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Bricker
01-16-2000, 12:03 PM
In one of the biographies of Richard Feynman, he told the story of trying to solve a thought problem by dint of actual experimentation. It was a funny story, but he never gave the answer!

Here is the thought problem:

Picture a hollow S-shaped tube, with a flexible joint in the middle, perpendicular to the surve of the S. If you pump water through the joint into the tube, it will come out the ends of the S. Pump it with enough pressure, and (because the joint is flexible) the S-shape will start to spin backwards, like a lawn sprinkler.

Now take this same apparatus and put it underwater. Instead of pumping water out, create a vacuum, and suck water in. Which way will the S shape turn?

He gives two arguments, one for each answer.

I don't wish to poison the well, so I won't give either of his arguments here.

Which way will the thing turn?

- Rick

Scylla
01-16-2000, 12:21 PM
Once it's filled up, with water, it won't spin at all as the water pressure will be pushing equally from all sides.

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Often wrong.... NEVER in doubt

Lord Jim
01-16-2000, 12:23 PM
Can I make a guess before the smart people get here?
It turn backwards since the inertia force is on the outside of the curve no matter which direction the water is going, in or out.

Bricker
01-16-2000, 06:47 PM
so no votes for "it would move forward, because you're sucking rather than squirting water?"

- Rick

hansel
01-16-2000, 07:27 PM
Actually, forward was my guess, but it seems so obvious that it must be wrong.

To be clear, by forward I mean that the openings at each end of the S would be pulled forward, in the opposite direction from squirting water through it.

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Never attribute to an -ism anything more easily explained by common, human stupidity.

Scylla
01-16-2000, 10:31 PM
It does nothing. This is why you don't need to anchor the fill side of a siphon while the empty side would wip around under sufficient pressure.

When you drain a fishtank for example, the inlet will remain still unless it is near an obstruction (like a pebble) that impedes the water flow enough to create a variance in pressure, in which case the inlet if it is not anchored will be drawn towards the low pressure zone (pebble.)

As you read this a column of air the height of the earth's atmosphere is pushing down on you. The reason you are not crushed like a bug by this is because it is also pushing you in all other directions including from the inside out. The pressure is exactly equal and you feel nothing.

Such is the case inside Feynman's tube once it is filled with water. The pressure is exactly equal.

Prove me wrong and we will all be instantly crushed by the weight of the earth's atmosphere!

So, argue at your own peril. ;)

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Often wrong.... NEVER in doubt

Bricker
01-16-2000, 10:48 PM
Wait a sec.

Are you saying that, even under water, Feynman's tube wouldn't spin backwards if it were squirting out more water?

I don't buy that at all.

I admit that parts of this thing defy my ability to reason, but that isn't one of them. I am convinced that, under water or not, if the tube is squirting, it will spin backwards.

So why is the pressure "unequal" in that case but "equal" if the tube is sucking water in?

- Rick

Scylla
01-16-2000, 10:57 PM
You do realize that by arguing this you're placing yourself under imminent peril of implosion, don't you?

Are you sure you want to procede?

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Often wrong.... NEVER in doubt

Bricker
01-16-2000, 11:08 PM
Scylla, frankly I'm more worried about having a poster named Charybdis come along and having to choose between your argument and his!

But I'm willing to risk that, as well as the crushing air pressure thing, because I am Truth's Servant.

- Rick

Scylla
01-16-2000, 11:25 PM
I only changed my name 3 days ago, and you're the first to make that reference in spite of the fact I've been "trolling" for it. Thanks for rising to the bait. :)

Here's why we would implode if I were wrong:

It is true that there is is an equal and opposite reaction to every reaction. The mistake made here is in assuming that that reaction would be focussed at the inlet.

In fact, the reaction is that the water level is dropping. In turn it is dropping because it is being pushed down by a column of air equal to one atmosphere.

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Often wrong.... NEVER in doubt

Bricker
01-16-2000, 11:30 PM
Yes, I'm easy - everyone says so.

But I don't understand your reasoning at all. What does the air pressure on the surface have to do with anything? Why isn't it a factor if the water is squirting out?

Why does the thing move for the squirting but not for the sucking?

Er... so to speak.

- Rick

Scylla
01-16-2000, 11:33 PM
meant equal and opposite reaction to every action.

The tube is not squirting. It's not doing anything. The pump or siphon is drawing water through it. The energy (action) is being expended at the pump in the former case, and the reaction is at the outlet.

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Often wrong.... NEVER in doubt

Scylla
01-16-2000, 11:36 PM
The water has more mass/volume than air

Lord Jim
01-16-2000, 11:37 PM
I'm going to jump back in here. Whether the tube is in the air or in the water, all the pressures cancel each other and the tube just sets there. The only pressure difference is from the middle of the tube to the ends. The only thing this will cause is the water to flow through the tube. The only thing that will create any motion by the tube itself is movement of the water.

Newton had a law of motion that that dealt with a body in motion tends to continue in the same direction unless another force is applied. The water will try to go straight, therefore banging it's head on the outside of the curved tube. So, for the water to move along the curved path the outer wall of the tube has to push it out of it's straight line. For every action there is an equal and opposite reaction, (I didn't make that up, somebody said it, Newton I think) so,if the water makes a left, then the tube makes a right.

It doesn't matter which direction the water is going, just that it isn't allow to go in a straight line.

Bricker
01-16-2000, 11:42 PM
I'm sorry to be so slow...

Jim, are you saying that the tube doesn't move in either case? What is the essential difference between the cases that makes the tube move when it squirts but not when it sucks? It seems to me in both cases the water is going around the curve in the tube.

Lord Jim
01-16-2000, 11:46 PM
No, Bricker, I've already said that the tube moves backwards and it doesn't matter which direction the water is going.

Scylla
01-16-2000, 11:48 PM
There is no net directional force in the S tube once it is filled with water. Take an empty garden hose and turn the water on. If you are holding the hose in a coil you can feel it torque as the water moves through it. Once it's filled all the force is at the outlet, and if there is a sprinkler head thingie on it, the thingie will spin.
There is no net force in the S tube when filled because once the water gets the through the S it pulls the water behind it

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Often wrong.... NEVER in doubt

Scylla
01-16-2000, 11:50 PM
Once the garden hose is full it doesn't jump around. See?

Lord Jim
01-17-2000, 12:26 AM
Wait, Scylla We've already established that the pressure of the water flowing through the tube would make the tube turn backwards. We were trying to establish the direction the tube would turn if suction was applied to draw water in instead of if pressure was applied to push the water out.

My hose will try to straighten itself out anytime pressure is applied, but you may have a very special hose that doesn't do that. But anyway, that part has already been established in the original statement.

I still don't see any difference in the direction the tube will turn based on which direction the water is going. It will still turn backwards in both cases.

Scylla
01-17-2000, 12:38 AM
Jimb:
Sorry if I'm not explaining myself well.

Both our hoses (no puns here) attempt to straighten themselves out as a result of the pressure inside them. They do this because the pressure inside the hose is pushing outward. Like blowing up a ballon.

