8. As shown in Figure 4, there are five circles that are externally tangent to each other in sequence, and all are tangent to lines $a$ and $b$. If the diameters of the smallest and largest circles are 18 and 32, respectively, then the diameter of $\odot \mathrm{O}_{3}$ is $\qquad$
8.24.

As shown in Figure 7, let the radii of the five circles $\odot O_{1}, \odot O_{2}, \odot O_{3}, \odot O_{4}, \odot O_{5}$ be $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$, respectively. Draw perpendiculars from points $O_{1}, O_{2}, O_{3}$ to line $a$, with the feet of the perpendiculars being $A_{1}, A_{2}, A_{3}$, respectively. Connect $O_{1} O_{3}$. Clearly, $O_{2}$ lies on $O_{1} O_{3}$. Draw $O_{1} B_{1} \perp O_{2} A_{2}$ at point $B_{1}$, and $O_{2} B_{2} \perp O_{3} A_{3}$ at point $B_{2}$.
Then $\triangle O_{1} O_{2} B_{1} \backsim \triangle O_{2} O_{3} B_{2}$
$$
\begin{array}{l}
\Rightarrow \frac{O_{1} O_{2}}{O_{2} O_{3}}=\frac{O_{2} B_{1}}{O_{3} B_{2}} \\
\Rightarrow \frac{r_{1}+r_{2}}{r_{2}+r_{3}}=\frac{r_{2}-r_{1}}{r_{3}-r_{2}} \Rightarrow \frac{r_{2}}{r_{1}}=\frac{r_{3}}{r_{2}} .
\end{array}
$$

Similarly, $\frac{r_{3}}{r_{2}}=\frac{r_{4}}{r_{3}}, \frac{r_{4}}{r_{3}}=\frac{r_{5}}{r_{4}}$.
Let $\frac{r_{2}}{r_{1}}=k$. Then
$$
r_{2}=k r_{1}, r_{3}=k^{2} r_{1}, r_{4}=k^{3} r_{1}, r_{5}=k^{4} r_{1} \text {. }
$$

Since $r_{1}=\frac{18}{2}=9, r_{5}=\frac{32}{2}=16$, we have
$$
k^{2}=\frac{4}{3} \text {. }
$$

Thus, $r_{3}=12$.
Therefore, the diameter of $\odot O_{3}$ is 24.