In mathematics, Viviani's curve, also known as Viviani's window, is a figure eight shaped space curve named after the Italian mathematician Vincenzo Viviani. It is the intersection of a sphere with a cylinder that is tangent to the sphere and passes through two poles (a diameter) of the sphere (see diagram). Before Viviani this curve was studied by Simon de La Loubère and Gilles de Roberval.[1] [2]
The orthographic projection of Viviani's curve onto a plane perpendicular to the line through the crossing point and the sphere center is the lemniscate of Gerono, while the stereographic projection is a hyperbola or the lemniscate of Bernoulli, depending on which point on the same line is used to project.[3]
In 1692 Viviani solved the following task: Cut out of a half sphere (radius
r
4r2
In order to keep the proof for squaring simple,
the sphere has the equation
x2+y2+z2=r2
the cylinder is upright with equation
x2+y2-rx=0
r/2
(r,0,0) .
Elimination of
x
y
z
The orthogonal projection of the intersection curve onto the
x
y
\left(x-\tfrac{r}{2}\right)2+y2=\left(\tfrac{r}{2}\right)2 .
x
z
x=-\tfrac{1}{r}z2+r .
y
z
z4+r2(y2-z2)=0 .
Representing the sphere by
\begin{array}{cll} x&=&r ⋅ \cos\theta ⋅ \cos\varphi\\ y&=&r ⋅ \cos\theta ⋅ \sin\varphi\\ z&=&r ⋅ \sin\theta -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2} , -\pi\le\varphi\le\pi , \end{array}
\varphi=\theta,
\begin{array}{cll} x&=&r ⋅ \cos\theta ⋅ \cos\theta\\ y&=&r ⋅ \cos\theta ⋅ \sin\theta\\ z&=&r ⋅ \sin\theta -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2} . \end{array}
One easily checks that the spherical curve fulfills the equation of the cylinder. But the boundaries allow only the red part (see diagram) of Viviani's curve. The missing second half (green) has the property
\color{green}\varphi=-\theta .
With help of this parametric representation it is easy to prove the statement: The area of the half sphere (containing Viviani's curve) minus the area of the two windows is
4r2
\iint | |
Ssphere |
r2\cos\thetad\thetad\varphi=r2
\pi/2 | |
\int | |
0 |
\theta | |
\int | |
0 |
\cos\thetad\varphid\theta=r2\left(
\pi | |
2 |
-1\right) .
2\pir2-4r2
2\pir2
4r2
The quarter of Viviani's curve that lies in the all-positive quadrant of 3D space cannot be represented exactly by a regular Bezier curve of any degree.
However, it can be represented exactly by a 3D rational Bezier segment of degree 4, and there is an infinite family of rational Bezier control points generating that segment.
One possible solution is given by the following five control points:
\boldsymbol{p0}=\begin{bmatrix}0\ 0\ 1\ 1\end{bmatrix} \boldsymbol{p1}=\begin{bmatrix}0\
1 | |
2\sqrt{2 |
The corresponding rational parametrization is:
\left(\begin{array}{c}
2\mu2\left(\mu2-2\left(2+\sqrt{2 | |
\right) |
\mu+4\sqrt{2}+6\right)}{\left(2(\mu-1) \mu+\sqrt{2}+2\right)2}\\
2(\mu-1)\mu\left((\mu-1)\mu-3\sqrt{2 | |
-4\right)}{\left(2 |
(\mu-1)\mu +\sqrt{2}+2\right)2}\\ -
(\mu-1)\left(\sqrt{2 | |
\mu |
+\sqrt{2}+2\right)}{2(\mu-1)\mu+\sqrt{2}+2}\\ \end{array} \right) \mu\in\left[0,1\right]
\varphi=c \theta .
Subtracting 2× the cylinder equation from the sphere's equation and applying completing the square leads to the equation
(x-r)2+y2=z2 ,
(r,0,0)
a) the intersection of a sphere and a cone and as
b) the intersection of a cylinder and a cone.