In reply to Horace Heffner's message of Thu, 31 Mar 2005 23:33:55 -0900: Hi Horace,
Thanks. I have now derived the formula for myself, so I understand where it comes from, and what the various constants mean. I have also applied the same derivation principle to an active vortex that it constantly being "topped up" to maintain a constant level. The result for a vortex with no initial angular velocity can be found at http://users.bigpond.net.au/rvanspaa/vortex-shape.mcd and http://users.bigpond.net.au/rvanspaa/vortex-shape.gif I'm still thinking about how to correctly introduce an initial angular momentum. I may try it with a fixed angular velocity at the rim, and see what happens. (This is what one would get with tangential addition of water as in one of your previous drawings). BTW the initial restriction imposed on the angular velocity by the radius needing to be less than the drain radius doesn't appear to be serious. IOW even an initial angular velocity that produces just a slight dip in the surface would already be sufficient to yield OU according to the first document I posted. However, I'm now having second thoughts about the validity of that first document (http://users.bigpond.net.au/rvanspaa/vortex.gif). >At 4:55 PM 4/1/5, Robin van Spaandonk wrote: > >>In short, is h the distance up from the bottom of the tank, or the >>distance down from the surface? > > >The variable h is the distance up from the bottom of the tank in the >equations I provided. However, I should note that the equation from >Feynman's *Lectures on Physics*, Vol II, 40-10 ff, namely: > > h = k/R^2 + h0 > >takes h in the downward direction. However, Feynman's equation and >derivtion are well defined in the reference. > >Regards, > >Horace Heffner > Regards, Robin van Spaandonk All SPAM goes in the trash unread.
