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This ain't yo Daddy's Problem Set
January 9, 2007 6:01 AM Subscribe

Poof and Foop.Can Metafilter tackle yet another horrible physics clusterfuck? Answers go here, or we can just argue it out right here at home.
posted by tehloki (136 comments total) 3 users marked this as a favorite

C. Not be moving.
posted by IronLizard at 6:07 AM on January 9, 2007

And also: D. Not be posting this crap.
posted by IronLizard at 6:08 AM on January 9, 2007 [1 favorite]

I agree with the "conservation momentum" camp, and say the can will not be moving. Every time I think I've got the answer infallibly right, though, I look at the problem again and it stops making sense. Argh.
posted by tehloki at 6:08 AM on January 9, 2007

shouldn't this be in askme?
posted by empath at 6:10 AM on January 9, 2007

It's simple. The only thrust (in a perfect scenario) is against the inside of the can, therefore canceling itself out.
posted by IronLizard at 6:11 AM on January 9, 2007

This reminds me of a story in one of Feynman's books. A bunch of physicists at Princeton argued for weeks about what would happen if you rigged a spinning sprinkler to suck water in and put it in a tub of water - would it spin in the opposite direction than regular, the same direction, or sit still. They argued theories back and forth for weeks with, as I recall, Feynman changing his mind several times. Feynman finally figured a way to conduct the experiment and found that it did not move at all.

I'll go with C.
posted by qldaddy at 6:14 AM on January 9, 2007

empath: d'oh! I knew I was forgetting about something...
posted by tehloki at 6:16 AM on January 9, 2007

Duh: the can will be sucked to the right.
posted by Tlogmer at 6:17 AM on January 9, 2007

I think it's sucked to the right. LIke our supreme court of late.
posted by mert at 6:18 AM on January 9, 2007 [2 favorites]

This is inane. The air sucking it to the right would push it straight to the left again at the same time as it pushed it to the right because it's hitting the inside of the damn can in the opposite direction. This is so intuitive it's not even funny. I'll explain better when I sober up sometime this evening.
posted by IronLizard at 6:23 AM on January 9, 2007 [1 favorite]

There's a similar problem:

You have a glass jar with a fly in it, sitting on the bottom of the jar. The jar sits upon a scale. If the fly starts flying around inside the jar without touching it, does the weight increase, decrease, or stay the same?
posted by tehloki at 6:26 AM on January 9, 2007

My favorite part of that post is the fact that Boing Boing, despite its position as the center of the blogosphere, still refuses to have comments. And as such, when they want to turn on commenting, they're forced to use a third-party system like Technorati or, in this case, fucking QuickTopic. Just goes to show you that you don't have to think ahead to become important.
posted by Plutor at 6:29 AM on January 9, 2007

Is the fly flying from left to right or right to left, also how did it get in there and are there airholes so it can breath cause it'll probably die soon and hit the scale pretty hard from asphyxiation and make the weight increase slightly, but you would need some insanely sensitive scale to be able to measure it.
posted by Skygazer at 6:30 AM on January 9, 2007

This question sucks
posted by srboisvert at 6:33 AM on January 9, 2007

Duh: the can will be pushed to the left.
just to round things out, ya know?
posted by carsonb at 6:37 AM on January 9, 2007

Yeah, but a more important question is, if nature's abhorrence of a vacuum were patented by Disney, could a diagram of the displacement of air caused by moving subway trains in Toronto still be put under a Creative Commons license?

And could Xeni Jardin podcast about it?

Naked?

On CNN?
posted by felix betachat at 6:43 AM on January 9, 2007 [1 favorite]

Fine. The can will explode. SPOILER WARNING.
posted by IronLizard at 6:46 AM on January 9, 2007

Wait, doesn't a perfect vacuum have infinite potential energy?
posted by tehloki at 6:49 AM on January 9, 2007

There is no can. There is only can't.
posted by cortex at 7:01 AM on January 9, 2007 [2 favorites]

Wouldn't the can already be squished?
posted by mkb at 7:03 AM on January 9, 2007

Well, are we observing the fly or not?
posted by Mikey-San at 7:09 AM on January 9, 2007

This is inane. The air sucking it to the right would push it straight to the left again at the same time as it pushed it to the right because it's hitting the inside of the damn can in the opposite direction. This is so intuitive it's not even funny. I'll explain better when I sober up sometime this evening.

