Useful term: separatrix. The “separatrix” is the imaginary closed surface that divides the moving-smoke part of the vortex from the “outside air” part of the vortex. Example: when a smoke-ring is travelling across the room, its separatrix is roughly sphere shaped, and all the air molecules inside the sphere travels forwards with the vortex-pattern as a whole.
Now the big question (which I asked before, if you read early messages)… WHY do we ignore the flow-pattern in the outside air? We need a very good reason to do so, otherwise it’s a totally arbitrary decision. As I said before, I WANT to ignore that backwards-flowing outside air, but I cannot do so without knowing some detailed physics which FORCES me to ignore that air whether I want to or not.
I cannot ignore that backwards-moving air just because I want to do so, or just because it makes the vortex seem to carry momentum. That would be backwards reasoning. I need some forwards reasoning that REQUIRES that we always ignore the momentum carried by the backwards-flow region.
If we look at a vortex we find that the pattern of flow is much more than just the smoke-filled core region. There is also a backwards-flow region immediately outside. There is no boundary between them. The name “vortex” refers to the whole pattern, not just to the smoke-filled region.
Let me try to make it clearer. In mathmatical terms, to calculate the momentum carried by the moving vortex-ring, we must integrate the momentum contained in each moving parcel of air associated with the vortex pattern. Simple? If we limit this integration to within the separatrix surface, then the vortex appears to carry forward momentum, but if we choose a boundary which encompasses the ENTIRE vortex pattern, then apparently the net momentum is zero. In not less that 500 words please justify the act of limiting the integration to within the separatrix surface.
One possible answer: the air molecules outside the separatrix do not follow the vortex pattern, therefore we should ignore them. In my opinion this is a lousy reason. Those molecules keep transferring their momentum to other molecules, so the “backwards flow” pattern moves with the vortex as a whole, and it really is a part of the entire pattern.
I think you’re wrong. Try doing a crude calculation with squared-off cylindrical flow regions as the vortex-pattern penetrates an imaginary plane. The flow rates through a plane must be equal. The flow velocities depend on the cross sectional areas of the forward-moving and backward-moving regions. I find that if the velocities are unequal, the moving masses are unequal in the opposite direction, so the net momentum carried by the vortex ends up being zero. Maybe I’m making a mistake somewhere.
If a small vortex hits a large flat cardboard target, the forward momentum gets dumped into the target. But where does the vortex’s rearward momentum go?
Also, if a vortex carries zero net momentum, why does a vortex-gun experience a kick as if it had deposited some forward momentum into the air? (I’m pretty sure it has a kick, but I haven’t sat down and measured it.)
I’m confused as to whether we’re talking about a vortex that is acting as a projectile or a stationary vortex. If it’s a stationary vortex then you are correct in that the entire system as a whole experiences no momentum; the only momentum exists within the vortex itself, in the individual particles. If we’re talking about a moving vortex (ala the Whammo gun projectile vortex), then I would think the airflow around the vortex is similar to that of a bullet in that particles are temporarily displaced as the vortex passes, then return to fill the void. In this case, the vortex is no different than say a bullet, in that there is a mass (of particles) moving forward at a measureable velocity. It matters not the relative densities, the only affect that has is on the mass of the vortex.
What makes you think so? It’s not true. Well, not true as far as I have been able to determine from my reading and from reasoning. For example, when a wing cuts through air, the air doesn’t just “open up” to let the wing pass. Instead the air really truly flows backwards, see:
If you’re right about this, it answers my original question. However, I guess it would be better for all of us to avoid making confident but unsupported statements. Instead, initially assume ignorance, then ask:
“IS there any ambient fluid rushing backwards past the object trying to fill the volume recently vacated by the forward-moving object?”
My reasoning with blocks travelling in fluid-filled pipes says that the backwards flow must exist. The diagram in the article above shows this backwards flow associated with aircraft wings. Maybe I can find a fluid-dynamics simulation online somewhere which shows the actual flows surrounding a moving vortex-ring.
