Thank you for your patience Robin. Here's one more try at a complete answer.
At 9:46 PM 3/31/5, Robin van Spaandonk wrote: >In reply to Horace Heffner's message of Sun, 13 Mar 2005 17:53:41 >-0900: >Hi Horace, > >I'm having some trouble understanding this formula. If it's meant >to give the relationship between the absolute height of the water >surface at any radius, then it seems to say that at w=0, h= h0, >i.e. h0 is the height of stationary water in the tank. The variable h0 is merely a constant that allows positioning the parabolic surface cross section at the right elevation. > >>Correction follows. Sorry! >> >>The shape of the final equilibrium surface is: >> >> h = (w^2/2g) x R^2 + h0 > >However when the water rotates, a dip forms at the middle, which >can drop right down to the floor of the tank at sufficiently high >w. However, according to the formula, for any w > 0, h > h0 for >all R, since the first term is always positive. The h0 above is negative. >Therefore, the >formula either doesn't represent what I thought it was meant to >represent, or it doesn't describe reality. The confusion here is my fault, partly due to the fact the value of h0 is negative. It is given by: h0 = - 2g/( w^2 x (R1)^2) where R1 is the radius of the drain hole. The surface given by rotating: h = (w^2/2g) x R^2 + h0 about the tank axis is only meant to represent a *final* state (or initial state) surface where no water is falling at all and the rotational velocity at every radius is a fixed constant w. This really only applies to the case where the water is in a tank which rotates. When considering such a tank viscosity is not important except maybe for considerations of how the vortex *reaches* this final state wherein the angular velocity is constant at every radius. Viscosity is not an issue when defining the initial state of a rotating tank assuming the angular velocity is desired to be constant at every radius in the initial state. Note that the above surface is only meaningful when h > 0. There is no water in the tank above radii where h <= 0. The hight of the water at the perifery of the tank where R = R2, is given by: h = (w^2/2g) x R^2 - 2g/( w^2 x (R1)^2) I provided this formula because it shows in a precise way why water may not go down the drain at all, and how a final state can arise in which some or all the water might not be able to go down the drain. The case for an "active" vortex tank that is well in progress of emptying, where the tangential speed is proportional to 1/r, and the vertical speed to 1/r as well, is shown in Feynman's *Lectures on Physics*, Vol II, 40-10 ff. Fig. 40-12 is a great drawing of such a tank, showing the surface contour of a sample vortex. In this case the water surface is a rotation of: h = k/R^2 + h0 which I noted earlier (different H0 though.) I've undoubtedly confused the situation by mixing assumptions regarding viscosities and angular velocity distributions, and looking at initial, intermediate, and final states separately. I thought I was making the concept simpler to follow by these assumptions. I see no way to make analysis of the intermediate states simple. The initial and final states are fairly easy to analyze though, assuming in those states angular velociy is constant at every radius. Regardless the function used to represent angular velocity, say w = F(R), a *final* surface results which has a perimeter height and a radius at h=0 that is equal to the hole radius. Given a function w = G(R) that represents the initial angular velocity as a function of R, there is a corresponding surface. If the radius of that surface at h=0 is larger than the drain radius then no water can go down the hole at all, and F(R) = G(R). You expressed concern about how it is water can not go down the drain when there is insufficient energy to obtain the assumed vortex inner angular momentum. This discussion hopefully helps understand why. When there is insufficient potetnial energy, the surface contour does not extend within the drain radius. If there is not sufficient energy from gravity (mgh) to make the water go down the drain, it can not go down the drain (assuming a rotating tank or zero viscosity.) If the water is viscous, however, and the tank does not spin too, the viscosity gradually reduces the angular velocity of water remaining in the tank and thus it all can go down the drain. However, the assumption that rotational energy plus heat increases by more than the potential energy being expended is then invalid. There is therefore no reason to think free energy is available. Good grief. I hope I got that all right for a change! 8^) I'm very busy with tax reporting so all the vortex activity of late came at a bad time for me. Regards, Horace Heffner