Intuitive != true.

Your example would only make sense if the air particles were moving infinitely fast.
posted by delmoi at 7:15 AM on January 9, 2007

Is the can on a treadmill?
posted by zsazsa at 7:22 AM on January 9, 2007

The air entering the can doesn't slam against the left end like a bunch of marbles. It would fill the can uniformly, with equal pressure on all sides.
On the outside of the can, however, the air on the right side would be in motion as it rushes toward the hole in the can, while the air on the left side would be relatively static. Since moving air has lower pressure than static air, there would be a net force on the can pushing it to the right.
posted by rocket88 at 7:23 AM on January 9, 2007 [1 favorite]

To me it's simple.

F = Pressure * surface area.

The pressure on the outside of the can is equal in all directions, but there is less surface area on the right side And on the outside of the can, air pressure from the right moves the can left, and air pressure from the left moves the can right. so the net outside force is positive in the right direction.

But what about the air on the inside of the can? Well, again with less surface area on the right side, the force of the air pressure is less on the right side. But the air on the inside of the can pushes in the other direction, that is air pressure on the right inside wall moves the can to the right, and air pressure on the left side of the wall moves the can left.

So, if the pressure on the inside and outside were equal, the can would not move. but the pressure is not the same. Until the pressure equalizes, the can will move to the right.

That's my analysis anyway.
posted by delmoi at 7:30 AM on January 9, 2007 [2 favorites]

Your example would only make sense if the air particles were moving infinitely fast.

Oh man, this makes no sense whatsoever. Velocity has nothing to do with it. I'm going to bed, hope the amateur physics association has fun till someone who knows what they're talking about shows up with a practical example.
posted by IronLizard at 7:32 AM on January 9, 2007

If the fly starts flying around inside the jar without touching it, does the weight increase, decrease, or stay the same?

D, all of the above. The weight of the fly pressing on the bottom of the jar is replaced by downward force of air pushed by its wings, causing the scale to oscillate.
posted by eddydamascene at 7:33 AM on January 9, 2007

And also: D. Not be posting this crap.

Huh? This is something interesting, on the web, that most of us haven't seen before. By definition it's an acceptable post. Me, I'm convinced by the rocket88/delmoi theory, but since I'm an ignoramus in these matters, I'll probably change my mind a couple more times in the course of this thread.
posted by languagehat at 7:40 AM on January 9, 2007

Is the can on a treadmill?

You can go to Hell.
posted by Mikey-San at 7:41 AM on January 9, 2007

B: it moves to the right,
but the movement is slight.

Conserve momentum.
Move some air to the left
Something must move right
The can thus moves right
However, not much air moves and it doesn't move far so only a little movement of the can to the right conserves the momentum
posted by caddis at 7:41 AM on January 9, 2007

d) get a vacuum can, punch a hole in it and see what happens ... or talk to someone who did

why guess when we could know?
posted by pyramid termite at 7:43 AM on January 9, 2007 [1 favorite]

The can won't move perceptibly. There's no mass being ejected from the can, so that rules out rocket action. The only unbalanced force exerted on the can, then, is due to the surface area subtracted from the right hand end by the puncture (force = pressure * surface area). But the puncture accounts for bugger-all surface area, so there's bugger-all unbalanced force and therefore bugger-all acceleration; and such unbalanced force as exists due to surface area difference will be pretty much countered by the drag force of the air rushing in past the edge of the hole. If you made the hole very big you might see a smidgen of movement to the right, but if you left it at rocket-nozzle proportions you'd get nowhere.

Coming at the thing from another angle: consider the case of a can fitted with some mechanism that suddenly detached the entire right-hand end and allowed it to piston inward to the left while maintaining the interior vacuum. In that case, the body of the can would accelerate to the right due to the unopposed force on its left end.

