This is a function, but I'm sure one can make an equation out of it.
It's for rectangular coordinates. Every angle here is in radians.
$n$ is the number of sides.
$r$ is the radius of the circle where the polygon is inscribed.
$\beta$ is the tilting angle, you may adjust it to make sure there aren't any vertical lines.
$$\alpha(k) = \frac {2\pi}nk + \beta$$$$\alpha_\beta(k) = \frac {2\pi}nk + \beta$$ The function $\alpha(k)$$\alpha_\beta(k)$ gets the angle of the vertex of order/index $k$, starting at $0$. $$k(x) = \left\lfloor\frac{\cos^{-1}(x/r) - \beta}{2\pi}n\right\rfloor$$$$k_\beta(x) = \left\lfloor\frac{\cos^{-1}(x/r) - \beta}{2\pi}n\right\rfloor$$ The function $k(x)$$k_\beta(x)$ gets the index of the first vertex (anti-clockwise) of the side of the polygon where $x$ belongs. The floor function $\lfloor \cdot\rfloor$ rounds the fractional result down so it snaps to integer order vertices.
$$\alpha(k(x))$$$$\alpha_\beta(k_\beta(x))$$ Combining the two functions gives us the angle of the first vertex (anti-clockwise) of the side that each $x$ belongs to. $$\mathrm P_{n,\ r, \ \beta}(x) = r\sin \alpha(k(x)) + (\cos \alpha(k(x))-x)\cot\left(\frac {\pi}n(2k(x) +1) + \beta\right)$$$$\mathrm P_{n,\ r, \ \beta}(x) = r\sin \alpha_\beta(k_\beta(x)) + (\cos \alpha_\beta(k_\beta(x))-x)\cot\left(\frac {\pi}n(2k_\beta(x) +1) + \beta\right)$$ This renders the upper "half" of the polygon centered at the origin. The other half is given by $-\mathrm P_{n,\ r,\ -\beta}(x)$.
Note that depending on $\beta$ there may be some lines extending out of the circle, nevertheless the polygon will be there.
This is a function, but I'm sure one can make an equation out of it.
It's for rectangular coordinates. Every angle here is in radians.
$n$ is the number of sides.
$r$ is the radius of the circle where the polygon is inscribed.
$\beta$ is the tilting angle, you may adjust it to make sure there aren't any vertical lines.
$$\alpha(k) = \frac {2\pi}nk + \beta$$ The function $\alpha(k)$ gets the angle of the vertex of order/index $k$, starting at $0$. $$k(x) = \left\lfloor\frac{\cos^{-1}(x/r) - \beta}{2\pi}n\right\rfloor$$ The function $k(x)$ gets the index of the first vertex (anti-clockwise) of the side of the polygon where $x$ belongs. The floor function $\lfloor \cdot\rfloor$ rounds the fractional result down so it snaps to integer order vertices.
$$\alpha(k(x))$$ Combining the two functions gives us the angle of the first vertex (anti-clockwise) of the side that each $x$ belongs to. $$\mathrm P_{n,\ r, \ \beta}(x) = r\sin \alpha(k(x)) + (\cos \alpha(k(x))-x)\cot\left(\frac {\pi}n(2k(x) +1) + \beta\right)$$ This renders the upper "half" of the polygon centered at the origin. The other half is given by $-\mathrm P_{n,\ r,\ -\beta}(x)$.
Note that depending on $\beta$ there may be some lines extending out of the circle, nevertheless the polygon will be there.
This is a function, but I'm sure one can make an equation out of it.
It's for rectangular coordinates. Every angle here is in radians.
$n$ is the number of sides.
$r$ is the radius of the circle where the polygon is inscribed.
$\beta$ is the tilting angle, you may adjust it to make sure there aren't any vertical lines.
$$\alpha_\beta(k) = \frac {2\pi}nk + \beta$$ The function $\alpha_\beta(k)$ gets the angle of the vertex of order/index $k$, starting at $0$. $$k_\beta(x) = \left\lfloor\frac{\cos^{-1}(x/r) - \beta}{2\pi}n\right\rfloor$$ The function $k_\beta(x)$ gets the index of the first vertex (anti-clockwise) of the side of the polygon where $x$ belongs. The floor function $\lfloor \cdot\rfloor$ rounds the fractional result down so it snaps to integer order vertices.
$$\alpha_\beta(k_\beta(x))$$ Combining the two functions gives us the angle of the first vertex (anti-clockwise) of the side that each $x$ belongs to. $$\mathrm P_{n,\ r, \ \beta}(x) = r\sin \alpha_\beta(k_\beta(x)) + (\cos \alpha_\beta(k_\beta(x))-x)\cot\left(\frac {\pi}n(2k_\beta(x) +1) + \beta\right)$$ This renders the upper "half" of the polygon centered at the origin. The other half is given by $-\mathrm P_{n,\ r,\ -\beta}(x)$.
