Viviani's curve

Floor plan, elevation and side plan

Elimination of ${\displaystyle x}$ , ${\displaystyle y}$ , ${\displaystyle z}$ respectively yields:

The orthogonal projection of the intersection curve onto the

${\displaystyle x}$-${\displaystyle y}$-plane is the circle with equation ${\displaystyle \;(x-{\tfrac {r}{2}})^{2}+y^{2}=({\tfrac {r}{2}})^{2}\ .}$

${\displaystyle x}$-${\displaystyle z}$-plane the parabola with equation ${\displaystyle \;x=-{\tfrac {1}{r}}z^{2}+r\;.}$

${\displaystyle y}$-${\displaystyle z}$-plane the algebraic curve with the equation ${\displaystyle \;z^{4}+r^{2}(y^{2}-z^{2})=0\;.}$

For parametric representation and the determination of the area

Representing the sphere by

${\displaystyle {\begin{array}{cll}x&=&r\cdot \cos \theta \cdot \cos \varphi \\y&=&r\cdot \cos \theta \cdot \sin \varphi \\z&=&r\cdot \sin \theta \qquad \qquad -{\tfrac {\pi }{2}}\leq \theta \leq {\tfrac {\pi }{2}}\ ,\ -\pi \leq \varphi \leq \pi \;,\end{array}}}$

and setting ${\displaystyle \;\varphi =\theta ,\;}$ yields the curve

• ${\displaystyle {\begin{array}{cll}x&=&r\cdot \cos \theta \cdot \cos \theta \\y&=&r\cdot \cos \theta \cdot \sin \theta \\z&=&r\cdot \sin \theta \qquad \qquad -{\tfrac {\pi }{2}}\leq \theta \leq {\tfrac {\pi }{2}}\ .\end{array}}}$

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 ${\displaystyle \;\color {green}\varphi =-\theta \;.}$

With help of this parametric representation it is easy to proof the statement: The area of the half sphere (containing Viviani's curve) minus the area of the two windows is ${\displaystyle 4r^{2}}$:

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:

${\displaystyle {\boldsymbol {p0}}={\begin{bmatrix}0\\0\\1\\1\end{bmatrix}}{\boldsymbol {p1}}={\begin{bmatrix}0\\{\frac {1}{2{\sqrt {2}}}}\\{\frac {1}{\sqrt {2}}}\\{\frac {1}{\sqrt {2}}}\end{bmatrix}}{\boldsymbol {p2}}={\begin{bmatrix}{\frac {1}{3}}\\{\frac {1}{2}}\\{\frac {1}{2}}\\{\frac {2}{3}}\end{bmatrix}}{\boldsymbol {p3}}={\begin{bmatrix}{\frac {1}{\sqrt {2}}}\\{\frac {1}{2{\sqrt {2}}}}\\{\frac {1}{2{\sqrt {2}}}}\\{\frac {1}{\sqrt {2}}}\end{bmatrix}}{\boldsymbol {p4}}={\begin{bmatrix}1\\0\\0\\1\end{bmatrix}}}$

The corresponding rational parametrization is:

${\displaystyle \left({\begin{array}{c}{\frac {2\mu ^{2}\left(\mu ^{2}-2\left(2+{\sqrt {2}}\right)\mu +4{\sqrt {2}}+6\right)}{\left(2(\mu -1)\mu +{\sqrt {2}}+2\right)^{2}}}\\{\frac {2(\mu -1)\mu \left((\mu -1)\mu -3{\sqrt {2}}-4\right)}{\left(2(\mu -1)\mu +{\sqrt {2}}+2\right)^{2}}}\\-{\frac {(\mu -1)\left({\sqrt {2}}\mu +{\sqrt {2}}+2\right)}{2(\mu -1)\mu +{\sqrt {2}}+2}}\\\end{array}}\right)\;\mu \in \left[0,1\right]}$

The area of the upper right part of Viviani's window (see diagram) can be calculated by an integration:

${\displaystyle \iint _{S_{sphere}}r^{2}\cos \theta \,\mathrm {d} \theta \mathrm {d} \varphi =r^{2}\int _{0}^{\pi /2}\int _{0}^{\theta }\cos \theta \,\mathrm {d} \varphi \mathrm {d} \theta =r^{2}({\frac {\pi }{2}}-1)\ .}$

Hence the total area of the spherical surface included by Viviani's curve is ${\displaystyle 2\pi r^{2}-4r^{2}}$ and

• the area of the half sphere (${\displaystyle 2\pi r^{2}}$) minus the area of Viviani's window is ${\displaystyle \;4r^{2}\;}$, the area of a square with the sphere's diameter as the length of an edge.