5 Reputation

10 years, 236 days

@Preben Alsholm  Thanks for your r...

Thanks for your reaction. Yes, Sol=Eq75. Sorry for the confusion..... The code should be:

> Eq75 := dsolve({dsys, ini}, {phix(t), phiy(t), phiz(t), thetax(t), thetay(t)}, numeric, output = listprocedure, range = 0 .. ts)

> odeplot(Eq75, [t, Z2], 0 .. ts, numpoints = 100)

The code solves rigid body motion. Z2 is the z coordinate of a point on the rigid body and can be expressed as Z2=f(t,phix(t), phiy(t), phiz(t), thetax(t), thetay(t)).

What I want to do is store the value Z2 at t=ts after running the simulation for t=0..ts. How can I call upon this value? The odeplot command can.......

desired value:= Z2(ts) doesnt work anyway...

Kind regards, Gerben

Solved!...

Hi Axel!

Thanks for your top tip! Using fsolve instead of solve provides a numerical solution for me. My problem is solved!

Best regards.

Solved!...

Hi Axel!

Thanks for your top tip! Using fsolve instead of solve provides a numerical solution for me. My problem is solved!

Best regards.

2 solutions...

Thanks for you advice rlopez. It helped me quite a bit. Still maple is struggling with providing a solution.

When I looked for a numeric solution, I gave a positive value to y1 and y2. Maple still gives a solution with RootOf and _Z because there are solutions for a >0 and for a< 0. I need the solution for a positive a. There is only one solution for this positive a. How do I get maple to solve this?

2 solutions...

Thanks for you advice rlopez. It helped me quite a bit. Still maple is struggling with providing a solution.

When I looked for a numeric solution, I gave a positive value to y1 and y2. Maple still gives a solution with RootOf and _Z because there are solutions for a >0 and for a< 0. I need the solution for a positive a. There is only one solution for this positive a. How do I get maple to solve this?

y1 and y2 are positive...

Ok, I know know that the RootOf is used because of the asymptotes in the arccosh. However y1 and y2 are positive, and thus ((y1+a)/a) > 1 and ((y2+a)/a). So we are working to the right of all asymptotes and there should be one solution. How do I let the solver take these assumptions?

Maple sais:

`> assume(a > 1);> solve(x, a);Warning, solve may be ignoring assumptions on the input variables.`
` `
`How do I get Maple to look at this one existing solution?!`
` `

I hope the formula is more readable this...

`> x := a*arccosh((y1+a)/a)+a*arccosh((y2+a)/a);                      /y1 + a\            /y2 + a\             a arccosh|------| + a arccosh|------|                      \  a   /            \  a   /> solve(x, a);                /       /y1 + _Z\          /y2 + _Z\\          RootOf|arccosh|-------| + arccosh|-------||                \       \  _Z   /          \  _Z   //`
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