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Possible to Isolate C?

  1. Sep 5, 2014 #1
    Hi, I'm not the best with algebra and I've been wrestling with a formula this afternoon. It is a formula to calculate a belt length based on 2 pulley sizes and the distance between the centres. I would like to be able to calculate the distance between the centres based on the pulley sizes and the belt length so instead of solving for L I need to solve for C. I would love to know what I'm doing wrong in isolating C. Anyway, here is the eqn.

    L=2C+∏ (D+d)/2 + (D-d)^2/4C

    L= Belt length
    C= Distance between pulley centres
    D= Big pulley Dia
    d= Small pulley Dia

    Thanks :)
     
  2. jcsd
  3. Sep 5, 2014 #2

    mathman

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    Multiply through by C. You then have a quadratic in C, which is readily solvable.
     
  4. Sep 5, 2014 #3
    I kinda know what you mean but I'm not really sure how to do that. As in multiply every element by c? Top and bottom of the fractions?
     
  5. Sep 5, 2014 #4

    HallsofIvy

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    Multiplying both sides of
    [tex]L=2C+∏ (D+d)/2 + (D-d)^2/4C[/tex]
    by C will give
    [tex]LC= 2C^2+ ∏ C(D+ d)/2+ (D- d)^2/4[/tex]
     
  6. Sep 7, 2014 #5
    For anybody still hung up on this, the next step is to reformulate the equation using the quadratic equation.
     
  7. Sep 8, 2014 #6
    Ok, looking good so far, but how come the second part (D-d)^2/4 isn't also multiplied by C and why are the denominators not multiplied by C?

    Ugh, I also suck at factoring. I need to go through my algebra textbook again and teach myself all this stuff from scratch.
     
  8. Sep 8, 2014 #7

    SteamKing

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    Here's a tip:

    [itex]L = 2C+\pi\frac{(D+d)}{2}+\frac{(D-d)^{2}}{4C}[/itex]

    use a lower case [itex]\pi[/itex] to indicate the ratio of the circumference of a circle to its diameter. The upper case [itex]\Pi[/itex] is generally used in mathematics to indicate the product of several different terms.
     
  9. Sep 8, 2014 #8
    Ok, So I'm guessing
    [itex]L = 2C+\frac{\pi(D+d)}{2}+\frac{(D-d)^{2}}{4C}[/itex]
    Should become
    [itex]LC = 2C^2+\frac{\pi C(D+d)}{2}+\frac{C(D-d)^{2}}{4C}[/itex]

    Is that correct?

    It still has me wondering why the C doesn't effect the denominator.
     
    Last edited: Sep 8, 2014
  10. Sep 8, 2014 #9

    HallsofIvy

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    It does! That was the whole point of multiplying by it. The C in the denominator cancels the C in the numerator leaving
    [tex]LC= 2C^2+ \frac{\pi C(D+ d)}{2}+ \frac{(D-d)^2}{4}[/tex]
    That is the same as
    [tex]2C^2+ \left(\frac{\pi(D+ d)}{2}-L\right)C+ \left[\frac{(D- d)^2}{4}\right]= 0[/tex]

    That is a quadratic equation. I wouldn't worry about "factoring"- use the quadratic formula.
     
  11. Sep 8, 2014 #10
    :P Right. I missed that you had canceled the C in the denominator.

    Ahhhh! The quadratic Eq'n! Forgot about that old chestnut. Alright, let's give this a try.

    [itex]C=\frac{-\left(\frac{\pi(D+d)}{2}-L\right)\pm \sqrt{\left(\frac{\pi(D+d)}{2}\right)^2-4(2)\left(\frac{(D-d)^2}{4}\right)}}{2(2)}[/itex]

    That's very interesting.... I need to try to jam all that into a CAD program now. I should probably attempt to check it first too....
     
    Last edited: Sep 9, 2014
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