Viviani's curve

File:Viviani_kromme.jpg

In order to keep the proof for squaring simple, suppose that the sphere and cylinder have the equations

: $x^2+y^2+z^2=r^2$

and

: $x^2+y^2-rx=0,$

respectively. The cylinder has radius $r/2$ and is tangent to the sphere at point $(r,0,0).$

File:Viviani-fenster-xyz.svg

Elimination of $x$, $y$, and $z$ respectively yields the orthogonal projections of the intersection curve onto the:

: $x$-$y$-plane is the circle with equation $\backslash left(x-\backslash tfrac\{r\}\{2\}\; \backslash right)^2+y^2\; =\; \backslash left(\backslash tfrac\{r\}\{2\}\; \backslash right)^2,$

: $x$-$z$-plane the parabola with equation $x=-\backslash tfrac\{1\}\{r\}z^2+r,$ and

: $y$-$z$-plane the algebraic curve with the equation $z^4+r^2(y^2-z^2)=0.$

File:Viviani-fenster-3.svg

Representing the sphere by

: $\backslash begin\{array\}\{cll\}$

x &=& r \cdot \cos \theta \cdot \cos \varphi \\

y &=& r \cdot \cos \theta \cdot \sin \varphi \\

z &=& r \cdot \sin \theta \qquad \qquad -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}, \ -\pi\le\varphi\le \pi,

\end{array}

and setting $\backslash varphi=\backslash theta,$ yields the curve

: $$

\begin{array}{cll}

x &=& r \cdot \cos \theta \cdot \cos \theta \\

y &=& r \cdot \cos \theta \cdot \sin \theta \\

z &=& r \cdot \sin \theta \qquad \qquad -\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 $\backslash varphi=-\backslash theta.$

With help of this parametric representation it is easy to prove that the area of the hemisphere containing Viviani's curve minus the area of the two windows is $4r^2$. The area of the upper-right part of Viviani's window (see diagram) can be calculated by an integration:

:$\backslash iint\_\{S\_\{sphere\}\}\; r^2\; \backslash cos\backslash theta\; \backslash ,\backslash mathrm\{d\}\backslash theta\backslash ,\; \backslash mathrm\{d\}\backslash varphi\; =r^2\; \backslash int\_0^\{\backslash pi/2\}\; \backslash int\_0^\backslash theta\; \backslash cos\backslash theta\; \backslash ,\backslash mathrm\{d\}\backslash varphi\; \backslash ,\backslash mathrm\{d\}\backslash theta=\; r^2\; \backslash left(\backslash frac\{\backslash pi\}\{2\}-1\; \backslash right).$

Hence the total area of the spherical surface included by Viviani's curve is $2\backslash pi\; r^2-4r^2$, and the area of the hemisphere ($2\backslash pi\; r^2$) minus the area of Viviani's window is $4r^2$, the area of a square with the sphere's diameter as the length of an edge.

The quarter of Viviani's curve that lies in the all-positive octant of 3D space cannot be represented exactly by a regular Bézier curve of any degree. However, it can be represented exactly by a 3D rational Bézier segment of degree 4, and there is an infinite family of rational Bézier control points generating that segment. One possible solution is given by the following five control points:

$p\_0\; =\; (0,\; 0,\; 1,\; 1)$

$\backslash begin\{array\}\{llllll\}$

p_0 & = & (0 & 0 & 1 & 1) & & \\

p_1 & = & (0 & 1 & 2 & 2) &/& (2\sqrt{2}) \\

p_2 & = & (2 & 3 & 3 & 4) &/& 6 \\

p_3 & = & (2 & 1 & 1 & 2) &/& (2\sqrt{2}) \\

p_4 & = & (1 & 0 & 0 & 1) & &

\end{array}

:$$

\boldsymbol{p0} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix}

\boldsymbol{p1} = \begin{bmatrix} 0 \\ \frac{1}{2\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}

\boldsymbol{p2} = \begin{bmatrix} \frac{1}{3} \\ \frac{1}{2} \\ \frac{1}{2} \\ \frac{2}{3} \end{bmatrix}

\boldsymbol{p3} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{2\sqrt{2}} \\ \frac{1}{2\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}

\boldsymbol{p4} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}

The corresponding rational parametrization is:

:$$

\left(

\begin{array}{c}

\displaystyle \frac{2 \mu ^2 \left(\mu ^2-2 \left(2+\sqrt{2}\right) \mu +4 \sqrt{2}+6\right)}{\left(2 (\mu -1)

\mu +\sqrt{2}+2\right)^2} \\

\displaystyle \frac{2 (\mu -1) \mu \left((\mu -1) \mu -3 \sqrt{2}-4\right)}{\left(2 (\mu -1) \mu

+\sqrt{2}+2\right)^2} \\

\displaystyle -\frac{(\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]

• The 8-shaped elevation (see above) is a Lemniscate of Gerono.
• Viviani's curve is a special Clelia curve. For a Clelia curve, the relation between the angles is $\backslash varphi=c\backslash theta.$

File:Viviani-kugel-kegel.svg

Subtracting twice the cylinder equation from the sphere's equation and completing the square leads to the equation

:$(x-r)^2+y^2=z^2,$

which describes a right circular cone with its apex at $(r,0,0)$, the double point of Viviani's curve. Hence, Viviani's curve can be considered not only as the intersection curve of a sphere and a cylinder but also as the intersection of a sphere and a cone, and as the intersection of a cylinder and a cone.

• Sphere-cylinder intersection

{{reflist}}

• Berger, Marcel: Geometry. II. Translated from the French by M. Cole and S. Levy. Universitext. Springer-Verlag, Berlin, 1987.
• Berger, Marcel: Geometry. I. Translated from the French by M. Cole and S. Levy. Universitext. Springer-Verlag, Berlin, 1987. xiv+428 pp. {{ISBN|3-540-11658-3}}
• {{springer|title=Viviani curve|id=Viviani_curve}}
• {{mathworld|urlname=VivianisCurve|title=Viviani's Curve}}

Category:Spherical curves