I've worked on this a little more, so I guess I'll take a stab at my own question. Here is our limacon again.
Notice what we begin with: a time parameter $t$ which runs in equal steps over the interval $[t_0,T]$, the cumulative distance $s$ traveled along the arc length at time $t$, and the Cartesian coordinates of the limacon $\gamma $ at time $t$. We set this much up in the question. What we want instead is an arc length parameter $s$ which runs in equal steps over the interval $[0,L]$, where $L$ is total arc length, together with the time $t(s)$ at $s$ and the Cartesian coordinates $\gamma (t(s))$.
Real quick, let's redo the cumulative distance $s$ traveled along the arc length with derivative (File Exchange link) so that we don't lose samples.
x_t = derivative(x);
y_t = derivative(y);
s = cumtrapz( sqrt(x_t.^2 + y_t.^2 ) );
We will be using the Matlab function interp1 in order to interpolate the timepoints $t(s_i)$ and coordinates $\gamma (t(s_i))$ which correspond to equally spaced steps $s_i$ along the arc length. Accordingly, we declare the arguments to interp1 with the variable names given in its documentation. The vector X is the original $s_i$, i.e. the distance traveled along the arc length at time $t_i$. The matrix V has column vectors whose rows are $t_i$ (our original time parameter in equal step size), and the coordinates $\gamma (t_i)$.
X = s.';
V = [ t.', x.', y.' ];
We need to interpolate to find the time points $t(s_i)$ corresponding to equally spaced arc length steps along the curve. These equally spaced steps run from $s_0=0$ to $s_N=L$, where $L$ is the total arc length.
L = s(end);
s0 = s(1);
N = length(s);
Xq = linspace(s0,L,N); % equally spaced indices
We can verify that s(N) == sum(diff(Xq)), as should be the case. We use the multidimensional piece-wise linear interpolation capability of Matlab's interpolation function interp1 to find interpolated values for time $t(s_i)$ and coordinates $\gamma (t(s_i))$ which correspond to the equally spaced steps of $s_i$ given in Xq.
Vq = interp1(X,V,Xq);
A call to diff(Xq) shows that we are moving in steps of constant arc length along $\gamma $, while a call to diff(Vq) shows that the time steps are no longer equal (first column). Thus, the result of our interpolation is:
- arc length $s_i$ in constant steps (
Xq)
- corresponding timepoints $t_i$, no longer in constant steps (first column of
Vq)
- coordinates $\gamma(t(s))$ as function of arc length (second and third columns of
Vq)
So the arc length parametrization of the limacon $\gamma (t(s))$ is expressed in Cartesian coordinates as xs and ys.
xs = Vq(:,2);
ys = Vq(:,3);
These vectors give the position of a particle moving along $\gamma $ as a function of $s$, where $s$ increments in steps of 0.1058. Here are the two parameterizations of the curve $\gamma $, namely $\gamma (t)$ in blue and $\gamma (t(s))$ in red.
I'll have to do some more work if I want to allow periodic solutions, since the current Matlab script does not in general. To see this, just look at the distance between the last and first sample of $\gamma (t(s))$ in red above. If we were to advance one more step of 0.1058 units along the limacon we would not land on a point we already visited, even though we know limacons are $2\pi$-periodic. I think this basically amounts to choosing a different step size for $s$ than 0.1058.
Best Answer
If we flip your parabola over and shift it to the left $$x' = x -\frac L2\\y' = H - y$$ your parabola becomes $$y' = \frac {4H}{L^2}x'^2$$. Since the transformation is rigid, we haven't changes the arclengths. Now the base of your parabola was at $(0,0)$ and $(L,0)$. These points have moved to $\left(\pm\frac L2, H\right)$. If we let $a = \frac{4H}{L^2}$ then we can use this calculation to obtain
$$s = \frac 1{4a}\left(2ax'\sqrt{4a^2x'^2 + 1} + \sinh^{-1} 2ax'\right)$$ for the arclength $s$ from the apex to the point $(x', y')$. Letting $x' = \frac L2$, we get the full arclength on one side to be $$S = \sqrt{H^2 + \frac{L^2}{16}} + \frac {L^2}{16H}\sinh^{-1}\frac{4H}L$$ Now you want to divide the full length of this parabola into $N$ parts. If $N$ is even, then set $d = \frac {2S}N$, and find the points that are a distance $kd$ away from the apex on the curve for $1 \le k < \frac N2$. If $N$ is odd, let $d = \frac SN$, and find the points that are a distance $(2k - 1)d$ from the apex on the curve, for $1 \le k <\frac N4$.
There is no nice closed form for the inverse of $s$, so you will need to use numerical methods to find the $x'$ values. Newton's Method should work fine, though the derivative is not nice.