When you turn on the hose it jumps around as a result of the water colliding against the walls of the hose during turns. It's trying to keep its momentum just as you described earlier. Once full it just bulges.

Scylla
01-17-2000, 12:41 AM
Basically we are talking about the equivalent of a rotating sprinkler except that the head is underwater, were drawing water through the hose instead of pushing it out, and the outflow is somewhere else that doesn't come into play. Am I correct?

Bricker
01-17-2000, 12:50 AM
Scylla, your summary of the situation is exactly correct.

Scylla
01-17-2000, 12:56 AM
Good, I was getting worried that I had it all wrong and was about to look very stupid.

The thing to understand is that the whole mass of water is pushing the S curve or sprinkler head equally in all directions, including from the inside out. The head is in fact trapped. It can't move unless another force is applied to cause it to do so.

Lord Jim
01-17-2000, 01:01 AM
Scylla, you seem to be arguing against the given portion on the problem. Plus against yourself. Forget the hose itself. Several things will influence whether it will move around, like the weight of the hose with water in it, friction against the ground. But, your sprinkler continues to turn even after it is full of water because the moving water tries to go in a straight line. The sprinkler changes the direction of the water and therefore is forced in the opposite direction. The S tube in the problem does the same thing.

The question is, though, if instead of applying water under pressure, you draw water thru by suction, does the tube (sprinkler) still move backwards?

KuaNalu
01-17-2000, 01:12 AM
it would spin backwards because the sucking of the water creates a vacuum, and for every action there is an equal but opposite reaction, so it would spin backwards.
It would be easier to show in a diagram, but I can't really draw anything, now can I?

by the way, JimB, we're all a part of the teeming millions....Cecil's the "smart person," if there, indeed, is only one. So feel free to make guesses as often as you like.

Scylla
01-17-2000, 01:18 AM
Jimb:

I maintain that the head does not rotate in any direction once it's under water and suction is applied.

It is not the water changing direction that causes the sprinkler head to rotate when we run water through it. It is the force of the water being expelled from the head.

Like a jet plane.

The water being expelled from the head is pushing it backwards. This is like standing on a sled on a frozen pond and throwing a snowball off. The sled moves in the opposite direction.

The effect of the directional change of the water is nil.

Scylla
01-17-2000, 01:31 AM
As for the hose, forget the other stuff. I'm just talking about the way it jumps or torques when you initially fill it, and then fails to once it's filled.

Once this occurs as the water is being pushed through the hose, it's also being pulled by the water ahead of it. Otherwise there would be a vacuum.

Scylla
01-17-2000, 01:33 AM
Bricker: Am I making ANY sense?

mr john
01-17-2000, 01:42 AM
Yeh backwards the same way as if it was squirting. Feynman's a small thinker anyway. Allatime wantin to make things little. Even the \$250,000 dollar prize is worth more than that.

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"Pardon me while I have a strange interlude."-Marx

Scylla
01-17-2000, 01:43 AM
Bricker:

One of your earlier questions was if the sprinkler would spin if water was being expelled, and the whole apparatus is underwater.

Absolutely.

I liked your pun so much I forgot to answer the question.

Scylla
01-17-2000, 01:59 AM
I can see where that might have confused things, and it's my fault, sorry.

Lord Jim
01-17-2000, 02:04 AM
Ok, let's take your jet engine example. When the plane is setting on the runway and the throttle is pushed forward. The thrust out the back pushes the plane forward. But, if the pilot drops those diverters across the back of the engines, the plane goes backwards. The air still comes in the front of the engine and out the back but since the diverters change the direction of the flow the final result is that the plane goes backwards.

To change the direction of any motion, there must be a force applied. When the direction changes, there is an equal and opposite reaction. So, it shouldn't matter why or which direction the water is moving, it still has to change directions to follow a S curve.

If there wasn't a tap in the middle and the flow went from end to end, then the flow and the forces would cancel since they would be making a hard left then a hard right. But with the tap in the middle the flow into both ends are making only hard lefts. They therefore add to each other rather than cancel each other.

Bricker
01-17-2000, 02:45 AM
Jim, I am with Scylla on this.

The jet engine example is good, because I can now tell you the difference between the squirting and the suction cases.

When the tube is squirting, the flow of water is in one direction: right out of the tube. This creates the strong push backwards that we all agree would happen.

BUT - when it's sucking in water, it's grabbing the water from right around the nozzle, from all different directions, so to speak. So there wouldn't be the same force.

How's this: picture the tube under water... and then we pump out water with green food coloring... we'd SEE the green water shoot directly out. But if we then immediately reversed and started sucking water in, the green water wouldn't be pulled directly back in. Instead, water from right around the mouth of the tube would be pulled in; the green water that was farthest away wouldn't be touched.

Wow.

That wasn't one of the options in the book. Feynman had arguments for forwards, and for backwards (the latter of which we have recreated perfectly, I might add) but never gives any answer.

I think this is it. None of the above. It doesn't move.

- Rick

RickG
01-17-2000, 02:50 AM
Scylla,

I'm going to have to weigh in to agree with others that you are making an incorrect argument. As far as I can tell, you are arguing a problem in dynamics with a statics argument. What you neglect is that, once some of the water is moving, there are dynamic pressure effects that may cause the sprinkler head to rotate (think Bernoulli). You seem to be arguing as if all the pressure forces were entirely static.

I am embarrassed to admit that, despite waaay too much education in physics, I can't remember the resolution of Feynman's problem. Frankly, I'm not sure if he ever came up with a satisfactory theoretical prediction, which may be why he resorted to experiments.

Rick (not the OP)

01-17-2000, 03:27 AM
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; The sprinkler head does not turn because it is shaped like an "S". It turns because water is moving out of it, and the force acting on the water acts also on the sprinkler. It acts in the opposite direction. The sprinkler is also acted upon by the force of the water moving out from its center, but that force remains static, since the center does not move. A straight pipe, capped on the ends, with holes on opposite sides of the horizontal about the axis of rotation would act the same. The static forces of pressure are not what causes the sprinkler to sprinkler to spin, it is the movement of water.

&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; The containing medium would not alter that characteristic, other than to modify the speed with which the rotation took place, because of friction, and momentum from mass. An air "sprinkler" would act the same, in air, or in water. Reversing the direction of the air would reverse the direction of the rotation. It would also involve additional frictional elements because the air drawn in is dynamically involved in the forces opposing rotation, unlike the converse case, where our water appeared at the center from outside of the system. Water propellant in a water environment would do so even more. None the less, it would tend to rotate in the opposite direction, albeit inefficiently.

&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Now, go out and get your garden hose, and a sprinkler, and go into your bathroom, fill the bathtub, and try it out. You will need a pump, for the reversed flow, unless you live on the top floor of a tall building, and have a very long hose.

