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Offline ElSid  
#21 Posted : 19 March 2018 20:01:21(UTC)
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Ahh. I missed that. FindRoot(ES1).smz (4kb) downloaded 27 time(s).
Originally Posted by: omorr Go to Quoted Post
This is a matter of third and fourth parameter of FindRoot() function (numerical precision). The solution will be different if you change them. Just look at the order of magnitude of the solutions. The second and third values are quite close to zero. Therefore, one might expect this behavior.


Offline ola_nicolas  
#22 Posted : 19 March 2018 20:43:43(UTC)
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Originally Posted by: Jean Giraud Go to Quoted Post
... I don't think so ! Check I recomposed correctly ...


I think it's just a speculation Jean Giraud. A case at one billion, or even the only case of an infinite number of possibilities. In any case, the variables B1 and B2 physically represent inductions through an electromagnetic core. Exponent n (or α if you prefer) is a strictly positive number as the formula is constructed. It is an empirical one, known as the Steinmetz relationship. That is why (with all due respect) I will never assign negative values to exponent n. Then (with all due respect) I think you have researched a lot to find out that negative value α = -1.23456789... (irrational number) very probably (perhaps certainly) the only one for which there is linear dependence relative to equations 1 and 3 in the system.

Edited by user 19 March 2018 21:05:10(UTC)  | Reason: Not specified

Offline omorr  
#23 Posted : 19 March 2018 21:45:35(UTC)
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Hello ola_nicolas,

My advice to you is to try changing the units of your known variables (if this can have a physical meaning, of course) in order to get the solution of unknown variables not so close to zero. And every time check the solution - surprises are always possible.

Regards,
Radovan
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Offline Jean Giraud  
#24 Posted : 19 March 2018 22:06:36(UTC)
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Originally Posted by: ola_nicolas Go to Quoted Post
negative value α = -1.23456789...

Was just to attract your attention. Now, plug 1.23456789 ... anything.
Maybe the equality system is zombie ?

Offline ola_nicolas  
#25 Posted : 20 March 2018 09:31:09(UTC)
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Hello everybody

Thanks Jean Giraud for the alert. Thanks Radovan for advice. Thank you all for your involvement. Without you it could not have been a convenient solution. Radovan is right. I have made ab initio the transformations in the units of measurement that I use in practice. I will do more tests and checks until I have established the usable form of the solution. However, from the point of view of the result, I find myself close to physical reality. Thank you once again.

Post scriptum: Please tell me what kind of files are those with the ending "msz". I have tried unsuccessfully to open the files posted by ElSid.

Edited by user 20 March 2018 09:42:41(UTC)  | Reason: Not specified

Offline Davide Carpi  
#26 Posted : 20 March 2018 13:48:04(UTC)
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Originally Posted by: ola_nicolas Go to Quoted Post
Post scriptum: Please tell me what kind of files are those with the ending "msz". I have tried unsuccessfully to open the files posted by ElSid.


It is a compressed format of the smath file format. I see no issues opening it in SMath Studio 0.98.6484 portable, except that you have to skip the warning messages (file created with more recent SMath Studio + Writer Region plugin requested; the second is in red because there isn't a 0.10 version for your SS, however you can skip the message and will be fully readable if you have the latest version of the Writer Region for your SS, otherwise Writer Regions will be displayed with a red cross)

Edited by user 20 March 2018 13:49:48(UTC)  | Reason: Not specified

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Offline ola_nicolas  
#27 Posted : 20 March 2018 15:57:17(UTC)
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Auguri Davide

Thank you for your answer. I do not have time now, but I will try later to see how I have to do it. Otherwise, all the best.
Offline Jean Giraud  
#28 Posted : 21 March 2018 07:43:37(UTC)
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Originally Posted by: ola_nicolas Go to Quoted Post
Exponent n (or α if you prefer) is a strictly positive number as the formula is constructed.


The other way around. As you give the system, a 3 conditions system, you need
a vector of initials [3]. Whereas 'n' can be anything, make 'n init =1'.
This will solve uniquely wrt dummy 'n' considering the [3] set of conditions
is ill posed for two parameters.
Offline ola_nicolas  
#29 Posted : 21 March 2018 11:14:06(UTC)
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Greetings to all

In attachment I intend to do a comparative study between solving the equation system using different units of measure. The first variant is the one already previously analyzed, with the measurement units dictated by practical considerations - [Hz] for frequencies, [T] for inductions and [W / mm3] for specific losses in the magnetic core. The following three variants correspond to the set of measuring units - [kHz] for frequencies, [T] for inductions and [kW / m3], used in ferrite cores producers data sheets. It is seen that depending on the set of three initial data, the result of the calculation is altered in variants 3 and 4. I also responded to Radovan's request to try to use different measurement units in order to avoid the zero convergence of the solutions for the variables η si λ. It can be seen that the use of this set of units of measure has advantages over the other. Considering the very wide range of ferrite cores currently produced, it is very likely that in some cases the calculation will yield vague results.

