The discharge through a Cipolletti weir is given by

[symbols have their usual meanings]

- 1
  
  $\frac{2}{3} C_{d} \sqrt{2 g} L H^{3 / 2}$
- 2
  
  $\frac{8}{15} C_{d} \sqrt{2 g} \tan \frac{θ}{2} H^{5}$
- 3
  
  $\frac{2}{3} C_{d} \sqrt{2 g} (L - 0.2 H) H^{3 / 2}$
- 4
  
  $\frac{2}{3} C_{d} \sqrt{2 g} L H^{5 / 2}$

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