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Itertive Calculation / Extending of current Situation / 2 Parameter Variation

sm�ller-2
11-Garnet

Itertive Calculation / Extending of current Situation / 2 Parameter Variation

Hey guys,

with the help of Werner i could manage my Iterative Calculation Routine.

Now i need to extend my work.

What i would need:

2 Parameter Variation of "F" and "ME" with a well defined Calculationstep of those 2 parameters.

For Example:

Calculate FAmax with ME=0    ---> Iterationstep for FA = FAmax/40

Calculate MEmax with FA=0 ---> Iterationstep for ME = Memax/40

Now those 2 Iterationsteps should applay to the "2 Parameter Variation"

The Results should then be written in a Matrix so that i can export the results to excel.

I think i only need a small add in to my Programm, what i have now.

Maybe anybody can help me

Calculating the Iterationstep should be possible with my mathcad program as i have now:

Calculate FAmax with ME=0    ---> Iterationstep for FA = Scalingfactor x FAmax

Calculate MEmax with FA=0 ---> Iterationstep for ME = Scalingfactor x MEmax

Scalingfactor should be choosable!

Next step: Mathcad should perform a Iterative Calculation for every combination of thoose steps. (Here Scalingfactor 5 %)

Calculation 1 100% Fa 0% Me     Results

Calculation 2 95% Fa 5% Me      Results

......

On Page 15 you can see the scalingfactor.

I only would need to know how to write it in Programcode

Best thx!

6 REPLIES 6

I already checked of failures. There shouldnd be any. But im not perfect hehe and someone is seeing some things what are not logicaly

especially im not 100 percent sure if my conditions are well defined:

Condition_OK_NOK_1 and Condition_OK_NOK_2

i did an actualisation but i still dont know where the failure is.

i did an actualisation but i still dont know where the failure is.

Tying to write a 2 Parameter Variation but dont know how

not logical Result here

-MFra-
21-Topaz II
(To:sm�ller-2)

I have given you two examples of iterations, to let you understand that when you work with iterative methods, the program must always do at each iteration (loop) a comparison between the result of the current calculation, and the result of the previous iteration in absolute terms . If the difference is lesser then a predetermined precision, we can say that the solution has been found, otherwise, one must repeat the cycle up to a certain limit, where the program exits from the loop without having found the solution. In your problem, I do not think there are real iterations, since the comparison is made with data values,  there isn't even defined a precision, I believe.

Moreover being your problem fundamentally mechanical, you should place it in the mathcad section devoted to the mechanics.

Greetings

F. M.

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