<P ALIGN="CENTER">Tris</P>
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"...it doesn't matter how beautiful your theory is, it doesn't matter how smart you are -- if it doesn't agree with experiment, it's wrong."
-- R.P. Feynman

douglips
01-17-2000, 03:40 AM
There seem to be a few arguments:
JimB: The centrifugal force of going around the curve is the same in the case of squirting or sucking, so it spins the same in either case.
hansel: The act of water being sucked in will suck the tube along, in exactly the time-reversed way that squirting makes it spin the other way. It turns out this is correct, though not because of the suction acting like reverse thrust theory.
Scylla: It doesn't go anywhere because the water going into the tube is coming from all directions and is not directed like it is in the case of a squirt.

First, lets cover the 'suction as a time-reversal of squirting' option. This can't be because fluid that is moving will willingly change directions when being sucked but not when being squirted. To see the difference, hold your hair dryer up and see how far away you can be and still feel the air blast. Now, hold your vacuum cleaner and see how far away you can feel the suction - it's a much smaller distance. As Scylla said, the water does go in from all directions. Picture someone exhaling a bunch of cigarette smoke. It shoots out in a fairly straight line, then diffuses. Now imagine seeing that film in reverse - all the smoke shooting straight into the guys mouth like a piece of spaghetti. Not gonna happen. Simpler experiment: Blow out a candle. Now, light it again and try to 'suck' it out. (danger :Don't do this if there is the remotest chance you could hurt yourself and sue me.)

So, the suction option doesn't fly. What about the centrifugal force? The argument about the jet engine is intriguing, though it is worth noting in passing that it demonstrates that the suction of the engine is not what provides thrust, lending more credence to the above logic.

But what about the change in direction (centrifugal force) vs. reaction thrust (rocket-engine type propulsion)? Without doing any math, let me try to convince you it is the thrust that matters most. If it is the change in direction that is important, a jet engine with reversers deployed and running in reverse would work just as well to stop the plane as with the engine running normally.

Let's just think about this. A jet engine running normally without the reversers achieves a lot of thrust without any direction changing of exhaust. Thus, the thrust must be important. If we now deploy the reversers and the thrust points in the other direction with 100% efficiency (not the case, but let's just say) then the engine is working just as well to slow the plane.

But wait! We get the extra bonus of the change of direction of the thrust bouncing off the reversers! That must be ADDED to the engine's thrust, giving a more efficient engine! So then, jet engines should be mounted backwards and fly with their reversers deployed to take advantage of this.

Nope, if Boeing could eke another 5% efficiency out of a jet engine they'd do it in a heartbeat. So, it must be the thrust that matters most, and since we can't 'suck' thrust, the forces on the inverse sprinkler are going to be smaller.

The final answer is, yes, the change in direction is a component of the forces on the sprinkler (or the jet engine) but it is small compared to the thrust component, and it is exactly balanced by the much smaller suction-anti-thrust component.

In fact, if you have a very good experimental setup, you will get the thing to go in the suction direction, but only slowly. The only reason that an inverse sprinkler goes in the opposite direction of the normal sprinkler is the angular momentum change that Scylla describes above with his jerky water hose. After the flow becomes steady, the sprinkler stays spinning slowly, as opposed to spinning quickly when squirting. And, if you add any friction, the thing will just sit there.

Not terribly helpful: MIT's Edgerton Center (http://web.mit.edu/edgerton/Corridor-Lab/feynman-sprinkler.html)
Much better: University of Maryland Physics (http://www.physics.umd.edu/deptinfo/facilities/lecdem/d3-21.htm)
Also good: http://www.wiskit.com/marilyn/sprinkler.html

douglips
01-17-2000, 03:44 AM
Grrr - blew it.
Hansel has the direction correct, but the reason for the direction wrong, and the effect is much smaller than in the normal sprinkler mode.
The final result is that it will spin in the opposite direction to the normal sprinkler mode, but much slower.

Bricker
01-17-2000, 04:03 AM
The MIT link claims if you do the experiment, it won't move.

The U of M link claims it will move forward.

Apparently they are both describing their actual experimental results.

NOW, I am well and truly baffled. This is clearly confounding the great minds of our time.

- Rick

Zor
01-17-2000, 07:53 AM
Hmm... I still can't believe I actually registered myself just so I could post this. I guess even my ego can have a field trip some days. Anyway, below is a simple argument as to why, under ideal circumstances, the S-shaped tube will stay still while it sucks in fluids.

Key Concept: Conservation of Linear Momentum

First, let's examine the case of your everyday lawn sprinkler. Just for sake of reference, let's place the S-shaped tube on a Cartesian plane with your standard x-y coordinates. Now since the system has symmetry, let's cut of the lower part of the S-shape tube, so we now have a C-shaped tube, with water entering from it's lower leg. Furthermore, we'll also assume the C-shape tube is a half circle.

1): When water enters the C-shaped tube perpendicular to the x-y plane, it has no momentum in the x nor y direction.

2): When water exits the C-shaped tube from the top, it now has momentum in the +x direction.

3): Net change in linear momentum is +x.

4): Since linear momentum of the system has to be conserved, the +x gained by the exiting water must be provided by a corresponding -x of the tube itself. This means the tube will spin counterclockwise.

Now with the above in mind, we can tackle Feynman's problem with ease.

1): When water is just beginning to get sucked in, it has no momentum in the x nor y direction.

2): When the water leaves the tube, it leaves perpendicular to the x-y plane. It therefore has no momentum in the x or y direction either.

3): Net change in linear momentum is 0.

4): No momentum change necessary for the tube, no movement for the tube.

I'd like to stress that any number of real life factors and imperfections in experimental technique and or equipment can throw the final result off. For example, the entrance and exit for the C-shaped tube must be perfectly parallel. You can probably see that even a slight discrepancy will force the tube to provide momentum in the y direction, thus inducing a net torque that turns the tube. Toss in local variations in pressure and the initial momentum of the fluid, it would be hard to perform this experiment with anything except your mind. I believe the reason why the freshmen physics project found the suction sprinkler to turn in reverse is due to such factors. The fact that they reported the suction sprinkler didn't work with air tends to confirm this suspicion (not enough momentum change to overcome friction due to the low mass of air).

The principal still stands however :)

Mr Thin Skin
01-17-2000, 09:15 AM
I just gotta throw in my \$0.02. After reading these posts, I kept waiting for Bernoulli to be used. He was mentioned, but not used. Here's my backwards rotation argument: Yes, in the reverse case the water is moving toward the entrance from all directions, but at the entrance to the tube, there will be water moving radially across the tube wall's face (tube has non-zero wall thickness). This moving water will be at a lower pressure that the water "behind" the tube (sum of vectors on back side of "S"). Because of the dynamic situation, the the will be pushed backwards. Furthermore, this should be a very slight effect. If cajoles, I can produce drawings (only if you promise not to laugh).

To clarifiy what I mean by radially moving water, image you are staring straight into the tube opening. The tube appears to you as a ring of non-zero thickness. This radially moving water moves from outside the ring, across the ring's thickness, and into the inner circle.

Mr Thin Skin
01-17-2000, 09:20 AM
I am sorry about all the typos. "that the," should be "then the", "the the" should read "The tube," and "cajoles," should read "cajoled."

Sheesh, I did reread, I just read what I thought I wrote.

