Villarceau circles

Torus: Villarceau circles
The image below was projected perpendicularly onto the cutting plane . The circles appear here undistorted.

{\ displaystyle \ varepsilon}

Torus with two families of Villarceau circles

Animation for generating Villarceau circles

Villarceau circles are pairs of circles in geometry on a torus , which are created by intersecting with suitable planes. They are named after the French astronomer Yvon Villarceau . It is obvious that there are two sets of circles on a torus: 1) A set (meridians) is created by the rotation of a circle when the torus is generated. 2) the second set (parallel circles) is created by cutting the torus with planes that are perpendicular to the axis of rotation. 3) +4) Two other less obvious families are made up of Villarceau circles. Villarceau circles are created in pairs by cutting the torus with planes that touch twice (see picture).

Description of the torus

A torus can always be represented by appropriate introduction of coordinates in such a way that the axis of rotation is the z-axis and the center is the zero point. If a meridian (circle) has the radius and the center points of the meridian circles have the distance from the axis of rotation, the torus can be expressed by the equation ${\ displaystyle r}$ ${\ displaystyle R}$

{\ displaystyle (x ^ {2} + y ^ {2} + z ^ {2} + R ^ {2} -r ^ {2}) ^ {2} -4R ^ {2} (x ^ {2} + y ^ {2}) = 0}

describe.

Double-touching plane of the torus

The plane containing the x-axis and touching the two meridians in the yz-plane (see picture) also touches the parallel circles through the two torus points (for reasons of symmetry) and is therefore a tangential plane of the torus. Since the torus touches in two points, it is called a double touching tangent plane . The following applies to the angle of inclination of the plane (see figure) . If you let it rotate around the z-axis, all double-touching tangential planes of the torus are created. ${\ displaystyle \ varepsilon}$ ${\ displaystyle \ varepsilon}$ ${\ displaystyle \ varepsilon}$ ${\ displaystyle \ alpha}$ ${\ displaystyle \ sin \ alpha = r / R}$ ${\ displaystyle \ varepsilon}$

Generation of the Villarceau circles

claim: The intersection of the plane (see above) with the torus (see above) consists of the two circles with the centers and the radius . ${\ displaystyle \ varepsilon}$ ${\ displaystyle M_ {1} = (r, 0,0), \; M_ {2} = (- r, 0,0)}$ ${\ displaystyle R}$

proof

For the proof, one rotates the coordinate system around the x-axis by the angle and then sets the new 3rd coordinate zero. ${\ displaystyle \ alpha}$ ${\ displaystyle \ zeta}$

Rotation and delivers

{\ displaystyle: \; x = \ xi, \; y = \ eta \ cos \ alpha + \ zeta \ sin \ alpha, \; z = - \ eta \ sin \ alpha + \ zeta \ cos \ alpha}

{\ displaystyle \ zeta = 0}

{\ displaystyle y = \ eta {\ sqrt {R ^ {2} -r ^ {2}}} / R, \; z = - \ eta r / R}

.

If you put this in the torus equation, you get the equation of the intersection curve:

{\ displaystyle (\ xi ^ {2} + \ eta ^ {2} + R ^ {2} -r ^ {2}) ^ {2} -4R ^ {2} (\ xi ^ {2} + \ eta ^ {2}) + 4r ^ {2} \ eta ^ {2} = 0 \ ;.}

Break the brackets and compare the solution with the solution of the equation

${\ displaystyle ((\ xi -r) ^ {2} + \ eta ^ {2} -R ^ {2}) \ cdot ((\ xi + r) ^ {2} + \ eta ^ {2} -R ^ {2}) = 0}$ ,

this is how we see: Both equations describe the same curve, i. H. the sectional figure consists of the two circles with the equations

{\ displaystyle (\ xi -r) ^ {2} + \ eta ^ {2} = R ^ {2}}

and .

{\ displaystyle (\ xi + r) ^ {2} + \ eta ^ {2} = R ^ {2}}

Parametric representations of the Villarceau circles

With the position vectors of the center points and the orthonormal basis of the cutting plane , the two cutting circles can be passed through ${\ displaystyle (\ pm r, 0,0) ^ {T}}$ ${\ displaystyle (1,0,0) ^ {T}, (0, \ cos \ alpha, \ sin \ alpha) ^ {T}}$ ${\ displaystyle \ varepsilon}$

