Machine characterization of $SAC^i$
Yes. Stack heights.  $\mathsf{SAC^1} = \mathsf{NAuxPDA}(\log n, \log n)$, that is, with $O(\log n)$ space and $O(\log n)$ stack height; this implies $\log n$ configurations and therefore $\log^2(n)$ bits.  We have 

$$\mathsf{SAC^k} = \mathsf{NAuxPDA}(\log n, \log^k n);$$

these machines will run in time $2^{\log^{k}(n)}$.  Without restriction on stack height, we will get exactly $\mathsf{P}$.  The result should follow from: W. Ruzzo, Tree-size bounded alternation. JCSS 1980.