Underwood equation

[technopersia]

Dear All,

I was wondering if you coluld help me with the methodology to calculate Θ = 1.645

Where q = 1

1- q = ((2.32*0.75)/(2.32-Θ))+((1.54*0.17)/(1.54-Θ))+((1*0.08)/(1-Θ))

Many thanks

 

[Deleted member 4993]

Dear All,

I was wondering if you coluld help me with the methodology to calculate Θ = 1.645

Where q = 1

1- q = ((2.32*0.75)/(2.32-Θ))+((1.54*0.17)/(1.54-Θ))+((1*0.08)/(1-Θ))

Many thanks

I would use Newton-Raphson iteration technique to solve for roots.

 

[technopersia]

iterations

Hi,

I was wondering if this is possible to solve without iterations, I can easily use the solver function on excel to solve it however I have no access to such tools in exams, and also I am curious.

Many Thanks

 

[technopersia]

circles

Hi Denis,

Thanks for the answer, however I am struggling to get to 2.0818x^2 - 5.597576x + 3.5728 = 0

What I get is this:

3.5728 - 2.32X-1.54X+X^2=(-0.08/(1-X))
3.5728-2.32X-1.54X+X^2(1-X)
3.5728-X(2.32+1.54+1)(1-X)
3.5728-4.86X+4.86X^2

Many Thanks

 

[technopersia]

3.5728 - 2.32X-1.54X+X^2=(-0.08/(1-X))
3.5728-2.32X-1.54X+X^2(1-X)
3.5728-X(2.32+1.54+1)(1-X)
3.5728-4.86X+4.86X^2

Please forget you have ever seen this.

I am ashamed to say I had completly forgotten about the straight forward method to do such problems . Thanks for the eequation it was extremly helpfull. Got exact same answer as you. Now I have a 6 component version of this to do!

Thanks for the help again.

 

[Deleted member 4993]

Ain't that simply "pretty"?!

Queens English.....Queens English.....Off with his head.....

 

[Deleted member 4993]

....but at least you read the Whig historians.

No...No... No - That's Queen of Hearts - Alice in Wonderland (or is it through the looking glass??)

 

[technopersia]

Duplicate post:

http://www.freemathhelp.com/forum/threads/74296-Solution-of-Non-linear-equation