On 26 Feb., 20:56, "Bruno Luong" <b.lu...@fogale.findmycountry> wrote: > "Torsten" wrote in message <kghobs$5q...@newscl01ah.mathworks.com>... > > Given a point on the spiral > > (t0*cos(t0)*Width,t0*sin(t0)*Width), > > the distance squared to another point on the spiral is given by > > d^2=(t*cos(t)*Width-t0*cos(t0)*Width)^2 + (t*sin(t)*Width-t0*sin(t0)*Width)^2. > > This is a quadratic equation in t you can solve analytically. > > I don't think it's quadratic in t. > > It is plausible to imagine a circle centered about a fixed point that might intersect the spiral in as much as many considered turns (take a very large circle, where the arc looks like almost a radial line). An equation that has more than 2 solutions cannot be a quadratic polynomial equation. > > Bruno
You are right ; I only saw the quadratic term and didn't notice that the linear term still contains trigonometric expressions in t. So I think there is no other way than to solve the above nonlinear equation using MATLAB's "fzero" for t.