Note that depending on $\beta$ there may be some lines extending out of the circle, nevertheless the polygon will be there.
This is a function, but I'm sure one can make an equation out of it.
It's for rectangular coordinates. Every angle here is in radians.
$n$ is the number of sides.
$r$ is the radius of the circle where the polygon is inscribed.
$\beta$ is the tilting angle, you may adjust it to make sure there aren't any vertical lines.
$$\alpha(k) = \frac {2\pi}nk + \beta$$ The function $\alpha(k)$ gets the angle of the vertex of order/index $k$, starting at $0$. $$k(x) = \left\lfloor\frac{\cos^{-1}(x/r) - \beta}{2\pi}n\right\rfloor$$ The function $k(x)$ gets the index of the first vertex (anti-clockwise) of the side of the polygon where $x$ belongs. The floor function $\lfloor \cdot\rfloor$ rounds the fractional result down so it snaps to integer order vertices.
$$\alpha(k(x))$$ Combining the two functions gives us the angle of the first vertex (anti-clockwise) of the side that each $x$ belongs to. $$\mathrm P_{n,\ r, \ \beta}(x) = r\sin \alpha(k(x)) + (\cos \alpha(k(x))-x)\cot\left(\frac {\pi}n(2k(x) +1) + \beta\right)$$ This renders the upper "half" of the polygon centered at the origin. The other half is given by $-\mathrm P_{n,\ r,\ -\beta}(x)$.
Note that depending on $\beta$ there may be some lines extending out of the circle, nevertheless the polygon will be there.
This is a function, but I'm sure one can make an equation out of it.
It's for rectangular coordinates. Every angle here is in radians.
$n$ is the number of sides.
$r$ is the radius of the circle where the polygon is inscribed.
$\beta$ is the tilting angle, you may adjust it to make sure there aren't any vertical lines.
$$\alpha(k) = \frac {2\pi}nk + \beta$$ The function $\alpha(k)$ gets the angle of the vertex of order/index $k$, starting at $0$. $$k(x) = \left\lfloor\frac{\cos^{-1}(x/r) - \beta}{2\pi}n\right\rfloor$$ The function $k(x)$ gets the index of the first vertex (anti-clockwise) of the side of the polygon where $x$ belongs. The floor function $\lfloor \cdot\rfloor$ rounds the fractional result down so it snaps to integer order vertices.
$$\alpha(k(x))$$ Combining the two functions gives us the angle of the first vertex (anti-clockwise) of the side that each $x$ belongs to. $$\mathrm P_{n,\ r, \ \beta}(x) = r\sin \alpha(k(x)) + (\cos \alpha(k(x))-x)\cot\left(\frac {\pi}n(2k(x) +1) + \beta\right)$$ This renders the upper "half" of the polygon. The other half is given by $-\mathrm P_{n,\ r,\ -\beta}(x)$.
Note that depending on $\beta$ there may be some lines extending out of the circle, nevertheless the polygon will be there.
This is a function, but I'm sure one can make an equation out of it.
It's for rectangular coordinates. Every angle here is in radians.
$n$ is the number of sides.
$r$ is the radius of the circle where the polygon is inscribed.
$\beta$ is the tilting angle, you may adjust it to make sure there aren't any vertical lines.
$$\alpha(k) = \frac {2\pi}nk + \beta$$ The function $\alpha(k)$ gets the angle of the vertex of order/index $k$, starting at $0$. $$k(x) = \left\lfloor\frac{\cos^{-1}(x/r) - \beta}{2\pi}n\right\rfloor$$ The function $k(x)$ gets the index of the first vertex (anti-clockwise) of the side of the polygon where $x$ belongs. The floor function $\lfloor \cdot\rfloor$ rounds the fractional result down so it snaps to integer order vertices.
$$\alpha(k(x))$$ Combining the two functions gives us the angle of the first vertex (anti-clockwise) of the side that each $x$ belongs to. $$\mathrm P_{n,\ r, \ \beta}(x) = r\sin \alpha(k(x)) + (\cos \alpha(k(x))-x)\cot\left(\frac {\pi}n(2k(x) +1) + \beta\right)$$ This renders the upper "half" of the polygon centered at the origin. The other half is given by $-\mathrm P_{n,\ r,\ -\beta}(x)$.