Here’s this AVI, but as usual we can’t tell what the fluid outside of the dyed region is doing. (It clearly shows the ellipsoidal “separatrix” volume though. A smoke ring is not just a ring, it’s a travelling blob.)
Let me go back to the OP and address some of your misconceptions. I hope this will help explain the heart of the issue. This will be long-winded, and perhaps a little too back to basics.
This is inaccurate. The oar is not a rocket, spewing out mass to trade momentum.
Consider how the oar works. It is a lever. A simple lever has a bar balanced on a fulcrum. Force in input to one end of the bar, the bar rotates around the fulcrum, and thus force is output at the other end of the lever. The relative forces at each end depend upon the respective lengths on each side of the fulcrum. However, the fulcrum is typically assumed to be immovable.
On the rowboat, the fulcrum is the oar pivot, which is mounted to the boat hull. One end is inside the boat, which the operator pulls on. Force input. The other end sticks into the water. Force output.
Imagine a boat rigidly attached to the bottom of the lake. The fulcrum is now fixed. Pulling on the oar handle makes the oar head push the water around. Easy.
Now let us modify our thought experiment. Instead of a liquid lake, let’s freeze it to ice. Let’s put our rowboat on runners (like skates), and instead of using a paddle, let’s make the end of the oar a sharp point downward. Now when the oar point digs into the ice, it is rigidly fixed at that location. Pulling on the oar handle still inputs force to the handle, but now the oar head has more friction than the boat hull. Ergo, the boat moves forward. This is the ideal best case motion you can get with the boat. There is no loss because the oar head is rigidly fixed to the ice during the motion.
Okay, but what if the point were to slip a little? Instead of having a series of holes from the point, you’d have a series of gouges as the tip moves a bit. But in this thought experiment, the slip is minor, and there’s still more friction between the tips and ice than the runners and ice, so the boat still moves forward. Are you with me?
That is what happens in the rowboat on water. The rowboat has less friction between the hull and the water than the oar heads and the water, but the oarheads slip because the water is not perfectly rigid. The water gives a little, so the boat motion is not a perfect translation of pull to boat forward motion.
Where do the vortexes come in? The oar head is pushing a “lump” of water. This is the loss to the boat motion, the water moving out of the way. This wants to create a high pressure in front of the oar head, and a subsequent low pressure behind. Because of the edges of the oar head, the low pressure behind is connected to the high pressure in front. The motion is circular because it wraps in an arc around the ends, and then the angular momentum keeps it swirling.
Note that the vortex is not what makes the boat move forward - it is a loss to the boat movement. (I keep repeating myself for emphasis.) The water moved to the side to get out of the way of the paddle (reduce the pressure build up), and then moved back in to fill the low pressure. This imparted a circular motion, angular momentum. The angular momentum is what keeps the vortex swirling. Note that not all the water moves directly around the oar head. Some (most) is pushed along. This is the resistance force on the end of the oar, which moves the boat. See how water slipping around the end of the oar is a loss to that?
Okay, the big picture that you are getting to is that water has to fill in behind the oar head, and that water has to come from somewhere. If it is not the water directly moved by the oar head, it still comes from somewhere else. If we’re in a closed lake, then eventually that water is linked in a circle back to the water that moves. Is that a fair restatement of the question you are having?
I think the answer is the difference between a localized effect and an environmental effect. A property of fluids is that they want to maintain constant pressure. The equilibrium, which assumes there are no other dynamic inputs. So we’re in our ideal lake with no other disturbances, just the one boat (and neglecting disturbances from the boat hull for simplicity). Let’s ignore the vortices for a moment. I have already shown why they are peripheral, so lets just pretend they are a boundary from the front to the back of the oar head. They represent a loss to the fluid moved by the oar head, but we’ll call it negligible. They contribute to the pressure behind the oar head returning to normal, but not completely, so we’ll call that negligible. There is still a pressure difference.
