Lacey's relation between R, Q and f

Lacey's regime relations for a channel are

$$V = \left(\frac{Qf^2}{140}\right)^{1/6}, \qquad R = \frac{5V^2}{2f}$$

where $V$ = mean velocity (m/s), $Q$ = discharge (m³/s), $f$ = silt factor and $R$ = hydraulic mean radius (m).

Derivation

From the second equation, $V^2 = \dfrac{2fR}{5}$. Squaring the first equation gives

$$V^2 = \left(\frac{Qf^2}{140}\right)^{1/3}$$

Equating:

$$\begin{aligned} \frac{2fR}{5} &= \left(\frac{Qf^2}{140}\right)^{1/3} \\ R &= \frac{5}{2f}\left(\frac{Qf^2}{140}\right)^{1/3} = \frac{5}{2}\cdot\frac{Q^{1/3}f^{2/3}}{140^{1/3}\,f} \\ &= \frac{2.5}{5.19}\left(\frac{Q}{f}\right)^{1/3} \end{aligned}$$ $$R \approx 0.48\left(\frac{Q}{f}\right)^{1/3}$$

(Many textbooks write the constant as 0.47 because the constants 140 and 5/2 in the original regime relations are rounded.)

Result: the regime hydraulic mean radius increases with the cube root of the discharge and decreases with the cube root of the silt factor. For example, $Q = 20$ m³/s and $f = 1.5$ give $R = 0.48\,(13.33)^{1/3} = 1.14$ m, which agrees with $R = 5V^2/2f$ computed directly.