{\ displaystyle {\ vec {x}} _ {\ pm} (\ varphi) = {\ begin {pmatrix} \ pm r \\ 0 \\ 0 \ end {pmatrix}} + R {\ begin {pmatrix} 1 \\ 0 \\ 0 \ end {pmatrix}} \ cos \ varphi + R {\ begin {pmatrix} 0 \\ - \ cos \ alpha \\\ sin \ alpha \ end {pmatrix}} \ sin \ varphi, \ quad 0 \ leq \ varphi <2 \ pi,}

describe (see ellipse ). The equation of the cutting plane is or, because of ${\ displaystyle: \ y \ sin \ alpha + z \ cos \ alpha = 0}$ ${\ displaystyle \ sin \ alpha = r / R \ ;:}$ ${\ displaystyle \ yr + z {\ sqrt {R ^ {2} -r ^ {2}}} = 0}$

An arbitrary pair of Villsarceau circles is obtained by rotation of the above circles around the z-axis by an angle : ${\ displaystyle \ sigma, 0 \ leq \ sigma <2 \ pi}$

${\ displaystyle {\ vec {x}} _ {\ pm} (\ sigma, \ varphi) = {\ begin {pmatrix} \ pm r \ cos \ sigma \\\ pm r \ sin \ sigma \\ 0 \ end {pmatrix}} + R {\ begin {pmatrix} \ cos \ sigma \\\ sin \ sigma \\ 0 \ end {pmatrix}} \ cos \ varphi + R {\ begin {pmatrix} \ cos \ alpha \ sin \ sigma \\ - \ cos \ alpha \ cos \ sigma \\\ sin \ alpha \ end {pmatrix}} \ sin \ varphi, \ quad 0 \ leq \ varphi <2 \ pi,}$

{\ displaystyle = {\ begin {pmatrix} \ pm r \ cos \ sigma \\\ pm r \ sin \ sigma \\ 0 \ end {pmatrix}} + R {\ begin {pmatrix} \ cos \ sigma \\\ sin \ sigma \\ 0 \ end {pmatrix}} \ cos \ varphi + {\ begin {pmatrix} {\ sqrt {R ^ {2} -r ^ {2}}} \; \ sin \ sigma \\ - { \ sqrt {R ^ {2} -r ^ {2}}} \; \ cos \ sigma \\ r \ end {pmatrix}} \ sin \ varphi, \ quad 0 \ leq \ varphi <2 \ pi,}

The cutting plane has the equation ${\ displaystyle: \; - xr \ sin \ sigma + yr \ cos \ sigma + z {\ sqrt {R ^ {2} -r ^ {2}}} = 0 \ ;.}$

Villarceau circles (magenta, green) through a given point (red). 4 circles go through each point.

Determination of the Villarceau circles by a torus point

If a torus point is given and one searches through the two Villarceau circles , one must determine the cutting plane from the above group of cutting planes, which contains, i. H. one has to determine that ${\ displaystyle P_ {0} = (x_ {0}, y_ {0}, z_ {0})}$ ${\ displaystyle P_ {0}}$ ${\ displaystyle P_ {0}}$ ${\ displaystyle \ sigma}$

{\ displaystyle -x_ {0} r \ sin \ sigma + y_ {0} r \ cos \ sigma + z_ {0} {\ sqrt {R ^ {2} -r ^ {2}}} = 0 \ ;. }

This problem can be understood and solved by the substitution in the - -plane as the intersection problem of the straight line with the unit circle (see intersection of a straight line with a circle ). In general, this results in two levels and a total of four Villarceau circles, of which only two contain the given point . In the special case that the two levels are the same, there is one of the two intersections of the two Villarceau circles. ${\ displaystyle {\ color {red} u} = \ cos \ sigma, \; {\ color {red} v} = \ sin \ sigma}$ ${\ displaystyle {\ color {red} u}}$ ${\ displaystyle {\ color {red} v}}$ ${\ displaystyle y_ {0} r {\ color {red} u} -x_ {0} r {\ color {red} v} + z_ {0} {\ sqrt {R ^ {2} -r ^ {2} }} = 0}$ ${\ displaystyle {\ color {red} u} ^ {2} + {\ color {red} v} ^ {2} = 1}$ ${\ displaystyle P_ {0}}$ ${\ displaystyle P_ {0}}$

literature

Ulrich Graf , Martin Barner : Descriptive Geometry. Quelle & Meyer, Heidelberg 1961, ISBN 3-494-00488-9 , p. 209.
Georg Glaeser : Geometry and its applications in art, nature and technology , Springer-Verlag, 2014, ISBN 9783642418525 , p. 216.
Hellmuth Stachel : Remarks on A. Hirsch's Paper concerning Villarceau Sections . In: Journal for Geometry and Graphics 6 (2002), No. 2, pp. 133-139.
Yvon Villarceau : Théorème sur le tore . In: Nouvelles annales de mathématiques. Journal des candidats aux écoles polytechnique et normal 7 (1848), pp. 345–347.

Web links

Mathworld: Villarceau circles