The localized effect of the oar head is to push some water backwards. We would prefer it not move at all, but some of it moves. This water pressure is then not in equilibrium with the surrounding water. The pressure build up was caused by our input of momentum, if you will (or input of energy). What causes the pressure to equalize is not from our energy, but entropy. Entropy drives the system from a lump of high pressure in one place to a more even pressure all around the lake.
The same thing is occurring on the negative side of the oar head. There is a low pressure point. Entropy drives the water to balance into the low pressure point. Thus water moves from next to the empty spot to fill it, and from the spot next to that, etc, till the lake returns to equal pressure.
So I guess the answer to the question of why you can neglect the momentum of the returning water (or air or whatever) is entropy. The cause of the returning fluid is entropy, vs. the cause of the leaving fluid being the force imparted by the oar or wing or whatever.
Consider the walking on sand analogy. When you walk, you propel yourself forward with the friction of your feet with the ground. But on sand, ground slips a little, some of the sand moving behind you, a divot opening under your foot as you step away. That moving sand is a loss to your push, which makes your walking a little more difficult. After you’ve walked away, the divot remains in the sand. Now natural forces work over time to fill the divot back in, until no trace remains. These natural forces may be wind or water from tides coming in, but the natural forces are not from your energy. That is the same with the oar in the water. Natural forces push the water in the lake to fill in the empty spot, but that is not energy from your input.
The same applies to the fish. Her tail pushes against the water and pushes her along. Because the water is not static, it moves, which is a loss to her motion. But it’s a small loss, and allows her tail to reset against another “lump” of water for the next push off. Then natural forces move other water to fill in the gap and equalize the pressure, creating the circular flow you speak of.
I have no idea what a Wham-O™ air blaster gun is or how it works, so I can only speculate. The discharge of air is somehow columnated or swirled so it doesn’t spread in a cone like a blast from a blank gun (or a shot gun without the pellets). This swirl is in effect a projectile. It moves through the atmosphere, which reacts as described above. The air in front of the swirl either moves to the side, or gets pushed into the shock wave in front. The air moving to the side returns to its location behind like the vortices on the oar, but the air column in front is the shock wave. It then spreads out through entropy, and more air fills in for the missing air from the rest of the environment because of entropy.
So what about airplanes? Air is displaced downward by the wings. This is both from the bottom of the wing (direct reflection) and by drag over the top of the wing. The downward air blast is the exchange for the plane staying in the air. What about the wing vortices? Those are just like the oars, losses to the system because of the ends of the wings allowing part of the air to swirl back into the opening, and angular momentum keeping the swirling going a bit. The rest of the fill in air comes not directly from the downward thrust air, but from entropy driving the air pressure back to equilibrium. Thus you can ignore the return momentum.
I’m probably hammering this point a little too hard now. But now you can see what is wrong with your gif model. The motion is not 1:1 movement around the disturbance.
micco said:
No, vortices are still losses. Yes, with an oar you want high drag. The ideal* would be have 100% drag, i.e. the water doesn’t move at all. This is impossible, because the water is a fluid. So you maximize your drag. However, the vortex is the effect of the end of the oar shape allowing water to slip around it. I’m not sure what you mean by energy in the vortex balanced by the drag.
Desmostylus said:
I think there’s a miscommunication here. bbeaty is speaking of circular including the atmosphere as a whole. Clearly when a bird flaps his wings and pushes air down, more air has to move into the gap left by the displaced air. That is the circularity that he means. This is confused because of the speak of vortices, and not recognizing the distiction between a localized vortex from slip around the wing and fill in air from the surrounding environment that is only incidentally connected to the air displaced.
Yeah, I picked that up as soon as bbeaty responded to that post. Since then, I’ve been trying to point out, apparently in vain, that while the “fill in” air has the same mass, it doesn’t have to have the same velocity or momentum.
