Apology for minor typo --- and further ramblings ... WFL Dear Ralph, Thanks for checking that, but you were too trusting --- the actual howler occurs before the first line [always the best place to conceal them, I find!] Your (Forsyth/Maxwell's?) cyclide equation as originally stated was (x^2 + y^2 + z^2 - m^2 - b^2)^2 - 4(a x - c m)^2 - 4 b^2 y^2 which I had attempted manually to simplify via b^2 -> a^2 - c^2 --- my very first equation should then actually have read (x^2 + y^2 + z^2 - m^2 - a^2 + c^2)^2 - 4(a x - c m)^2 - 4(a^2 - c^2)y^2 = 0 and the remainder follows from that correctly as before. Incidentally, this exercise of my inimitable clerical skills prompted the observation that your / Maxwell's "b" is what I earlier denoted "r": the radius of the circle where a generator plane touches the cyclide, a Laguerre invariant. And I've since discovered that my alternative choice for the third parameter --- the Moebius invariant sine of the cone semi-angle "s" --- is what John Knapman (1987) calls the "eccentricity". Reassuring to find that both these were already recognised as significant. Now Groebner bases expose this simple relation, bilinear between their squares --- q^2 b^2 s^2 - p^2 q^2 s^2 + p^2 b^2 + p^2 q^2 = 0 . Since pairs of cyclide nodes are apparently symmetric, it's a bit of a shock that this expression is instead invariant under p -> q, q -> p, s -> 1/s, b -> i b; the exact geometrical interpretation of this subtle wrinkle in the symmetry at present eludes yours truly, Fred Lunnon On 7/29/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
Following my recent ignominious extraction from the Dupin node quicksands with the assistance of a flying-squad who do know some algebraic geometry, and muster a computer algebra system which can solve polynomial equations, I remounted my cyclide and pedalled off shakily to further adventures in neglected corners of this subfusc mathematical suburb. Among the customary extensive selection of technical lamposts, potholes and manhole-covers lying subsequently in wait, there lurked the following conundrum.
To recap a recent post by Ralph, a Dupin cyclide in canonical pose may be specified by three parameters (due apparently to Maxwell, rather than to Cayley as earlier hazarded). The radius "a" specifies a central circle at the origin; with centre on this, a perpendicular cross-sectional circle sweeps out a tube with mean radius "m"; while in the course of a single revolution, the actual tube radius is offset from m-c through m+c and back to m-c again.
In terms of c,a,m, the implicit equation is then a Cartesian quartic (x^2 + y^2 + z^2 - m^2 - a^2 + c^2)^2 - 4(a x - c m)^2 - 4(a^2 - c^2)y^2 = 0 . The general shape of the surface will be "horned" when 0 < m < c < a ; "ring" when 0 < c < m < a ; "spindle" when 0 < c < a < m ; with various special cases at boundaries between. For example, when the differential offset vanishes c = 0 we have a torus; when in addition the tube radius equals the central radius m = a we have a double sphere centred at the origin (and giving the average graphics surface plotter a furry tongue).
Now consider the special case where offset equals central radius c = a. It's easily established that the equation factors as the product of (x - 2a)^2 + y^2 + z^2 - (m - 2a)^2 , with (x + 2a)^2 + y^2 + z^2 - (m + 2a)^2 ; the surface comprises two tangent spheres with centres on the x-axis (by the way, as oriented spheres properly tangent, not anti-tangent).
'Ang abaht tho' --- one definition of a Dupin cyclide is the envelope (unique when spheres are oriented) swept out by a sphere moving tangent to 3 fixed spheres. For what 3 spheres (including planes and points) could such an envelope possibly be a pair of tangent spheres?
Fred Lunnon