To figure out what would happen when the detached right end slammed into the inside of the left end, we need several principles:

1. Momentum is conserved in a collision.

2. Momentum is mass times velocity.

3. Velocity, from a standing start, is acceleration times time.

4. Acceleration is force divided by mass.

Let's call the mass of the can M, and the mass of the detached end m (because M is bigger than m).

From (4), we see that for the entire time the bits are travelling toward each other, the acceleration of the detached end will be M/m times the acceleration of the can body; actually -M/m, because the accelerations are in opposite directions.

Because they're both travelling for the same amount of time before they collide, the velocity of the detached end will likewise be -M/m of that of the can body any time before impact.

Multiplying the mass of each body by its velocity, then, will give us equal but opposite momenta; the momenta sum to zero. At impact, the system will have still have zero momentum, and therefore zero velocity.

The net effect would be that the can jumps to the right a little way (less than its own length) and then stops. The energy of motion on impact would all be converted to heat and noise.

The less massive you make the detached end, the faster this all happens, and the less time the can gets to jump in. In the limit, if you make the detached end infinitesimally massive (i.e. turn it into a puncture the entire size of the can end) it travels up the can infinitely fast and the can doesn't jump at all.

So at the lower limit of hole size, you get stuff-all movement; and at the upper limit of hole size you get stuff-all movement. There may be some optimum value in between where you get a bit of a jump out of it but I wouldn't bank on it.
posted by flabdablet at 7:44 AM on January 9, 2007 [1 favorite]

On lack of preview: yeah, there will be a little pocket of low-pressure air adjacent to the hole, but it will be quite small (air can enter the region from all directions quite slowly and still make up for the tiny stream rushing down the hole). And I still reckon that the net force due to pressure and/or area differences will be pretty much balanced up by drag on the edges of the hole as the air rushes in.

I'll bet a dollar it doesn't move.
posted by flabdablet at 7:52 AM on January 9, 2007

Caddis's argument doesn't necessarily work either, because there is no guarantee that all the air that ends up in the can was originally on its right hand side.
posted by flabdablet at 7:55 AM on January 9, 2007

There is no can. There is only can't.

Actually, I prefer "There is no can. There is only Kant."
posted by mosk at 7:58 AM on January 9, 2007

And I'll bet a hundred dollars it won't be moving after the vacuum is filled.
posted by flabdablet at 7:59 AM on January 9, 2007

Am I missing something here? Isn't this really a question about friction?

The call of the question is: After the vacuum is filled the can will...

Aren't many of us answering what will happen while the vacuum is filling? While it is filling, the can may or may not move to the right, depending on what kind of surface the can is sitting on. If it is on no friction-causing surface, it will move to the right in some large or small amount. If it is on a surface with some friction, it may or may not move depending on how much drag there is.

Once the vacuum is filled, the can's momentum will either continue to carry it to the right, or not, depending on the surface.
posted by saladpants at 8:03 AM on January 9, 2007

Dude, I am like one joint toke away from full on crainial explosion right now.
posted by The Straightener at 8:08 AM on January 9, 2007

Everyone's missing the point here. If you look closely, you'll see little motion lines on the first can, indicating that it is moving.

On the second can, there are no motion lines. Therefore is it not moving.
posted by chrismear at 8:14 AM on January 9, 2007 [1 favorite]

I think for thought experiments like this it is best to assume a frictionless environment.
posted by caddis at 8:14 AM on January 9, 2007

Dude, I am like one joint toke away from full on crainial explosion right now.

After your head explodes, will you be moving forwards, or backwards?
posted by obvious at 8:22 AM on January 9, 2007 [1 favorite]

In a perfectly frictionless environment nothing would happen as the vacuum would also exist outside the can.

But lets say in earth atmosphere, irregardless of the amount of friction (but not zero):

The can will move neither left or right, but will move down ↓ over time

and here is my reasonong:

If nature abhors a vacuum, than nature's state is a non-vacuum.

