How DO rowboats work?

Factual Questions
81

It’s also wrong.

Here’s your mistake: the moment you imagine that the water is frozen, or even is slightly putty-like, then you fundamentally change the problem we’re discussing, and all your reasoning no longer applies to my original question.

Specifically you are trying to impose low-Reynolds dynamics on a system having very high Reynolds number. You’re visualizing water as being like putty, as if the forces applied by the paddle can communicate (via shear) with fluid parcels at great distances from the paddle. This “putty effect” does occur in nature, but it becomes insignificant in situations involving high Reynolds numbers (such as rowboat paddles.) If the viscosity effects of water did dominate over the inertial effects, then pushing a paddle through water would drag ALL the water along, not just the water local to the paddle. But then there would also be no ring-vortices which carry big blobs of water along for many tens of yards after the rowboat paddle has come to the end of each sweep.

If water WAS like putty then my question would be answered, because then the long-range shear forces would be very significant, and the paddle would push upon the entire volume of liquid as a whole. Water becomes “putty-like” at the scale of individual microorganisms, and bacterial cilia do behave like levers as you say. At the scale of bacteria, a propellor screws itself forward like an auger drilling into mud. The situation is utterly different at the scale of rowboats, ship propellors, fish tails, etc. For insights into the differences between high and low Reynolds number physics, read Vogel’s excellent book “Life in moving fluids.”

If rowboat paddles are causing disagreements, then just forget about them. My original question involves the propulsive force experienced by a vortex-launcher, and I used the rowboat paddle as a typical example of vortex-launcher. I didn’t expect this to cause a big distracting argument.

So instead just explain how this device can propel itself forward by repeatedly launching ring-vortices backwards:

The Airzooka
http://www.arborsci.com/images/airzookblueT.gif
http://www.arborsci.com/Products_Pages/Pressure&Fluids/Pressure&FluidsBuy2.asp

Wrong, but I’m not going to try to shake your confidence, since it leads away from my original question about vortices.

Another possible source of confusion: when I say “vortex”, possibly you hear the phrase “swirling motion?” If so, then we’re miscommunicating. A “ring vortex” is much more than just a swirling motion of fluid about a circular core-filament. Besides swirling internally, the whole volume enclosed by the separatrix moves forward! And it carries mass as it moves, as if it was a linearly translating blob of fluid. Yes, I agree with you that the swirling motion contained within the separatrix surface is wasted energy. But the forward motion of a massive “chunk” of fluid is not wasted.

But you’re wrong each time you repeat this. I know for a fact that vortex mechanics explains “unsteady flight” or flapping-wing propulsion of all kinds, including fish tails, insect and bird wings, and any kind of periodic motion of any paddles in fluids when Reynolds number is large. This isn’t PhD-level aerodynamics here (but it certainly isn’t covered in the undergrad texts either.)

82

In aerodynamics it is standard to assume that incompressible fluids are used. This is done because the changes in the density of air are on the order of a few percent where wings are concerned, and if we use water instead, the density changes become totally insignficant. That’s basic aerodynamics 101. Now PRESSURE is different. If you deflect some mass-bearing parcels, or pass a high-velocity flow through a constriction, then you experience large pressure differences.

Remember, if you look at a flow diagram, you can’t see the
pressures.

83

I was thinking about this “concentration” at one point long ago, but then I realized that a relatively small ring-vortex can knock over a relatively wide cardboard target.

If a travelling ring-vortex pattern has forward momentum in its center, and is surrounded by a region of backwards momentum, then when it strikes a flat surface I would imagine that BOTH regions of momentum would interact with the surface. (If not, then where does the backwards momentum go?)

This all still bothers me. A vortex-launcher sends out a pattern with ZERO net momentum? Yet the vortex-launcher experiences a kick! Doesn’t that violate conservation of momentum? Where did the backwards momentum carried by the vortex come from? And then when the travelling pattern finally hits a wall, it delivers a kick to the wall… yet the travelling vortex contained no net momentum.

I wish I still had a swimming pool. I’m tempted to build a motorized piston device which sends out underwater ring-vortices. Those vortices should be very heavy, so the engine should produce a large thrust. Sort of like a motorboat prop, but instead of shooting out a screw-thread jet of water, it throws out a jet which looks like a stack of pancakes!

