Re: vortex mystery
In reply to Horace Heffner's message of Tue, 5 Apr 2005 16:46:12 -0800: Hi, [snip] 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 Uh, if there is no initial angular velocity the water merely runs down the drain. Not sure what you mean here. What I mean is that the body of water as a whole has no initial angular velocity, though a small portion adjacent to the drain may have. IOW just enough to get the vortex kicked off, so that it can grow from the centre out. Ok, I have now resolved the whole thing. The complete model with freely chosen tank rim parameters (including injection velocity), can now be found at http://users.bigpond.net.au/rvanspaa/vortex-shape+.mcd with gif at http://users.bigpond.net.au/rvanspaa/v-shape.gif Perhaps needless to say, we missed out on that free lunch again! :) Regards, Robin van Spaandonk All SPAM goes in the trash unread.
Re: vortex mystery
At 4:52 PM 4/6/5, Robin van Spaandonk wrote: Perhaps needless to say, we missed out on that free lunch again! :) Nuts! I had no other plans. Regards, Horace Heffner
Re: vortex mystery
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.
Re: Vortex mystery
The interesting series of posts regarding this subject is fascinating. Anyone sitting in the middle of a tornado or hurricane can testify that the forces generated are awesome and certainly didn't come from the effect of gravity of falling water. A water vortex performs an interesting " reverse" flow similar to a hurricane. The evidence of this reverse flowing condition is from the micro "twisters" spun off from the main vortex. The twisters are described as microbursts by the weathermen and account for the strange and specific damage during a hurricane. The twisters produce a short durationwhistling sound easily remembered by anyone that has spent hours located in the path of a hurricane as it passes. We can produce many variations of a vortex in our glass test tanks, The short duration vortexes spun off from the center vortex can be vertical, horizontal or diagonal and visible in the water tank because of entrained air. As for the mathematics of a vortex, my old professor used to state.. one can perform wonders with numbers while eating cucumbers. Richard Blank Bkgrd.gif
Re: vortex mystery
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
Re: vortex mystery
Greetings to all members Horace Heffner's message of Thu, 31 Mar 2005 13:43:54 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. A non-physcist 's visualization of solitonic vortices is at URL: http://lewfh.tripod.com/themindthingthegiftofvisualization/ With regards Lew Horace Heffner wrote: At 4:55 PM 4/1/5, Robin van Spaandonk wrote: In reply to Horace Heffner's message of Thu, 31 Mar 2005 13:43:54 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. If h = (w^2/2g) x R^2 + h0 and h0 is negative, then for w=0, h=h0 and is thus also negative. How does one end up with a negative height? As I stated in the last post, the above surface is only meaningful when h 0. There is no water in the tank above [at] radii where h = 0. If w=0 then h=0 everywhere because no water will stay in the tank. No angular momentum is involved. Any water in the tank is not in equilibrium as assumed - it will all run out. Please note again that the coreolis force is ignored throughout. Or should the original formula perhaps be: h = h0 - (w^2/2g) x R^2 ? (Since the second term in this version is positive, the height becomes less for higher w and also for smaller R, both of which make sense). 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. When h=0 or h=0 then there is no water above the radius at which h is computed. The variable h at final equilibrium is a function of R, R1, and w, where R1 is the radius of the drain hole. The initial or final surface, assumed to be in equilibrium with w constant at every radius, is concave upwards. The coefficient of R^2 is thus positive. In the initial condition, h0 can be anything depending on how much water is in the rotating tank. This h0 does not affect the *curvature* of the surface, however, which is only a function of w, g, and R, assuming the drain hole is plugged, and w is constant over all radii. The variable h0 changes as the water drains from the tank. The equation describing the water surface changes as well, but the final surface should return to the form h = (w^2/2g) x R^2 + h0 for a rotating tank, assuming that viscosity forces w to be uniform across all radii, where in the *final* equilibrium: h = (w^2/2g) x R^2 - 2g/( w^2 x (R1)^2) and w is the final angular velocity of the water and tank, R1 is the drain radius, and R is a given radius. Regards, Horace Heffner
