Make "too large to display" a function so I can use it in plot?? mathcad attached

[ptc-4451268]

I need help!

I have I long equation, and I am trying to solve for one of the variables in the equation:

K=f(a,W,K) --> solve for K, to make K=f(a,W).

I have attached the mathcad.

The solution for my solve function is too big, but I need to use this solution to create a plot. How do I do this??

Thank you!!!

[Werner_E]

without units and using a substitution Mathcad is able to show the result, but I guess I won't make you happy. Due to the nature of the solution it cannot assigned a function. Details see attached.

Are you sure that solve should find a compact, unique solution?

[zhulaojianke]

I try to substitute it, but it can not work, even I try to substitute the simple one, it can not work too.

[Werner_E]

Look if the attached (based on the idea in a post of 朱 欣研 which seemed to have been deleted) makes sense.

Caution - takes quite a while to evaluate

[zhulaojianke]

hehe, because I think the method of exhaustion maybe no efficency on this problem, so I have deleted the reply...

I'm use another method, simplify it by hand, and use W=1, plot K(a,W)--a/W, seemly it only one point on it...Maybe wrong.

[Werner_E]

Not that much of value without the sheet (as mine, as I forgot to include it).

Look at the range for a and K in my solution and throw that in yours. Maybe you will see more than one point then.

After all - if you are going for a sulution using the "brute force" way with those nested for loops, it would not be necessary to make that long winded manual simplification and substitution process anyway.

[zhulaojianke]

Ah... I only want this function to be easily look and operate, and that big expression really annoying me to think about it. My simplify method go false on this mission.

[Werner_E]

朱 欣研 wrote:

Ah... I only want this function to be easily look and operate, and that big expression really annoying me to think about it. My simplify method go false on this mission.

Unfortunately the necessary resubstitution will undo most of the simplification.

BTW you would not need B if you assign the expression directly a function with formal parameter K.

[Werner_E]

Sorry, forgot to include the worksheet - here it is

[jkowalski]

Better look

[Werner_E]

Jan Kowalski wrote:

Better look

But much less precision. For a better look AND correct values you would need

• a decreased tolerance value (and not one increased to 0.1)
• a decreased step value at least for K
• an increased number of cups of coffee to bridge the waiting time

[Werner_E · Accepted solution]

If the goal is plotting and tables of values, there are resonable numerical solutions using "root" or "Find"available. The reason for those "artefacts" we see when using root for a values greater than 2*W/3 are due to the fact that root delivers real roots for selected single values where it should return a complex number, I guess. So I think the graph produced with the sollve block is the correct one and function K1 should be rewritten to return NaN if a is greater than 2*W/3.

Another question. How can I add the errorhandling to K2 as I did it with K1. Do I have to define another function which is then asigned to K2?

• Using the "root" function

• Using a solve block with "Find"

Edited: Changed attached file, include modified solution with root (as decribed above) now

[zhulaojianke]

The herizontal ordinate of the plot is a/w, thus K(a,1.5), is a/1.5, and k(a,2), is a/2, OK?

[Werner_E]

If you plot with a/W on the ordinate and K(a,W) on the abscissa you will get a lot of graphs which cover/overlap/are identical with the red one (for W=1). So nothing new. This of course only applys if its possible at all to make the function dependable of the ratio a/W alone, which does NOT seem to be the case here as the graphs show. So either K is not supposed to be dependent of the ratio a/W or the original poster should check the formulas in my file against the one in his sheets.

[ptc-4451268]

Wow you guys are great. I never expected this much help!!

K is only a functon of sigma, a, and W. I am given the instruction to plot K vs. a/w. The plot I am creating is supposed to prove that the K function does not rely on the first bit:

W-2(a+rp)/W

This is why, I believe, we are getting the different scales for a/W.

I believe this is exactly what I need! Thank you!!!

Also - I should have noted this before - the domain is only between a/w = 0-0.5