Viviani on Sphere parametrization

Question

How should parametrization of the 2 parameter surface of a sphere (latitude u, longitude

v) be changed to result in 1 parameter curve of Viviani?

Answer 1

The parametrization of the 2-sphere is given by $$\vec{x}(u,v) = (\cos(v)\sin(u),\sin(v)\sin(u),\cos(u)).$$

The equation for a cylinder of radius $\frac{1}{2}$ and center at $\left(\frac{1}{2},0,0\right)$ is $$ \left(x-\frac{1}{2}\right)^2+y^2=\frac{1}{4}.$$

Thus the intersection of the sphere and the cylinder is determined by $$ \left(\cos(v)\sin(u)-\frac{1}{2}\right)^2+(\sin(v)\sin(u))^2 = \frac{1}{4}$$ and we derive $$ \sin(u) = \cos(v).$$

We therefore have a parametrization $(u(t),v(t)) = \left(t,\frac{\pi}{2}-t\right)$ for $0\leq t\leq 2\pi$. We have $$\vec{x}(u(t),v(t)) = (\sin(t)^2,\sin(t)\cos(t),\cos(t)).$$

Answer 2

I was asking for changes needed in:

$$\vec{x}(u,v) = (\cos(v)\sin(u),\sin(v)\sin(u),\cos(u)).$$

to directly get after derivation:

$$\vec{x}(u(t),v(t)) = (\sin(t)^2,sin(t)cos(t),cos(t)).$$

OK, the asked change/substitution simply is:

$$ u = v , ( = t, here) $$

$(\pi/2 - u = latitude , v= longitude)$

This is so because the Viviani curve has this standard parameter link property of

spherical coordinates:

$$ latitude = longitude. $$

This is a little known property, imho.