My question is based around comparing the physics of a tornado against, what I imagine/assume to be the marine equivalent of a tornado, that is a whirlpool located either at sea, in a tidal region or in an estuary.


My assumptions are:

  • They both have the same basic cause, opposing streams of fluid, (air versus water) meet and an "inward" velocity vector arises, creating the inital spinning motion, which then, temporarily, becomes self perpetuating, dragging in more fluid and increasing the angular momentum of the system.

  • I assume, from video footage of tornados that I have seen, that tornados generally seem to be created at the height of the cloudbase, which can, it seems to me, range from 100 to 500 metres in vertical height.

  • I would assume that, due to the much greater density of water relative to air, a whirlpool could not descend very far, as the forces required to maintain it are far greater that it's atmospheric equivalent. However, I don't know enough about fluid dynamics to estimate how far down it could actually descend before it dissipates. My (pretty obvious, I admit) guess would be on the order of metres, i.e. far less than tornado height.

In other words, how good is my analogy of picturing a whirlpool as the marine equivalent of a atmospheric tornado and where does it break down?

  • 1
    $\begingroup$ From what I learned in cartoons, if you pull the big plug out of the bottom of the ocean, the whirlpool will extend all the way down $\endgroup$
    – Jim
    Aug 27, 2015 at 17:36
  • $\begingroup$ I would guess the right answer contains (in)compressibility. $\endgroup$
    – Bernhard
    Aug 27, 2015 at 17:51
  • $\begingroup$ Interesting question. Can water spouts create deep whirlpools? $\endgroup$
    – Alex
    Aug 27, 2015 at 19:24
  • 1
    $\begingroup$ People have survived going over Niagara Falls in a barrel. Given that, I won't say people couldn't survive a whirlpool in a barrel. Don't try this at home though $\endgroup$
    – Jim
    Aug 27, 2015 at 20:41
  • 1
    $\begingroup$ @Jim if I could fit a barrel in my bathtub I'd probably attempt it, just saying. $\endgroup$
    – Asher
    Aug 27, 2015 at 21:36

1 Answer 1


According to Heimholz second theorem it goes all the way. It even can't end in the fluid. You of course mean just the air, but it's merely a matter of how the situation is developing. If the flow conditions goes over Froude number 1 you will allways have the connection. It's explained here; Air core Vortex; Physical explanation of the "air Entrainment Hook" at $F_{co}=0.7$ -on my answer.

And according to experience gathered at hydropowerplants it really goes "all the way." Here is a nice study about the issue see ie. picture in page one.

enter image description here

Note that the pressure in intake is not the same, and "all the way" is the connection between two pressures.

  • $\begingroup$ Yes, it was only the honest structure of your question which motivated me to answer. There is a really nice video about the water analogy of the vortex, in youtube "13. Secondary Flow" in the channel of Barry Belmont, the point I mean is around 11 minutes from beginning, -if a I remember right. $\endgroup$
    – Jokela
    Oct 14, 2015 at 16:24

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