Itertive Calculation / Extending of current Situation / 2 Parameter Variation

[sm�ller-2]

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!

[sm�ller-2]

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

[sm�ller-2]

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

Condition_OK_NOK_1 and Condition_OK_NOK_2

[sm�ller-2]

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

[sm�ller-2]

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

Tying to write a 2 Parameter Variation but dont know how

[sm�ller-2]

not logical Result here

[-MFra-]

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.