Given a 2D-curve arc $\mathscr{C}$, I would like to be able to easily compute a subset of $n$ points belonging to $\mathscr{C}$, so that the points are separated by equal-length curve arcs.
For that purpose, I'm trying to find a convenient parametric representation of the curve $f : t \mapsto (x(t),y(t))$ (for my purposes, $t \in [0,1])$ such that $\forall t, t'>t$, $\exists v$ such that the length of the arc between $f(t)$ and $f(t')$ is $k(t'-t)$. In other words, if $t$ is time, $f$ describes the trajectory of a material point moving at constant speed $v$ along $\mathscr{C}$.
Thus, my initial problem is solved by the values of $f$ for equally-spaced values of $t$ in [0,1], e.g. if $n = 3$, $\{ f(0), f(\frac{1}{2}), f(1) \}$ is the solution.
In the case of segment [AB], such a parametric curve is easy to find: $f:t \mapsto A + t(\vec{AB})$ works (with $v = AB$). Circle arcs are also pretty nice to me: $t \mapsto r(cos(t), sin(t))$ is such a parametric curve for origin-centered $r$-radius circle, with $k = r$.
I am well-aware that the length of the arc between points $f(t_0)$ and $f(t_1)$ is $ L = \displaystyle \int_{t=t_{0}}^{t_{1}} \sqrt{\displaystyle \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt$, and my definition of a convenient parametric function must imply that $\sqrt{\displaystyle \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$ is constant and equals $v$.
I'm happy to be able to solve my problem well for circle arcs and segments, but I want more. Segments and circles are pretty simple, I would like to find other classes of curves for which the explicit form of their convenient parametric representation can be known... I've tried setting $x(t)$ to some polynomial function of $t$ and computing the corresponding $y(t)$, turns out there's no such y(t) when x(t)'s degree is stricly more that one... Any suggestions ?