Another of nature's default states is one of entropy (or decay).

Entropy and unvacuumness are default states that define nature.

When the air rushes into the can, the force produced by the state of unvacuumness will not move the can. A vacuum has a kinetic value no greater than the weight and mass of a thing itself. In other words, nature would not contradict itself.

The only movement is over time and that movement is down ↓
posted by Skygazer at 8:25 AM on January 9, 2007

After your head explodes, will you be moving forwards, or backwards?

That's...whoa.
posted by The Straightener at 8:26 AM on January 9, 2007

Whether the can actually moves or not is subject to factors not specified in the problem, relating to the can's mass, the hole size, friction on the base and, in the case of falling over, the shape of the can (a tuna can wouldn't fall over, a hairspray can might). But the question seems to be one of net forces and it does seem to me that the net force is to the can's right, irrespective of whether anything really moves.
posted by George_Spiggott at 8:30 AM on January 9, 2007

I say B, on general intuition.

I tried to imagine in using my cup of coffee, and it just seemed to "make sense" that it would move to the right. I don't see it "rocketing" towards the right, but kind of jerking suddenly to the right and then coming to rest again.
I spilled some coffee, but it's ok. It's for science.
posted by CitrusFreak12 at 8:32 AM on January 9, 2007

er. "This" not "in"
posted by CitrusFreak12 at 8:32 AM on January 9, 2007

If nature abhors a vacuum, than nature's state is a non-vacuum.

But a vacuum is better than most of the stuff Nature fills it with.
posted by GuyZero at 8:34 AM on January 9, 2007

On the second can, there are no motion lines. Therefore is it not moving.

QEmotherfuckingD. Case closed, people. Go watch football or something.
posted by cortex at 8:35 AM on January 9, 2007

Let's see if I keep thinking about this will I be moving towards keeping myself afloat or getting fired? Hmmmm.....

*Twirls thumbs*
posted by Skygazer at 8:40 AM on January 9, 2007

This question is stupid. It all depends on whether God wants the can to move or not.
posted by LordSludge at 8:48 AM on January 9, 2007

Wait, doesn't a perfect vacuum have infinite potential energy?

What? No. Who told you that?
posted by bshort at 8:55 AM on January 9, 2007

Oh man, this makes no sense whatsoever. Velocity has nothing to do with it. Velocity has nothing to do with it. I'm going to bed, hope the amateur physics association has fun till someone who knows what they're talking about shows up with a practical example.

Amateur physics association? Which you're not a member of? Whatever. You're both arrogant and wrong.
posted by delmoi at 8:56 AM on January 9, 2007

I think Rocket88 and delmoi have it right, if the explanation of jet propulsion my high school physics professor gave us was right.
posted by nanojath at 8:58 AM on January 9, 2007

I'm arrogant and correct. You're arrogant and wrong. There, that should settle it.
posted by IronLizard at 8:59 AM on January 9, 2007 [1 favorite]

Forget what happens inside the can, because whatever it is, it is in roughly spherical direction. As the air on one side enters it creates lower pressure where it left from, until the pressure has time to equalize its distribution. So it should be sucked to the right into that relative vacuum. And you can't just discount that distribution time factor like you are expected to do with for instance the friction of the gas entering the hole, because if the pressure change was instantaneous the whole question would not make sense anymore.
posted by StickyCarpet at 9:00 AM on January 9, 2007

if the explanation of jet propulsion my high school physics professor gave us was right.

That explanation applies to jet propulsion. Not internalized suction. Anyone who has worked with compressed air and vacuum pumps to a large degree will understand the difference.
posted by IronLizard at 9:01 AM on January 9, 2007

Just to clarify, if there was a pressure differential inside a sealed can for a time, it would'nt make the can move.
posted by StickyCarpet at 9:04 AM on January 9, 2007

mmmmmm....

on first glance my intuition is that the can will not move, but I have no math to back that up.

What follows is a stream of consciousness, so caveat emptor.

