Thread Subject: Linear Simlutaneous Equations

Hi all,

I have the following six equations:

1. Reaction-Y1 + Force13*sin(A1) + Force12*sin(A3) = 0
2. Reaction-X1 + Force13*cos(A1) + Force12*cos(A3) = 0
3. Reaction-Y2 + Force21*sin(A1) + Force23*sin(A2) = 0
4. Reaction-X2 + Force21*cos(A1) + Force23*cos(A2) = 0
5. Reaction-Y3 + Force31*sin(A2) + Force32*sin(A3) = 0
6. Reaction-X3 + Force31*cos(A2) + Force32*cos(A3) = 0

I have calculated the Reaction values for all 6 of them, and all the angles of each of them.

How would I substitute these equations into a Matrix, so that I can solve for the forces, using the form of

AX = B

Where A is a square matrix, X is a vector of unknowns (Forces) and B is a vector of knowns, i.e, the Reactions forces.

Any help will be appreciated. Thanks.

Damien

Damien wrote:
> Hi all, I have the following six equations:
> 1. Reaction-Y1 + Force13*sin(A1) + Force12*sin(A3) = 0
> 2. Reaction-X1 + Force13*cos(A1) + Force12*cos(A3) = 0
> 3. Reaction-Y2 + Force21*sin(A1) + Force23*sin(A2) = 0
> 4. Reaction-X2 + Force21*cos(A1) + Force23*cos(A2) = 0
> 5. Reaction-Y3 + Force31*sin(A2) + Force32*sin(A3) = 0
> 6. Reaction-X3 + Force31*cos(A2) + Force32*cos(A3) = 0
>
> I have calculated the Reaction values for all 6 of them, and all the
> angles of each of them. How would I substitute these equations into a
> Matrix, so that I can solve for the forces, using the form of
> AX = B
> Where A is a square matrix, X is a vector of unknowns (Forces) and B
> is a vector of knowns, i.e, the Reactions forces.
> Any help will be appreciated. Thanks. Damien

Just rewrite is as a matrix/vector equation the same way that you stated it.

[sin(A3) sin(A1) 0 0 0 0 ] [Force12] [Reaction-Y1]
[cos(A3) cos(A1) 0 0 0 0 ] [Force13] [Reaction-X1]
[ 0 0 sin(A1) sin(A2) 0 0 ]*[Force21]= [Reaction-Y2]
[ 0 0 cos(A1) cos(A2) 0 0 ] [Force23] [Reaction-X2]
[ 0 0 0 0 sin(A2) sin(A3) ] [Force31] [Reaction-Y3]
[ 0 0 0 0 cos(A2) cos(A3) ] [Force32] [Reaction-X3]


                                              [A]*[X] = [B]

X = inv(A)*B if the inverse exists.
Otherwise, using the pseudoinverse to give you a leastsquare solution.
X = A\B

Note the wrap-around pattern of A3-A1,A1-A2,A2-A3

Freelance Embedded Systems Engineer <g9u5dd43@yahoo.com> wrote in message <48EE355C.9000106@yahoo.com>...

> X = inv(A)*B if the inverse exists.

Better use this
X = A \ B


> Otherwise, using the pseudoinverse to give you a leastsquare solution.
> X = A\B
>

No, use X = pinv(A)*B or better still X=lsqr(A,B)

Example:

A=[1 2; 2 4]
B = [3; 6]

>> A\B
Warning: Matrix is singular to working precision.

ans =

   NaN
   NaN

>> pinv(A)*B % Use lsqr

ans =

    0.6000
    1.2000

Bruno