84

I don’t think it propels itself forward by launching ring vortices backwards. I think forward moving ring vortices would be a side-effect (losses/friction) of air flowing forward…propelled in that direction by the Newtonian “kick” you say exists.

Since you know the facts, why ask the question? I’m afraid I’m part of the repetitive crowd here – vortices in rowing are an undesirable side effect. In fact, I can’t think of anything (of course that doesn’t say much…) where vortices are desirable. They are losses caused by setting in motion part of a fluid while the rest is at rest.

85

I haven’t seen the vortex cannons you talk about, but I suspect that much of the “kick” you feel is mechanical. If the cannon contains a piston on a spring, released by a trigger, then the piston is probably heavier than the air in the barrel. The kick is probably mostly from the piston’s sudden acceleration.

The motion of the piston would still impart some momentum to the air in the barrel. I’m just suggesting that it’s probably much less than you might think from the force you feel. You could only make a propulsion system out of this if you had a separate intake (and then you’ve just built a water-jet engine like a jet-ski has); otherwise, you’d lose all of your momentum sucking air back into the barrel after creating the vortex. (I don’t think I can speak intelligently about knocking down the cardboard targets without seeing one in action or playing with it, so I won’t give my thoughts on that.)

I think the disagreement here is because of a confusion between two sizes of vortices. The little vortex-whirlpools which curl off the end of an oar are undesirable; it took energy to make them, but there’s no net momentum transferred to your boat. But your boat, moving through a lake, can also be thought of as part of a pair of large, counterrotating vortices, one on either side of the boat. This, as the OP points out, has to be the case; the water is essentially incompressible, and it is constantly being displaced from the bow and replaced at the stern, so there must be a net backwards flow of water around the boat.

A vortex large enough to hold the whole boat can impart momentum to the boat. There’s nothing magic about this: the vortex system as a whole (boat plus water) has no momentum, but by clever positioning of vortices you can ensure that the boat is in a forward-moving part of the vortex at all times. (Think of Chronos’ experiment, with a plug in a cylinder. If the cylinder has a ring vortex, the plug may move along the center of the vortex even though the total momentum of the system remains zero.)

86

As I said once before, my original question involved the finer points of ring-vortex motion and momentum. It wasn’t about rowboats. I just used a rowboat as a real-world example of a craft which is propelled by periodically launching individual vortices. (I didn’t expect objections to rowboats to take over the thread.)

I say again, forget the rowboats. Imagine a Wham-O vortex launcher gun which is motorized so it sends out repeated ring- vortices. Its internal diaphragm moves back and forth, yet it drives itself forward. And any flat object placed in the “exhause stream” of ring-vortices is driven backwards. Yet if we place an imaginary UNMOVING box around each ring vortex (where the box is big enough to enclose all the significant flows associated with each ring-vortex) then we find that each ring vortex has net zero flow.

Possible answer:

Maybe a flow-pattern with zero net flow is allowed to carry non-zero momentum? If so, then it’s an extremely weird idea (since eseentially only the momentum flows, while mass doesn’t!) And in that case I’m after some sort of simple thought-experiment where the net flow is easily proved to be zero yet the net momentum being transported is easily proved to be non-zero. I tried to simplify a ring-vortex to make the calculations easy, but that didn’t work.

87

You can always draw a control volume large enough that nothing interesting happens, but if you really want to analyze what’s going on, why not pick a more useful CV? If you set your CV such that the moving vortex crosses the boundary, then you can do a more useful momentum balance. That’s what you have to do if you want to look at the vortex striking an object anyway. Your vortex generator is injecting energy into the system, translating mechanical energy in a spring or something into motion in the fluid. Why does it seem so strange that if you use a mechanical device to dump energy into the fluid, the vortex it creates can carry momentum?

88

Why large? I want the edge of the cube to be maybe 10x larger than the ring diameter, so that it intercepts the significant flows in the way that a cardboard target would. To intercept ALL the flows, the volume would have to have infinite extent. Imagine the ring-vortex to be a bar magnet. The magnet’s field drops to a very low level at distances a few times larger than the magnet length.