To solve this problem you just need to know the rate at which the can will be refilled with air. If we include no friction, there is no loss of energy (momentum will still be conserved). We need to know the potential energy of the system before the puncture. But then again, if this is a reversible system then we should just be able to reverse time and recreate the motion. I suspect this is NOT a reversible situation (entropy of the system increases as the gas escapes the cannister) When the can is "refilled" with air, it will NOT be pressurized as before.

Ah ha now I think I see it. OK the analogous situation would be piercing a can filled with air at exactly the same pressure as the surrounding environment. Nothing would happen, the system is already in equilibrium. So the reverse situation would be filling a can of air until it reaches equilibrium with its surrounding. All we have to do is time reverse the situation I described above.

No movement I say (assuming no air friction)
posted by ozomatli at 9:06 AM on January 9, 2007

"Nature abhors a vacuum" is part of Aristotle's explanation of intertia. The idea is that, if you throw an object then a vacuum is left behind it. Air rushes in to fill that vacuum and the pushes the object along a bit more. I'm not an expert in the field, but I don't think this is the model of inertia that is in currently in favor at MIT or Cal Tech.

Having done some noodling around with pneumatic cylinders (for Halloween props) I can tell you that if you plug one opening in a two way cylinder and then force the piston (so that it's drawing a vacuum or compressing the air) it will definitly push back. The difference is that with compressed air, you can go, more or less, as high as you want/can. With a vacuum, the maximum difference you are going to have between your high and low pressure reservoirs is less than 15 psi. FYI, 15 psi will not fling a PVC and foam rubber corpse out of a plywood coffin with sufficient vigor to impress the Trick-or-Treaters.

As I see it, there is 15 psi pushing on the can in all directions. If one end suddenly vanished there would be 15 psi pusing from all directions except for the end where the air is racing in to fill up the vacuum. Once that vacuum is filled no additional acceleration will occur. Of course that is assuming no friction, the can being rigid enough not to be crushed and that, somehow, we can remove an end (or open a hole or whatever) and not impart any vector on our can. Oh, and I want a pony.
posted by Kid Charlemagne at 9:17 AM on January 9, 2007

Anyway, Where exactly is this can? If it's sitting on a table, it probably wouldn't move much. Is it sitting on a frictionless table? Is it floating in a zero-g environment?
posted by delmoi at 9:18 AM on January 9, 2007

I vote for "not the best question" and "not moving."

I suspect that this question is really about conservation of momentum. In the top case, the net momentum of the system is 0. In the bottom case, the net momentum should also be zero. Assuming idealized cans and gasses of course.

My favorite part of that post is the fact that Boing Boing, despite its position as the center of the blogosphere, still refuses to have comments.

Of course, the central caveat of that graph is Daily Kos was arbitrarily selected as the center. Which is one of the big problems with creating social network graphs.
posted by KirkJobSluder at 9:20 AM on January 9, 2007

A rocket engine works because the high pressure has been "trapped" inside a container with small opening. The forces on the lateral walls are balanced, but the force on the nozzle end escapes, allowing the upward force on the nose to prevail. The difference in the "foop" case is that the pressure on the outside of the can is effectively infinite (not trapped), so any imbalance of force on the outside of the can is instantly replaced by the earth's atmospheric pressure. The effect would be the same as if you submerged an empty pop can in a swimming pool. There would be some movement caused by turbulence around the opening, but no "rocket" effect.
posted by weapons-grade pandemonium at 9:22 AM on January 9, 2007

Imagine an astronaut. While floating in space, he throws a ball away from him. He is propelled in the opposite direction of the ball. Now imagine him throwing a ball directly at his chest, but doing it with inhuman precision. He doesn't move, momentum is transferred from him to the ball and right back to him equally. Same goes for the can and the air (again, in a perfect scenario with a valve instead of a forced puncture, to make it simpler to envision). Everything else is just obfuscation.
posted by IronLizard at 9:22 AM on January 9, 2007

ironlizard is right. the rest of you will have to repeat this course.