Re: vortex mystery
Greetings to all members A non-physcist 's visualization of solitonic vortices is at URL: 1. http://lewfh.tripod.com/themindthingthegiftofvisualization/ 2. http://lewfh.tripod.com/coloursarecodedfrequenciesinphotonicbandgapcrystalstructures/id4.html With regards Lew FHLew wrote: Greetings to all members Horace Heffner's message of Thu, 31 Mar 2005 13:43:54 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. A non-physcist 's visualization of solitonic vortices is at URL: http://lewfh.tripod.com/themindthingthegiftofvisualization/ With regards Lew Horace Heffner wrote: At 4:55 PM 4/1/5, Robin van Spaandonk wrote: In reply to Horace Heffner's message of Thu, 31 Mar 2005 13:43:54 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. If h = (w^2/2g) x R^2 + h0 and h0 is negative, then for w=0, h=h0 and is thus also negative. How does one end up with a negative height? As I stated in the last post, the above surface is only meaningful when h 0. There is no water in the tank above [at] radii where h = 0. If w=0 then h=0 everywhere because no water will stay in the tank. No angular momentum is involved. Any water in the tank is not in equilibrium as assumed - it will all run out. Please note again that the coreolis force is ignored throughout. Or should the original formula perhaps be: h = h0 - (w^2/2g) x R^2 ? (Since the second term in this version is positive, the height becomes less for higher w and also for smaller R, both of which make sense). 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. When h=0 or h=0 then there is no water above the radius at which h is computed. The variable h at final equilibrium is a function of R, R1, and w, where R1 is the radius of the drain hole. The initial or final surface, assumed to be in equilibrium with w constant at every radius, is concave upwards. The coefficient of R^2 is thus positive. In the initial condition, h0 can be anything depending on how much water is in the rotating tank. This h0 does not affect the *curvature* of the surface, however, which is only a function of w, g, and R, assuming the drain hole is plugged, and w is constant over all radii. The variable h0 changes as the water drains from the tank. The equation describing the water surface changes as well, but the final surface should return to the form h = (w^2/2g) x R^2 + h0 for a rotating tank, assuming that viscosity forces w to be uniform across all radii, where in the *final* equilibrium: h = (w^2/2g) x R^2 - 2g/( w^2 x (R1)^2) and w is the final angular velocity of the water and tank, R1 is the drain radius, and R is a given radius. Regards, Horace Heffner
Re: vortex mystery virus alert
Mr. Lew, and everyone else: I just now went to the first URL listed below in your posting - apparently on your own website. I immediately got a couple of pop-ups followed by several Trojan virus alerts / blocked hits. Please be careful. NR --- FHLew [EMAIL PROTECTED] wrote: Greetings to all members A non-physcist 's visualization of solitonic vortices is at URL: 1. http://lewfh.tripod.com/themindthingthegiftofvisualization/ 2. http://lewfh.tripod.com/coloursarecodedfrequenciesinphotonicbandgapcrystalstructures/id4.html __ Do you Yahoo!? Yahoo! Sports - Sign up for Fantasy Baseball. http://baseball.fantasysports.yahoo.com/
Re: vortex mystery virus alert
Thanks Nick. I will be careful. With regards Lew Nick Reiter wrote: Mr. Lew, and everyone else: I just now went to the first URL listed below in your posting - apparently on your own website. I immediately got a couple of pop-ups followed by several Trojan virus alerts / blocked hits. Please be careful. NR --- FHLew [EMAIL PROTECTED] wrote: Greetings to all members A non-physcist 's visualization of solitonic vortices is at URL: 1. http://lewfh.tripod.com/themindthingthegiftofvisualization/ 2. http://lewfh.tripod.com/coloursarecodedfrequenciesinphotonicbandgapcrystalstructures/id4.html __ Do you Yahoo!? Yahoo! Sports - Sign up for Fantasy Baseball. http://baseball.fantasysports.yahoo.com/
Re: vortex mystery
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. 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. Therefore, the formula either doesn't represent what I thought it was meant to represent, or it doesn't describe reality. Regards, Robin van Spaandonk All SPAM goes in the trash unread.
