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Thread: Trying to solve for specific variable

  1. #1
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    Trying to solve for specific variable

    Hello all,

    It's been a few(several) years ago I studied algebra and find myself trying to re-write a 4-variable formula to solve for a specific one.
    Here's what I have;

    L = 2C + pi((D2+D1)/2) + ((D2+D1)^2/4C)

    A picture would probably be better...

    Trying to solve for specific variable-belt-length-required.jpg

    Thank you for any assistance
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  2. #2
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    Re: Trying to solve for specific variable

    Quote Originally Posted by RustyAlgebra View Post
    Hello all,

    It's been a few(several) years ago I studied algebra and find myself trying to re-write a 4-variable formula to solve for a specific one.
    Here's what I have;

    L = 2C + pi((D2+D1)/2) + ((D2+D1)^2/4C)

    A picture would probably be better...

    Click image for larger version. 

Name:	belt-length-required.jpg 
Views:	5 
Size:	60.2 KB 
ID:	38632

    Thank you for any assistance
    What variable are you trying to solve for?

    -Dan
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  3. #3
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    Re: Trying to solve for specific variable

    Guess that would probably help huh? Doh

    I need to solve for C

    Thank you
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  4. #4
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    Re: Trying to solve for specific variable

    You have $\displaystyle L = 2C + \pi((D2+D1)/2) + ((D2+D1)^2/4C)$ and want to solve for C. The first thing I notice is that "4C" in the denominator of a fraction. Just because I don't like fractions, I would multiply both sides of that equation by 4C:
    $\displaystyle 4CL= 8C^2+ 2\pi(D2+ D1)C+ (D2+ D2)^2$

    We can write that as $\displaystyle AC^2+ BC+ D= 0$ with A= 8, $\displaystyle B= 2\pi(D2+ D1)- 4L$ and $\displaystyle D= (D2+ D1)^2$.

    Now, by the quadratic formula, $\displaystyle C= \frac{-B\pm\sqrt{B^2- 4AD}}{2A}$.
    Last edited by topsquark; Apr 1st 2018 at 02:40 PM.
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  5. #5
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    Re: Trying to solve for specific variable

    Quote Originally Posted by RustyAlgebra View Post
    L = 2C + pi((D2+D1)/2) + ((D2+D1)^2/4C)
    Simplifying a bit: P = pi, A = D2+D1 ; equation becomes:
    L = 2C + P(A/2) + A^2/(4C) : I'm assuming the "4C" you show should be "(4C)"
    Multiply by 4C:
    4CL = 8C^2 + 2CPA + A^2
    Re-arrange:
    8C^2 + 2CPA - 4CL + A^2 = 0
    8C^2 + (2PA - 4L)C + A^2 = 0

    That's a quadratic: a = 8, b = 2PA - 4L, c = A^2 : solve it!

    EDIT: sorry Halls, me no see yours...
    anyhoo.....mine's better!!
    Last edited by DenisB; Apr 1st 2018 at 01:18 PM.
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  6. #6
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    Re: Trying to solve for specific variable

    Thank you both for the help!
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