Vector Solve Block Simple Equation

Forum|Forum|7 years ago
February 20, 2019
9 replies
3352 views

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

1_14.zip

Mechanical_Engineering

L

LucMeekes

23-Emerald IV

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

D

DJF

16-Pearl

Try this.

1_14-(3).zip

T

terryhendicott

21-Topaz II

Hi,

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

Cheers

Terry

1_190220-14.zip

D

DJF

16-Pearl

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.