Note that depending on $\beta$ there may be some lines extending out of the circle, nevertheless the polygon will be there.
This is a function, but I'm sure one can make an equation out of it.
It's for rectangular coordinates. Every angle here is in radians.
$n$ is the number of sides.
$r$ is the radius of the circle where the polygon is inscribed.
$\beta$ is the tilting angle, you may adjust it to make sure there aren't any vertical lines.
$$\alpha(k) = \frac {2\pi}nk + \beta$$ The function $\alpha(k)$ gets the angle of the vertex of order/index $k$, starting at $0$. $$k(x) = \left\lfloor\frac{\cos^{-1}(x/r) - \beta}{2\pi}n\right\rfloor$$ The function $k(x)$ gets the index of the first vertex (anti-clockwise) of the side of the polygon where $x$ belongs. The floor function $\lfloor \cdot\rfloor$ rounds the fractional result down so it snaps to integer order vertices.
$$\alpha(k(x))$$ Combining the two functions gives us the angle of the first vertex (anti-clockwise) of the side that each $x$ belongs to. $$\mathrm P_{n,\ r, \ \beta}(x) = r\sin \alpha(k(x)) + (\cos \alpha(k(x))-x)\cot\left(\frac {\pi}n(2k(x) +1) + \beta\right)$$ This renders the upper "half" of the polygon. The other half is given by $-\mathrm P_{n,\ r,\ -\beta}(x)$.
Note that depending on $\beta$ there may be some lines extending out of the unit circle, nevertheless the polygon will be there.
This is a function, but I'm sure one can make an equation out of it.
It's for rectangular coordinates. Every angle here is in radians.
$n$ is the number of sides.
$r$ is the radius of the circle where the polygon is inscribed.
$\beta$ is the tilting angle, you may adjust it to make sure there aren't any vertical lines.
$$\alpha(k) = \frac {2\pi}nk + \beta$$ The function $\alpha(k)$ gets the angle of the vertex of order/index $k$, starting at $0$. $$k(x) = \left\lfloor\frac{\cos^{-1}(x/r) - \beta}{2\pi}n\right\rfloor$$ The function $k(x)$ gets the index of the first vertex (anti-clockwise) of the side of the polygon where $x$ belongs. The floor function $\lfloor \cdot\rfloor$ rounds the fractional result down so it snaps to integer order vertices.
$$\alpha(k(x))$$ Combining the two functions gives us the angle of the first vertex (anti-clockwise) of the side that each $x$ belongs to. $$\mathrm P_{n,\ r, \ \beta}(x) = r\sin \alpha(k(x)) + (\cos \alpha(k(x))-x)\cot\left(\frac {\pi}n(2k(x) +1) + \beta\right)$$ This renders the upper "half" of the polygon. The other half is given by $-\mathrm P_{n,\ r,\ -\beta}(x)$.
Note that depending on $\beta$ there may be some lines extending out of the unit circle, nevertheless the polygon will be there.
This is a function, but I'm sure one can make an equation out of it.
It's for rectangular coordinates. Every angle here is in radians.
$n$ is the number of sides.
$r$ is the radius of the circle where the polygon is inscribed.
$\beta$ is the tilting angle, you may adjust it to make sure there aren't any vertical lines.
$$\alpha(k) = \frac {2\pi}nk + \beta$$ The function $\alpha(k)$ gets the angle of the vertex of order/index $k$, starting at $0$. $$k(x) = \left\lfloor\frac{\cos^{-1}(x/r) - \beta}{2\pi}n\right\rfloor$$ The function $k(x)$ gets the index of the first vertex (anti-clockwise) of the side of the polygon where $x$ belongs. The floor function $\lfloor \cdot\rfloor$ rounds the fractional result down so it snaps to integer order vertices.
$$\alpha(k(x))$$ Combining the two functions gives us the angle of the first vertex (anti-clockwise) of the side that each $x$ belongs to. $$\mathrm P_{n,\ r, \ \beta}(x) = r\sin \alpha(k(x)) + (\cos \alpha(k(x))-x)\cot\left(\frac {\pi}n(2k(x) +1) + \beta\right)$$ This renders the upper "half" of the polygon. The other half is given by $-\mathrm P_{n,\ r,\ -\beta}(x)$.
Note that depending on $\beta$ there may be some lines extending out of the circle, nevertheless the polygon will be there.