[math-fun] Rational Parametric Polys vs Conic Sections vs (Un)?Solvable Quintics
AIUI, in order to form conic setion curves using rational parametric polynomials such that the (Cartesian) coordinates of the loci, the radii and the coefficients are all in Q, the polymonials have to be at least quintic. Otherwise, at least one of the loci, radii or coefficients are radicals. Presuming that is true, is there any relation between that and the fact that quintics are the lowest degree polys which are not always solvable in terms of radicals (given coefficients in Q)? -JimC -- James Cloos <cloos@jhcloos.com> OpenPGP: 1024D/ED7DAEA6
I didn't understand this question at all, I'm afraid. Where you wrote "loci", I think you probably meant "foci". The "radii" presumably refers to major and minor radius of an ellipse. The coefficients are presumably those of the parameters in the coordinate functions. A circle is rationally parameterised by [w,x,y] = [1+t^2, 1-t^2, 2t], the Cartesian coordinates of the conic then being [x/w, y/w]. Why should this not disprove your assertion? A reference to the relevant background might be of some assistance! Fred Lunnon On 8/5/11, James Cloos <cloos@jhcloos.com> wrote:
AIUI, in order to form conic setion curves using rational parametric polynomials such that the (Cartesian) coordinates of the loci, the radii and the coefficients are all in Q, the polymonials have to be at least quintic. Otherwise, at least one of the loci, radii or coefficients are radicals.
Presuming that is true, is there any relation between that and the fact that quintics are the lowest degree polys which are not always solvable in terms of radicals (given coefficients in Q)?
-JimC -- James Cloos <cloos@jhcloos.com> OpenPGP: 1024D/ED7DAEA6
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"Fl" == Fred lunnon <fred.lunnon@gmail.com> writes:
Fl> I didn't understand this question at all, I'm afraid. Fl> Where you wrote "loci", I think you probably meant "foci". Yes. An uncaught typo. Fl> The "radii" presumably refers to major and minor radius of an ellipse. Of course. And the logical equivs for parabolae and hyperbolae. Fl> The coefficients are presumably those of the parameters in the Fl> coordinate functions. As I explain below I was thinking of Bernstein basis polys, but I Fl> A circle is rationally parameterised by [w,x,y] = [1+t^2, 1-t^2, 2t], Fl> the Cartesian coordinates of the conic then being [x/w, y/w]. Fl> Why should this not disprove your assertion? Fl> A reference to the relevant background might be of some assistance! So perhaps then it only holds for bezier and b-splines? Or, since cubics are the norm for graphics, I may have just forgotten to consider quadradics.... :-/ The standard cubic NURB-Spline for a circle with radius=1 centered at [0,0] has four arcs with (normalized) knot vector: [ 0 0 0 1/4 1/4 1/2 1/2 3/4 3/4 1 1 1 ] and coefficient matrix: x = [ 1 sqrt(1/2) 0 -sqrt(1/2) -1 -sqrt(1/2) -0 sqrt(1/2) 1 ] y = [ 0 sqrt(1/2) 1 sqrt(1/2) 0 -sqrt(1/2) -1 -sqrt(1/2) 0 ] w = [ 1 sqrt(1/2) 1 sqrt(1/2) 1 sqrt(1/2) 1 sqrt(1/2) 1 ] for a circle with radius=1 centered at [0,0]. Whereas that same circle can be drawn with a single arc NURB-Spline using quintics with knot vector [ 0 0 0 0 0 0 1 1 1 1 1 1 ] and coefficient matrix: x = [ 0 4 2 -2 -4 0 ] y = [ -5 -1 3 3 -1 -5 ] w = [ 5 1 1 1 1 5 ] I don't beleive that one can reduce that to a quartic and retain the lack of radicals. -JimC -- James Cloos <cloos@jhcloos.com> OpenPGP: 1024D/ED7DAEA6
Well then, to answer your original question, I have not the slightest doubt that there's no connection with quintic solvability whatsoever! [But of course, I could be wrong ...] Exercises like this are excellent training, but it's worth bearing in mind that generally single patches which depart a long way from linearity are rarely satisfactory in practice; also high degree polynomials suffer computational costs in both time and accuracy. Have you tried looking at triangular patching (for surfaces in 3-D)? It's never caught on, and I think one reason is that few people can actually understand it. Its great advantage is that the degree does not multiply up in the same way as the Cartesian product used for NURBS surfaces does. One reference is @incollection{[Sei92], author = {Seidel,~H-P.}, title = {Polar Forms and Triangular {B}-Spline Surfaces}, editor = {D-Z.~Du and F.~Hwang}, booktitle = {Computing in Euclidean Geometry}, publisher = {World Scientific Publishing Co. Pte. Ltd.}, year = 1992, } WFL On 8/6/11, James Cloos <cloos@jhcloos.com> wrote:
"Fl" == Fred lunnon <fred.lunnon@gmail.com> writes:
Fl> I didn't understand this question at all, I'm afraid. Fl> Where you wrote "loci", I think you probably meant "foci".
Yes. An uncaught typo.
Fl> The "radii" presumably refers to major and minor radius of an ellipse.
Of course. And the logical equivs for parabolae and hyperbolae.
Fl> The coefficients are presumably those of the parameters in the Fl> coordinate functions.
As I explain below I was thinking of Bernstein basis polys, but I
Fl> A circle is rationally parameterised by [w,x,y] = [1+t^2, 1-t^2, 2t], Fl> the Cartesian coordinates of the conic then being [x/w, y/w].
Fl> Why should this not disprove your assertion?
Fl> A reference to the relevant background might be of some assistance!
So perhaps then it only holds for bezier and b-splines?
Or, since cubics are the norm for graphics, I may have just forgotten to consider quadradics.... :-/
The standard cubic NURB-Spline for a circle with radius=1 centered at [0,0] has four arcs with (normalized) knot vector:
[ 0 0 0 1/4 1/4 1/2 1/2 3/4 3/4 1 1 1 ]
and coefficient matrix:
x = [ 1 sqrt(1/2) 0 -sqrt(1/2) -1 -sqrt(1/2) -0 sqrt(1/2) 1 ] y = [ 0 sqrt(1/2) 1 sqrt(1/2) 0 -sqrt(1/2) -1 -sqrt(1/2) 0 ] w = [ 1 sqrt(1/2) 1 sqrt(1/2) 1 sqrt(1/2) 1 sqrt(1/2) 1 ]
for a circle with radius=1 centered at [0,0].
Whereas that same circle can be drawn with a single arc NURB-Spline using quintics with knot vector [ 0 0 0 0 0 0 1 1 1 1 1 1 ] and coefficient matrix:
x = [ 0 4 2 -2 -4 0 ] y = [ -5 -1 3 3 -1 -5 ] w = [ 5 1 1 1 1 5 ]
I don't beleive that one can reduce that to a quartic and retain the lack of radicals.
-JimC -- James Cloos <cloos@jhcloos.com> OpenPGP: 1024D/ED7DAEA6
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