Derive the balancing depth of a canal

Balancing depth is the depth of excavation (depth of the bed below ground level) at which the volume of earth cut from the canal equals the volume of earth needed to form the banks, so that no earth has to be borrowed or spoiled. It gives the most economical section.

Notation

$B$ = bed width, $d$ = depth of the bed below ground level (balancing depth), $n_1$ = side slope in cutting ($n_1$H:1V), $n_2$ = side slope of the banks, $b$ = top width of bank, $h$ = height of bank above ground level (for a bank-on-both-sides canal).

        b                      b
      |---|                  |---|
      ____                    ____
     /    \                  /    \    h
 ---/      \__    GL       __/      \---
              \            /   d
               \__________/
                    B

Derivation (per metre length of canal)

Cutting area (trapezoid of bottom width $B$, depth $d$, side slope $n_1$):

$$A_c = (B + n_1 d)\,d$$

Filling area of one bank (top width $b$, height $h$, slope $n_2$ on both faces):

$$A_f = h\,(b + n_2 h)$$

For two banks, $A_f = 2h\,(b + n_2 h)$.

For balancing, cutting = filling:

$$(B + n_1 d)\,d = 2h\,(b + n_2 h)$$ $$n_1 d^2 + B d - 2h(b+n_2 h) = 0$$ $$d = \frac{-B + \sqrt{B^2 + 8 n_1 h\,(b + n_2 h)}}{2 n_1}$$

When the bank height depends on d

For a canal of full supply depth $D$ and free board $F$, the top of the bank is at $D + F - d$ above ground level, so $h = D + F - d$. Substituting in the equation above gives a quadratic in $d$ whose smaller positive root is the balancing depth. If a service bank of top width $b_s$ is provided on one side, then fill $= h(b + n_2 h) + h(b_s + n_2 h)$.

If the actual depth is greater than $d$, earth is surplus (spoil banks are needed); if it is less, borrow pits are needed.