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ptc-53413201-Visitor
1-Visitor
October 24, 2013
Solved

Another "variable is undefined" question

  • October 24, 2013
  • 2 replies
  • 2610 views

Hi I am very new to MatchCAD (and to PTC so I am sorry if I am in the wrong part). I keep getting variable is undefined at the bottom of my find block. I will attach my file so that someone can look at it. It isn't very complex, I just can't figure out why it won't work.

Best answer by AlanStevens

The real problem here lies in your fifth equation, where the factor multiplying B is x5Ac instead of x6Ac.

Alan

2 replies

W
Werner_E25-Diamond I
25-Diamond I
October 24, 2013

I keep getting variable is undefined at the bottom of my find block

Thats the only error message you will get from a solve block, no matter what the real reason may be.

Richt click on the error message and use trace Error to go to the first. Mathcad will tell you that it found no solution.

You have six equations in your solve block but just five variables.This is likely to have no solution.

If you look closely at the top three equations you notice that the only deal with three variables, so you could solve for them. They are so simple you won't need a solve block to do so, but you can.

The last three equations also deal only with three variables. Variable F is in both blocks of three and chances are the results for F are different in both solve blocks which is the reason your big block throws the error.

I guess your last equations are wrong - you are demanding the impossible. Fill in your nuimbers and look what kind of equations you get. The only solution possible is when all variables are zero.

P
ptc-53413201-VisitorAuthor
1-Visitor
October 25, 2013

Thanks for looking at that. My assignment requires that I only use one given/find solve block though. I traced the error and it comes up as F has no solution, but of course it does. I will attach the matrix methods version of the same problem. Our assignment was to use matrix methods and a given/find solve block to solve for these variables.

W
Werner_E25-Diamond I
25-Diamond I
October 25, 2013

Here it works as you only have five equations for your five variables, not six.

A
AlanStevens19-TanzaniteAnswer
19-Tanzanite
October 25, 2013

The real problem here lies in your fifth equation, where the factor multiplying B is x5Ac instead of x6Ac.

Alan

W
Werner_E25-Diamond I
25-Diamond I
October 25, 2013

The real problem here lies in your fifth equation, where the factor multiplying B is x5Ac instead of x6Ac.

Ah, yes - well spotted. And so now the last three equations are linear dependend, the sum of last two giving the fourth, so you may omit any one of them.

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