Sorry Again

Bricker
01-17-2000, 11:20 AM
I'm not quite sure that what you describe is exactly what Bermoulli was talking about.

While it's very true that the Bernoulli principle involve pressure differential on two sides of an object causing the object to move in the direction of the lower pressure -- I thought the key element Bernoulli explains was why there is lower pressure (airplane wing shape and distance for flowing air to travel being greater on top than on the bottom). So I don't think the pipe represents the Bernoulli principle so much as it it MIGHT just represent differential pressure. (I say MIGHT only because I was all set to buy into the standing-still theory until you messed it up for me again!)

- Rick

Scylla
01-17-2000, 11:40 AM
Bricker:

In the experiments, did they apply a LOT of suction in one case (more than one atmosphere and enough to cause cavitation.) If they did, it might very well move forward.

Also, if the S were immersed in a relatively small pool of water, it might create a whirlpool effect that would move the sprinkler.

1. It doesn't move because God doesn't want it to.

2. According to the Heisenberg Uncertainty principle, it is possible to either know the sprinkler heads location or spin, Never both. Since we know the sprinkler is underwater, it is impossible to determine spin.

3. The sprinkler head won't move until you observe it.

4. When you put the sprinkler head under water the universe splits in three, in order to actualize all possibilities.

5. Feynman is screwing with us.

I'm glad this debat brought new people!

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Often wrong.... NEVER in doubt

Scylla
01-17-2000, 11:44 AM
Bernoulli has nothing meaningful to do with this

douglips
01-17-2000, 11:58 AM
Re the experiments: The UMD experiment was more skillfully designed, and goes along with an article referenced on one of the pages I linked to - I didn't find a link to the article, but it was published in a physics journal so it is findable and believable.

Zor, your linear momentum argument is good, but you are missing the angular momentum factor - this is why the sprinkler goes backwards if you remove friction. It only goes fast enough to preserve angular momentum, not as fast as if you were squirting water out.

Scylla
01-17-2000, 12:11 PM
Douglips:

I Checked links. I still say Sprinkler doesn't move, and if it does we all die.

Marilyn was already defeated by Cecil in another column, and as such lacks the infallibilty of the Dopers.

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Often wrong.... NEVER in doubt

Scylla
01-17-2000, 12:22 PM
I would also mention that the experiment exploded in one of those links.

As I warned at the outset this is an extremely dangerous avenue of inquiry.

Perhaps there are some things man was not meant to know.

Heh. heh. heh.

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Often wrong.... NEVER in doubt

Mr Thin Skin
01-17-2000, 12:35 PM
Let me try again, please. Below is an small little gif I created illustrating my argument. If you don't like the image, I'll happily kill it.

The red arrows illustrate the flow across the tube's thickness. I maintain that this flow results in a lower pressure area.

Mr Thin Skin
01-17-2000, 12:39 PM
Dammit, I followed the FAQ to the letter. The image is here:

Scylla
01-17-2000, 12:42 PM
Mr. Thin Skin:

I'm sorry, but I can't see your GIF.

In an earlier reference I made to low pressure underwater and the movement of a siphon towards an obstacle I misspoke.

The currents pull the pebble and siphon together, not the pressure.

The foundation of hydraulics is that water is not compressible.

The pressures remain exactly the same despite the currents.

Mr Thin Skin
01-17-2000, 12:42 PM

I give up.

Scylla
01-17-2000, 12:44 PM
I get "The page you select is not currently available."

Scylla
01-17-2000, 12:49 PM
MIT and physicists worldwide disagree. We are being thwarted by forces beyond are comprehension.

Perhaps this experiment represents a glitch in the software of life.

We need to appeal to a higher power.

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Often wrong.... NEVER in doubt

douglips
01-17-2000, 02:16 PM
Scylla writes:I Checked links. I still say Sprinkler doesn't move, and if it does we all die.
Your 'column of air will destroy us' is both amusing and pointless to this discussion. Either you did not check links, or I did not provide enough. I will remedy that below. The fact is that if the experiment is done with little friction, the sprinkler does revolve backwards (opposite squirt mode.)

Scylla again:Marilyn was already defeated by Cecil in another column, and as such lacks the infallibilty of the Dopers.Which is exactly why I linked to a page explaining why she was wrong.

The experiment has been done, and was published in the American Journal of Physics (http://ojps.aip.org/journals/doc/AJPIAS-home/top.html), a well respected scientific journal. If you go to the AJP's search page (http://ojps.aip.org/journal_cgi/search?KEY=AJPIAS) you can search for 'Feynman AND sprinkler' (here is a link to the results, but it is long and may not work: Try it (http://ojps.aip.org/AIPcgipath/doc/search-AJPIAS/disk3/journals/journal_cgi/search?origquery=&vdk_query=&page=1&chapter=0&docdisp=0&smode=strresults&query_type=search&KEY=AJP IAS&CURRENT=NO&ONLINE=NO&SMODE=strsearch&possible1=Feynman&possible1zone=article&bool1=and&possible2=sprinkler&possible2zone=article&fromyear=&frommonth=&fromday=&toyear=&tomonth=&today=&fromvolume=&tovolume=&fromissue=&toissue=&sort=rel&maxdisp=10&threshold=0&%5BSearch%5D.x=58&%5BSearch%5D.y=12))

You will find 5 documents that have been published over the years regarding this issue.

The definitive article(s) are (links to abstracts may work)
American Journal of Physics -- July 1989 -- Volume 57, Issue 7, pp. 654-657
The Feynman inverse sprinkler problem: A demonstration and quantitative analysis (http://ojps.aip.org/AIPcgipath/doc/tomonth/disk3/journals/journal_cgi/search?KEY=AJPIAS&CURRENT=NO&ONLINE=NO&smode=display&docnum=4&coden=AJPIAS&vol=57&iss=7&page=1&chapter=0 &fpage=654&seqno=1&issuedate=July%2B1989&journal=American+Journal+of+Physics&numret=5&sort=rel&maxdisp=10&docnum=4&origquery=%28%28Feynman%29++%3Cand%3E+sprinkler%29+&threshold=0&a bscall=strresults&logid=YES&allprl=0&pjournals=&pyears=)
Richard E. Berg and Michael R. Collier
Lecture–Demonstration Facility, Department of Physics and Astronomy, University of Maryland, College Park, Maryland 20742

American Journal of Physics -- April 1991 -- Volume 59, Issue 4, pp. 349-355

The Feynman inverse sprinkler problem: A detailed kinematic study (http://ojps.aip.org/AIPcgipath/doc/tomonth/disk3/journals/journal_cgi/search?KEY=AJPIAS&CURRENT=NO&ONLINE=NO&smode=display&docnum=2&coden=AJPIAS&vol=59&iss=4&page=1&chapter=0 &fpage=349&seqno=1&issuedate=April%2B1991&journal=American+Journal+of+Physics&numret=5&sort=rel&maxdisp=10&docnum=2&origquery=%28%28Feynman%29++%3Cand%3E+sprinkler%29+&threshold=0& abscall=strresults&logid=YES&allprl=0&pjournals=&pyears=)

Michael R. Collier, Richard E. Berg, and Richard A. Ferrell

Department of Physics and Astronomy, University of Maryland, College Park, Maryland 20742

Excerpt from the abstract of the first article:...The inverse sprinkler moves in a direction opposite to that of the normal sprinkler ... Recent experiments designed to test this effect may have produced a null result because of friction associated with the systems...

