Five. (Full marks 20 points) There is a quantity $W$, after "modeling" the relationship is given by
$$
W=\frac{1}{c}\left(\frac{3 a}{\sqrt{1-u^{2}}}+\frac{b}{\sqrt{1-t^{2}}}\right),
$$

where $a, b, c, u, t$ are all positive, $u<1, t<1$, and satisfy $a t+b u=c, a^{2}+2 b c u=b^{2}+c^{2}$. Please design a method to find the minimum value of the quantity $W$.
Given $a^{2}=b^{2}+c^{2}-2 b c u$. All $a, b, c, u$ are positive, and $ub^{2}+c^{2}-2 b c=(b-c)^{2}$.

From this, we get $|b-c|<a<b+c$.
Therefore, the positive numbers $a, b, c$ can be the lengths of the three sides of a triangle. Let the vertices opposite these sides be $A, B, C$, respectively. By transforming the constraint, we get
$$
u=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\cos A.
$$

Substituting into the other constraint, we get
$$
a t+b \cos A=c.
$$

Using the projection theorem $c=a \cos B+b \cos A$, we get
$$
t=\cos B.
$$

Since $u$ and $t$ are positive, $\angle A$ and $\angle B$ are acute angles, so,
$$
\begin{array}{l}
\sqrt{1-u^{2}}=\sqrt{1-\cos ^{2} A}=\sin A, \\
\sqrt{1-t^{2}}=\sqrt{1-\cos ^{2} B}=\sin B.
\end{array}
$$

Let the circumradius of $\triangle ABC$ be $R$, then
$$
\begin{array}{l}
a=2 R \sin A, b=2 R \sin B, c=2 R \sin C. \\
\begin{aligned}
\therefore W & =\frac{1}{c}\left(\frac{3 a}{\sqrt{1-u^{2}}}+\frac{b}{\sqrt{1}-\overline{t^{2}}}\right) \\
& =\frac{1}{2 R \sin C}\left(\frac{6 R \sin A}{2}+\frac{2 R \sin B}{\sin B}\right) \\
& =\frac{8 R}{2 R \sin C}=\frac{4}{\sin C} \geqslant 4.
\end{aligned}
\end{array}
$$

The equality holds if and only if $\angle C$ is a right angle, i.e., $a^{2}+b^{2}=c^{2}$. Therefore, the minimum value of $W$ is 4.