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. 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. In 1692, Viviani solved the following task: Cut out of a hemisphere (radius r {\displaystyle r} ) two windows, such that the remaining surface (of the hemisphere) can be squared; that is, a square with the same area can be constructed using only ruler and compass. His solution has an area of 4 r 2 {\displaystyle 4r^{2}} (see below).
Equations 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 {\displaystyle x^{2}+y^{2}+z^{2}=r^{2}}
and
x 2 + y 2 − r x = 0 , {\displaystyle x^{2}+y^{2}-rx=0,}
respectively. The cylinder has radius r / 2 {\displaystyle r/2} and is tangent to the sphere at point ( r , 0 , 0 ) . {\displaystyle (r,0,0).}
Properties of the curve
Floor plan, elevation, and side plan
Elimination of x {\displaystyle x} , y {\displaystyle y} , and z {\displaystyle z} respectively yields the orthogonal projections of the intersection curve onto the:
x {\displaystyle x} - y {\displaystyle y} -plane is the circle with equation ( x − r 2 ) 2 + y 2 = ( r 2 ) 2 , {\displaystyle \left(x-{\tfrac {r}{2}}\right)^{2}+y^{2}=\left({\tfrac {r}{2}}\right)^{2},}
x {\displaystyle x} - z {\displaystyle z} -plane the parabola with equation x = − 1 r z 2 + r , {\displaystyle x=-{\tfrac {1}{r}}z^{2}+r,} and
y {\displaystyle y} - z {\displaystyle z} -plane the algebraic curve with the equation z 4 + r 2 ( y 2 − z 2 ) = 0. {\displaystyle z^{4}+r^{2}(y^{2}-z^{2})=0.}
Parametric representation
Representing the sphere by
x = r ⋅ cos θ ⋅ cos φ y = r ⋅ cos θ ⋅ sin φ z = r ⋅ sin θ − π 2 ≤ θ ≤ π 2 , − π ≤ φ ≤ π , {\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}}}
… excerpt ends here. Continue reading the full article.