The weirdness is because the vortex carries no net mass through a surface, yet it carries significant momentum, as if “pure momentum” was flowing. I want to understand how this can happen (or perhaps discover a more common and easily visualized situation where a similar thing occurs.)

If I throw a baseball through a large metal ring so that it strikes a target, significant mass passes through the ring, and so does significant momentum. As I understand it, if I fire a ring-vortex through a large metal ring, zero net mass passes through that ring, yet the momentum passing through the ring is significant. Huh? How can this be?!!! (Or is it somehow wrong?) If the answer lies in velocity vectors rather than fluid speed as someone mentioned earlier, that’s fine, but that’s only a hint and not an answer to my question.

89

Bear with me if I’m just completely missing something here, but I just don’t understand something.

The vortex is a coherent structure. If it passes through your CV surface, it carries mass and momentum. We started this discussion using the vortices shed by oars, which travel perpendicular to their axis and now we’re talking about ring vortices which travel in the same direction as their axis, but in both cases they’re coherent structures made up of specific rotating fluid elements. They are not like a wave which passes through fluid; a vortex is fluid. This is easily seen in smoke rings where the smoke in the ring moves. The ring structure doesn’t just pass out of the smoke into the ambient fluid and leave a stationary blob of smoke behind. I don’t mean to belabor this point since we probably don’t disagree about it, but I just wanted to make it clear: the vortex is a coherent moving mass of fluid.

Now, if the vortex is a moving mass of fluid, it carries momentum. There are a lot of rotational velocity components which sum to zero, but if the vortex is moving as a whole instead of just rotating in place, then there is a translational velocity component superimposed on the rotational components. This translational component does not cancel within the vortex (if it did, the vortex wouldn’t move). In this respect, the momentum balance on a vortex as a whole is no different than the momentum balance on a puff of air, a packet of fluid moving through an ambient fluid. Of course, the rotational components of the vortex allow it to remain coherent and resist diffusion so it persists much longer than a puff would, but at any given moment the puff and the moving vortex are analogous as far as balancing momentums.

In both cases (moving vortex and puff) there is a motion in the ambient fluid as a result of the structure’s motion. Fluid directly in front of the structure is pushed “forward” (in the direction of motion) and “out” (away from the centerline path of the motion). Fluid behind and beside the structure is entrained in the wake moving forward and “in” (toward the path). All of these movements have momentum, but they do not cancel the forward momentum of the vortex itself. In fact, depending on where you draw your CV, there may be no “backward” flow whatsoever (except those rotational components in the vortex, which we’ve already said are irrelevant).

Again, I apologize if I’m missing something. This thread has gone on so long that I think someone (possibly me) has a fundamental misunderstanding of what we’re discussing. But I’ve done a lot of fluid dynamics both experimentally and analytically and I don’t understand why you say a moving vortex carries no net mass.

90

Take a little baloon with very low density, and give it an aerodynamic shape. Fill it with helium. Launch it at a paper target. Target is knocked over.

The helium baloon is lighter than air. Does this mean it carries negative momentum? No, I didn’t think so. Launching it from side A of a room to side B of a room will nonetheless move net mass from side B of a room to side A of a room - this is just entropy.

Someone said earlier that the air moving “backwards” throughout the flight of the air vortex would net out to have as much momentum as the air moving forward in the vortex. This is bullshit. Momentum is calculated in the moment, not over time. Momentum summed up during a set period of time is effect, not momentum. Thus, if the air moving forward has more m*v than the air moving backward, it carries momentum.

What’s the frickin’ problem?

91

Well, I meant to say that it carries momentum anyway (like my helium balloon example should show), but it carries net momentum if it has more m*v than the air moving backward. Preview twice, post once. Doh.

92

bbeaty, you commented on several parts of my post, but ignored the essence of the response that answers your question.

You say you want to know what accounts for the momentum balance. For example:

And this:

This suggests to me that you are not looking at the vortex itself (everything inside the separatrix), but the overall flow, the surrounding media returning to the empty space. Looking at your gifs, I see the same thing:


Red vortex flowing through a green medium. Is this a correct description of your position? If not, explain how not. If so, agree so I know we are discussing the same thing.