Re: vortex mystery
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. 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. Therefore, the formula either doesn't represent what I thought it was meant to represent, or it doesn't describe reality. It apparently doesn't mean what you think it means, and that's my fault. The surface given by: h = (w^2/2g) x R^2 + h0 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 so 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 this defines the initial state of a rotating tank assuming the angular velocity is desired to be constant at every radius in the initial state. I provided this because it shows precisely 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 surface is a rotation of: h = k/R^2 + h0 which I noted earlier. 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 w = K*R for some constant K. 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. This describes why. 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, 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. Regards, Horace Heffner
Re: vortex mystery
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. 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. Therefore, the formula either doesn't represent what I thought it was meant to represent, or it doesn't describe reality. I think I misunderstood the problem you note! The error is not in the above formula, but in stuff you snipped. It is merely a sign error. Following is a repeat derivation of H0 with corrections: The shape of the final equilibrium surface is: h = (w^2/2g) x R^2 + h0 where h is height, w is angular velocity, g = 9.80665 m/s^2, and R is radius. Using R1 as the radius of the hole, R2 as radius of of tank, we have h = 0 at the radius R1 when equlilbrium is established and no more water can go down the hole. So: 0 = (w^2/2g) x (R1)^2 + h0 h0 = - 2g/( w^2 x (R1)^2) = Note the sign change! The final height Hf of the water at the edge of the tank is thus: Hf = (w^2/2g) x (R2)^2 - 2g/( w^2 x (R1)^2) Note subtraction! The above assumes that the initial angular velocity is small. If the angular velocity of the initial condition is high then the initial condition integration of angular momentum and energy also has to be outside the boundary established by: h = (w^2/2g) x R^2 + h0 It is possible that in the initial condition all the water will be located outside the radius of the hole, and thus no water can go down the hole at all. Regards, Horace Heffner
Re: vortex mystery
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
Re: Vortex mystery
and another link on vortex http://homepage.ntlworld.com/ufophysics/vortex.htm Richard Blank Bkgrd.gif
Re: vortex mystery
In reply to Horace Heffner's message of Thu, 31 Mar 2005 13:43:54 -0900: Hi Horace, [snip] 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. If h = (w^2/2g) x R^2 + h0 and h0 is negative, then for w=0, h=h0 and is thus also negative. How does one end up with a negative height? Or should the original formula perhaps be: h = h0 - (w^2/2g) x R^2 ? (Since the second term in this version is positive, the height becomes less for higher w and also for smaller R, both of which make sense). In short, is h the distance up from the bottom of the tank, or the distance down from the surface? Regards, Robin van Spaandonk All SPAM goes in the trash unread.
Re: vortex mystery
At 4:55 PM 4/1/5, Robin van Spaandonk wrote: In reply to Horace Heffner's message of Thu, 31 Mar 2005 13:43:54 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. If h = (w^2/2g) x R^2 + h0 and h0 is negative, then for w=0, h=h0 and is thus also negative. How does one end up with a negative height? As I stated in the last post, the above surface is only meaningful when h 0. There is no water in the tank above [at] radii where h = 0. If w=0 then h=0 everywhere because no water will stay in the tank. No angular momentum is involved. Any water in the tank is not in equilibrium as assumed - it will all run out. Please note again that the coreolis force is ignored throughout. Or should the original formula perhaps be: h = h0 - (w^2/2g) x R^2 ? (Since the second term in this version is positive, the height becomes less for higher w and also for smaller R, both of which make sense). 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. When h=0 or h=0 then there is no water above the radius at which h is computed. The variable h at final equilibrium is a function of R, R1, and w, where R1 is the radius of the drain hole. The initial or final surface, assumed to be in equilibrium with w constant at every radius, is concave upwards. The coefficient of R^2 is thus positive. In the initial condition, h0 can be anything depending on how much water is in the rotating tank. This h0 does not affect the *curvature* of the surface, however, which is only a function of w, g, and R, assuming the drain hole is plugged, and w is constant over all radii. The variable h0 changes as the water drains from the tank. The equation describing the water surface changes as well, but the final surface should return to the form h = (w^2/2g) x R^2 + h0 for a rotating tank, assuming that viscosity forces w to be uniform across all radii, where in the *final* equilibrium: h = (w^2/2g) x R^2 - 2g/( w^2 x (R1)^2) and w is the final angular velocity of the water and tank, R1 is the drain radius, and R is a given radius. Regards, Horace Heffner
Re: Vortex mystery
In reply to RC Macaulay's message of Sat, 12 Mar 2005 20:34:11 -0600: Hi Richard, [snip] I saw a pic of the collapse of a tiny bubble in a SL experiment. A vortex was visible extending into center of the sphere at the moment of collapse..hmm. I think the picture you saw was the one from Stringham/George. However that bubble collapsed *asymmetrically* against a solid surface, not in the centre of a spherical flask. Regards, Robin van Spaandonk All SPAM goes in the trash unread.
