Re: [math-fun] equilateral elliptical n-gons
After a deep discussion with Macsyma, M convinced me that this recursive procedure works, and does indeed produce points whose coordinates are the solutions of polynomials with coefficients in the rational field extended by a & b. Furthermore, these polynomials are Galois-solvable by towers of square roots. Computing these points is really no more difficult than computing the coordinates of the 2^n-th roots of unity, except that a & b keep entering into the computation. If a=b=1, then the computations are exactly the same as those for calculating the coordinates of the 2^n-th roots of unity. Basically, given 2 points already on the ellipse, compute the equation of the perpendicular bisector line through those two points. This line intersects the ellipse in two places, _both_ of whom become new additional points (by symmetry). The equation set can be produced by resultant(ellipseeqn,bisectoreqn,y) [a quadratic in x] for the two x coordinates, and resultant(ellipseeqn,bisectoreqn,x) [a quadratic in y] for the two y coordinates. A slightly different approach can produce the points w/o any ambiguity about which x coordinate goes with which y coordinate.
Date: Thu, 18 Nov 2010 07:29:26 -0800 From: Henry Baker <hbaker1@pipeline.com>
I guess one way to inscribe an equilateral n-gon in an ellipse of semimajor axis a and semiminor axis b would be to start with the 4 points
(a,0), (0,b), (-a,0), (0,-b)
forming a diamond inscribed in the ellipse, and then recursively constructing a little triangle on each of these diagonals which intersects the ellipse. In this way, we can construct an equilateral 2^n-gon within the ellipse. I assume that calculating the point of intersection of the perpendicular bisector of an existing side with the ellipse is the solution of a polynomial equation ??
At 06:47 AM 11/18/2010, Henry Baker wrote:
I was looking for an equi _lateral_ n-gon inscribed in an ellipse. It is more important that the (straight line segment) sides be equal, than that the vertices occur at equal curve segments around the edge. In an ellipse, it won't be possible to do both at the same time.
At 06:40 AM 11/18/2010, Michael Kleber wrote:
Wait, Henry, do you really mean you want "equal spacing along the perimeter of the ellipse"? I thought you wanted equal side lengths, and didn't care about the length of the ellipse arc they subtended.
--Michael
On Thu, Nov 18, 2010 at 9:30 AM, Henry Baker <hbaker1@pipeline.com> wrote:
I think that by using the term "regular" instead of "equilateral", I confused everyone.
I'm trying to find a simple way to generate n-vertex equilateral ellipses. It is particularly important for the application I have in mind that the vertices are at equal spacing along the perimeter of the ellipse, and most importantly, that the sequence closes after one trip around -- i.e., there can't be any gap.
The reason is that I want to use these little vectors as accelerations, and if they don't sum up exactly to zero, then the velocities & positions don't have any hope of closing up.
The resulting polygon won't be equilateral. It works for octagons, but the next stage will produce a 16-gon with alternating blocks of 4 equal edges, in the pattern aabbbbaaaabbbbaa. If you need a numerical answer, guess a side length, start at some point on the ellipse, calculate the next intersection, repeat. After N sides, calculate the under/overshoot, add 1/Nth to your guessed side, repeat. If you want an algebraic number answer, let your side^2 be X; calculate the next intersection point -- likely doubling the degree? -- N/2 times in each direction around the ellipse, and equate the final points. This hints that the polynomial degree could be pretty big, even for N=10. When N is even or a multiple of 4, you could start from both ends of the ellipse, or the four "corners", to get the degree down. Rich ------- Quoting Henry Baker <hbaker1@pipeline.com>:
After a deep discussion with Macsyma, M convinced me that this recursive procedure works, and does indeed produce points whose coordinates are the solutions of polynomials with coefficients in the rational field extended by a & b. Furthermore, these polynomials are Galois-solvable by towers of square roots.
Computing these points is really no more difficult than computing the coordinates of the 2^n-th roots of unity, except that a & b keep entering into the computation. If a=b=1, then the computations are exactly the same as those for calculating the coordinates of the 2^n-th roots of unity.
Basically, given 2 points already on the ellipse, compute the equation of the perpendicular bisector line through those two points. This line intersects the ellipse in two places, _both_ of whom become new additional points (by symmetry). The equation set can be produced by resultant(ellipseeqn,bisectoreqn,y) [a quadratic in x] for the two x coordinates, and resultant(ellipseeqn,bisectoreqn,x) [a quadratic in y] for the two y coordinates. A slightly different approach can produce the points w/o any ambiguity about which x coordinate goes with which y coordinate.
Date: Thu, 18 Nov 2010 07:29:26 -0800 From: Henry Baker <hbaker1@pipeline.com>
I guess one way to inscribe an equilateral n-gon in an ellipse of semimajor axis a and semiminor axis b would be to start with the 4 points
(a,0), (0,b), (-a,0), (0,-b)
forming a diamond inscribed in the ellipse, and then recursively constructing a little triangle on each of these diagonals which intersects the ellipse. In this way, we can construct an equilateral 2^n-gon within the ellipse. I assume that calculating the point of intersection of the perpendicular bisector of an existing side with the ellipse is the solution of a polynomial equation ??
At 06:47 AM 11/18/2010, Henry Baker wrote:
I was looking for an equi _lateral_ n-gon inscribed in an ellipse. It is more important that the (straight line segment) sides be equal, than that the vertices occur at equal curve segments around the edge. In an ellipse, it won't be possible to do both at the same time.
At 06:40 AM 11/18/2010, Michael Kleber wrote:
Wait, Henry, do you really mean you want "equal spacing along the perimeter of the ellipse"? I thought you wanted equal side lengths, and didn't care about the length of the ellipse arc they subtended.
--Michael
On Thu, Nov 18, 2010 at 9:30 AM, Henry Baker <hbaker1@pipeline.com> wrote:
I think that by using the term "regular" instead of "equilateral", I confused everyone.
I'm trying to find a simple way to generate n-vertex equilateral ellipses. It is particularly important for the application I have in mind that the vertices are at equal spacing along the perimeter of the ellipse, and most importantly, that the sequence closes after one trip around -- i.e., there can't be any gap.
The reason is that I want to use these little vectors as accelerations, and if they don't sum up exactly to zero, then the velocities & positions don't have any hope of closing up.
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