If we are on the same page, then let me address your problem. The second gif seems to show the red vortex is directly responsible for taking the green particles from in front and depositing them behind, in a one-for-one exchange. This is not a correct representation. Nevermind the spin within the vortex - I think we agree on that. The vortex itself is a moving mass of air. That is what carries the momentum.

So how does the air get from in front to behind? Entropy. The vortex is not trying to replace the air, and the energy from the vortex is not doing the work of replacing the air behind it. That is why it can be ignored for the momentum balance against the Wham-O gun, and why it can be ignored for the impact against the target.

The vortex flowing through the medium is pushing whatever is in the way out of the way. This creates localized pressure increases (in front) and pressure decreases (behind). But what balances out those localized imbalances is not the energy you put in via the vortex cannon. You shoved a chunk of air, and that air is a coherent group so it keeps together, so it keeps moving. What drives the pressure imbalances to even out is entropy. When you input energy, you created an localized decrease in entropy - or added order if you want to think of it that way. Entropy is what makes the molecules all spread out, vs. staying clumped in localized pressure imbalances. Once the energy you input is used/gone, then the entropy increases again, and the molecules all spread back out evenly. The momentum return is not caused by your momentum imput.

93

There’s your problem. You’re assuming that this is in air. Yet it involves fluids, e.g. gas and liquid. In fluid mechanics it’s normal to assume an incompressible fluid, that way you arrive at correct general explanations and not just explanations which apply only to gases.

Perform all of this stuff underwater. If it only works in air, then it’s not an explanation.

Explanations have implications. Your explanation doesn’t work underwater, nor does it work for a perfectly incompressible fluid, yet ring-vortices are easy to produce underwater (and they are common in fluid flow computer simulations which use incompressible fluid.)

Your explanation is wrong. I have to look elsewhere.

94

Would this actually happen? To make the same effect much stronger that with a helium balloon, let’s use a hollow glass sphere made from very thin glass and submerge it within a water tank in a weightless environment so that it doesn’t try to rise from buoyancy. Now fling the sphere with your hand underwater. Does it coast along? Or does it stop instantly when you stop pushing? After all, the hollow sphere carries almost no mass inside (zero mass if we pump out the air while also ignoring the mass of the very thin glass layer.)

The sphere cannot carry momentum inside itself. So, can it coast through a fluid environment and then knock over a distant target? If so, then this suggests that the mass carried by a ring-vortex is irrelevant, and it’s the flows outside the separatrix surface which control everything.

That’s your belief. It doesn’t matter how strongly you make the assertion, if not backed up by reasoning or evidence it remains purely a belief. Why not come up with a simple thought-experiment that shows that IT IS bullshit? The only experiments I can imagine end up proving the very opposite.

Please, just show that the air moving forwards DOES have more m*v than the air moving backwards. If you can do this, I’d have my answer. Whenever I try to do it, I find that I get lots of m and tiny v on one side of the equation, but small m with lots of negative v on the other side. They are equal and opposite.

95

That’s what I’m trying to show. But I fail! I want to explain how a ring-vortex can deposit momentum into a target.

Yet when I construct easily-analyzed situations, as with small cubical or cylindrical pistons moving through a large closed-ended pipe, I find that the momentum of the fluid immediately surrounding the piston is equal and opposite to the momentum stored in the piston. Example: for a large piston with a small gap between piston and pipe, we have lots of piston-mass moving forward, and less fluid-mass moving backward through the gap. But the fluid in the gap moves fast backwards, and it’s just fast enough so that mv = -m’v’. The piston carries momentum forward, but it’s surrounded by a fluid flow which carries equal and opposite momentum. This fluid flow pattern follows the piston as it moves. I don’t quite know what happens when such a piston strikes the closed end of the pipe.

(Don’t forget that the density of the piston is that of fluid, since the “piston” is supposed to model fluid carried forward by the moving ring-vortex.)

OK, so maybe a piston moving through a closed-ended pipe is somehow fundamentally different than a small ring vortex moving through a large tank of water. As I make my closed-ended pipe larger and larger, does something suddenly change? If I could grasp the single small fact that makes the pipe and the tank totally different, I think I would finally understand. Yet I don’t find any sudden change, and I only succeed in convincing myself that the closed-ended pipe is the same as the water tank, and the moving piston is the same as the moving ring-vortex.