Does that satisfy you? I feel no impending implosion from a column of miffed atmosphere.

douglips
01-17-2000, 02:19 PM
The links to the abstracts and search results above are in fact broken. To read the abstracts or order the articles for your very own selves, go to the AJP Search page (http://ojps.aip.org/journal_cgi/search?KEY=AJPIAS) and search for Feynman AND sprinkler.

Enjoy.

douglips
01-17-2000, 02:41 PM
I think I figured out how to link to these darn things.
The Feynman inverse sprinkler problem: A demonstration and quantitative analysis[b] (http://ojps.aip.org/journal_cgi/getabs?KEY=AJPIAS&cvips=AJPIAS000057000007000654000001)
Richard E. Berg, Michael R. Collier
American Journal of Physics, Volume 57, Issue 7, pp. 654-657
[b]The Feynman inverse sprinkler problem: A detailed kinematic study (http://ojps.aip.org/journal_cgi/getabs?KEY=AJPIAS&cvips=AJPIAS000059000004000349000001)
Michael R. Collier, Richard E. Berg, Richard A. Ferrell
American Journal of Physics, Volume 59, Issue 4, pp. 349-355

douglips
01-17-2000, 02:42 PM
I think I figured out how to link to these darn things.
The Feynman inverse sprinkler problem: A demonstration and quantitative analysis (http://ojps.aip.org/journal_cgi/getabs?KEY=AJPIAS&cvips=AJPIAS000057000007000654000001)
Richard E. Berg, Michael R. Collier
American Journal of Physics, Volume 57, Issue 7, pp. 654-657
The Feynman inverse sprinkler problem: A detailed kinematic study (http://ojps.aip.org/journal_cgi/getabs?KEY=AJPIAS&cvips=AJPIAS000059000004000349000001)
Michael R. Collier, Richard E. Berg, Richard A. Ferrell
American Journal of Physics, Volume 59, Issue 4, pp. 349-355

douglips
01-17-2000, 02:43 PM
Looks like I need to go the remedial HTML farm. :(

Scylla
01-17-2000, 03:59 PM
I did what you suggested, and found brief summaries for 5 or so articles, But no real data.

Some say it moves backwards, some say it don't.

I'd say it's unresolved.

In a small resevoir, or an immense amount of suction you might get the head to rotate due to a whirlpool effect.

Cavitation in the latter case may also have effects.

Both of these effects (neither of which has been addressed,) as well as the friction problem, leave the issue unresolved. Quite simply we would need ot order the articles and see for ourselves.

Im not kidding about the column of air and ambient pressure (though it is amusing.)

Sorry you disagree.

------------------
Often wrong.... NEVER in doubt

Scylla
01-17-2000, 04:04 PM
Douglips:

Just saw the choose thing you put in the phaser discussion. Very cool, how'd you do that?

Scylla
01-17-2000, 04:14 PM
Also, one of the summaries seemed to imply that depending on very minute differences in how the experiment is set up, you get different results.

Deep underwater, with no currents, cavitation, or nearby masses the head doesn't move.

Unfortunately, the links as they stand show as much evidence as if I said "Hey, I tried it in my bathtub and it didn't spin."

This is not enough to change my mind.

With no offense to physicists, this wouldn't be the first time an experiment was conducted in a flawed manner.

Due to the disagreement in your links, it's obvious that this has occured.

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Often wrong.... NEVER in doubt

douglips
01-17-2000, 04:22 PM
Unfortunately, I'm not willing to pay \$30 to get the article which I couldn't publish here anyway. From what I've seen, my conclusion is that the non-spinning experiments are due to friction, and that better designed experiments show the spin.

Anyway, i think this has been done to death.

As for the select thingie, use the HTML select tag, something like this:
&lt;select&gt;&lt;option&gt;Option 1&lt;option&gt;Option 2&lt;/select&gt;

<select><option>Option 1<option>Option 2</select>

ZenBeam
01-17-2000, 04:51 PM
I think it will move, if there is low enough turning friction. In normal operation, the sprinkler recoils due to the water shooting out in a jet. If the water shot out in all directions equally, it wouldn't move. When the sprinkler operates in reverse, if it sucked in water from all directions, it wouldn't move, but it doesn't suck water in equally from all directions. In a spherical region centered on the tip of the sprinkler, water can't come from
the region occupied by the tube. It will come preferentially from in front of the tube, breaking the spherical symmetry of water flow. Since there is more water flowing toward the nozzle from the front, the sprinkler will turn slowly in the opposite direction to its normal sprinkling direction. Those experiments where the sprinkler doesn't turn can be explained by there being too much rotational friction. Arguments which approximate the flow of fluid as being spherically symmetric have approximated away the effect which breaks the symmetry and causes the motion.

------------------
It is too clear, and so it is hard to see.

Scylla
01-17-2000, 04:52 PM
Agreed. It either do, or it don't.

I ain't spending \$30.00 either.

Thanks for the tip.

Stephen
01-17-2000, 05:32 PM
Bricker said....
I thought the key element Bernoulli explains was why there is lower pressure (airplane wing shape and distance for flowing air to travel being greater on top than on the bottom).

Sorry, but the "pathlength difference" explanation of Bernoulli's effect, over an aircraft wing, assumes physical laws which don't exist.

Scylla said....
The foundation of hydraulics is that water is not compressible.

Water contains gasses in solution. Gasses are compressible. The foundation of hydraulics is not that water is not compressible, but that fluids are not compressible. If water were totally incompressible, we could freely dive to the deepest depths of the ocean. As it is, there's a pesky 1Atm of pressure increase, for every 33 feet of depth.

------------------
Stephen
Stephen's Website (http://stephen.fathom.org)
Satellite Hunting 1.1.0 visible satellite pass prediction
Satellite Hunting (http://stephen.fathom.org/sathunt.html)

Stephen
01-17-2000, 06:55 PM
I said...
If water were totally incompressible, we could freely dive to the deepest depths of the ocean. As it is, there's a pesky 1Atm of pressure increase, for every 33 feet of depth.

I see the error in the above statement. A better analogy...
A hydraulic car jack, filled with tap water, would be less efficient than one filled with oil (or even deoxygenated water), since some of the effort would be spent compressing the gasses in the water.

The water at the bottom of the ocean is still under more pressure than the water near the surface (a measured volume of water at depth contains more mass than the same volume near the surface), but even if that weren't the case, a diver would be subjected to the same forces as the gasses. So, an ocean of incompressible fluid, would be no more hospitable to divers than water. My mistake.

------------------
Stephen
Stephen's Website (http://stephen.fathom.org)
Satellite Hunting 1.1.0 visible satellite pass prediction
Satellite Hunting (http://stephen.fathom.org/sathunt.html)

Scylla
01-17-2000, 06:58 PM
Arrg!

