Help with Setting Up Solver

[mechowl]

If Y is known, how can this be setup to solve for l?

Thank you.

 

[GregLocock]

Loopy l is more like an index than a measurement I think, and given that p=1 that equation is not very frightening at all.

I don't think Solver is the right approach, just loops.


Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[GregLocock]

Use a while loop containing a for loop for the summation. That's a funny old function, what is it?

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[mechowl]

Thanks Greg!

I must admit my knowledge in this area has not been used for many many years, and to add I'm very new to MathCAD.

I am playing with the FOR and WHILE loops right, and despite the examples online I'm a bit lost. Can you elaborate how to set this up?

This is from a paper that proposes a procedure for determining the capacity of a surge bin between two machines based on their rates, availability, etc.

 

[GregLocock]

In pseudo code, since I don't have mathcad
Y=user input
L=1
P=1
thisE=Y
while thisE<Y
L=L+1
thisE=longcomplicatedfunctionof P and L using a for loop

end while

print(L)

end

The problem I have with it is that unless I have mistranscribed your equation, E is always negative






Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[GregLocock]

So here's how I did it in Smath, a very similar program to Mathcad. You can see why I am confused- the grayed out section would give the L you want, but since all values of E were negative it is no good. You need to reset L to 0 or delete the active while loop to use the second one.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[Stress_Eng]

Have you tried graphing the equation? Keeping E as a function of L, with increasing L, the graph =<0.E(2)= -6.373, after which a step is seen, giving -14.273 and, with increasing L, the curve goes negative rapidly. Looks as though the solution is highly dependent on your given Y.

 

[mechowl]

Thank you to all for your commentary and input. If either of you are interested in seeing the paper that describes this procedure please email me at mechowl1980@gmail.com

Maybe it will be helpful to understand this equation. I have two identical papers printed by two separate publishers and the equation in one of them is missing a ( which leads me to believe there is a small possibility the equation is incorrect. Anyways I'm stuck too.

Thanks again.

 

[GregLocock]

Unfortunately neither paper presents a derivation of that equation.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?