Hi,
I am really new to Mathcad.
I have a problem to find a "d" variable in constraint with L as the dependent factor. The problem is, I want to get "d" with range variables of L from 0 to 1000 mm. So I get range value of "d" based on the L inputs.
Could we do this by that way with solve block? Appreciate your guy's help
Yes, you could. But I think it's much simpler if you first solve, like this:
That is, you define a function d, where the definition uses a symbolic solve. That 'll get you best accuracy, and speed.
In your case, R=0.85.
Now if you make M a function of z, and z a function of L, D and t, you can call:
d(S.F,S.flow,t,M(z(L,D,t)),0.85)=...
And you can plot d(...) versus L.
Success!
Luc
Hi,
Along the same line of thinking paramatise the solve block in terms of L
Cheers
Terry
I did mine the way I did as I believe it's faster. As I understand things, once mathcad starts iterating it will call M(L), and hence z(L), for each iteration. Not a big deal here, but it can slow things down in big programs. Also the solver isn't terribly robust imho and gets confused with 'constants' inside of a solve block. I've found the 'program' internal to the solve block to be very effective. But that's probably more than our poster needs to know at this point.
Thanks for the hints DJF
Terry
Thanks DJF, it is also another way of thinking that solve my problem as well.
Cheers!!
In other words, simply like this:
Success!
Luc
You made my day simpler. Really appreciate it
Thanks,
Here is a way which follows closely your approach and still avoids multiple calculation of M(L) during iterations:
But I agree with Luc that in case of your rather simple equation it would be better to solve it symbolically and go for the most direct approach:
Worksheet in Format Prime5 attached