No, you’ve got the idea. Essentially you’re saying that the momentum of the flows outside the separatrix surface enclosing the ring-vortex don’t cancel out the momentum of the moving mass within the separatrix. Either you are wrong about this and they do cancel… or we are very close to where my own mistake lies.

Might you have a simple thought experiment which shows why a closed-ended pipe is fundamentally different than a tank of water, and that a cubical piston is fundamentally different than a moving ring-vortex? Or maybe you have a simplified situation where the sum of momenta in all the moving parcels associated with the “ring vortex” do in fact add up to a particular non-zero number?

96

OK. I missed it.

But you’re not going to just come out and tell me what this essence is? :slight_smile:

Yes, you’ve got it. We can simply pretend that the ring-vortex is actually a moving balloon. To avoid misunderstandings, call it a water-filled balloon submerged in a large tank of water. Now fling it across the tank.

If I fling a water balloon through the air or through a vacuum, I have no problem. It’s when I fling some water while under water that I run into troubles.

Right. We could just as well say that the backwards-moving beads keep transferring their energy to other beads at their tail end (ahead of the red block) whenever the backwards-moving beads suddenly stop, and this would cause more beads to start moving backwards whenever some other beads suddenly halted.

So we agree that the outside fluid carries some backwards momentum, and the fluid inside the balloon carries some forward momentum? And this pattern of flows moves along as the balloon moves along?

I don’t understand your explanation. Why does Entropy tell us to ignore the backwards momentum? I’m looking for a simple and detailed explanation, not just a statement “it’s because of entropy.”

More important: where does this backwards momentum go when the vortex strikes a target and gets disrupted? If the forward momentum goes into the target and knocks it over, the remaining backwards momentum can’t just vanish. If it goes racing off into the distance somewhere, this “racing off” process needs to be explained too.

Actually the pressures around a moving sphere are high in front and high behind (these mostly cancel out, so the sphere feels no gigantic drag). And the low pressure is in a band around the equator of the sphere (the low pressure is radially outward, so it has no effect on the trajectory of the moving sphere.)

But entropy can’t knock over a target! Momentum is required. If the water balloon plus the surrounding flow together carry zero momentum, then after the balloon and the flow have both stopped (after the solid target has stopped them,) how can there be any momentum deposited into the target? Saying that entropy does it, to me that seems like saying a magic word to make a problem go away. The true magic words would be the words in a “simple explanation.”

97

Attacking the problem from another front.
Suppose I use a machine gun to shoot at a steel plate target. Suppose the bullets bounce off and come back at me. Suppose I hold up another steel plate and reflect the bullets back at the target. Then I can put away the gun and let the bullets keep bouncing. They apply a force to my plate and to the target.

Isn’t this situation very close to the vortex-ring momentum problem? After all, the outgoing stream of bullets has equal and opposite momentum to the return stream, so as a whole the streams have zero momentum. The bullets reverse course at each metal plate, so they’re depositing momentum as they reverse. If we consider the bullets as a whole to be a “fluid jet”, then the fluid has zero momentum, since the momentum carried by the incoming stream is equal and opposite to that carried by the outgoing stream.

This situation resembles one where an underwater hose blasts a water jet against a distant target, and then a water pump collects some ambient water to power the underwater hose.

But where this hose is concerned, there’s a continuous fluid stream connecting the “launcher” to the “target.” The vortex-ring situation is similar, but the vortex is a brief pulse instead of a constant stream. (Wouldn’t a series of moving ring-vortices stacked up like pancakes be exactly the same as an underwater jet of water?!)

98

The vortex produced is represantive of the inefficiency of the oar pushing itself through the water. Since the water being pushed “east” as a result of the oar pushing it eastward has the opportunity to also exert perssure upward, downward, south and north, the surrounding water responds my moving in those directions.
Also remember that the oar is basically flat and it 's thrust angle changes respective to the intended direction of the boat as the oar makes it’s limited arc through it’s radius. Note that the racing boats (you know, those rich guys) have very long oars. (and probably other things) This reduces the angular changes, so the result is more constant, and the mechanical advantage is such that the paddle end of the oar travels a great distance with respect to the handle end. Shape the paddle differently and the result should be different, but when you look at a primitave oar, it looks pretty much like the hi-tech ones available now.
BUT, look at a propeller and get into the real high tech fluid dynamics.
Bernoulli doesn’t mean too much in a lake or in the air. Works real well in a pipe. Check out the Bernoulli effect on a Cessna 150 at take off speed. With Bernoulli only, I think it would drop like a rock

99

AHA!!! I THINK I GOT IT! And it has nothing to do with anything I was thinking before. I was making a big mistake which I just couldn’t see.