Re: vortex mystery
At 7:39 AM 3/14/5, Robin van Spaandonk wrote: In reply to Horace Heffner's message of Fri, 11 Mar 2005 17:22:03 -0900: Hi, [snip] Given the complexity of the equations for ASCII representation, I have placed a Mathcad document (24 kB) and a gif version thereof (36 kB), for readers without Mathcad, on my web page at http://users.bigpond.net.au/rvanspaa/vortex.mcd and http://users.bigpond.net.au/rvanspaa/vortex.gif respectively. Most of the variables at the top of the page are irrelevant (inherited from my standard template). This is great stuff! I hope, for the sake of the archives, one of us takes the time to post this in ascii when all is done, and tabularly descirbe each variable. I suppose I can insure that happens. This could be a useful analysis to refer to for various things. I also hope escribe continues in existance. I see you are opting to analyze the zero viscosity fluid version, which is very handy. So, to obtain the energy or momentum of what went down the drain you need only integrate inside the boundary consisting of the edge of the tank or the surface of the remaining water, whichever is smaller at a given height. To obtain the energy and angular momentum of the water that remains you need only intergate those values outside that surface and inside the tank. You might find the following analysis useful for your next phase. The shape of the final equilibrium surface is: h = (w^2/2g) x R + h0 where h is height, w is angular velocity, g = 9.80665 m/s^2, and R is radius. Using R1 as the radius of the hole, R2 as radius of of tank, we have h = 0 at the radius R1 when equlilbrium is established and no more water can go down the hole. So: 0 = (w^2/2g) x R1 + h0 h0 = 2g/( w^2 x R1) The final height Hf of the water at the edge of the tank is thus: Hf = (w^2/2g) x R2 + 2g/( w^2 x R1) Bravo and keep cooking! Regards, Horace Heffner
Re: vortex mystery
I assumed in the prior analysis that the initial angular velocity is small. If the angular velocity of the initial condition is high then the initial condition integration of angular momentum and energy also has to be outside the boundary established by h = (w^2/2g) x R + h0 It is possible that in the initial condition all the water will be located outside the radius of the hole, and thus no water can go down the hole at all. Regards, Horace Heffner
Re: vortex mystery
Correction follows. Sorry! The shape of the final equilibrium surface is: h = (w^2/2g) x R^2 + h0 where h is height, w is angular velocity, g = 9.80665 m/s^2, and R is radius. Using R1 as the radius of the hole, R2 as radius of of tank, we have h = 0 at the radius R1 when equlilbrium is established and no more water can go down the hole. So: 0 = (w^2/2g) x (R1)^2 + h0 h0 = 2g/( w^2 x (R1)^2) The final height Hf of the water at the edge of the tank is thus: Hf = (w^2/2g) x (R2)^2 + 2g/( w^2 x (R1)^2) The above assumes that the initial angular velocity is small. If the angular velocity of the initial condition is high then the initial condition integration of angular momentum and energy also has to be outside the boundary established by h = (w^2/2g) x R^2 + h0 It is possible that in the initial condition all the water will be located outside the radius of the hole, and thus no water can go down the hole at all. Regards, Horace Heffner
Re: Vortex mystery
Been following this thread with interest. I remain amazed at the statue of this group and the insight expressed on such a range of subjects. I'm waiting on the next step.. that being ..explain the energy unleashed in a tornado or hurricane. Water drains, hurricanes, tornados, galaxies, black holes, etc. make me cosider an atom may be vortex shaped. I saw a pic of the collapse of a tiny bubble in a SL experiment. A vortex was visible extending into center of the sphere at the moment of collapse..hmm. Richard Blank Bkgrd.gif
Re: vortex mystery
At 4:16 PM 3/11/5, Robin van Spaandonk wrote: In reply to Horace Heffner's message of Wed, 9 Mar 2005 00:31:22 -0900: Hi, [snip] mv1r1 = mv2r2, i.e. v2 = v1 x r1/r2. (m is the same before and after, because we are dealing in both cases with the identical chunk of water). Yes, you are certainly right about this. Momentum is conserved. The speed of a chunck is thus not constant in a vortex, as I had assumed. In fact, I think in your imaginary tank the tangential speed is proportional to 1/r, i.e. v2 = v1 x r1/r2 Where r2 is your r, and V1 x r1 is the proportionality constant. It is a highly idealized picture. This describes the motion of an idealized chunk unaffected by its surroundings other than the force