[Edit: earlier version was even more messy] I just watched a video of a jellyfish caught in a diver's 'air ring' - a torus blown for the sake of watching it rise. The jellyfish gets drawn into the torus and spun like a top as the bubble rises. I want to know how fast, in rpm, the jellyfish is spinning at a given distance from the surface. I think I need to know a formula for the circumference of the torus section at any given point as it rises , and a formula for the speed of rise.

My assumptions: - air torus is released at 10m, 2 atmospheres. - the exhaled volume at 10m is 500mls, so that'll double to 1000mls at sea-level - the jellyfish is isobuoyant, so is the same mass as the seawater it displaces - so can be treated as water. -the radial rotation of the torus is arbitrary (ie due to the mechanics of how the diver blew the bubble and is at a normal to the axial rotation), so can be ignored.


[Apologies if this is completely the wrong forum for these questions - in which case signposting to other places appreciated]

  • $\begingroup$ "torus" is a donut. "taurus" is a bull. This is a nice question, and I don't have any idea how to tackle it. Maybe there's a fluid dynamics whiz on SE who can help. $\endgroup$
    – S. McGrew
    Jul 7, 2019 at 3:15
  • $\begingroup$ Thanks - have edited, for ease of reading, not because I'm vein ;-) $\endgroup$
    – Laurence
    Jul 7, 2019 at 4:11
  • $\begingroup$ @S.McGrew What I don't understand is how the jellyfish is drawn in to the torus. How can there be a flow towards it with no outflow? Perhaps the fluid dynamics whiz could comment on that, or it could be another Question. $\endgroup$ Jul 7, 2019 at 4:12
  • $\begingroup$ Keith, I think it's a 'simple' pressure effect, the moving water creates a lower pressure, Bernoulli's principle, and so the jellyfish is drawn in - if my assumption about mass of jelly being close to seawater is correct, the jelly is showing us what is happening to all the water around the torus. $\endgroup$
    – Laurence
    Jul 7, 2019 at 4:23
  • $\begingroup$ " if my assumption about mass of jelly being close to seawater is correct, the jelly is showing us what is happening to all the water around the torus." You mean mass density, but your assumption and conclusion are both correct. @KeithMcClary 's question is interesting though: it appears that there is inflow to the torus without outflow. But I suspect that only an extended object like the jellyfish gets pulled to the long circular axis of the torus. $\endgroup$
    – S. McGrew
    Jul 7, 2019 at 5:14

1 Answer 1


Bernoulli's Principle provides a sufficient explanation. In the toroidal flow, water moves faster near the long circular axis. Water is not flowing toward the axis; but flows faster around the axis as you get closer. Bernoulli's Principle indicates that the pressure, then, decreases as the distance to the axis decreases. An extended body will experience greater pressure on the side away from the axis and lesser pressure on the side toward the axis, so will be pulled toward the axis. When the extended body is on the axis, the pressure is the same on all sides so it gets trapped there, spinning with the fluid flow.


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