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Yet another solve block issue

awiest
1-Visitor

Yet another solve block issue

New MathCAD user here. I have spent the last day going through other solve block issues and see if someone else made a similar mistake as myself, to no avail.

I am trying to solve a 3 variable system of equations with solve block and get and error. Any guidance is appreciated

ACCEPTED SOLUTION

Accepted Solutions
Werner_E
25-Diamond I
(To:awiest)

The problem is the equal sign after Find.

Find returns a vector and your three solved for variables have different units (more precisely: dimensions). Mathcad 15 and below has static unit checking and does not allow for a matrix consisting of elements with different dimensions.

This limitation is not present in Prime, as Valery has shown. While Prime has many (and I mean really many!) drawbacks and is not one but quite a lot of steps backward in evolution, this sure is one of the very few improvements made. Not that it would be a reason to switch from MC15 to Prime, but fair is fair, its an improvement.

There is an easy solution to your problem, though. While you cannot evaluate(=) Find, as this would force MC to display a vector with different units, you can assign the result of Find to variables which you put into a vector and display each one separately (and of course you have to delete the evaluation equal sign after Find):

BTW,  you don't have to use the same variable names when assigning.

WE

View solution in original post

5 REPLIES 5

May be problems with units in solve block Mathcad 15.

err.png

All ok in Prime:

Spans.png

Werner_E
25-Diamond I
(To:awiest)

The problem is the equal sign after Find.

Find returns a vector and your three solved for variables have different units (more precisely: dimensions). Mathcad 15 and below has static unit checking and does not allow for a matrix consisting of elements with different dimensions.

This limitation is not present in Prime, as Valery has shown. While Prime has many (and I mean really many!) drawbacks and is not one but quite a lot of steps backward in evolution, this sure is one of the very few improvements made. Not that it would be a reason to switch from MC15 to Prime, but fair is fair, its an improvement.

There is an easy solution to your problem, though. While you cannot evaluate(=) Find, as this would force MC to display a vector with different units, you can assign the result of Find to variables which you put into a vector and display each one separately (and of course you have to delete the evaluation equal sign after Find):

BTW,  you don't have to use the same variable names when assigning.

WE

Werner!

I am glad to your back on the forum!

We needed You!

For example for the old task about the catenary with a load.

I hope we solved it.

See please Catenary with load

We needed you and... one more (9-th) equation.

Check please!

It is new type of systems - algebra-programming system of equation!

Valery Ochkov wrote:

Werner!

I am glad to your back on the forum!

Not really back as I used to be, I guess. But it looks like a combination of too much spare time and a computer with Mathcad still installed is tempting 😉

Its sad to see that nothing has improved here. Neither the devastated Forum software, nor Mathcad or Prime itself. From what I read from Stuart it looks like MC15 M040 is rather buggy and not an improvement. Seems to be the PTC way to deal with tasks.

As far as i remember the question in that old monster thread where I played the role of the Advocatus Diaboli ,basically was, if the rope with point load can be described in an exact analytical form by two ordinary cosini hyperbolici which degenerate in two straight lines in the one extreme (infinite point load) or are identical in the other (no load).

Is this question now already answered (symbolically)? I still see just a numeric solve block, not a set of two equations 😉

WE

Werner Exinger wrote:

As far as i remember the question in that old monster thread where I played the role of the Advocatus Diaboli, basically was, if the rope with point load can be described in an exact analytical form by two ordinary cosini hyperbolici which degenerate in two straight lines in the one extreme (infinite point load) or are identical in the other (no load).

Is this question now already answered (symbolically)? I still see just a numeric solve block, not a set of two equations 😉

WE

The answer is simple!

If we change the load to one fixed point - why we must  create new equation for this two part of catenary.

A catenary is catenary - with one load or with one fixed point.

CatCat.png

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