Recently, Richard Webb and myself gave a polynomial-time algorithm for computing the asymptotic translation length of a mapping class $$ \ell(h) = \lim_{n \to \infty} d(x, h^n(x)). $$ This appears as Algorithm 6 of the paper and relies on being able to compute geodesics in the curve complex. Assuming this, the key observation is that a midpoint $c'$ of a curve $c$ and its image under a large enough power of $h$ this lies close to the tight axis of $h$. Hence $d(c', h^n(c')) \approx n \ell(h)$. Therefore, since $\ell(h)$ is a rational number with bounded denominator, if we take $n$ large enough then $$ \ell(h) = \frac{1}{n} d(c', h^n(c')). $$

If you fix a finite generating set $\langle X \rangle = \textrm{Mod}(S)$ then the running time of this algorithm is a polynomial function of the word length of the mapping class $|h|_X$ since this is the running time of our algorithm for computing a geodesic from $c$ to $h(c)$ (Algorithm 4).

Recently, Richard Webb and myself gave a polynomial-time algorithm for computing the asymptotic translation length of a mapping class $$ \ell(h) = \lim_{n \to \infty} d(x, h^n(x)). $$ This appears as Algorithm 6 of the paper and relies on being able to compute geodesics in the curve complex. Assuming this, the key observation is that a midpoint $c'$ of a curve $c$ and its image under a large enough power of $h$ lies close to the tight axis of $h$. Hence $d(c', h^n(c')) \approx n \ell(h)$. Therefore, since $\ell(h)$ is a rational number with bounded denominator, if we take $n$ large enough then $$ \ell(h) = \frac{1}{n} d(c', h^n(c')). $$

If you fix a finite generating set $\langle X \rangle = \textrm{Mod}(S)$ then the running time of this algorithm is a polynomial function of the word length of the mapping class $|h|_X$ since this is the running time of our algorithm for computing a geodesic from $c$ to $h(c)$ (Algorithm 4).