Yes, there is some relatively small, very localized “backward” acceleration over the top of the wing. However, do you notice all those forward-pointing arrows on the right side of the image you linked? That’s the bulk motion of the fluid in the direction of the wing caused by the passing of the wing. A bit of fluid which passes over the top of the wing may (depending on relative velocities) have a period of acceleration toward the rear while it’s directly over the wing, but that will immediately be overcome by the forward motion induced in the main fluid. See the vectors in your linked image numbered 1-5? Those represent fluid that is briefly accelerated “backward” and then reversed.
The air immediately over the wing is accelerated toward the back of the wing by the pressure gradients, but that is a very localized effect. You’re right that the air doesn’t just part to let the wing pass and remain unaffected in the wake, but it doesn’t end up with a persistent motion opposite the wing either. The wing induces an overall motion in the air in the same direction as the wing. Have you ever stood next to the road close enough to feel a car pass or stood on a train platform when a train passed? Which way did the “wind” blow? The bulk motion of the air is in the same direction as the vehicle and the same is true with a wing. The passing of a train does not induce a “rushing backward” flow in the ambient fluid any more than a moving vortex does.
I think you’re letting these small localized velocities confuse the issue when you try to do momentum balances on the large-scale effects. In your plug flow example (baseball in the pipe) there is, of course, some “backward” flow to fill the void left by the initial position of the plug. This is a pressure-driven flow pulled into the wake region. However, The fact that you’re limiting your example to a cylinder with the plug initially at one end may be confusing things because in this case the main body of fluid does move entirely “backward”. If you looked instead at a very long pipe with the plug initially in the center or, better yet, a channel open to a reservoir behind the plug, you’d see that the fluid to fill the plug’s initial position was pulled in (predominantly) from behind and there was no bulk motion “backward”. This situation is a better comparison for your original question about vortices moving in water because the fluid to fill the wake will be pulled from behind and beside the vortex because the fluid right next to the vortex is essentially stationary (aside from the small induced motion from the rotation of the vortex and small Bernoulli-effect accelerations like on the wing discussed above), not rushing backward to fill the wake.
OK. I think I’m closer to the misunderstanding that is happening.
Thus far, we’ve entirely ignored density in all these discussions. The gif animation shows two things that are wrong…
an implicit assumption that the density of the forward moving air is the same as that of the surrounding air…
it ignores the fact that localized pressure differences can and do exist.
As an object (vortex or anything else) moves through a fluid, the fluid in front is compressed creating an area of higher density/pressure, and an area of low pressure is created behind. But the flow is not simply from the high to the low.
The high pressure area is higher than ALL surrounding fluid…therefore there is flow out in all directions. Similarly with the low pressure area behind. Flow into that area comes from all directions.
Eventually, when the object is long passed, and the fluid is at rest (ok, it never is…) there will have been some net back flow of fluid. This would be an equal volume, but not necessarily an equal mass, to fill the area behind the moving object.
Although Irishman’s post is mostly accurate, I can’t stop myself from responding to this little bit. Airflow over the top of a wing is accelerated; if there was drag over the top, the flow would be slowed.
Also, and I apologize for bringing it up, the implication that wings deflect air to stay up is over-simplified and incorrect. As a pilot and a sailor (both highly reliant on fluid dynamics) it’s hard not to want to get this understood. I will admit that it is easier to think of as deflection, but easy to understand doesn’t make it right (hey, just ask the quantum physicists…). Thankfully, those who design the wings of the planes I fly in are willing to endure the complexities.
If you define the system boundary as much larger than the forward moving vortex (eg the whole room with a smoke ring moving through it) then:
we agree that the volume and hence mass of air that moves backward is the same as that of the smoke ring (discount smoke particles and density differences please)
is the total velocity of all the air that moves behind the smoke ring equal to the velocity of the smoke ring? This is the vector addition of all the air along the x axis (direction of movement). I would think yes. This is even if the air comes from anywhere in the room.
However on a localised scale, there is a concentration of momentum inside the separatrix (great word), because of a localised velocity. This is the momentum that imparts a force on knocking over a paper target.
Thanks irishman. Ditto what you said.