momentum is conserved. when a pressurized can is punctured on the right, air shoots out and keeps going, all the way to the horizon and over. the air has momentum, which is mass times velocity (mv). the air particles bump into other air particles and contribute their momentum to the newly hit particles, so over a short time you have a lot higher m with a lot less v. the can has a much higher m than the air particles, so it will be propelled to the left with a proportionally tinier v.

when you puncture a vacuum can on the right, the air rushing in doesn't have an open path all the way to the horizon and over, it quickly runs into the interior walls of the can and stops. in the ground state of this situation, there was no momentum in the system. the end state of this situation has no air moving because the can is full of air now. since there is no mv for the air, the mv for the can will be zero, which means the v has to be zero. the can doesn't move!
posted by bruce at 9:24 AM on January 9, 2007

A punctured can of compressed air moves because of the high pressure that it contains, which is a lot of potential energy. A can that contained only ambient air pressure wouldn't move at all when punctured. So why shoud ambient air pressure move a punctured vacuum canister?
posted by mr_crash_davis at 9:26 AM on January 9, 2007

on preview, several others got it right too, hat's off to you.
posted by bruce at 9:28 AM on January 9, 2007

This is just viral advertising for the newest innovation in personal air travel: The Suck Pack.

Oh hell, here's one: Why doesn't your vacuum cleaner's hose go anywhere (now now, it can't be next to something it'll begin sucking on that's cheating.) ? Ok, now put it in reverse. Oh? So it can keep a soccer ball in the air in reverse but not propel itself upwards on 'suck' mode?

So why shoud ambient air pressure move a punctured vacuum canister?

Ack. That's not quite it either. A can of vacuum still has potential energy and it can be realized by puncturing, the key is getting usable energy out of it by not having it work against itself, as it is in this scenario. A -14 psi can of air has the same potential energy ad a +14 psi can of air.
posted by IronLizard at 9:32 AM on January 9, 2007

Some people have mentioned that the question actually asks what happens after the can stops. Well, if the can is moving while its filling, and no other force acts on it, it will continue to move after it's finished filling. So immediately after it finishes filling it will begin to slow down in the air, just as a rocket, in air, slows down when it finishes firing.

The difference is that a rocket has far more pressure difference then an empty can.
posted by delmoi at 9:34 AM on January 9, 2007

As the iris on the vacuum can opens, two things will happen: greater force on the LHS of the can will cause it to accelerate to the right, and unopposed gas from the RHS of the can will move into the can, giving the gas near there an overall momentum to the left.

The dynamics shortly after that get complicated, depending on a lot of variables we just don't have access to, but steady state pretty much has to be stationary. Even if we're dealing with an ideal gas, made up of a bunch of tiny "marbles" that don't interact except by collision, moving through it would still (assuming uniform temperature, density and pressure) generate a net force against the can opposing its movement. I suspect you can almost get away with treating the filling process as an inelastic collision between the gas and the can...

So, the can has to end up at rest relative to the gas, and the only way conservation of momentum can be satisfied is to have both at rest.

On preview, I type (and, though I often hesitate to admit it, think) too slowly, and other folks have beaten me to it.
posted by MadDog Bob at 9:34 AM on January 9, 2007

The vacuum cleaner is a bad example, and I apologize if I've been abrasive. Still can't sleep.
posted by IronLizard at 9:38 AM on January 9, 2007

ha! I have successfully started an internet fight. My work here is done. Dance, mefites, dance!
posted by tehloki at 9:38 AM on January 9, 2007

Oh hell, here's one: Why doesn't your vacuum cleaner's hose go anywhere

Unless the vacuum cleaner's hose isn't attached to anything and sitting on a frictionless surface, the situation isn't analogous. Obviously if we're talking about a can sitting on a table or a can attached to a piece of household equipment, with a thick plastic tube, it won't move.
posted by delmoi at 9:39 AM on January 9, 2007

I just want to point out that the OP of this has an obligation... no no, a duty, to post the follow up answer to this question. If nothing else, some egos may need deflation.
posted by Holy foxy moxie batman! at 9:40 AM on January 9, 2007

Rocket88:
The air entering the can ~~doesn't~~ does slam against the left end like a bunch of ~~marbles~~ molecules.