required to reduce its radius. and the vertical speed to 1/r as well. This is shown in Feynman's *Lectures on Physics*, Vol II, 40-10 ff. Fig. 40-12 is a great drawing of your tank, showing the surface countour of a sample vortex. The reason kinetic energy is increased is that work is done on the chunck as it moves inward. In the case of a cylinder of water the work is just the work of falling, m*g*h. The work is completely analogous to the work This is what I also initially assumed, and it must be the case for a tank in which the water is essentially initially motionless, and is eventually brought into motion by the vortex action spreading from the centre. Assuming that some random minor current in the water near the drain starts the vortex rotating. This is almost never the case though. The vortex is not started by some random motion, but rather by residual angular momentum from the filling of the tank. Did you try an experiment? If the water *were* completely still, then it would be started rotating by the coreolis force, but this is a nominal force, and would not exist if the experiment were done on the equator. If the water were completely still, and located on the equator, then it would flow radially only down the drain. In this case, the velocity at the rim of the tank will eventually assume a value determined by the gravitational energy gain, call this velocity V0, with matching AM m x V0 x R0 (where R0 is the radius of the rim of the tank). The water at the exact rim of the tank always has tangential velocity zero. The water part way between the rim and the hole carries most of the initial angular momentum. The gravitational component of the final velocity can't exceed sqrt(2 x g x h), where h is the initial height of the chunk of water, and g is the Earth's gravitational acceleration at the surface. This implies a maximum rim velocity due to gravity of V0 = sqrt(2 x g x h) x Rd/R0. This is not true in general I think. You are attempting to apply conservation to a single chunk. Conservation of angular momentum and energy only applies in the aggregate. The above relation might apply to water on the surface of the vortex which is falling along the surface into the drain, but not to water deeper down. It falls along the water-air surface having a cross section curve h = k/r^2 + h0, where h is the height, and r the radius. Water falling inside the surface falls from a lessor height and contracts from a lessor radius. The pressure and work required to accelerate the chunks is done by water falling which is closer to the rim and above the chunk. The problem arises, when one starts with a tank full of water that is already rotating, especially where the initial velocity at the rim V0. If we now follow this down the drain, we find that the energy gain of the water is greater than would ensue purely from gravity. I think maybe you are ignoring the fact that the water to go down the drain last has less angular momentum and energy than the water going down the drain earlier. The energy of the rim water falling is translated into increased kinetic energy and angular momentum of the water nearer the drain. The water at the bottom and near the rim doesn't even move much toward the drain until the very end, and then its height is low and it has low initial angular momentum as well. The energy *gain* per unit mass is: 1/2 x V0^2 x ((R0/Rd)^2 -1) which clearly is a function of V0! (R0 and Rd being constants). IOW by choosing a larger V0, we gain more energy, which implies that either gravity is not the only source, or the water doesn't flow, or part of the angular momentum is passed off, such that the velocity can remain constant. This was the reasoning that led to my initial question. If a large part of the angular momentum is not passed to the Earth, then where does the energy come from? It appears the problem is you don't have a good estimate of conditions throughout the tank, and thus you don't have valid integration of the energy and angular momentum that goes down the drain. A visualization that may be helpful is a tank that rotates, and is initially rotating. In that case the boundary conditions of the tank are very different, more like you imagine. If
Re: vortex mystery
