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## Better match of results

How can I better display 3D results in matrix equations? An example in the appendix. I have tried to separate everything but it is very tedious. It is very important to me that all the results are 3D.

1 ACCEPTED SOLUTION

Accepted Solutions
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## Re: Better match of results

Two problems:

1. Up in the top, you have defined a value for F.02. The GPa as its unit is a type 'Constant' instead of 'variable'. But potentially that's not a problem for Mathcad 15.

2. Down at the bottom, you provide a guess value for lambda. That must have the correct unit, Pa^3. If you add that, the solve block at least finds the lambda value for x and y both 1 m. But for other values of x and y (guided by the CreateMesh function) no suitable lambda was found. You may need to provide a better guess value for lambda. This is what I get with a mesh of 10x10 points, and a guess value of -10 MPa^3 for lambda: And with a guess value of -50 MPa^3 and 20x20 points, it shows that there are 2 points where lambda produces totally different values: I'll leave it to you to investigate that.

Success!
Luc

17 REPLIES 17
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## Re: Better match of results

That's quite a big file, impressive!

But it doesn't make clear what you are trying to achieve.

I see no 'Appendix' in the attached file, but I see a lot of equations with undefined variables, and plot's that are red.

Can you create a small example to illustrate your question?

Success!
Luc

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## Re: Better match of results

Thanks for interesting of problem. I'm going to show in points what we need to do. This problem will be in attachment.

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## Re: Better match of results

It still isn't clear to me what you are trying to accomplish.This is what I see:

Point 1 defines a set of 3 expressions.

Point 2 appears to assign those expressions to 3 separate functions in x and y, but the expressions contain no x and no y, so the functions are constants.

Point 3 defines an equation, involving the 3 functions and three variables capital Pi.something (which are undefined). Then it tries to find the root of a function f in lambda, but f is undefined.

Point 4 attempts to create a plot of one of the functions defined under point 2, using the two parameters of that function (those that the function does not 'have').

You'll need to be more precise in what you want.

Success!
Luc

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## Re: Better match of results

Look at sample. We need solve equation for all x,y.

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## Re: Better match of results

The notation s(x,y) in Mathcad defines a function to be evaluated for any value(s) of x and y.

As Luc has observed, your symbolic results do not contain any x or y terms.  (They also, note the red, contain terms that the symbolic engine assumes are constants but are not defined to have values.)  As a result, these functions cannot be evaluated by the numeric engine; if you defined values for each of the constants, then the functions would return the same value for ALL values of x and y.

For example: Highlighted

## Re: Better match of results

Look at sample. We are looking for lambda and lambda can't to be named.

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## Re: Better match of results

You've got one equation and one unknown (lambda)?

Make sure you supply lambda as a parameter to each of the signa functions, then you should be able to solve lambda as a function of x, y and all other parameters.

If you want/need my help, save and attach the worksheet as a mathcad 11 (.MCD) file.

Success!
Luc

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Here you are:)

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## Re: Better match of results

Unfortunately your equation cannot be solved symbolically, not by Mathcad, and I doubt if it can be solved symbolically by any other math program.

I did the following: I'm only showing the left part of these. They're running a few pages wide though.  To show you that this is the right way to do it, I've (over)simplified the sigma functions: For which I can solve, symbolically, the following equation: Note that if I make one sigma function slightly more complicated: The resulting solution of the equation becomes: So what can you do?

Abandon the idea of finding a symbolic solution to this complicated equation, and go for numeric.

With the exact same set of sigma functions, the equation can be put in a solve block, the result of which can be assigned to a function of x and y. Note that you have to supply the (numeric) value for each of the parameters (c, q, P11, P12 etc).

Success!

Luc

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## Re: Better match of results

Look at my first solusion (appendix). Thanks for spend your time on this. I suppose we need solve it by newton raphson method.But I know this metod but unfortunately I do not know how to modeling it in mathcad. can we do it for all x,y??

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## Re: Better match of results

You shouldn't need to worry about the method, like Newton Raphson.

Just use a solve block. Here's an example.

First you define your functions (if needed). Like: Make sure you define all known variables numerically. Like: Provide a guess value for the variable that you want to solve for. Then follows the solve block. It consists of the statement "given" followed by the constraint(s) and finally a keyword so the solver knows what to do, in this case that's 'find''.

To make a function of it, the 'find' result is assigned to a function: Now you can use that function to calculate the solution for any set of x and y values: The file is attached, you can play with it. Note that due to the symmetry of my simple sigma functions, not every combination of x and y values leads to a solution. I hope that is not the case for your functions.

Oh. If you need to control the solution method, right-click on the 'find' statement and see what options you have.

Success!
Luc

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## Re: Better match of results

I would like to thanking you for spending your time for me. I've tried to get your solution to real sample. Something is wrong in my calculation. I wanted plot it but something is wrong too.

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## Re: Better match of results

Two problems:

1. Up in the top, you have defined a value for F.02. The GPa as its unit is a type 'Constant' instead of 'variable'. But potentially that's not a problem for Mathcad 15.

2. Down at the bottom, you provide a guess value for lambda. That must have the correct unit, Pa^3. If you add that, the solve block at least finds the lambda value for x and y both 1 m. But for other values of x and y (guided by the CreateMesh function) no suitable lambda was found. You may need to provide a better guess value for lambda. This is what I get with a mesh of 10x10 points, and a guess value of -10 MPa^3 for lambda: And with a guess value of -50 MPa^3 and 20x20 points, it shows that there are 2 points where lambda produces totally different values: I'll leave it to you to investigate that.

Success!
Luc

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## Re: Better match of results

That's it. Great job Luc. That is what I wanted to see. I have only one quastion: can we put if? That means if we have -0,001 this is 0 and if we have +0*10-20 or +0,002 this solusion is +0,002 and on the graps need to be the same showing.

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## Re: Better match of results

I don't understand what you mean:

0* 10- 20= -20

Luc

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## Re: Better match of results

My mistake. Correct is 1*10 to-20 in simplification 0,00000000000000000000001=0

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## Re: Better match of results

At least you have a unit problem in your functions.

For example in sigma.xyp... Either lambda is not unitless but has the dimension Pressure^3 or you add 1*1/Pa^-3 instead of just 1, or ....

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