Does this help?
posted by IronLizard at 9:42 AM on January 9, 2007

As the OP, I can definitively state that through all my exhaustive research, I haven't found a single definitive -explanation- behind the answer. However, the consensus in the "people who actually understand physics" portion of the boingboing posters (and Richard Feynman, on the similar "sucking sprinkler" problem seems to be that the can will not be moving (after the intake of air has stopped).
posted by tehloki at 9:44 AM on January 9, 2007

Oh... errr, you could always buy the book the problem is from, but that would be cheating, wouldn't it?

does anybody know where I can get a pirated copy of the thinking physics eBook?
posted by tehloki at 9:49 AM on January 9, 2007

My short answer: c, will not be moving. Judders but stops once pressure has equalized.

Reasoning. It's a mistake to think that conservation of momentum means the can must move. The air around, above and below the can is rushing right as the air rushes left, so that could balance the momentum alone.

First moment: can moves slightly to the right as air rushes in. This is because the pressure on the right is now slightly less due to the hole.

Then, remember the initial air molecules through the hole are all blasting leftwards. Once they hit the left wall, they jolt the can hard left.

However, they then bounce back the other way. They hit incoming air molecules and shove them towards the right wall. That jolts the can right again, with slightly less momentum due to the hole. This impact is will take place over a longer space of time as the incoming air softens it.

So basically, the can judders, but stays where it is by the end.
posted by TheophileEscargot at 9:52 AM on January 9, 2007

Finally, just to be clear/pedantic, I make no claim about the position of the can afterward, only that it will be at rest relative to the gas around it.
posted by MadDog Bob at 10:10 AM on January 9, 2007

A decidedly non-amateur physics club shows that the Feynman Sprinkler will move in the opposite direction. After a brief look around, seems that friction is the key. Assuming that this is analogous to the can question, (and I think the principle is the same) I'm changing my answer to moves the other way.
posted by qldaddy at 10:13 AM on January 9, 2007

Clearly the universe moves a little to the left, as it's sucked into the can.
posted by RollingGreens at 10:25 AM on January 9, 2007

qldaddy> This reminds me of a story in one of Feynman's books. A bunch of physicists at Princeton argued for weeks about what would happen if you rigged a spinning sprinkler to suck water in and put it in a tub of water - would it spin in the opposite direction than regular, the same direction, or sit still. They argued theories back and forth for weeks with, as I recall, Feynman changing his mind several times. Feynman finally figured a way to conduct the experiment and found that it did not move at all.

If I remember correctly, Feynman blew up a pump when he tried this experiment back at Princeton, so "did not move at all" is correct only in the most literal sense.
posted by UrineSoakedRube at 10:52 AM on January 9, 2007

I knew he blew up something! But, I think the experiment ran long enough before the explosion that he felt good about his "does not move" result. As noted above, more recent experiments show otherwise.

BTW, I love it that physicists spend time on stuff like this.
posted by qldaddy at 11:55 AM on January 9, 2007

Rocket88:
The air entering the can ~~doesn't~~ does slam against the left end like a bunch of ~~marbles~~ molecules.

Does this help?

No, it doesn't. Expanding gases tend to expand uniformly in all directions. The molecules would strike all interior surfaces of the can.
posted by rocket88 at 11:59 AM on January 9, 2007

Why would "all the way to the horizon" matter? How does the air in the pressurized can know what it has in front of it at t=0? If there has to be a clear path all the way to the horizon, isn't all that intervening air between here an the horizon going to cause problems? And if you have to have air to push against how can rocket propulsion work in space?

If the air entering the can is going to move it back the way it came, why wouldn't the air outside the can push the can back the way it came in the pressurized can model?

Like I said before - the big difference is that we are used to the bigest vacuum possible being 15 psi away from atmospheric pressure, but I can run out and get a 2000 psi oxygen cylinder at my local welding supply place.
posted by Kid Charlemagne at 12:18 PM on January 9, 2007

Some people have mentioned that the question actually asks what happens after the can stops [filling].

