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Math Help - Rearranging an equation. A big equation.

  1. #1
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    Rearranging an equation. A big equation.

    Rearranging an equation. A big equation.-mathforum.jpg


    Hello,

    I am working on a project, and I have an equation that i need to rearrange in terms of x. The equation:

    z = x (r sqrt(1-x^2/(4 r^2))-r+y)+0.5 r^2 (2 sin^(-1)(x/(2 r))-sin(2 sin^(-1)(x/(2 r))))
    Which is the area of the given shape (basically a rectangle with a sector spliced on top).

    Tried to input in to Wolframalpha, but ran out of computational time, and I have since ran out of talent!

    Any help would be much appreciated!

    Regards

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  2. #2
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    Re: Rearranging an equation. A big equation.

    It's very doubtful there's a closed form solution for x in terms of z, r, and y.

    What are you actually trying to do? Can't you just solve it numerically?
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  3. #3
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    Re: Rearranging an equation. A big equation.

    Quote Originally Posted by romsek View Post
    It's very doubtful there's a closed form solution for x in terms of z, r, and y.

    What are you actually trying to do? Can't you just solve it numerically?
    We are standardising a design document for shape such as that. The aim is to maintain a given area without compromising the y value.

    The aim is to dictate that if z,r and y are known, we can use an equation to find what x should be.

    Thanks for taking a look though. Might need to change my approach.

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  4. #4
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    Re: Rearranging an equation. A big equation.

    Quote Originally Posted by adcb View Post
    We are standardising a design document for shape such as that. The aim is to maintain a given area without compromising the y value.

    The aim is to dictate that if z,r and y are known, we can use an equation to find what x should be.

    Thanks for taking a look though. Might need to change my approach.

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    I might have something for you.

    If you redo the drawing a bit.

    Let $x_2$ be the height of your shape up to the point of the circular sector.

    Then let

    $h=y-x_2$

    According to this

    The area of the circular segment is given by

    $a_{seg}=r^2 \cos ^{-1}\left(\dfrac{r-h}{r}\right)-(r-h) \sqrt{2 h r-h^2}$

    The rectangular area is just

    $a_{rect}=x x_2$ so the total area is

    $\large z=a_{rect}+a_{seg}=r^2 \cos ^{-1}\left(\frac{r-h}{r}\right)-(r-h) \sqrt{2 h r-h^2}+ x x_2$

    Solving this equation in terms of x I get

    $x=\frac{z+r^2 \left(-\cos ^{-1}\left(\frac{r+x_2-y}{r}\right)\right)+r \sqrt{-(x_2-y) (2 r+x_2-y)}-y \sqrt{-(x_2-y) (2 r+x_2-y)}+x_2 \sqrt{-(x_2-y) (2 r+x_2-y)}}{x_2}$

    I'd double and triple check this but it's possible that it will work.
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  5. #5
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    Re: Rearranging an equation. A big equation.

    Quote Originally Posted by romsek View Post
    I might have something for you.

    If you redo the drawing a bit.

    Let $x_2$ be the height of your shape up to the point of the circular sector.

    Then let

    $h=y-x_2$

    According to this

    The area of the circular segment is given by

    $a_{seg}=r^2 \cos ^{-1}\left(\dfrac{r-h}{r}\right)-(r-h) \sqrt{2 h r-h^2}$

    The rectangular area is just

    $a_{rect}=x x_2$ so the total area is

    $\large z=a_{rect}+a_{seg}=r^2 \cos ^{-1}\left(\frac{r-h}{r}\right)-(r-h) \sqrt{2 h r-h^2}+ x x_2$

    Solving this equation in terms of x I get

    $x=\frac{z+r^2 \left(-\cos ^{-1}\left(\frac{r+x_2-y}{r}\right)\right)+r \sqrt{-(x_2-y) (2 r+x_2-y)}-y \sqrt{-(x_2-y) (2 r+x_2-y)}+x_2 \sqrt{-(x_2-y) (2 r+x_2-y)}}{x_2}$

    I'd double and triple check this but it's possible that it will work.
    Thanks, I'll give it a try. I'll let you know of the outcome

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  6. #6
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    Re: Rearranging an equation. A big equation.

    Quote Originally Posted by romsek View Post
    I might have something for you.

    If you redo the drawing a bit.

    Let $x_2$ be the height of your shape up to the point of the circular sector.

    Then let

    $h=y-x_2$

    According to this

    The area of the circular segment is given by

    $a_{seg}=r^2 \cos ^{-1}\left(\dfrac{r-h}{r}\right)-(r-h) \sqrt{2 h r-h^2}$

    The rectangular area is just

    $a_{rect}=x x_2$ so the total area is

    $\large z=a_{rect}+a_{seg}=r^2 \cos ^{-1}\left(\frac{r-h}{r}\right)-(r-h) \sqrt{2 h r-h^2}+ x x_2$

    Solving this equation in terms of x I get

    $x=\frac{z+r^2 \left(-\cos ^{-1}\left(\frac{r+x_2-y}{r}\right)\right)+r \sqrt{-(x_2-y) (2 r+x_2-y)}-y \sqrt{-(x_2-y) (2 r+x_2-y)}+x_2 \sqrt{-(x_2-y) (2 r+x_2-y)}}{x_2}$

    I'd double and triple check this but it's possible that it will work.
    Is there a method of finding $x_2$ or $y$ without first referencing $x$?
    Or am I just being dense?

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    Re: Rearranging an equation. A big equation.

    Quote Originally Posted by adcb View Post
    Is there a method of finding $x_2$ or $y$ without first referencing $x$?
    Or am I just being dense?

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    it's your shape. You said you are given $y$. I assume you know where the rectangular portion of your shape ends and where the circular sector part begins. That height is $x_2$.

    If $x_2$ is not known ahead of time it can be calculated from $x$ and $r$.
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  8. #8
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    Re: Rearranging an equation. A big equation.

    Take partial derivatives with respect to $r$, $y$, and $z$, then use some approximation method (maybe Newton's Method might work) to find an approximation of $x(r,y,z)$.
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  9. #9
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    Re: Rearranging an equation. A big equation.

    Quote Originally Posted by romsek View Post
    it's your shape. You said you are given $y$. I assume you know where the rectangular portion of your shape ends and where the circular sector part begins. That height is $x_2$.

    If $x_2$ is not known ahead of time it can be calculated from $x$ and $r$.
    I think I might not have been clear in my brief. The aim was to derive the $x$ dimension from the known values of $y$,$r$ & $z$.
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  10. #10
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    Re: Rearranging an equation. A big equation.

    Is r both the arc length and the radius of the corresponding circle?
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  11. #11
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    Re: Rearranging an equation. A big equation.

    Quote Originally Posted by JeffM View Post
    Is r both the arc length and the radius of the corresponding circle?
    r Is just the arc radius. Chord length is equal to x, arc length is equal to r, being the internal angle.

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