Water, fluids, same difference! I was not about to launch into a discussion of hydraulics in general, and we we're talking about water. Your technicality of a correction is noted, and the point is conceded.

Your statement about being able to dive to the depths of the if it were not for the solubility of gases in water is 100% false.

Sorry about that, but it's the truth.

When you dive down 33 feet (assuming you started at sea-level,) Your body is at exactly 2 atmospheres of pressure. One atmosphere is from the weight of the column of air above the water, the other is from the weight of the water above you.

At this point, if you have held your breath then the volume of air in your lungs is exactly 1/2 what it was at the surface. If you have not performed the Valsalva manuever by now, your eardrums and sinuses have probably ruptured.

For our purposes the volume of air in the 33ft of water above you, and it's weight are insignifact compared to the weight of the water.

This is why your ears don't pop when you walk up the stairs.

If you were to dive 1000+ feet into a vat of hydraulic fluid with no soluble gas in it you would be crushed and probably die, just as quickly as you would in the ocean.

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Often wrong.... NEVER in doubt

Scylla
01-17-2000, 07:00 PM
Stephen:

Thanks for correcting yourself.

:)

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Often wrong.... NEVER in doubt

NightGirl44
01-17-2000, 11:16 PM
Jeepers Creepers!![/i]

A little applied physics here. Forcing air OUT of it creates a backward spin due to thrust conversion /action v/s reaction.

SUCKING [b]IN water creates a reverse action provided the central pivot joint is loose enough and the resistance v/s foot pounds per second of the suction v/s volume of water intake (too little water will NOT cause a reaction) which means that the S will move in the direction directly TOWARDS of the suction (or the reverse of the expulsion of volumes of air). In essence, it will be pulling itself along instead of responding to a thrusting action.

AGAIN, this depends on the force of the suction verses the density of the water. Too little suction power means that the S will remain stationary. Insufficient data was given for this problem. :D

Bricker
01-18-2000, 01:04 AM
Actually, this is a good point.

There have been reasoned, erudite analyses for both the "stays still," "moves forward," and "moves backwards," camps.

Perhaps if Ed is reading this, he could bring it to Unca Cecil's attention. Since MIT and U of M seem stymied, we clearly must appeal to a higher authority.

- Rick

Scylla
01-18-2000, 01:09 AM
We must now all bow our heads, and invoke the ritual chant:

Ahem,

Repeat after me.

"I will send you a sawbuck. I will send you a sawbuck. I will send you a sawbuck."

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Often wrong.... NEVER in doubt

Rocket
01-18-2000, 01:42 AM
This is an interesting series of explanations. My initial reaction when reading the problem statement was that the device would not move, and I have not read anything that would change my opinion on that.

I have no explanation of this but here are my thought processes:

First, imagine the "normal" sprinkler spraying water from the nozzles into the air. Now imagine air is being sprayed through the system instead of water. The sprinkler would still spin in the same direction although probably slower (depending on the pressure used). This made me think it was a friction problem not a suction vs. pressure problem. the pressure of the water spraying from the ends overcomes the friction of the apparatus itself and the air it is moving through. If you reverse the flow under these conditions, the movement will stop (just as the one of the previous posters described using the blow dryer/vaccuum cleaner example).

Now take this a step further to the underwater scenario. It seems if you pump water through the system, it will only move if the pressure overcomes the friction resistance of the apparatus and the water it is moving through (probably will move slowly in this case). Now reverse the flow, but not by suctioning through the device, but by increasing the pressure on the pool it is in (assuming it is a closed system, except for the outlet from the apparatus). It seems obvious to me in this case that the apparatus will not move.

Feel free to shoot this entire post full of holes.

My God, it's full of stars.

moriah
01-18-2000, 03:37 AM
I find myself agreeing with Zenbeam again!

Sure, the conservation of angular motion argument would indicate that the nozzle doesn't turn, however, for as long as the thing is pumping, it is creating a 'vacuum' (read: lower density region) in front of the nozzle. The nozzle gets 'pulled' into that lower denstity region. (It's the spaghetti slurping problem all over again!)

And Scylla, please lay off the column of air argument. If everything 'equalized' then the sprinkler nozzle wouldn't even move backward if air were pumped through it. Even though we have so many pounds of air pushing on all sides, the pressure varies in regions for varying reasons. You know, like wind. IOW, the nozzle moving does not mean we get crushed.

Peace.

Pasta
01-18-2000, 09:57 AM
I hope this proves useful:

The key in both cases is how much angular momentum is being lost via the escaping water. First the normal case, then its counterpart. Let the z-axis be the rotation axis (i.e., the S-curve lies in the x-y plane).

Normal: Let's create a set of initial conditions. We'll say simply that the water has been flowing for some time now and the sprinkler head is currently not spinning. If you'd like, this can be established by running the sprinkler for a bit and then grabbing the head while the water continues to flow. At this time (t=0), the system has a certain angular momentum (L). Since the head is not spinning, the only thing contributing to L is the water in the tubes. Now if we increment t just a bit, we notice that some momentum-carrying water has come out of the end of the tube. Momentum conservation requires that this "lost" momentum be accounted for in the system. Pressure, etc., has dictated the speed of the water flowing around in the tube (relative to the tube) and this will not change. The only thing that can make up for the momentum is the sprinkler head itself. So, the small bit of momentum (dL) carried away by the water during this infinitessimal bit of time has been made up by a change of -dL in the sprinkler head's angular momentum -- leaving total angular momentum at L for our micro-universe (which, of course, includes the expelled water.)

Underwater: Set up the same sort of initial conditions -- flow has been going for some time, and at t=0, the head is stationary. THE KEY TO LIFE: The only way the sprinkler head will start to spin is if some momentum escapes. Remember that even though there is non-zero angular momentum in the water in the arms of the tube, it was there at t=0 -- there has been no net change to be made up for. The only way for momentum to leave: down the center escape tube. Thus, if the water goes down the tube in a sort-of spiral way (like your toilet water), then there is a finite dL/dt (momentum escaping per unit time). This must be accounted for if we want our micro-universe's L to be conserved, so the sprinkler head will have to start spinning.

This sheds light on the experimental inconsistencies -- the L lost down the center tube is highly geometry-dependent. Some geometries will allow the water to transfer most of it momentum back to the head before leaving on its merry way. In such a system, the head would not spin up very much.

I should say, for completeness, that everywhere in the above that I have used angular momentum, I could have (and really should have) used only the z-component of angular momentum. Recall that the three components are independently conserved, and it is only L_z that we need here.

This also makes it clear why the two cases are so different. In the normal case, the water is escaping in a manifestly momentum-carrying way. In the sucking case, the amount of momentum carried away is not so manifest.

-P

Pasta
01-18-2000, 04:22 PM
CurtC,

You said:
I'm surprised at the length of time it's taking to arrive at an answer to this one. It seems to me there is a simple explanation...

