Re: Oblique torus slice: intersecting circles

In article
<c6f3ce75-aacd-4051-b6bb-b7bf3ace69a8@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Jim Ferry <corklebath@xxxxxxxxxxx> wrote:

Apparently a certain oblique slice of a torus yields a pair of
intersecting circles. This is pictured at
http://johnbanks.maths.latrobe.edu.au/Torus/

Can anyone offer a simple explanation of this?

Here is an elementary explanation, which is not very elegant but gets
the job done.

To begin with, let the coordinate system be such that the torus has
rotational symmetry about the Z axis, and also has mirror symmetry in
the plane Z=0. Let the inner and outer radii of the torus be R-r and
R+r, so that the torus may be constructed by taking a circle with center
at (R, 0,0) and radius r, rotated about the Z axis. (I will later call
this the "generating circle.")

The intersection of the torus with a plane at distance z above the XY
plane is two circles, with radii R \pm \sqrt{ r^2 - z^2} .

I am using TeX and :LaTeX notation, but not very much. \pm is "plus or
minus", ^ introduces superscripts or exponents, and {} is for TeX-level
grouping. So, if I had a complicated superscript of subscript, I could
encluse it all in braces. Happily, I don't need to for this article.

Anyhow, points (x,y,z) on the torus satisfy the equation

x^2 + y^2 = (R \pm \sqrt{r^2 ­ z^2} )^2.

Expand out the right-hand side of this equation; it is

R^2 + r^2 ­ z^2 \pm 2 R \sqrt{r^2 ­ z^2} .

Let's quarantine that sqrt subexpression on one side of the equal sign.
We get

x^2 + y^2 + z^2 ­ R^2 ­ r^2 = \pm 2 R \sqrt{r^2 ­ z^2}.

Now we can square both sides of the equation, to get a polynomial
equation satisified by the coordinates of a point on the torus:

( x^2 + y^2 + z^2 ­ R^2 ­ r^2) ^2 = 4 R^2 (r^2 ­ z^2) .

Now, the two circles that Jim Ferry asked about are on a certain plane.
Let O be the origin of coordinates (and the center point of the torus);
let P be the point (R,0,0), which is the center of the generating
circle; let Q be the point where the circle is touched by the tangent
line. Then OQP is a right triangle, with OP = R, PQ = r, and OQ = \sqrt
{ R^2 - r^2} . The line OQ and the Y axis define the plane, which I
shall prove intersects the torus in two circles.

On this plane we need a coordinate system. y will do fine as one
coordinate; the other, which I shall call u, is measured obliquely,
along the line OQ. After some messing about with similar triangles, we
find that the x and z coordinates of a point on the plane are determined
by u as follows:

x = (\frac{\sqrt{R^2 ­ r^2} } {R}) u
z = (r/R) u.

(More comments about TeX: in my equation for x, the \sqrt expression is
the numerator of a fraction, and "R" is the denominator. The whole
built-up fraction is enclosed in parentheses.)

Now we substitute into the equation for the torus. We get that a point
with coordinates (u,y) on that plane and also on the torus satisfies

((u^2 + y^2) ­ (R^2 + r^2))^2 = 4 r^2(R^2 ­ u^2).

I expand out the square in the left-hand side:

(u^2 + y^2)^2 ­ 2(u^2 + y^2)(R^2 + r^2) + (R^2 + r^2)^2 =
4 r^2 R^4 ­ 4 r^2 u^2.

Now I ranspose everything to one side of the equals sign, and do a
little rearrangement.

(u^2 + y^2)^2 ­ 2(R^2 - r^2)u^2 ­ 2(R^2 + r^2)y^2 + (R^2 - r^2)^2 = 0.

I do another rearrangement, of adding 4r^2 y^2 to both sides. This is
the trickiest part of the whole argument. Everything up to now has been
just plodding along, with a little attention to keeping the argebraic
expressions as compact as possible. The only justification for adding 4
r^2 y^2 is "because it works."

(u^2 + y^2)^2 ­ 2(R^2 ­ r^2)u^2 ­ 2(R^2 ­ r^2)y^2 + (R^2 ­ r^2)^2 =
4 r^2 y^2.

Now I can simplify by grouping to terms in which (R^2 ­ r^2) is a factor:

(u^2 + y^2)^2 ­ 2(R^2 ­ r^2)(u^2 + y^2) + (R^2 ­ r^2)^2 = 4r^2 y^2.

Hey look! The left hand side is a perfect square!

((u^2 + y^2) ­ (R^2 ­ r^2))^2 = 4 r^2 y^2.

That is to say,

((u^2 + y^2) ­ (R^2 - r^2))^ 2 ­ 4(ry)^2 = 0.

The left-hand side is a difference of two squares, so it factors:
A^2 ­ B^2 = (A+B)(A ­ B).

For a point on the intersection, one of those factors must equal 0. If
you write out the two factors separately, you will find that each one,
equated to 0, gives the equation of a circle. The centers are at u=0, y
= \pm r; the common radius is R.

--
Christopher J. Henrich
chenrich@xxxxxxxxxxxx
http://www.mathinteract.com
"A bad analogy is like a leaky screwdriver." -- Boon
.

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