4. Through the focus $F$ of the parabola $E: y^{2}=2 x$, two lines $l_{1}, l_{2}$ are drawn with the product of their slopes being $-\frac{1}{2}$. Line $l_{1}$ intersects $E$ at points $A, C$, and line $l_{2}$ intersects $E$ at points $B, D$. Then the minimum value of $|A C|+2|B D|$ is $\qquad$.
Let $l_{1}: x=2 n y+\frac{1}{2}, l_{2}: x=-\frac{1}{n} y+\frac{1}{2}$,
Solving $\left\{\begin{array}{l}x=2 n y+\frac{1}{2} \\ y^{2}=2 x\end{array} \Rightarrow y^{2}-4 n y-1=0 \Rightarrow|A C|=\sqrt{4 n^{2}+1} \cdot \sqrt{16 n^{2}+4} \right.$;
Solving $\left\{\begin{array}{l}x=-\frac{1}{n} y+\frac{1}{2} \\ y^{2}=2 x\end{array} \Rightarrow y^{2}+\frac{2}{n} y-1=0 \Rightarrow|B D|=\sqrt{\frac{1}{n^{2}}+1} \cdot \sqrt{\frac{4}{n^{2}}+4}\right.$.
Thus, $|A C|+2|B D|=2\left(4 n^{2}+1\right)+4\left(\frac{1}{n^{2}}+1\right)=8 n^{2}+\frac{4}{n^{2}}+6 \geqslant 8 \sqrt{2}+6$, equality holds when $n^{2}=\frac{\sqrt{2}}{2}$. Therefore, the minimum value of $|A C|+2|B D|$ is $8 \sqrt{2}+6$.