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From: Ali on 3 Jun 2010 17:38
Okay so I kind of figures it out!

Here is the code I used:

syms m1 m2 k1 k2 c1 c2


solve([(m2*k1)+(c2*c1)-73.6225,
(k1*k2)-2.656e6,
(m1*m2)-0.000575,
(c2*m1)+ ((c2+c1)*m2)-0.01065,
(m1*k2)+ (k1*m2)+(k2*m2)+(c1*c2)-1.134625000000000e+02,
(c2*k1)+(c1*k2)-604.5]);


as you can see some of the numbers changed, this is due to some typing mistakes but either way... Since this is an exact solution the exact equations needed to be provided... Lets say I don't have the exact number (less digits lets say) so the exact solution really doesn't exist and we'll need to use some sort of iterative method? How would I go about doing that!?
From: Walter Roberson on 3 Jun 2010 18:08
Ali wrote:
> Okay so I kind of figures it out!
> Here is the code I used:
>
> syms m1 m2 k1 k2 c1 c2
>
>
> solve([(m2*k1)+(c2*c1)-73.6225,
> (k1*k2)-2.656e6,
> (m1*m2)-0.000575,
> (c2*m1)+ ((c2+c1)*m2)-0.01065,
> (m1*k2)+ (k1*m2)+(k2*m2)+(c1*c2)-1.134625000000000e+02,
> (c2*k1)+(c1*k2)-604.5]);
>
>
> as you can see some of the numbers changed, this is due to some typing
> mistakes but either way... Since this is an exact solution the exact
> equations needed to be provided... Lets say I don't have the exact
> number (less digits lets say) so the exact solution really doesn't exist
> and we'll need to use some sort of iterative method? How would I go
> about doing that!?

solve() is still going to do the best job that can be done with the
information you do have. About the only thing you could do better would be to
do some interval arithmetic calculations to see the range of values that could
be output based upon the precision of the input values. For example, you could
add +T to the end of the last equation and solve for m1 m2 k1 k2 c1 c2, and
then you would see how the solution changes. k1 turns out to be particularly
sensitive -- e.g., a change of 4/100 in the 604.5 constant (i.e., would round
to the same value) makes a difference of 3200 in k1.



[{c1 = .1500000000, c2 = .1500000000, k1 = 3200., k2 = 830.,
m1 = .2500000000E-1, m2 = .2300000000E-1},

{c1 = -.1500000000, c2 = -.1500000000, k1 = -3200., k2 = -830.,
m1 = -.2500000000E-1, m2 = -.2300000000E-1},

{c1 = -3.078422704-29.01028502i, c2 = -.1070289711+1.152755637i,
k1 = 8112.510827-1069.414581i, k2 = 321.8034898+42.42106441i,
m1 = .1168663341-.1673147690E-1i, m2 = .4821328277E-2+.6902581763E-3i},

{c1 = 3.078422704+29.01028502i, c2 = .1070289711-1.152755637i,
k1 = -8112.510827+1069.414581i, k2 = -321.8034898-42.42106441i,
m1 = -.1168663341+.1673147690E-1i, m2 = -.4821328277E-2-.6902581763E-3i},

{c1 = -3.078422704+29.01028502i, c2 = -.1070289711-1.152755637i,
k1 = 8112.510827+1069.414581i, k2 = 321.8034898-42.42106441i,
m1 = .1168663341+.1673147690E-1i, m2 = .4821328277E-2-.6902581763E-3i},

{c1 = 3.078422704-29.01028502i, c2 = .1070289711+1.152755637i,
k1 = -8112.510827-1069.414581i, k2 = -321.8034898+42.42106441i,
m1 = -.1168663341-.1673147690E-1i, m2 = -.4821328277E-2+.6902581763E-3i}]
From: Steven Lord on 3 Jun 2010 21:55

"Ali " <sabti(a)mie.utoronto.ca> wrote in message
news:hu97bs$rfm$1(a)fred.mathworks.com...
> Okay so I kind of figures it out!
> Here is the code I used:
>
> syms m1 m2 k1 k2 c1 c2
>
>
> solve([(m2*k1)+(c2*c1)-73.6225,
> (k1*k2)-2.656e6,
> (m1*m2)-0.000575,
> (c2*m1)+ ((c2+c1)*m2)-0.01065,
> (m1*k2)+ (k1*m2)+(k2*m2)+(c1*c2)-1.134625000000000e+02,
> (c2*k1)+(c1*k2)-604.5]);

That's quite a change from the equations you originally posted -- the value
you subtracted from the last equation changed from 2.656e6 to -604.5. The
fact that one of your values is still really large compared to the others
could still cause problems -- either k1 or k2 have to be really large or
they both need to be medium. When you combine that with the first and fifth
equations, m1 and m2 are probably going to have to be fairly small, which
means c1 and c2 will need to be large to counteract that in the fourth
equation.

Still, if you use OPTIMSET to increase the number of function evaluations
allowed for FSOLVE, you can get reasonably small residuals after a few
thousand iterations, which (due to the simplicity of the operations being
performed) is farily quick -- a second or two. When I allow 2000 function
evaluations, the residuals I receive are:

fval =

Columns 1 through 3

2.32177902148578e-006 -0.000115446280688047 -0.00133763329396516

Columns 4 through 6

0.011010895224216
3.9505012864538e-005 -2.60158685705392e-008

All but the residuals for 3 and 4 are close to 0, and even those resdiuals
aren't that bad.

--
Steve Lord
slord(a)mathworks.com
comp.soft-sys.matlab (CSSM) FAQ: http://matlabwiki.mathworks.com/MATLAB_FAQ
To contact Technical Support use the Contact Us link on
http://www.mathworks.com


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