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?
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
Ad
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.
Ad
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.