How to solve these two equations for 'tau' and 'b'? All the other symbols are constants. Please help??

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Anik Faisal 2018 年 8 月 7 日
コメント済み: Anik Faisal 2018 年 8 月 9 日
I've tried using the sysms command but not sure I'm getting the input formatting correct

4 件のコメント

Anik Faisal 2018 年 8 月 7 日
Yeah I did. I think the image is attached now
Stephan 2018 年 8 月 7 日
you should also show us what you have done so far by using syms...
Anik Faisal 2018 年 8 月 7 日
syms m b nu R rc tau g b0 u0
eqn1=(m*b/2)*((2-nu)/(1-nu))*R*log(R/rc)-tau*pi*R^2+(g*pi^2*R^2/b0)*sin(2*pi*(u0+b)/b0)==0;
eqn2=(m*b^2/4)*((2-nu)/(1-nu))*(1+log(R/rc))-2*pi*tau*R*b-g*pi*R*sin(pi*b/b0)*sin(pi*(2*u0+b)/b0)==0;
eqns=[eqn1 eqn2];
vars=[b tau];
[solv,solu]=solve(eqns,vars)

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採用された回答

Stephan 2018 年 8 月 7 日
Hi,
i used the isolate function instead of solve and found:
sol_tau =
tau == -(2778046668940015*R*g*sin((2*pi*(b + u0))/b0) + 281474976710656*b*b0*m*log(R/rc) - 2778046668940015*R*g*nu*sin((2*pi*(b + u0))/b0) - 140737488355328*b*b0*m*nu*log(R/rc))/(281474976710656*R*b0*pi*(nu - 1))
sol_b =
b == 0
by using these lines of code:
syms m b nu R rc tau g b0 u0
eqn1=(m*b/2)*((2-nu)/(1-nu))*R*log(R/rc)-tau*pi*R^2+(g*pi^2*R^2/b0)*sin(2*pi*(u0+b)/b0)==0;
eqn2=(m*b^2/4)*((2-nu)/(1-nu))*(1+log(R/rc))-2*pi*tau*R*b-g*pi*R*sin(pi*b/b0)*sin(pi*(2*u0+b)/b0)==0;
sol_tau = simplifyFraction(isolate(eqn1,tau))
eqn2_new = simplifyFraction(subs(eqn2,tau,rhs(sol_tau)));
sol_b = simplifyFraction(isolate(eqn2_new,b))
I did not check if you have typos in your equations...
Best regards
Stephan

11 件のコメント

Anik Faisal 2018 年 8 月 9 日
could you suggest any other technique for solving this?
Walter Roberson 2018 年 8 月 9 日
Assuming that your initial code was correct in https://www.mathworks.com/matlabcentral/answers/413950-how-to-solve-these-two-equations-for-tau-and-b-all-the-other-symbols-are-constants-please-help#comment_597512 then, No, there is no closed form solution for this, aside from
tau = g*Pi*sin(2*Pi*u0/b0)/b0
b = 0
The other solutions involve the roots of something that is similar to a degree 4 polynomial but involving terms that are also trig.
Well, correction: there are also a couple of analytic solutions if the other parameters happen to have special relationships, such as if
-8*R*g*Pi^2*(nu-1)*sin(2*Pi*u0/b0)/(b0^2*m*(nu-2)*(3*ln(R/rc)-1))
happens to be an integer
Anik Faisal 2018 年 8 月 9 日
Hi Walter,
I got the same
tau = g*Pi*sin(2*Pi*u0/b0)/b0
b = 0
I am curious about the technique for other analytical solutions you're speaking of.

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