Viviani's curve
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
) two windows, such that the remaining surface (of the half sphere) can be squared, i.e. a square with the same area can be constructed using only compasses and ruler.
In order to keep the proof for squaring simple, and The cylinder has radius
respectively yields: The orthogonal projection of the intersection curve onto the Representing the sphere by and setting
But the boundaries allow only the red part (see diagram) of Viviani's curve.
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
The area of the upper right part of Viviani's window (see diagram) can be calculated by an integration: Hence the total area of the spherical surface included by Viviani's curve is
) minus the area of Viviani's window is
, 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 quadrant 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: The corresponding rational parametrization is: Subtracting 2× the cylinder equation from the sphere's equation and applying completing the square leads to the equation which describes a right circular cone with its apex at
