Lacey's P = 4.75 sqrt(Q) derivation

Lacey's regime equations (basic equations) for a channel in regime:

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

where $V$ = mean velocity (m/s), $Q$ = discharge (m$^3$/s), $R$ = hydraulic mean radius (m), $f$ = silt factor ($f = 1.76\sqrt{d_{mm}}$), and the wetted perimeter $P$ in m.

Derivation of $P = 4.75\sqrt{Q}$

Step 1. Raise (1) to the 6th power, then take the square root:

$$V^6 = \frac{Q f^2}{140}\;\Rightarrow\; V^3 = f\sqrt{\frac{Q}{140}}\qquad (3)$$

Step 2. Continuity: $Q = A V$ and $A = P R$, so

$$Q = P\,R\,V$$

Step 3. Substitute $R$ from (2):

$$Q = P\left(\frac{5V^2}{2f}\right)V = \frac{5}{2f}\,P\,V^3$$

Step 4. Substitute $V^3$ from (3):

$$Q = \frac{5}{2f}\,P\,f\sqrt{\frac{Q}{140}} = \frac{5\,P\sqrt{Q}}{2\sqrt{140}}$$

Step 5. Solve for $P$ (note the silt factor $f$ cancels):

$$P = \frac{2\sqrt{140}}{5}\sqrt{Q} = 4.733\sqrt{Q}$$

Lacey rounded the coefficient to 4.75, so

$$\boxed{P = 4.75\sqrt{Q}}$$

Remarks: the wetted perimeter depends only on the discharge, not on the silt factor. Other Lacey relations follow: $A = Q/V$, $S = \dfrac{f^{5/3}}{3340\,Q^{1/6}}$.