I posted this to stackexchange, and after some hours got a comment that was so pessimistic about finding some neat orderly solution, that I'm posting it here too. (In case anyone cares, this is related to this questionthis question, which I posted here earlier.)
I'd like $\alpha,\beta,\gamma$ as functions of $t$, satisfying the following conditions: $$ \begin{align} \alpha+\beta+\gamma & = 0 \\ \sin^2\alpha + \sin^2\beta + \sin^2\gamma & = c^2 \\ \left| \frac{d}{dt}(\sin\alpha,\sin\beta,\sin\gamma)\right| & = 1 \end{align} $$ I'm thinking of $c^2$ as small. At the very least that means $<2$, and intuitively it means $\ll 2$. Some geometry shows that there is a qualitative change in the nature of the solutions when $c^2$ goes from $<2$ to $>2$.
Later edit: The question above asks for a parametrization by arc-length. Here's an ugly parametrization by something quite remote from arc length: $$ \beta = \frac{\arccos\left(\frac{1 + \sin^2\alpha - c^2}{\cos\alpha}\right)-\alpha}{2} $$ And then $\gamma = \pi - \alpha - \beta$. In order to get the whole curve, you'd need a multiple-valued arccosine and then you'd pick the right value for the particular point on the curve. One thing that fails to be obvious to me just from the way the function above is written, giving $\beta$ as a function of $\alpha$, is that that function is its own inverse.
So here's a less demanding question that the one above: Is there some nice pleasant way of parametrizing the curve that, if not by arc-length, at least treats $\alpha$, $\beta$, and $\gamma$ equally, so that it's perfectly self-evident from the way it's written that the whole expression is symmetric in $\alpha,\beta,\gamma$?