I was actually making TWO mistakes, but their effects cancel out, leaving behind a cloud of confusion and unanswered questions. (A similar thing happens in software debugging, when two bugs cancel each other out but still create residual odd effects that are almost impossible to trace to any source.) I was also both right and wrong when making my GIF animations, so they didn’t help me locate my errors.

I have to say thanks to everyone here. Without your counterarguments I would just keep making the same mistake. I would have remained confused forever.

I was ignoring something important. When a vortex-launching device throws out a ring-vortex, the device receives (I hope!) some negative momentum while the center of the ring-vortex pattern receives equal positive momentum. The center of the ring-vortex and the launcher-device then start moving in opposite directions, with the total momentum remaining zero. The same thing happens when an astronaut in free fall throws a brick. But the momentum contained in surrounding fluid gives added complexity. So what happens with the fluid flows if they’re confined to a tank?

The center of the ring-vortex moves forward and carries forward momentum. And as I’ve been insisting, the “surround” of the ring-vortex moves backwards and carries backwards momentum… for a total momentum of zero.

But look at the vortex launcher device! The device itself moves backwards, carrying negative momentum. But the fluid surrounding the vortex launcher, it moves forwards, carrying forward momentum, for a net momentum of zero (assuming for simplicity that the device is made of hardened fluid having the same density as the liquid environment.) Same problem as the ring-vortex: zero total momentum.

If they’re inside a very large pipe, then MAYBE the backwards-flowing “surround” of the vortex ring can expand to a very large volume, with very slow backwards flow (yet still carrying constant negative momentum as it expands.) Does the flow get big like this? It seems to be true of the elementary Doublet flows in fluid dynamics, see http://www.idra.unige.it/~irro/elementari5_e.html . They have forward flow in the middle, and backwards flow in the surrounding fluid.

If the pipe is very large, then the same thing happens with the vortex-launcher device: the device carries backwards momentum, while a LARGE volume of the surrounding fluid carries equal forwards momentum, and the momenta sum to zero.

But look! The moving “surround” of the ring-vortex, and the moving “surround” of the launcher device… they’re huge, and the two flow patterns superpose… and they almost entirely cancel out!

ARRRRRRRRG! That was it! All these months I’ve been dinking with this, that’s where the backwards momentum of the vortex-ring was going. It was greatly expanding in volume and being eaten by the corresponding forwards momentum which surrounds the vortex-launcher. (I think! I still have to try putting them both inside a tiny pipe.)

How the heck did I come up with this? I was half-asleep this morning, which makes it very easy to think, since dreamlike intuitive flashes were still appearing like some Greek Chorus from the subconscious. I drew the small pipe and drew the piston. But suddenly I went backwards for no good reason, and called the piston “region inside the ring” with some forward-moving fluid contained inside the piston. Flash! The entire small closed pipe now constitutes an UNMOVING vortex ring. It has forward flow in the middle (the piston) and equal backwards flow outside (the back-flowing fluid.)

Flash! Put the entire closed-pipe-assembly inside a larger closed-ended pipe and move the small pipe forward within it. The net flow and the net momentum still remain zero, since this larger outer pipe and it’s entire contents are NOT moving forward. The small pipe containing the entire vortex moves forward, but it is surrounded by some back-flow which carries exactly equal and opposite momentum. Still a problem!

Here’s the key. Do the same thing again and again, putting the larger pipe inside a still-larger pipe, and we finally obtain a large water tank, where the water is not moving ahead on average, yet it is composed of a tiny hunk of forward-moving water in the center, surrounded by a huge volume of backwards-moving water which contains exactly opposite momentum.