On the other hand, in connection with Radovan's fear that the use of very small subunits can affect the calculation, I think it is unreasonable. I made a SMath spreadsheet to see what result returns my program to 1/10^308. It was 10^-308. So in numerical representation, the program can distinguish between extremely close to zero results. By comparison, MathCAD has set the (infinite) level of 10^307. At the same time, however close to zero, the result does not in any way affect my initial values ​​set for the variables. The difference between the value returned by calculating of two variables, and their initial value, will always be less than the unit. For the value of exponent n, in practice we find values ​​of the order of the result from this calculation. So it seems to me that the advantage of the first option is obvious. Otherwise, all the best.

FindRoot comparative.jpg
Offline Jean Giraud  
#30 Posted : 21 March 2018 22:37:16(UTC)
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Originally Posted by: ola_nicolas Go to Quoted Post
In attachment I intend to do a comparative study between solving the equation system using different units of measure.

Totally useless, it has nothing to do with the maths handled by the solver.
Up to this point, the project is brute force solved in the attached.
It could be refined from there by finishing outside of the solver.
It might be possible to manually optimized [η,λ] and let 'n=2',
this can't be predicted ... you may have to play [n,η,λ].
Could this finishing be done done automatically ?
Answer is NO ... why ? because of the X^Y power function involved.
Under the hood, X^Y is a combination exp(,), ln(,)
Can't help more, from this lecture. Lot more difficult/impossible
projects were done, well done from patience.

FindRootsRefexive.sm (16kb) downloaded 12 time(s).

FindRootsReflexive.PNG
Offline mkraska  
#31 Posted : 21 March 2018 23:41:03(UTC)
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Originally Posted by: ola_nicolas Go to Quoted Post
Greetings to all

In attachment I intend to do a comparative study between solving the equation system using different units of measure.



Using different units for the equations in FindRoot(2) essentially means to use different relative convergence criteria if these aren't adjusted in the FindRoot() call.
Martin Kraska

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Offline ola_nicolas  
#32 Posted : 22 March 2018 15:51:40(UTC)
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Hello

Originally Posted by: mkraska Go to Quoted Post
... use different relative convergence criteria ...

Can you detail (possibly with examples) please?

Edited by user 22 March 2018 15:53:20(UTC)  | Reason: Not specified

Offline mkraska  
#33 Posted : 22 March 2018 16:40:18(UTC)
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FindRoot() internally converts the numbers to base units (meter). The default tolerance of 0.0001 for variable and function is to be interpreted in the base unit of the given quantity.

In the first example with volume and height being close to unity in base units, the default tolerance is appropriate.
If the dimensions are scaled by factor 0.001 (mm instead of m) then the tolerance is nearly of the same magnitude as the variable. The result is complete rubbish. The problem is fixed by adjustment of the tolerances.

2018-03-22 14_31_05-SMath Studio - [findroot.sm].png
findroot.sm (30kb) downloaded 13 time(s).
Martin Kraska

Pre-configured portable distribution of SMath Studio: https://smath.com/wiki/SMath_with_Plugins.ashx
thanks 1 user thanked mkraska for this useful post.
on 22/03/2018(UTC)
Offline Jean Giraud  
#34 Posted : 22 March 2018 17:29:40(UTC)
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Originally Posted by: ola_nicolas Go to Quoted Post
Can you detail (possibly with examples) please?


Maybe Martin has some trick.
That's what I meant about NON-linear model wrt X^Y,
Great Mathcad failed on that one [I guess it toke me > one day]
Maybe the project is not properly constructed, can it be made implicit ?

Difficult.PNG

Solve Algebraic.sm (47kb) downloaded 21 time(s).

Offline ola_nicolas  
#35 Posted : 23 March 2018 12:48:19(UTC)
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Hi

Originally Posted by: mkraska Go to Quoted Post
... If the dimensions are scaled by factor 0.001 (mm instead of m) then the tolerance is nearly of the same magnitude as the variable. The result is complete rubbish. The problem is fixed by adjustment of the tolerances ...


All right, I think I understand. Corroborating Martin's information with Jean Giraud's example (using graphical representations) I understand it is possible that the function passes through the points I choose to apply the FindRoot procedure, while the rest of the points are quite far from what they should results using the negligible tolerances in relation to the magnitude of variables. I will study much more carefully these numerical procedures for solving equations and / or systems of equations. Thank you all.
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