Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Reading about storm surge, I found it fascinating that the gradual slope of the Gulf Coast of Florida resulted in a much higher storm surge but much lower energy release in breaking waves.

Is there a critical bathymetric profile to the ocean floor that would result in virtually all wave energy (of various frequencies) being converted into surge elevation rather than breakers? (I'm assuming ideal waves -- all parallel to each other and orthogonal to the profile. Also, to further simplify, assume that there is no change in total power with time -- so there is, presumably, virtually no back-flow.)

share|improve this question

1 Answer 1

At a glance, I'd guess that the flatter the approach to the shore, the longer the wave would go before it breaks. A breaking wave occurs, as I understand it, because the components of the wave don't move at the same speed. When a wave hits a coast, the forward and lower parts tend to move slower, while the back and top tends to move faster and pile up above the surface of the water. There's a broad circular motion to the overall wave and anything floating in it. As a result a wave's crest tends at a critical point to spill over the front resulting in the breaking of the wave. My thinking is that if the approach rises slowly enough, then the internal viscosity of the wave would have time to transfer energy among the various parts of the wave, both delaying onset of wave breaking and dissipating energy overall.

The latter would also help delay breaking of the wave because, again as I understand it, less energy means a shorter in height wave which in turn can go into shallower water before it threatens to break.

Now where would the energy go? I imagine some of it would go into heat and some into lifting water some amount.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.