So why can a ring-vortex knock over a cardboard target? It’s because the region of backwards momentum, IF IT EXISTED, would be much larger than the target. In other words, my GIF animations are both right and wrong:


They’re right in that the net momentum associated with a “lone ring-vortex” really is zero. They’re wrong in that the backwards flow is not as shown. It’s not like a narrow backwards stream, instead it exists as a slow flow which occupies a larger volume. Yet the net momentum of the whole thing would still be zero, but only if I don’t take into account the equal and opposite flow-pattern caused by the vortex-launcher.

This answers an earlier question too. As the fluid opens up in front of the vortex, and closes up behind, it doesn’t NEED to flow from the front to the back. Why? Elsewhere there is a vortex-launcher device with fluid opening up ahead of it and closing up behind. The two large patterns combine together, leaving an opening/closing flow but no backwards flow. The excess water doesn’t have to compress to get out of the way. It flows outward, but any backwards motion is cancelled out by forces from an identical outward flow associated with the distant vortex-launcher device which is also moving.

Where does the negative momentum go when the vortex-ring hits the target? In reality, that region of backwards momentum doesn’t exist at all! It can only exist for any “lone vortex-ring” which was somehow inserted into the fluid “by god” without using a physical vortex-launcher device. That’s what my GIF animations show: a lone vortex-ring “inserted by god.” They depict an “ideal” vortex ring which was created mentally, not a real one which was created by a vortex-launcher.

Now I have to modify my animation so it shows TWO hunks of fluid being sent in opposite directions through a larger pipe. That should let me move the black beads in such a way that the backwards flows cancel out.

Again, those backward flows WOULD be there if we didn’t have a vortex-launcher moving one way while the ring-vortex moved the other way.
Wow! I think I FINALLY understand some of the breakthroughs physicists have been having about fish tails and flapping wings in the last five years. I was missing the important fact that the fish or bird itself is part of a larger “ring-vortex.” Someone else mentioned it earlier regarding rowboats, but it didn’t connect. To move faster, the animal must throw out some small opposite-spinning vortex rings. The fluid within the separatrix does carry significant momentum, and can act as “rocket exhaust.”

100

I’m not sure I followed all of that last post, but I think you’re wrong somewhere. However, you are close to the answer. Yes, the vortex gun is the key.

However, the negative momentum is not spread into the fluid to generate counter reverse flows. Note that the gun is “rigidly”* held by the operator, who is “rigidly” fastened to the ground. The reverse momentum is absorbed by the rigid fixtures holding the gun, not returned to the fluid flow. It is like a car slamming into a wall, or a bullet hitting a clay pot. Any of those perfectly inelastic cases shows the velocity of both components after impact equal to zero.

But that is the key to your problem. You are wondering where the input momentum comes from that affects the target, if the fluid momentum all balances out. Answer: from the energy input to the system from the gun. The vortex gun adds the momentum by creating and launching the vortex. The fluid flow does not add any momentum, and only reduces the momentum somewhat via friction losses. The target also absorbs momentum, the momentum added by the vortex gun.

bbeaty said:

Here is a link to a discussion of entropy. It’s several pages, but they’re short and easy to read.

The key is to look at where the momentum came from - the motivation driving the movement. The forward momentum of the vortex comes from energy added to the system from the vortex gun. There is no momentum in the vortex gun to begin with, yet at the end the vortex leaves with momentum. Where did it come from? Answer: the energy input from the gun. That is what drives the vortex forward. The vortex driving forward does not directly drive the return momentum. Rather, it creates a condition of decreased entropy. As the link I provided discusses, entropy can be reduced by the use of energy. Energy is removed from the moving vortex through the vortex pushing the fluid in front of it out of the way. This creates localized pockets of decreased entropy - high and low pressure zones. Once the vortex has gone by, the system tries to return to a higher entropy state, since a higher entropy state is available. Thus entropy drives the system to balance the pressures, and fills in the missing pocket behind the vortex.

Your biggest fundamental mistake is looking only at the momentum. You can’t isolate momentum, but must also look at the energy and entropy of the system.

Here’s another link to a description of entropy and how it affects the world. http://www.secondlaw.com/two.html

  • While a human is not perfectly rigid, the gun could easily be mounted on a rigid frame. This would negate the negative momentum to the gun plunger getting pushed into the surrounding atmosphere.