A circular Clifford torus (radius at flat circle = h, section radius $ a , a<h $ ) is cut by a plane at an angle $ \cos \alpha = a/h \tag{1} $

centrally to the symmetry axis, the line of intersection is a Villarceau Circle.

Show how, or by what Law or Theorem the the Circle is a Loxodrome, i.e., makes a constant

$$ \psi = \alpha \tag {2} $$

to torus meridians, that appears to be a situation of intersections as indicated here, agreeing with result from Colonel Schölcher mentioned in Wiki article of reference.

Joachimsthal's theorem ( line of curvature of either surface cutting at same angle) does not seem to be of relevance here.

I have checked it to be so by numerical calculation, and also that the Villarceau circle has curvature

$ \kappa= 1/h \, \tag{3} $ Posted in SE Mathematica recently.

I seek your help in being able to appreciate it more directly by analytical derivation.

The loxodrome situation can be understood, hopefully around the new three common intersections.

The Villarceau Circles are lines of concurrent intersection between three surfaces. Torus,Plane and Sphere. Image is uploaded.

The sphere is not given in Wikipedia on the topic of Villarceau circles as being one of the generating intersection surfaces. Also I have not seen it in this context.


$$ z = y \,\tan \alpha \tag{4} $$


$$ (x,y,z) = ( ( h + a \cos u) \sin v ,( h + a \cos u) \cos v, a \sin u ) \tag{5} $$

where $ u $ is toroid latitude, $v$ longitude.

A relation between $u,v$ of Torus surface to obtain the Villarceau Circle intersection line that I obtain like spherical

$$ (u,v) = ( \phi ,\theta) \tag{6} $$

$$ \sec \alpha = \frac{\sin u \sin v }{\cos u - \cos v} \tag{7} $$

The above result is obtained by assuming *constant $\alpha = \psi $ angles between Villarceau Circle and Torus meridian for Loxodrome *, so the question still remains.

I have no access to the original article or the other references in:

Villarceau Circles Wiki


$$ (x,y,z)= (\cos u \sin v, \cos u \cos v \color {red} {+a } , \sin u) \tag{8} $$

where $ u $ is sphere latitude, $v$ longitude.

Common Villarceau Circle intersection

$$ (x,y,z)= (\pm (a-h) \cos u , \sqrt{h^2-a^2} \sin u, a \sin u) $$

where $u$ is rotation angle starting from y-axis around torus center.

The motivation of this post was to draw a noticeable parallel of intersections on the one hand between sphere/ eccentric cylinder defining the Viviani curve and and on the other torus/ eccentric sphere intersection curve in same way defining Villarceau Circle now noticed here.

Further, if $ \phi $ is meridional rotation around tube section and prime denotes differentiation with respect to arc length $s$, the Villarceau circle possesses a very simple differential relation:

$$ \phi^{'} = \frac {1}{h} \tag{11} $$

Arc length

$$ s_{max} = 2 \pi h \tag{12} $$

Villarceau Circle Plane, Torus, Sphere

Orthogonal projections are uploaded for a Villarceau Circle $ h=5, a= 3. $

The projection of planar intersection is not centered , but appears as an off-centred ellipse in 3D.



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