Date: Jan 20, 2013 10:40 AM
Subject: Re: create polygon coordinates
"Roger Stafford" wrote in message <firstname.lastname@example.org>...
> "Jessica" wrote in message <email@example.com>...
> > Thanks for the tip! Do you mind explaining how I could test whether a coordinate falls within an ellipse using just the equation of an ellipse and not converting it to polygon coordinates?
> - - - - - - - - - -
> Just in case you decide to use the general expression
> A*x^2+B*x*y+C*y^2+D*x+E*y+F = 0
> for defining an ellipse, you can use the following code to generate n points along it. For B not equal to zero this is an ellipse whose major and minor axes are not aligned with the x and y axes.
> % Calculate various useful parameters
> x0 = (B*E-2*C*D)/(4*A*C-B^2);
> y0 = (B*D-2*A*E)/(4*A*C-B^2);
> f = (A*E^2-B*D*E+C*D^2)/(4*A*C-B^2)-F;
> d = sqrt((A-C)^2+B^2);
> a = sqrt(2*f/(A+C+d));
> b = sqrt(2*f/(A+C-d));
> t2 = 1/2*atan2(B,A-C);
> s = sin(t2); as = a*s; bs = b*s;
> c = cos(t2); ac = a*c; bc = b*c;
> % Generate the ellipse
> t = linspace(0,2*pi,n); % <-- You choose n
> x = x0 + ac*cos(t) - bs*sin(t);
> y = y0 + as*cos(t) + bc*sin(t);
> % Test it against the original expression and plot it
> z = A*x.^2+B*x.*y+C*y.^2+D*x+E*y+F;
> axis equal
> In order to ensure a valid ellipse with the above expression, the following inequalities for its coefficients have been assumed:
> A > 0,
> C > 0,
> 4*A*C > B^2, and
> A*E^2-B*D*E+C*D^2 > F*(4*A*C-B^2).
> The third of these makes it an ellipse, as opposed to a parabola or hyperbola. The fourth prevents it from being a degenerate single-point ellipse or non-existent. (Also if they are not true, some of the above square roots produce imaginary values.)
> Roger Stafford
Thank you very much for all of these suggestions!