]
eq(3) = cross(r3-r2,rr2) + cross(r3-r1,rr1) + cross(r3-COM,mg) + cross(r3-r4,rr4) == [0,0,0]
eq(4) = cross(r4-r2,rr2) + cross(r4-r3,rr3) + cross(r4-COM,mg) + cross(r4-r1,rr1) == [0,0,0]
j = fsolve(eq,[R1,R2,R3,R4])
im trying to solve the above system of linear equations for R1 R2 R3 & R4
the error is
Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.
L_tilde2 = builtin('subsasgn',L_tilde,struct('type','()','subs',{varargin}),R_tilde);
C = privsubsasgn(L,R,inds{:});

Torsten on 26 Jun 2022
Edited: Torsten on 26 Jun 2022
Seems your equations are not independent.
syms R1 R2 R3 R4
COM = [-6,0,10];
r1 = [-54.2,-27.5,-16.4];
r2 = [-54.2,27.5,-16.4];
r3 = [42.9,-27.5,-16.4];
r4 = [42.9,27.5,-16.4];
rr1 = [0,0,R1];
rr2 = [0,0,R2];
rr3 = [0,0,R3];
rr4 = [0,0,R4];
mg = [0,0,980.65*16631];
%eq(1) = cross(r1-r2,rr2) + cross(r1-r3,rr3) + cross(r1-COM,mg) + cross(r1-r4,rr4) == [0,0,0]
%eq(2) = cross(r2-r1,rr1) + cross(r2-r3,rr3) + cross(r2-COM,mg) + cross(r2-r4,rr4) == [0,0,0]
%eq(3) = cross(r3-r2,rr2) + cross(r3-r1,rr1) + cross(r3-COM,mg) + cross(r3-r4,rr4) == [0,0,0]
%eq(4) = cross(r4-r2,rr2) + cross(r4-r3,rr3) + cross(r4-COM,mg) + cross(r4-r1,rr1) == [0,0,0]
%j = fsolve(eq,[R1,R2,R3,R4])
cross(r1-r2,rr2) + cross(r1-r3,rr3) + cross(r1-COM,mg) + cross(r1-r4,rr4)
ans =
cross(r2-r1,rr1) + cross(r2-r3,rr3) + cross(r2-COM,mg) + cross(r2-r4,rr4)
ans =
cross(r3-r2,rr2) + cross(r3-r1,rr1) + cross(r3-COM,mg) + cross(r3-r4,rr4)
ans =
cross(r4-r2,rr2) + cross(r4-r3,rr3) + cross(r4-COM,mg) + cross(r4-r1,rr1)
ans =
A = [0 -55 0 -55;55 0 55 0; 0 0 97.1 97.1;-97.1 -97.1 0 0];
rank(A)
ans = 3
b = [3588021833/8;-3588021833/8;-1648577405738025/2097152;836259700628521/1048576];
sol = A\b;
Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 3.144153e-17.
R1 = sol(1)
R1 = -8.5635e+06
R2 = sol(2)
R2 = 3.5012e+05
R3 = sol(3)
R3 = 4.0891e+05
R4 = sol(4)
R4 = -8.5047e+06
Dhanush Srikanth on 26 Jun 2022
Thanks a lot I got it now.

Walter Roberson on 26 Jun 2022
those cross() calls each return a vector, and you compare the vector to [0 0 0] giving a vector result. You then try to store the vector in a scalar location.
If you were to store the entire vector, then when you got to the fsolve you would have four sets of 3 equations, for a total of 12 equations. And you are trying to fsolve the 12 for four variables. That is unlikely to succeed.
Dhanush Srikanth on 26 Jun 2022
Right, so should I store the result from each equation seperately and then pick any 4 equations and then use them in solve?

R2021b

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