The problem is tricky (and therefore time-consuming) because it is misleading; there are many obvious and simple, yet incorrect, explanations. And, as yours is right there after mine, I will address its error. But, first...

Bricker: Yeah, I skirted that issue. The main reason is that the only way I can think to succinctly explain the error in the "sucking" argument is with diagrams. Hmm... Maybe I'll whip some up later in MS Paint.

Back to CurtC...

Yes, your setup is equivalent to Feynman's. Unfortunately, you got lost in your own handwaving (happens to us all). You state:

The fluid inside the container is under a higher pressure than outside (why else would it squirt out?), and that pressure on all sides except one, so the pressure moves it to the left.

I assume that final "it" has as its antecedent "the container." Your handwaving begins when you claim that "the pressure moves [the container] to the left." But really it doesn't. The pressure difference causes some fluid to be pushed out to the right (in your diagram) and, by momentum conservation, the container and its contents move left.

You then magically (and invalidly) reverse all the forces. A good way to think about this case (when the fluid in the container is at the lower pressure) is as follows: (Incidently, this is a good way to think about _any_ problem involving pressure since pressure is nothing but a statistical concept.)

Take a fluid with a pressure gradient in it. For simplicity, say that on one side of an imaginary plane we have one pressure (P1) and on the other side we have a different pressure (P2). Deep within the P1 (or P2) side, a given molecule is bouncing around without any particular care of direction. At the boundary, however, a molecule has a better chance of travelling toward the lower pressure side (i.e., it has a higher mean free path in that direction). No forces, no pulling. Simply, the lower pressure side invites over molecules from the higher pressure side if they happen to be headed that way.

Your container just sits there. Molecules from outside the container tend to move into it, but nothing at all is tugging on the container.

If this is still unsatisfying, take the pressures to the limit: Say the pressures are _so_ low that there are, on average, five molecules inside the container and twenty molecules inside the same volume of space _outside_ the container. As all these molecules bounce around, the ones near the boundary will have a relatively easy time making their way into the container since there is really nothing there for them to hit. Thus, molecules tend to make their way into the container. No forces on the container. Certainly nothing pulling it towards the open end. The result is simply a net fluid flow into the container.

Further, if you can somehow expel these from the back end as fast as they come in, the container just sits there -- the Feynman inverse sprinkler.

This is not to say it doesn't move: see my explanation a few posts back of how it _does_ move. I've simply shown that it is not due to sucking on fluid.

I must apologize for the length of this post. Thanks to those who have taken the time to read all the way through it. :)

Bricker: Hmm. It seems I have sort of done what I said I needed pictures for. I do have another way of thinking about it that definitely does require pictures; I'll share if some of you still feel unsatisfied.

