# Maximum speed of a toroidal vortex?

Consider a toroidal vortex, such as the iconic smoke ring or bubble rings that dolphins make at play.

What is the maximum speed of such a ring? Since they are a fluid flow phenomenon it seems that the speed of sound comes into play in determining an absolute hard limit. But, most rings demonstrated are rather slow in cruising (the translational movement of the entire ring structure), but that could be an artifact of how they are produced. Is there a fixed relationship between the (maximum) cruising speed and the toriodal speed of the particles making up the ring?

Also, is this partially a wave phenomena in which the disturbance propagates, but the particles making up the ring may exchange with the medium its passing through? In demonstrations involving a tracer substance being caught up within the ring, it's unclear, and the substantially heavier smoke particles may behave differently anyway. But in the case where the ring is made up of a different material than the surrounding media, is there a difference in the maximum cruising speed of a ring that delivers the specific particles vs a ring disturbance that leaves those original particles along the route but still delivers a vortex?

• Motion of a vortex is not a wave phenomenon. In an inviscid fluid (subject to some additional conditions) vortex lines move with the fluid (see Kelvin's circulation theorem: whoi.edu/fileserver.do?id=9349&pt=2&p=12248). Viscosity acts to diffuse vorticity, just like it diffuses momentum. Searching for "speed of vortex ring" throws up theoretical work that give (complicated) expressions for translation speed of vortex rings of small cross-section in a viscous incompressible fluid. I have not gone through them myself.
– Deep
Jun 12, 2017 at 5:06

Also, is this partially a wave phenomena in which the disturbance propagates, but the particles making up the ring may exchange with the medium its passing through? ... But in the case where the ring is made up of a different material than the surrounding media, is there a difference in the maximum cruising speed of a ring that delivers the specific particles vs a ring disturbance that leaves those original particles along the route but still delivers a vortex?

The "ring" to which you elude is a vortex resulting from a driven Kelvin-Helmholtz instability or KHI.

The linear speed of any point on the vortex is dependent upon the size of the vortex, the density of the media, the speed of the shear flow, and the viscosity of the fluid.

For instance, suppose we have an interface of two fluids (subscripts $1$ and $2$ to discriminate) flowing past each other with slightly different densities, $\rho_{j}$, and different speeds, $U_{j}$, where $j$ = 1 or 2 and we assume $\rho_{1} \geq \rho_{2}$ and $U_{1} > U_{2}$. If we include surface tension, $\sigma$, then the angular frequency, $\omega$, of a disturbance resulting from this shear flow is given by: $$\omega\left( k \right) = \frac{ \rho_{1} \ U_{1} + \rho_{2} \ U_{2} }{ \rho_{1} + \rho_{2} } \ k \pm \sqrt{ - \frac{ \rho_{1} \ \rho_{2} \ \left( U_{1} - U_{2} \right)^{2} }{ \left( \rho_{1} + \rho_{2} \right)^{2} } \ k^{2} + \frac{ \rho_{1} - \rho_{2} }{ \rho_{1} + \rho_{2} } \ g \ k + \frac{ \sigma }{ \rho_{1} + \rho_{2} } \ k^{3} } \tag{1}$$ where $g$ is the acceleration of gravity and $k$ is the wavenumber of the resulting mode.

What is the maximum speed of such a ring?

This depends upon the viscosity and compressibility of the fluids and what you mean by "speed of such a ring" (i.e., are you asking about the linear speed at the outer edge or the rate of rotation).

From Equation 1 one can estimate a linear speed using $V_{t} \simeq \omega \ r$, where $r$ is the radial distance from the center of the vortex. Since $\omega$ is dependent upon $k$ and the inverse of the wavenumber is a spatial scale length, i.e., $k^{-1} \propto L$, then you can make a very rough approximation that $V_{t} \sim \omega \ k^{-1}$.

Note that $\omega$ here is also loosely related to vorticity, $\nabla \times \mathbf{U}$, which actually varies as: $$\nabla \times \mathbf{U} \sim 2 \ k \ U \ A \ \cos{\left[ k \left( x - \frac{ \lambda }{ 4 } \right) \right]} \ e^{k \ U \ t} \tag{2}$$ where $\lambda$ is the wavelength, $t$ is time, $x$ is some spatial coordinate, and $A$ is some amplitude constant. We have also assumed, for simplicity, $\rho_{1} = \rho_{2} \equiv \rho$, $U_{1} = - U_{2} \equiv U > 0$ in Equation 2 (technically this is the real part of the vorticity, which also has an imaginary part corresponding to growth/decay).

Dividing both sides of Equation 2 yields an approximate relationship between linear speed of the outer edge of each vortex and the properties of the system, or $V_{t} \propto F\left( k, U, A, t \right)$.

Ultimately, the maximum speed will be limited by driving force (i.e., roughly $A$), speed of resulting shear flow (i.e., $U$), and viscosity of the fluid. The viscosity has not been explicitly included because it's usually small until gradient scale lengths become comparable to the mean free path for compressible fluids like the Earth's atmosphere at STP (though there are some similarities or weak relationships through surface tension, e.g., see discussions at Viscosity and surface tension). However, if you "push" a fluid hard enough viscosity will start to matter and will become one of the greatest limiting factors to the maximum speed. For instance, in collisional media like Earth's atmosphere viscosity is the primary dissipation mechanism that damps shock waves. However, it only becomes important during and after shock initiation when the gradient scale lengths become comparable to or smaller than the mean free path.

• “it depends…” so does that mean you could throw a vortex ring at supersonic speed (through normal air)? Jul 24, 2017 at 4:28
• @JDługosz - I am not sure I follow. Are you asking whether the guiding center of the vortex (i.e., the "eye") can move at supersonic speeds? The short answer is kind of, but not because we "throw" one. The turbulent eddies behind supersonic aircraft would satisfy this in some cases but for the most part you cannot "throw" a vortex like a frisbee. The speeds to which I referred in my answer were the linear speeds at the edge of the vortex (i.e., the "spinning" speeds). Jul 24, 2017 at 13:29
• I think OP means throwing ring vortex from a vortex cannon, like this: youtube.com/watch?v=GHiTDsFTFQQ May 25, 2018 at 10:21