1 ACCEPTED SOLUTION
Accepted Solutions
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
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 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.