-P

fLoWeRcHiLdGenerationY
01-18-2000, 05:13 PM
[QUOTE]Originally posted by KuaNalu:
it would spin backwards because the sucking of the water creates a vacuum, and for every action there is an equal but opposite reaction, so it would spin backwards.
It would be easier to show in a diagram, but I can't really draw anything, now can I?

~~~~
yeah okay, i don't get it - explain to me, someone why it wouldn't move or spin backwards???? i totally agree with KuaNalu....for every action there is an equal but opposite reaction that's the law, rite?? so, if the tube has become a vacuum and is sucking water in, it would have to move in the opposite direction (backwards) to make up for the direction that the water is flowing, rite??

Look, I asked my mom and dad about this, and they read Feynman (have plenty of his books) and my parents back me up on the explination above - my dad even said we could build a model and try it out in our pool (when winter's passed, of course!!)

I still say it'll spin backwards. If not, someone email me and tell me why.
-hayley

------------------
I'm not weird, I'm just Gifted...okay, so I'm weird too...
~I'm 15, people, but don't doubt my intelligence~
*fLoWeR cHiLd, 2nd generation...
"Im not opinionated, Im just always right." (c)Me

Scylla
01-18-2000, 05:20 PM
Sorry to jump in, especially as I am envious of Pasta's remarkably cogent and well put together explanation.

I still have a problem with the explanation though.

It is my understanding that the pressure in a body of liquid is uniform (discounting any gases dissolved into it,) and there can be no low and high pressure action within that body of liquid moving the sprinkler. (Hence my poorly received implosion joke.)

Water is not compressible so where is the differential, and how is it being created?

There is also no angular momentum because once the water has made the turn in the sprinkler it pulls the water behind it with exactly the same force with which it resisted the turn.
If the suction is great enough to overcome this you have cavitation (recall that one experiment blew up because of this very thing)If we discount the volume of air in the water as I believe the experiment is intended, then we have a pure exercise in hydraulics where the presure is uniform throughout the fluid.
The only thing that could move the head is a current, which would be faulty experimental setup.
In air, I can see how the sprinkler head might rotate due to low pressure but in a fluid?
Maybe I'm being pissy about the pressure changes, and I should just accept that dissolved gases are creating a pressure differentiall that pulls the sprinkler. Even if that's the case we are talking about immense suction.

I know that the angular momentum argument is bullshit (no offense intended,) as does anybody else who has ever replaced the seals on a hydraulic cylinder. The hoses twist every which way, but remain motionless under incredible pressure. Inside the cylinder, the fluid is directed along an occasionally convoluted path by VERY fragile parts that can be broken with a slip of a screwdriver.
They hold up under the pressure because it is exactly equal from all sides, including inside out.

It seems to me that the "angular momentum" described above would rip my hydraulic cylinder apart if such momentum existed. It doesn't, so it don't

This is why I've been arguing about it not rotating underwater since post one.

There is angular momentum up till the very moment that the sprinkler is filled. (my hose analogy.) After that, it just ain't there.

If somebody can show me that I'm wrong here I'd appreciate it. I would like to know the answer to this question.

The only things that can possibly move the &*%\$ing sprinkler head are air pressure, cavitation, or currents. Any of these is experimental error. Or am I going crazy?

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Often wrong.... NEVER in doubt

Pasta
01-18-2000, 07:00 PM
In regards to my post at 01-18-2000 08:57 AM:

I just want to reiterate the conclusion I tried to draw (as I sense some misinterpretation.) The sprinkler can spin backwards underwater, and the cause is independent of any pressures at the openings. The torque felt by the sprinkler head is identically equal to (minus) dL/dt = the angular momentum carried away by the draining water per unit time.

That is, if the water is still spinning as it leaves, the sprinkler will feel a torque. If the system is designed carefully such that the water does not have any angular momentum as it leaves, no torque. And, if I dare, one could design a sprinkler head whose arms turn back on themselves in a way to give the draining water angular momentum in the direction _opposite_ what we have all been thinking, implying that one could have it spin the same way as the normal case!

The abstraction brought in by the angular momentum argument is necessary to keep us from integrating (in the calculus sense) all the time. If one wishes to think of the physics involved from a purely force/torque standpoint, he can: just as the water drifts into the suction, it is still motionless. If it leaves the center drainpipe with some motion around the center axis, something along the way must have given it that motion. That something is the sprinkler head, and this gets us to Newton's action/reaction ==> the sprinkler spins. If the geometry of the tubes are such that the water bounces here and bounces there and leaves without any motion around the center axis, this just means that the sprinkler head pushed it one way for a bit and then another way for a bit and in the end netted zero total torque applied. (Insert "force" for "torque" if it makes you happier.)

Scylla: Water _is_ compressible, and this is where pressure comes from. But, it is compressible on a magnitude that makes it useless to think of any sort of volume change. That is really why pressure is useful at all -- the mathematical tools have been developed to work with this statistical concept, even though it is somewhat separated from the physical world it describes. Macroscopically, it's ususally best to think of some entity called pressure moving about in a liquid while the liquid maintains its continuous, incompressible nature. Sometimes, this view is not sufficient (as it is only an approximation.)

-P

Pasta
01-18-2000, 07:08 PM
Sorry for the double post. I should have re-read for clarity.

I said

...if the water is still spinning as it leaves, the sprinkler will feel a torque.

More correctly: ...if the water is still spinning as it leaves, then the sprinkler felt a torque a little bit ago, namely the "reaction" torque due to the "action" torque it gave to the water.

Maybe it was clear from the context of the paragraph that followed anyway....

-P

Scylla
01-18-2000, 07:23 PM
Pasta:

Thanks, it kinda makes sense now. Of course, if the water is still spinning, that force will be imparted to the head. duuuh! Why didn't I think of that? If my little hydraulic fittings don't spin the water, then they're okay, right

The compressible water thing sounds fishy to me, but I'll take your word on it.

So the whole thing depends on how the Fricken experiment is set up? Oh well.

Thanks again Pasta. You got a good noodle there.

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Often wrong.... NEVER in doubt

Bricker
01-19-2000, 01:06 AM
Pasta,

That is a remarkably coherent explanation.

While it disposes of angular momentum as a reason for the rig to spin underwater, it doesn't address the 'localized vacuum/cavitation' argument: that somehow, by sucking water, a small low pressure area is crated in front of the tub which will 'pull' the nozzle forward.

- Rick

CurtC
01-19-2000, 01:30 AM
I'm surprised at the length of time it's taking to arrive at an answer to this one. It seems to me there is a simple explanation, which also agrees with the published experimental result, that the sprinkler moves in the opposite direction.

First, consider the case where water is squirting out. The reason the sprinkler turns in this case is that there is pressure on all sides of the container except on the open end. If you have any closed container jetting out a fluid it can be reduced to this:

------------------------------------------
| /\
| pressure
|
| less
| <-- pressure pressure -->
|
|
| pressure
| \/
------------------------------------------

The fluid inside the container is under a higher pressure than outside (why else would it squirt out?), and that pressure on all sides except one, so the pressure moves it to the left.

If you reverse this, there is less pressure inside the container now, with more outside, and all the forces are reversed. The container now moves to the right.

Now mount two of these containers at the end of sprinkler arms, and you exactly have the problem as stated by Feynman.

RickG
01-19-2000, 03:35 AM
Originally posted by Scylla:
It is my understanding that the pressure in a body of liquid is uniform (discounting any gases dissolved into it,) and there can be no low and high pressure action within that body of liquid moving the sprinkler. (Hence my poorly received implosion joke.)

Water is not compressible so where is the differential, and how is it being created?

This is not a solution to the inverse sprinkler problem, but I wanted to give my take on correcting Scylla's misconception:

Just because a fluid is (ideally) incompressible doesn't mean it has no pressure gradients (it means it has no density gradients, which is not the same thing). For a trivial example, on the surface of the Earth, a container of fluid has a static pressure gradient that opposes gravity.

But even in the absence of gravity, pressure gradients are associated with any motion of a fluid. Bernoulli's Theorem essentially states that energy is conserved along a streamline in an incompressible flow (if we can ignore viscous forces). This energy is generally expressed as an energy density which is the sum of two components (three, if there is gravitational potential energy)--the pressure (which is, statistically, an energy density), and the kinetic energy per unit volume of the fluid. Since the sum of these two quantities is conserved along a streamline, if the velocity (thus kinetic energy) of the fluid increases as it follows its trajectory, the pressure decreases, and vice versa.

This is the source of the lift on airplane wings--the fluid on top of the wing is moving faster than that on the bottom, so the pressure is less and voila, we have net lift. Note that for the overall lift calculations, air may generally be treated as incompressible and inviscid (having zero viscosity). Compressibility and viscosity become essential factors only in the boundary layers around the wing.

Any motion of the fluid in the inverse sprinkler case implies pressure gradients. Whether you treat these as causing the motion or resulting from the motion is a matter of point of view, I think. On the other hand, the only way we have to suck the fluid into the sprinkler is to set up a pressure gradient in the first place (which is what a pump does), so perhaps it is reasonable to consider the pressure gradient as cause of the flow in this case.

As Pasta points out, pressure is a statistical concept, and one could argue the case with Conservation of Angular Momentum or with pressure forces equally well. One implies the other.

Rick (ex-Fluid Dynamicist)

Scylla
01-19-2000, 11:52 AM
Thanks for the help Rick. I learn more here than in college.

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Often wrong.... NEVER in doubt

Bricker
01-19-2000, 02:08 PM
1. I never dreamed, when posting this question, that it would garner so many... um... varied responses.

2. I have since acquired and read What Do You Care What Other People Think? This is the companion volume to Surely You're Joking, Mr. Feynman, from which mighty tome I got the puzzle. In What Do You Care, Feynman claims the thing doesn't move.

3. For whatever that's worth.

- Rick

Scylla
01-19-2000, 02:12 PM
Does he say why?

frolix8
01-19-2000, 02:12 PM
Another twist on this experiment is to consider if the same apparatus could be used to pump fluid in both directions if it was spun by an outside force. Many pumps use this principle, and I believe you'll find that they are far more efficient in the "squirt" mode but will also function in the "scoop" mode.

CurtC
01-20-2000, 01:55 AM
Pasta,

Sorry my ASCII diagram didn't come out just right - I forgot that HTML is rendered without multiple spaces unless you use &amp;nbsp;. Several of the things in it should have been shifted to the right, including the word "less" should have been near the right side where there is less pressure. You seemed to understand it anyway.

I don't completely buy that you can't look at the problem from the standpoint of pressures, and must use conservation of momentum. The pressure that I think is key here is on the container walls - on the left-side wall there is a pressure difference, which equates to a force, that is not balanced by a force from the open end. In the squirting-out case, the pressure is higher on the right side of the wall than on the left, so it and the whole container move to the left. In the opposite case, the pressure is higher outside the container, so the force pushes the left-side wall to the right, turning the sprinkler backwards. The container doesn't have an up-down force because the forces on the top and bottom walls counter each other.

Just because water is incompressible, that doesn't mean that there are no differences in pressure.

And if you would rather think of the pressure in terms of statistical events of moving molecules, just think of how the molecules on the left side of the wall hit it with more momentum than on the right, so the conservation of momentum wants to move this wall to the right. But as you said, on the open end there is no pushing or pulling, so the force on the left wall is not balanced.

I still say it moves in the opposite direction when sucking.