Re: [math-fun] Re: Simplest Ovals
Fascinating and surprisingly long-lived thread! My "natural" choice would be to make ovals by interpolating or blending two ellipses (perhaps from too much computer graphics): Standardize the graphs of the ellipses by aligning them with their major axis coincident with the X axis and scale them so their ends are at x=0 and x=1. These graphs form a family indexed by "eccentricity parameter" [e]: y = ellipse[e](x) Now define oval[e0,e1] := (1-x) ellipse[e0](x) + x ellipse[e1](x) (having "eccentricity vector" [e0,e1]) So if ellipse(x) is quadratic in x then oval(x) will be some cubic. Alternatively, we might define oval[e0,e1] := ellipse[(1-x) e0 + x e1](x) For what parameterizations of eccentricity are these the same?
Extending my earlier post... Standardizing so that the diameter is the unit interval, graph the circle y with y = sqrt(x(1-x)) Scale by c to get an ellipse y[c] with "circularity" c y[c](x) := c sqrt(x(1-x)) where c=1 is circular, c=0 is a line. This seems a natural parameterization since interpolating between the graphs of two ellipses is equivalent to interpolating their circularities: (1-x)y[c0](x) + x y[c1](x) = y[(1-x)c0 + x c1](x) Then without loss of generality we can set c0=1 to get a standard oval z[c] z[c](x) := (1 + x(c-1)) sqrt(x(1-x)) which is circular with c=1, a teardrop with c=0 and ovaltine for c between. The value c=0.5 is reasonably eggy; c=0.618... outputs an auric anser.
Of an epicyclic synthesis of Moss's egg, I sent:
Note the surprising presence of counterrotors, presumably due to my lazy choice of traversal speeds: constant dtheta/dt on all four arcs, meaning instant ac/deceleration at each arc [endpoint].
The constant arcspeed rendition of Moss's egg is oo (sqrt(2) - 11) n pi ==== ' cos(-------------------) \ i n t n 17
e (- 1) (------------------------
/ 4 n + sqrt(2) - 6 ==== n = - oo (2 sqrt(2) - 5) n 1 (sqrt(2) + 1) cos((----------------- - -) pi) 17 4 - ---------------------------------------------) 4 n - 2 sqrt(2) - 5 ----------------------------------------------------, n (8 n + sqrt(2) - 6) (after enormous simplification*). (Input form: (-1)^n*(cos((sqrt(2)-11)*%pi*n/17)/(4*n+sqrt(2)-6)-(sqrt(2)+1)*cos(%pi*((2*sqrt(2)-5)*n/17-1/4))/(4*n-2*sqrt(2)-5))/(n*(8*n+sqrt(2)-6))*cis(n*t).) We now have n^-3 convergence, but lingering (surprisingly large) counterrotors. The harmonics > .002, 0.00532 0.03246 0.09996 i t 2 i t 3 i t ------- - ------- + ------- + e + 0.00711 e + 0.03168 e 3 i t 2 i t i t e e e 4 i t 5 i t 6 i t 7 i t - 0.01495 e + 0.0044 e - 0.00297 e + 0.00246 e , come within a pixel (modulo slight scaling and translation), showing that Moss's curvature discontinuites are imperceptible. Just the four terms > 1%, 0.09996 i t 3 i t 4 i t ------- + e + 0.03168 e - 0.01495 e , i t e draw an egg only slighly plump. We didn't beat n^-3 because the acceleration vector, though of continuous direction, has step function magnitude. To get the n^-arbitrary Gene described it seems we must come to a full, ultrasmooth (like exp(-1/t) ?) stop at each junction in order to match all derivatives.
E.g.g. Ouch! Eggregious. Oeuf with his head. --rwg STAGGERY EGGTRAYS *The original expression was O(n^18)/O(n^19) and ran on for pages.
I Remezed the epicyclic, b i t 3 i t 4 i t z = ---- + e + c e + d e i t e and polar i t 2 2 r = e ((cos(2 t) + b) + (c cos(t) + d) ) approximations to Moss's egg, minimaxing (very nearly) the signed distance measured (somewhat arbitrarily) along the ray from the center of its semicircular arc. This proved difficult and surprising, due to a chicken-and-egg conundrum that does not occur with polynomial and rational Remez: To get started, you need an approximate set of turning points good enough for the equal-ripple equations to have a real solution, which in turn must be good enough to have as many turning points as degrees of freedom. Then, when I finally got enough ripples, both iterations developed an *extra* ripple for which there is no parameter to control. Then you have to guess which ripple to leave unconstrained (disrupting the alternating sign pattern of your equations), and hope it winds up smaller than the ones you minimaxed. I only eyeballed the turning points instead of computing them, and I didn't try every possible omission pattern, but for the epicyclic, b = 0.17108, c = - 0.03844, d = - 0.0484, with six ripples of .02 and a seventh of .008. (Slightly < 2% of the egg minor axis.) (The other three degrees of freedom are translation, scale, and ripple height.) For the polar, b = 3.83214, c = 1.41969, d = 5.42912, (vs my original 3,1, and 5), with six ripples of 3% and a seventh of 2%. Now the surprise: There is a better (2.5%) minimax, without the seventh ripple!: b = 5.40849, c = 1.81059, d = 6.07139 . I would have bet against the possibility of multiple solutions, and strongly against the superiority of fewer ripples. But all three are aesthetically competitive with Moss's egg, using only three buggerfactors. --rwg MONASTERIAL ANAL-EROTISM AMELIORANTS
I blathered
and polar i t 2 2 r = e ((cos(2 t) + b) + (c cos(t) + d) )
No, that's z(t) = cis(t) r(t), which should've said 2 2 r = (cos(2 t) + b) + (c cos(t) + d) . I guess this means nobody tried to plot it for b = 0.17108, c = - 0.03844, d = - 0.0484, or b = 3.83214, c = 1.41969, d = 5.42912 . For the former, there's a shift and scale which intersects Moss's egg fourteen times, and twelve for the latter, yet the latter is a better fit. --rwg
I spazzed again!
I guess this means nobody tried to plot it for b = 0.17108, [...] or b = 3.83214, [...]
Sorry!! That's b = 3.83214, c = 1.41969, d = 5.42912, (and) b = 5.40849, c = 1.81059, d = 6.07139 .
For the former, there's a shift and scale which intersects Moss's egg fourteen times, (http://gosper.org/egg14.gif) and twelve for the latter, (http://gosper.org/egg12.gif) yet the latter is a better fit. --rwg
participants (2)
-
Marc LeBrun -
R. William Gosper