Looks like I need to go the remedial HTML farm. 
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
Douglips:
Just saw the choose thing you put in the phaser discussion. Very cool, how’d you do that?
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.
Often wrong… NEVER in doubt
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:
<select><option>Option 1<option>Option 2</select>
<select><option>Option 1<option>Option 2</select>
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.
Agreed. It either do, or it don’t.
I ain’t spending $30.00 either.
Thanks for the tip.
Bricker said…
Sorry, but the “pathlength difference” explanation of Bernoulli’s effect, over an aircraft wing, assumes physical laws which don’t exist.
Scylla said…
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
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I said…
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
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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.
Often wrong… NEVER in doubt
Stephen:
Please disregard previous post.
Thanks for correcting yourself.
Often wrong… NEVER in doubt
*Jeepers Creepers!!
A little applied physics here. Forcing air OUT of it creates a backward spin due to thrust conversion /action v/s reaction.
SUCKING 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. 
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.
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,
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
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.
CurtC,
You said:
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:
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
[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)MeSorry 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?
Often wrong… NEVER in doubt
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