On Wed, May 22, 2013 at 8:00 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Tue, May 7, 2013 at 1:28 AM, Bill Gosper <billgosper@gmail.com> wrote:
DanA>On the plane, assume the foci are at (-1,0) and (1,0):
The ellipse is the locus of points whose sum of distances to the foci is a constant.
The ovals of Cassini is the locus of points whose product of distances to the foci is a constant. (See < http://en.wikipedia.org/wiki/File:Line_of_Cassini.svg >.)
Can you guess the evolution of the locus of points whose exponentiation of their distances to the foci is a constant, as that constant increases from 0 to, say, 2 ???
(I certainly didn't!)
--Dan
You mean distance1^distance2 = constant, right? (You left out the hyperbola from distance1 - distance2 = constant.)
Re Cassini: Although it has disappeared completely from my personal mail archive (?!!?), the math-fun
archive retains a three week(!) discussion, starting ~ 14 Feb 07, on constructing convincing ovals. Amazingly, of all the methods discussed (piecewise circular, Fourier series, polar fitting, rounding triangles, bending steel, the "7-11 curve", sections of surfaces, ...), no one brought up Cassini's
ovals! E.g., from sectioning a torus. Nobody thinks about two eggs at a time?
But it would make a nice eXcerise (I resisted!) to find the Cassini parameters that best approximate Moss's Egg, by various criteria. Make that two Moss's Eggs.
Before actually computing the Cassini parameter best fitting the Moss egg, I eyeballed ContourPlot[(-Sqrt[1 - x^2 - Sqrt[0.9` - 4 x^2]] + y) (-Sqrt[1 - x^2 + Sqrt[0.9` - 4 x^2]] + y) == 0, {x, -3/4, 3/4}, {y, 0, 3/2}] Then, Neil's little sister brought me some actual eggs from the fridge. They don't look anything like this. Or Moss's. Very nearly mere prolate spheroids. Grr:-{
So what is the area of a Cassini oval? Eqn (10) in http://mathworld.wolfram.com/CassiniOvals.html is garbled. (12) claims to apply only to b>a, but is wrong except when b=a. While I was still struggling to make a plot, Neil figured out that total enclosed area = 2 Re[b^2 EllipticE[ArcSin[b^2/a^2], a^4/b^4]] == 2 Re[b^2 EllipticE[a^4/b^4]] . Without the realparts, the lhs works for the two-loop, b<a case, and the rhs handles b>a. (a:= tube radius, b^2:=invariant distance product.) --rwg
Oops, no, 2a:= interfocal distance. a is the major radius of a torus with tube radius r:=b^2/(2a) sectioned by a plane at distance r from the torus center perpendicular to the plane of the torus, revealing the oval(s) under discussion. Again, Mathworld seems garbled on this. --rwg
Man, that green racetrack case sure looks flat. But I checked: it's just ±y = Sqrt[-1 - x^2 + 2 Sqrt[1 + x^2]] ~ 1-x^4/8+O(x^6). --rwg