In reply to Horace Heffner's message of Fri, 11 Mar 2005 07:45:11 -0900: Hi, [snip] direction the water goes down the drain. Any angular momentum exhibited by the vortex must be there initially, and some of that is lost by transfer to the tank bottom. In all cases the overall angular momentum is conserved. [snip] Correct, but not really the issue. I had already assumed that such was the case. The issue is where does the *energy* come from? Put simply, 1 kg of water that changes it's radius by a factor of 10, undergoes roughly[1] a ten fold velocity increase. This implies a 100 fold kinetic energy *increase*. Where does that energy come from? It can't all come from gravity, because gravity can only supply a fixed amount, equal to m x g x h, whereas the increase in energy is a function of the initial velocity, and is therefore different for each initial velocity, and consequently not always equal to m x g x h. You pointed out that angular momentum goes down the drain with the water. This is something I hadn't really considered, but it only serves to sharpen the problem. It is precisely the angular momentum that goes down the drain, which is *not* passed out of the system before the drain is reached. [1] In fact it will be slightly less, due to friction as you pointed out. Making the tank deeper will reduce this loss on average, because friction with the bottom will apply to a smaller fraction of the water. Or if you prefer, we can fill it with liquid helium instead, so that there is no friction. Regards, Robin van Spaandonk All SPAM goes in the trash unread.
Re: vortex mystery
At 11:29 AM 3/12/5, Robin van Spaandonk wrote: In reply to Horace Heffner's message of Fri, 11 Mar 2005 07:45:11 -0900: Hi, [snip] direction the water goes down the drain. Any angular momentum exhibited by the vortex must be there initially, and some of that is lost by transfer to the tank bottom. In all cases the overall angular momentum is conserved. [snip] Correct, but not really the issue. I had already assumed that such was the case. The issue is where does the *energy* come from? The COAM statement above was just an afterthought. My main point earlier was *what energy?*. You don't have an accounting of the energy on a system basis. Put simply, 1 kg of water that changes it's radius by a factor of 10, undergoes roughly[1] a ten fold velocity increase. This implies a 100 fold kinetic energy *increase*. There again is an example of treating a part of the system like it was the whole system! You just can't do that. The short simple answer is the 1 kg of water, if that's all you have, can't all go down the drain at a 10 fold increase in angular velocity. The kinetic energy of parts of the system remain either as (1) water which can't go down the drain (frictionless version, or rotating tank) due to centrifugal force or (2) water which depletes its kinetic energy before it goes down the drain and transfers its momentum in part to the earth via the tank and viscosity (normal verison). Where does that energy come from? What energy? You haven't intergated anything. You are treating the entirety of the water as a single lump at fixed radius. You just can't do that. You don't know from these particulars (1) what the initial velocities are (on a finite element basis), (2) the total system angular momentum, (3) the total initial starting kinetic energy of the system, (4) what angular momentum and kinetic energy is actually dumped down the drain (though it is clearly not nearly as much as you estimate unless the initial total momentum is very small), (5) how much kinetic energy and momentum remains in water which can not go down the drain due to centrifugal force or (6) what kinetic energy is dumped into heat and corresponding momentum dumped to the earth in order to allow all the water to go down the drain. It can't all come from gravity, because gravity can only supply a fixed amount, equal to m x g x h, whereas the increase in energy is a function of the initial velocity, and is therefore different for each initial velocity, and consequently not always equal to m x g x h. The m x g x h integrated throughout the mass may well be way too *much* energy. Almost all that energy could easily be converted to a combination of vertical velocity, water heat, and tank heat, provided the initial system angular velocity is very small. If m x g x h provides too little energy, then there is water which cannot go down the hole until its angular velocity is reduced by turning it into heat. The excess angular momentum in this case is indeed transfered to the earth as its corresponding energy is reduced to heat. You pointed out that angular momentum goes down the drain with the water. This is something I hadn't really considered, but it only serves to sharpen the problem. It provides a perspective to solve the problem. All the energy ends up as heat, kinetic energy going down the drain, or (in a frictionless version or rotating