STODOLA-VIANELLO itteration problem

Ronitmaddy · **Joined:** Mon, 26 Dec 2011 07:35:15 UTC **Posts:** 3

I don't know where to post this calculation problem in this forum,So I m posting it here.
I hope you people would have a look at it and would help me to understand it.

The following problem is part of the Stodola-Vianello iteration method.After doing some calculation the following three equations are found,from which four variables are to be calculated.

The above image shows the calculation which has been carried out on above 3 equations ( these three equations has got 4 unknowns ,they are Wn^2 ,a1n,a2n and a3n )

In order to simplify we can represent the above three equation as follows

x*p = (95049*x) - (47524.5* y)
y*p = - (47524.5*x) + (95049*y) - (47524.5*z)
z*p = - (47524.5*y ) + (47524.5*z)

where p = Wn^2
x = a1n
y = a2n
z = a3n

I have tried to solve these 3 equation but failed to get any result,infact I got struck in the middle of calculation.So I m asking you people to help me in solving this problem.

For the calculation ,the author of this example has taken "x" as 1 (i.e. a1n = 1) as assumption
and has tabulated the result ,but in table I found that all values x,y and z are assumed (at first cycle) except Wn .

If any understand this procedure or know how to do this calculation ,then please explain me.

Thanking you in advance
Regards
Ronitmaddy

outermeasure · **Posted:** Mon, 26 Dec 2011 08:50:30 UTC

Moved from Proposed Problems to Differential Equations.

outermeasure · **Posted:** Mon, 26 Dec 2011 11:32:02 UTC

Ronitmaddy wrote:

I don't know where to post this calculation problem in this forum,So I m posting it here.
I hope you people would have a look at it and would help me to understand it.

The following problem is part of the Stodola-Vianello iteration method.After doing some calculation the following three equations are found,from which four variables are to be calculated.

The above image shows the calculation which has been carried out on above 3 equations ( these three equations has got 4 unknowns ,they are Wn^2 ,a1n,a2n and a3n )

In order to simplify we can represent the above three equation as follows

x*p = (95049*x) - (47524.5* y)
y*p = - (47524.5*x) + (95049*y) - (47524.5*z)
z*p = - (47524.5*y ) + (47524.5*z)

where p = Wn^2
x = a1n
y = a2n
z = a3n

I have tried to solve these 3 equation but failed to get any result,infact I got struck in the middle of calculation.So I m asking you people to help me in solving this problem.

For the calculation ,the author of this example has taken "x" as 1 (i.e. a1n = 1) as assumption
and has tabulated the result ,but in table I found that all values x,y and z are assumed (at first cycle) except Wn .

If any understand this procedure or know how to do this calculation ,then please explain me.

Thanking you in advance
Regards
Ronitmaddy

No! What you have is an iterative scheme.

Start with
$\begin{aligned} a_{1n}\omega_n^2 & = 95049 a_{1n} - 47524.5 a_{2n}\\ a_{2n}\omega_n^2 & = -47524.5 a_{1n} + 95049 a_{2n}-47524.5 a_{3n}\\ a_{3n}\omega_n^2 & = -47524.5 a_{2n}+47524.5 a_{3n} \end{aligned}$
and the normalisation $a_{1n}=1$ . We use the starting condition $a_{2n}^{(0)} = -1.3$ , $a_{3n}^{(0)}=0.7$ to iteratively find
$\begin{aligned} (\omega_n^2)^{(k+1)} & = 95049 a_{1n} - 47524.5 a_{2n}^{(k)}\\ a_{2n}^{(k+1)} & = \frac{1}{(\omega_n^2)^{(k+1)}} \left( -47524.5 a_{1n} + 95049 a_{2n}^{(k)} - 47524.5 a_{3n}^{(k)}\right)\\ a_{3n}^{(k+1)} & = \frac{1}{(\omega_n^2)^{(k+1)}} \left( -47524.5 a_{2n}^{(k+1)} + 47524.5 a_{3n}^{(k)}\right) \end{aligned}$
(maybe $a_{2n}^{(k)}$ in the last equation instead of the updated value $a_{2n}^{(k+1)}$ ).

Ronitmaddy · **Joined:** Mon, 26 Dec 2011 07:35:15 UTC **Posts:** 3

Thanks for the quick reply.
Yes it is iterative scheme.

Will you please tell me why

and

why not any other values like 2,3,etc.Why -1.3 and 0.7
can we use any other value,will the final result change after iteration based on new values for a2n and a3n
thanks for the reply,i hope you will reply and explain me again.

outermeasure · **Posted:** Mon, 26 Dec 2011 13:01:50 UTC

Ronitmaddy wrote:

Thanks for the quick reply.
Yes it is iterative scheme.

Will you please tell me why

and

why not any other values like 2,3,etc.Why -1.3 and 0.7
can we use any other value,will the final result change after iteration based on new values for a2n and a3n
thanks for the reply,i hope you will reply and explain me again.

The starting values -1.3 and 0.7 are chosen to be close to the exact solution, so that the scheme will hopefully converge to the exact solution (with prescribed tolerance) within a few iterations.

Ronitmaddy · **Joined:** Mon, 26 Dec 2011 07:35:15 UTC **Posts:** 3

thanks,it was very helpful.

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STODOLA-VIANELLO itteration problem

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