Suppose $T$ is superstable. Let $\leq$ denote the fundamental order on complete types, $\lbrack p\rbrack$ the class of the bound of $p$, and $U(--)$ Lascar's foundation rank (see [LP]). We prove THEOREM 1. If $q < p$ and there is no $r$ such that $q < r < p$, then $U(q) + 1 = U(p)$. THEOREM 2. Suppose $U(p) < \omega$ and $\xi_1 < \cdots < \xi_\kappa$ is a maximal descending chain in the fundamental order with $\xi_\kappa = \lbrack p\rbrack$. Then $k = U(p)$. That the finiteness of $U(p)$ in Theorem 2 is necessary follows from THEOREM 3. There is an $\omega$-stable theory with a type $p \in S_1 (\phi)$ such that (1) $U(p) = \omega + 1$, and (2) there is a maximal descending chain of proper extensions of $\lbrack p\rbrack$ which has order type $\omega$.