tank version only) as kinetic energy of water remaining in the tank and which can not go down the drain. All the initial momentum of the system ends up either going down the drain or being transferred to the earth in a process where the corresponding kinetic energy is reduced to heat. It is precisely the angular momentum that goes down the drain, which is *not* passed out of the system before the drain is reached. That angular momentum is still in the water if that water has not gone down the drain. If there is not enough kinetic energy available to make its radius reduce, then that radius remains fixed - it can not go down the drain, it can not reduce its radius of rotation unless some other energy M x g x h comes from some other M in the system. In a friction world, however, the angular momentum of water which can not go down the drain is eventually reduced due to viscosity and heat generation, so that water can all eventually go down the drain. It just doesn't carry all the angular momentum that you would estimate based on the simplistic model. [1] In fact it will be slightly less, due to friction as you pointed out. Making the tank deeper will reduce this loss on average, because friction with the bottom will apply to a smaller fraction of the water. Or if you prefer, we can fill it with liquid helium instead, so that there is no friction. The case for a frictionless liquid is similar to the case for the rotating tank. The mass containing the angular momentum for which there is no energy available to put down the drain remains behind. Superfluids, though,
Re: vortex mystery
In reply to Horace Heffner's message of Wed, 9 Mar 2005 00:31:22 -0900: Hi, [snip] mv1r1 = mv2r2, i.e. v2 = v1 x r1/r2. (m is the same before and after, because we are dealing in both cases with the identical chunk of water). Yes, you are certainly right about this. Momentum is conserved. The speed of a chunck is thus not constant in a vortex, as I had assumed. In fact, I think in your imaginary tank the tangential speed is proportional to 1/r, i.e. v2 = v1 x r1/r2 Where r2 is your r, and V1 x r1 is the proportionality constant. and the vertical speed to 1/r as well. This is shown in Feynman's *Lectures on Physics*, Vol II, 40-10 ff. Fig. 40-12 is a great drawing of your tank, showing the surface countour of a sample vortex. The reason kinetic energy is increased is that work is done on the chunck as it moves inward. In the case of a cylinder of water the work is just the work of falling, m*g*h. The work is completely analogous to the work This is what I also initially assumed, and it must be the case for a tank in which the water is essentially initially motionless, and is eventually brought into motion by the vortex action spreading from the centre. Assuming that some random minor current in the water near the drain starts the vortex rotating. In this case, the velocity at the rim of the tank will eventually assume a value determined by the gravitational energy gain, call this velocity V0, with matching AM m x V0 x R0 (where R0 is the radius of the rim of the tank). The gravitational component of the final velocity can't exceed sqrt(2 x g x h), where h is the initial height of the chunk of water, and g is the Earth's gravitational acceleration at the surface. This implies a maximum rim velocity due to gravity of V0 = sqrt(2 x g x h) x Rd/R0. The problem arises, when one starts with a tank full of water that is already rotating, especially where the initial velocity at the rim V0. If we now follow this down the drain, we find that the energy gain of the water is greater than would ensue purely from gravity. The energy *gain* per unit mass is: 1/2 x V0^2 x ((R0/Rd)^2 -1) which clearly is a function of V0! (R0 and Rd being constants). IOW by choosing a larger V0, we gain more energy, which implies that either gravity is not the only source, or the water doesn't flow, or part of the angular momentum is passed off, such that the velocity can remain constant. This was the reasoning that led to my initial question. If a large part of the angular momentum is not passed to the Earth, then where does the energy come from? The only other possibility I can see, is that we have a situation where the centrifugal force is so large at the drain, that the inner radius of the vortex is in fact larger than the drain, and no water goes down at all, which means that the water eventually slows down due to internal friction, until a point is reached that the weakening centrifugal force allows it flow. Regards, Robin van Spaandonk All SPAM goes in the trash unread.
