Need some Math help from math guru!
Ok, i got a problem that i need to solve using a 6x6 matrix.
basically a *[6x1]=[6x6][6x1]
my equation are
A = 30 degree
.3(Pcos(A)) - 100(9.81)(0.1) = 0
Raz(.8) - 100(9.81)(.5) + P(.4) - .3(Psin(A)) = 0
Raz - 100(9.81) + Rbz-P = 0
(.3)Pcos(30) - Ray(.8) = 0
Rax - P = 0
Ray - Rby = 0
for the *[6x1] =
P
Rby
Rbz
Ray
Raz
Rax
Can anyone write out this matrix equation for me?
I can solve the problem easy without a matrix but i have to put it in the matrix form as given and solve it for a range of degree from 0-30
basically a *[6x1]=[6x6][6x1]
my equation are
A = 30 degree
.3(Pcos(A)) - 100(9.81)(0.1) = 0
Raz(.8) - 100(9.81)(.5) + P(.4) - .3(Psin(A)) = 0
Raz - 100(9.81) + Rbz-P = 0
(.3)Pcos(30) - Ray(.8) = 0
Rax - P = 0
Ray - Rby = 0
for the *[6x1] =
P
Rby
Rbz
Ray
Raz
Rax
Can anyone write out this matrix equation for me?
I can solve the problem easy without a matrix but i have to put it in the matrix form as given and solve it for a range of degree from 0-30
I started to work on this, but your notation is whacko: in the equations you have P, Rza, Rzb, Ryz, Rxa, Rya, and Ryb,
whereas in the vector you have: P, Rby, Rbz, Ray, Raz, and Rax.
Are Rby and Ryb the same variable? How about Ray and Rya? Where's Rxy in your vector?
This is an easy problem if I know the variables. It's impossible if I don't.
whereas in the vector you have: P, Rby, Rbz, Ray, Raz, and Rax.
Are Rby and Ryb the same variable? How about Ray and Rya? Where's Rxy in your vector?
This is an easy problem if I know the variables. It's impossible if I don't.
If I read this correctly, your system can be written as:
0.3*cos(A)*P + 0*Rby + 0*Rbz + 0*Ray + 0*Raz + 0*Rax = 100*9.81*0.1
(0.4 - 0.3*sin(A))*P + 0*Rby + 0*Rbz + 0*Ray + 0.8*Raz + 0*Rax = 100*9.81*0.5
-1*P + 0*Rby + 1*Rbz + 0*Ray + 1*Raz + 0*Rax = 100*9.81
0.3*cos(A)*P + 0*Rby + 0*Rbz + -0.8*Ray + 0*Raz + 0*Rax = 0
-1*P + 0*Rby + 0*Rbz + 0*Ray + 0*Raz + 1*Rax = 0
0*P + -1*Rby + 0*Rbz + 1*Ray + 0*Raz + 0*Rax = 0
If we let x be the vector:
[P]
[Rby]
[Rbz]
[Ray]
[Raz]
[Rax],
A be the 6 x 6 coefficient matrix:
[0.3*cos(A) 0 0 0 0 0]
[(0.4 - 0.3*sin(A) 0 0 0 0.8 0]
[-1 0 1 0 1 0]
[0.3*cos(A) 0 0 -0.8 0 0]
[-1 0 0 0 0 1]
[0 -1 0 1 0 0], and
b be the vector:
[100*9.81*0.1]
[100*9.81*0.5]
[100*9.81]
[0]
[0]
[0],
then the system can be written as:
Ax = b.
The solution is:
x = (A^(-1))b
(i.e., A-inverse times b).
I leave it to you to compute A-inverse.
0.3*cos(A)*P + 0*Rby + 0*Rbz + 0*Ray + 0*Raz + 0*Rax = 100*9.81*0.1
(0.4 - 0.3*sin(A))*P + 0*Rby + 0*Rbz + 0*Ray + 0.8*Raz + 0*Rax = 100*9.81*0.5
-1*P + 0*Rby + 1*Rbz + 0*Ray + 1*Raz + 0*Rax = 100*9.81
0.3*cos(A)*P + 0*Rby + 0*Rbz + -0.8*Ray + 0*Raz + 0*Rax = 0
-1*P + 0*Rby + 0*Rbz + 0*Ray + 0*Raz + 1*Rax = 0
0*P + -1*Rby + 0*Rbz + 1*Ray + 0*Raz + 0*Rax = 0
If we let x be the vector:
[P]
[Rby]
[Rbz]
[Ray]
[Raz]
[Rax],
A be the 6 x 6 coefficient matrix:
[0.3*cos(A) 0 0 0 0 0]
[(0.4 - 0.3*sin(A) 0 0 0 0.8 0]
[-1 0 1 0 1 0]
[0.3*cos(A) 0 0 -0.8 0 0]
[-1 0 0 0 0 1]
[0 -1 0 1 0 0], and
b be the vector:
[100*9.81*0.1]
[100*9.81*0.5]
[100*9.81]
[0]
[0]
[0],
then the system can be written as:
Ax = b.
The solution is:
x = (A^(-1))b
(i.e., A-inverse times b).
I leave it to you to compute A-inverse.
Originally Posted by nMeOnE,Aug 1 2010, 12:21 PM
Ok, i got a problem that i need to solve using a 6x6 matrix.
In what context does this problem arise?



