8. In the tetrahedron $S-ABC$, $SA=4, SB \geqslant 7, SC \geqslant 9, AB=5, BC \leqslant 6, AC \leqslant 8$, then the maximum volume of the tetrahedron is . $\qquad$
As shown in the figure, $\cos \angle S A B \leqslant \frac{4^{2}+5^{2}-7^{2}}{2 \cdot 4 \cdot 5}=-\frac{1}{5}$, then $\angle S A B \geqslant \pi-\arccos \frac{1}{5} \Rightarrow \sin \angle S A B \leqslant \frac{2 \sqrt{6}}{5}$ $\Rightarrow S_{\triangle S A B} \leqslant \frac{1}{2} \cdot 4 \cdot 5 \cdot \frac{2 \sqrt{6}}{5}=4 \sqrt{6}$,
and the height of the tetrahedron is no greater than the side length, i.e., $h \leqslant B C \leqslant 6$, at this time all other side lengths also meet the requirements.
Therefore, $V_{S-A B C}=\frac{1}{3} S_{\triangle S A B} \cdot h \leqslant 8 \sqrt{6}$.