Vector Solve Block Simple Equation

[intermoxil]

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

[LucMeekes]

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

[DJF]

Try this.

[terryhendicott]

Hi,

Along the same line of thinking paramatise the solve block in terms of L

Cheers

Terry

[DJF]

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.

[terryhendicott]

Thanks for the hints DJF

Terry

[intermoxil]

Thanks DJF, it is also another way of thinking that solve my problem as well. 

Cheers!!

[LucMeekes]

In other words, simply like this:

Success!

Luc

[intermoxil]

You made my day simpler. Really appreciate it

Thanks, 

[Werner_E]

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