Prove R = 1.35 (q^2/f)^(1/3)

For a very wide channel the wetted perimeter is almost equal to the width. Since $A = PR$, the discharge per unit width is $q = Q/P = VR$ (the regime flow per metre width).

Lacey's regime equation connecting $R$, $V$ and $f$:

$$R = \frac{5V^2}{2f} \quad\Rightarrow\quad V = \sqrt{\frac{2fR}{5}}$$

Proof

$$\begin{aligned} q &= V R = \sqrt{\frac{2fR}{5}}\;R = \sqrt{\frac{2f}{5}}\;R^{3/2} \\ R^{3/2} &= q\sqrt{\frac{5}{2f}} \\ R^{3} &= \frac{5}{2}\cdot\frac{q^2}{f} \\ R &= \left(\frac{5}{2}\right)^{1/3}\left(\frac{q^2}{f}\right)^{1/3} = 1.357\left(\frac{q^2}{f}\right)^{1/3} \end{aligned}$$ $$R \approx 1.35\left(\frac{q^2}{f}\right)^{1/3}$$

where $R$ is the normal scour depth (m), $q$ the discharge per metre width (m³/s/m) and $f$ the silt factor. This is Lacey's scour depth formula. It is used to fix the depth of cutoff walls, guide bunds and aprons, with $R$ multiplied by a factor for the maximum scour (about 1.27 for a straight reach, up to 2-2.75 at the nose of guide bunds).