Re: vortex mystery
It would be pretty neat to have a long time do this experiment right on the equator to see if it is possible to obtain a vortex flow with only radial motion visible. Regards, Horace Heffner
Re: vortex mystery
Let me try this one more time. At 4:47 PM 3/7/5, Robin van Spaandonk wrote: The question is, where does the energy come from to increase the velocity of the water? If the water is stationary to start with, then it comes from the change in height of the water as it leaves the tank, and the velocity at the edge automatically adjusts itself accordingly. Water in a tank, once spinning, does not stop for an amazingly long time. It is generally thought water goes down the drain a differing direction in the Southern Hemisphere than the Northern Hemisphere. However, if you check it a day or two after filling the tank it usually forms a vortex in the vortex direction that resulted when the tank was filled. I recall an article about this in the early 1960's - maybe in Scientific American or Popular Science. The rotation that results is a function of both gravity and initial angular momentum. Some of the gravity potential energy converts to angular velocity via the coreolis force. If the initial angular velocity of the water is zero, then your assertion that the energy comes entirely from the water falling to the drain is true, and the angular momentum of the vortex comes from the coreolis force. However if the water is already rotating before the plug is pulled, then it has to end up going faster than can be accounted for by gravity. Yes, except when the coreolis force is in a direction opposing the initial rotation, and its effect exceeds that of the inital angular momentum. If the radius decreases by a factor of ten before the water reaches the drain, then the velocity has to increase 10 fold, and the energy per unit mass must increase 100 fold. So where does the energy come from, or for some reason, does it simply not happen, and if not, then what does happen? Ignoring the coreolis force for a moment, the fallacy in the above statement is the assumption that the instantaneous speed v of some small chunk of the water changes as it approaches the drain. The speed of the chunk remains constant at all times, except for the speed added by converting gravitational potential energy PE to kinetic energy. Thus the instantaneous linear kinetic energy KE = 1/2 m v^2 of the chunck remains constant except for speed added by falling in the gravitational field. The angular velocity w increases however, because w = v/r. Now, you might say that for a rotational system KE = 1/2 I w^2, and w is increasing with reduction in radius, so where does the free energy come from? Well, the answer is that in a vortex the moment of inertia I of a chunk is not constant. We have I = m R^2, and w = v/R, so when we substitute these into KE = 1/2 I w^2 and we have: KE = 1/2 (m R^2) (v/R)^2 = 1/2 m v^2 which is constant, except for the PE converted to KE by falling down hill. Since the KE of every chunk remains constant the energy of what remains in the tank is the original gravitational potential energy PE plus KE less the KE of what went down the drain. I hope that is all a bit more correct. Regards, Horace Heffner
Re: vortex mystery
a large amount of the energy that forms is from the friction between the water and the air bubbling up. as well as the coriolis force (though theres barely any there). in fact, if you perform this experiment in near vacuum, with teh water at a stand still, it will NOT form a vortex. On Mon, 07 Mar 2005 16:47:13 +1100, Robin van Spaandonk [EMAIL PROTECTED] wrote: Hi, Suppose one has a tank of water, with a plugged drain in the centre at the bottom. Let the water be rotating in the tank. Remove the plug. A vortex forms as water exits the drain. Conservation of angular momentum (well established for a vortex) ensures that the velocity of the water changes as the inverse of the radius. Since all water eventually runs down the drain, all the water increases in velocity, with the water from the edge having increased the most. The question is, where does the energy come from to increase the velocity of the water? If the water is stationary to start with, then it comes from the change in height of the water as it leaves the tank, and the velocity at the edge automatically adjusts itself accordingly. However if the water is already rotating before the plug is pulled, then it has to end up going faster than can be accounted for by gravity. If the radius decreases by a factor of ten before the water reaches the drain, then the velocity has to increase 10 fold, and the energy per unit mass must increase 100 fold. So where does the energy come from, or for some reason, does it simply not happen, and if not, then what does happen? Regards, Robin van Spaandonk All SPAM goes in the trash unread. -- Monsieur l'abbé, I detest what you write, but I would give my life to make it possible for you to continue to write Voltaire
