I have taught open-channel hydraulics at San Diego State University for nearly three decades. I use Ven T. Chow's well-respected textbook, complemented with my own material on unsteady flow.

One formula that particularly intrigued me was Eq. 15-1 in Chow's book. This equation states that the efficiency of the hydraulic jump, i.e., the ratio of the specific energy after the jump to that before the jump, is equal to:

E₂/E₁ = [(8F₁² + 1)^3/2 - 4 F₁² + 1] / [8F₁²(2 + F₁²)]

The proof of this formula is particularly challenging. In the early 1980s, having failed to solve the problem, I decided to pose the problem to the class, in the hope that someone would find the solution. Eventually one of our students accomplished the seemingly daunting task. Indeed, it was an intricate job of algebra, sure to test the skills and patience of most people.

A hydraulic jump downstream of Tinajones reservoir, Chiclayo, Peru.