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S
sm�ller-21-Visitor
1-Visitor
November 15, 2019
Solved

Using "Vorgabe" Function in Iteration Calculation

  • November 15, 2019
  • 5 replies
  • 2944 views

Hallo guys,

 

i have a small problem. Actually i need to build an Iterative Calculation procedure.

I have just started the process to build it up. I got help from you guys already so i have a template.

 

Now i have a complicated equation which i cannot transform to the desired searched value.

So i use the mathcad "Vorgabe" Function in order to recalculate the value. 

Now i dont know how to write it correctly into my Iterative procedure.

 

I think i only need to know the correct Syntax in order to be able to proceed my work

 

Thank you,

 

BR Stefan

Best answer by -MFra-

Hi,

I attach the file with my changes.

Regards

5 replies

L
LucMeekes23-Emerald IV
23-Emerald IV
November 15, 2019

Stefan,

Du kannst die 'Vorgabe...Suchen()' Struktur nicht in einem Programm benützen.

LM_20191115_Iteration.png

Du kannst aber das Resultat einer solchen Struktur als Funktion schreiben, und die Function in einem Programm gebrauchen. Aber warscheinlich kommst du hier besser weg wen du einfach die root() Funktion gebrauchst um den Dp1 Wert zu finden.

Dann noch, du sollst keine Definition, wie z.B. "Dp2<-", schwischen 'Vorgabe' und 'suchen()' stellen.

Und wass ist der Fehler mit Beziehung der '2' (das Quadrat)?

 

Viel Erfolg!
Luc

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-MFra-21-Topaz II
21-Topaz II
November 16, 2019

Hi,
I don't think it necessary to use the Find operator because Dp2 is given by the "Lambert" function, see the image:

lambert1.jpg

lambert.jpg

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-MFra-21-Topaz II
21-Topaz II
November 16, 2019

Hi,
I don't think it necessary to use the Find operator because Dp2 is a "Lambert" function, see the image:

lambert1.jpglambert.jpg

In the first definition of ITER the parameters are 8, while in the next one the parameters are 9.

S
sm�ller-21-VisitorAuthor
1-Visitor
November 18, 2019

MFra,

thank you,

 

here its not working. Can you send me both mathcad files (the code in the second black circle and the yellow part lambert code. Thank you

br

Stefan

L
LucMeekes23-Emerald IV
23-Emerald IV
November 16, 2019

A simpler implementation of the LambertW function is:

LM_20191116_LambertW.png

It may be less powerfull than Stuarts implementation,

but it works for an argument range from slightly below 0 to 500 at least.

 

Success!
Luc

    S
    sm�ller-21-VisitorAuthor
    1-Visitor
    November 18, 2019

    this fomula does not work with the LambertW Dp2 calculation? Can you help me another time plz?

    S
    sm�ller-21-VisitorAuthor
    1-Visitor
    November 18, 2019

    I encounter another problem.

    For this equation there are 2 solutions:

     

    The Lambert Function looses 1 solution and with the lambert function i get the solution of 219,19898mm

     

    But the correct one is 98,81192mm

     

     

     

     

     

     

     

     
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