Community Tip - You can Bookmark boards, posts or articles that you'd like to access again easily! X
Dear Mathcad community,
I have a problem when plotting a piecewise function.
The error is "all evaluations resulted in either an error or a complex result".
I would really appreciate if anyone can help me!
Thank-you very much.
Best Regards.
Enrico
Solved! Go to Solution.
The evaluation of:
F.input_rod(a)=
results in an error message "Must be scalar".
Tracing further back I find that in your definition of F.input_rod(a) you are comparing a with 10 m/s^2.
Note that a is an array, you cannot compare an array with a scalar value.
Your way out is simple: vectorize the F.input_rod(a) so that the function acts on each value, but still produces all results.
See attached.
Success!
Luc
Can't see your sheet, unless you save as Mathcad 11..
Here's some advice to help you debug:
Evaluate the variable you want to plot. If you plot e.g. y(x) and have x defined (either as a range variable or as a vector), type
y(x)=
and
x=
To see the value(s) each has. There may be NaN's in there, or imaginary numbers. Or another error occurs.
In case of an error, select the evaluation and, using the righthand mouse button, select 'trace error', and trace that error back to where it occurs.
Success!
Luc
Dear Luc,
Thank-you for your reply.
Attached you find the file in .mcd format.
The evaluation of:
F.input_rod(a)=
results in an error message "Must be scalar".
Tracing further back I find that in your definition of F.input_rod(a) you are comparing a with 10 m/s^2.
Note that a is an array, you cannot compare an array with a scalar value.
Your way out is simple: vectorize the F.input_rod(a) so that the function acts on each value, but still produces all results.
See attached.
Success!
Luc
Just a quick look (in Smath), you have some unit balance issues.
Where?
I don't see them. Mathcad (11) also not.
Luc
Salve,
non puoi confrontare il vettore a con uno scalare. Il confronto con lo scalare va fatto elemento per elemento .
Buona giornata,
Ecco di cosa si occupa la vettorializzazione.
Successo!
Luc