Quantguy. Do this (thought) experiment. Fisk tank, filled with water, with a verticle wall down the centre that has a circular hole in the middle. Put dye on one side. Put rubber membrane over the dyed side. Tap membrane. Does a dye ring come out? Water is insignificantly compressible.
What is wrong with saying air is deflected over the wing? Not trying to pick an argument, but just want to be informed. I thought, the deflection (ie change in direction) of the flow of air above and below the wing is what produces a force. I understand that it is not the air hitting the wing that pushes it up, but I did think that the deflection of flow around the shape of the wing (top and bottom) created the delta V which leads to lift. Like the back of a spoon under a stream of tap-water.
Consider an aircraft with a loop wing, that is, circular looking at it from in front. If the wing profile was shaped appropriately it would fly.
There would be little of the turbulence you speak of, since the wash from the wing is radial and the wing has no tip. Any turbulence would be around a torroidal shape.
If this plane flys, then wouldnt that prove your hypothesis wrong?
is not correct. It’s not the vortex that does it. You’re firing a bullet; it’s just made out of air instead of lead. The vortex is a side effect, a loss in the process.
Not sure the point…yes, a dye ring comes out. You literally push it out when you hit the membrane. When you relax pressure on the membrane, fluid flows back the other way to fill the void. What does this demonstrate?
The argument isn’t whether or not the path of air changes. It is about the mechanics of flight. The “deflection” argument is that air hits the wing, reflecting downward, imparting an equal & opposite force upwards, keeping said wing aloft.
I’m not trying to say the path of air doesn’t change…of course it does.
I know - it does not matter whether it is vortex or puff of air (analogy I used at the start of the thread). The momentum only belongs to the puff of air, BUT overall there is no momentum.
It demonstrates that compression of surrounding air is not an issue.
OK I think I get you - just depends on the definition of ‘deflect’ - bounces off or flow around.
Poor word choice on my part. I was referring to the Coanda effect.
From subsequent statements, I think I was saying the same thing you are. It is not just the reflected air, but the air deflected from above the wing. That’s what I was trying to say, anyway. (Reflection is a deflection, but not all deflection is reflection. Capice?)
So we shouldn’t be scared of guns, because overall, there’s no momentum??? If you include the deer and ground inside the system definition, then no momentum gets transferred outside the system, yes.
But when you draw the boundaries at this scale, you’re ignoring what’s happening, not studying it. It’s like saying there’s no net flow of water on Earth and in its atmosphere. Maybe true, but not terribly inciteful.
My apologies if this comes across too strong or sarcastic.
Not to hijack, but this malapropism struck me as funny in context. I think you meant “insightful” since “incite” means to stir things up.
BTW, I agree completely with your point. Properly constructing your control volume is the first step in learning anything in fluid dynamics. Too large or too small and the analysis is pointless.
Not at all, since the “upper” and the “lower” sections of such a loop-wing would each still deflect air downwards, and a region of downwards-deflected air always forms itself into a vortex-pair. There would still be a tip-vortex. It would be large and fuzzy: same total vorticity but with the vorticity spread out in a flat region rather than concentrated in a narrow filament.
The wash from such a wing is only radial if the wing is deflecting air outward in all directions (and therefore generating zero lift.) To generate lift, the wing has to create high pressure underneath and low pressure above, and there are vertical flows associated with this.
It helps to imagine the loop-wing as being an old-style biplane with vertical fins connecting the tips of the wings to form a rectangular tube. Both horizontal wings try to generate tip-vortex patterns. They sum together to give one big fuzzy vortex-pair pattern. They do not cancel out.
Beware of this topic! It’s the “Bernoulli vs. Newton” controversy, and it extends to the highest levels of aerodynamics. University professors will disagree with your above statement. But then other university professors will disagree with the disagreement, resulting in months-long simmering flamewars on physics education list servers. Whole textbooks have been written from one or the other viewpoint. It’s a classic “Swiftian” controversy, where the Big Endians want to fight to the death with the Little Endians over which is the one best way to “crack an egg.”