Yes, and I need to do a better job reading the question. As the can is filling it should move ever so slightly to the right as the air moves left into it but after the can is done filling and the air has stopped inside the can, the can is stopped. So I amend my answer to C.
posted by caddis at 12:30 PM on January 9, 2007

If nothing else, some egos may need deflation.

But will they move to the left, right or not at all ?
posted by Pendragon at 12:41 PM on January 9, 2007

Adiabatic expansion of an ideal gas against a vacuum does not do any work. The can does not move.

The reverse-sprinkler experiment doesn't help much since water is not an ideal gas it's a Newtonian liquid.
posted by JoJoPotato at 1:18 PM on January 9, 2007

Aside from pageviews what's the point of garnering (mostly incorrect) postulations. The whole point of the internet is that the answer should follow rather closely to question whenever the answer is already known.
posted by furtive at 1:26 PM on January 9, 2007

Adiabatic expansion of an ideal gas against a vacuum does not do any work. The can does not move.

No, an adiabatic expansion has no heat transfer. They absolutely can do work in the thermodynamic sense.
posted by UrineSoakedRube at 1:32 PM on January 9, 2007

Yes! Adiabatic expansion of a gas can do work.... Unless it is expanding against a vacuum.
posted by JoJoPotato at 1:51 PM on January 9, 2007

Poof and Foop

Poop.
posted by qvantamon at 2:15 PM on January 9, 2007

Yes! Adiabatic expansion of a gas can do work.... Unless it is expanding against a vacuum.

Okay, take a cylinder divided into two regions by a circular plate. The top region of the cylinder is at half an atmosphere of pressure. The bottom region of the cylinder is under vacuum. The circular plate is free to move up, but there are hard stops which prevent it from moving down. Put a small mass on the plate (inside the top region). And let's assume all of this takes place on the surface of the earth (in other words, there is a gravitation force in effect).

Now let's say there's a valve at the bottom of the bottom region of the cylinder, and outside of the cylinder the pressure is 1 atmosphere. Open the valve.

1) Will the circular plate move up?

2) If so, isn't the expansion doing work by lifting the mass?
posted by UrineSoakedRube at 2:22 PM on January 9, 2007

...therefore I clearly can't drink the wine in front of you!
posted by carsonb at 2:23 PM on January 9, 2007 [2 favorites]

Also, I'm not sure that you can say that this is an adiabatic expansion, unless you take the entire system -- the container that encloses the vacuum can and the gas around it -- as being thermally isolated from some other system. In that case, dW for the system wouldn't necessarily change as a result of the expansion, but that wouldn't necessarily mean that the can couldn't move.
posted by UrineSoakedRube at 2:42 PM on January 9, 2007

So now that I read my previous post, my thought experiment wouldn't be an adiabatic expansion (although it would do work), but it's not at all clear to me that this problem ("Poof and Foop") is that of an adiabatic expansion in any non-trivial sense.
posted by UrineSoakedRube at 2:47 PM on January 9, 2007

Conservation of momentum. If the can has momentum to the right, something must have momentum to the left. What is it?
posted by KirkJobSluder at 3:19 PM on January 9, 2007

http://web.mit.edu/Edgerton/www/FeynmanSprinkler.html
posted by pullayup at 3:36 PM on January 9, 2007

(oops, sorry)
feynman sprinkler
posted by pullayup at 3:37 PM on January 9, 2007

Conservation of momentum. If the can has momentum to the right, something must have momentum to the left. What is it?
posted by KirkJobSluder at 3:19 PM PST on January 9 [+]
[!]

The air.
posted by UrineSoakedRube at 3:39 PM on January 9, 2007

Would somebody please just call Adam and Jamie and get them on this?
posted by Mr.Encyclopedia at 4:33 PM on January 9, 2007

What it's doing while it's filling is complex but, by definition, when the can and gas eventually attain equilibrium the pressure on all surfaces of the can will sum to zero. Hence the rms velocity of the gas molecules is the same on all surfaces, so the can has no mean relative velocity. (I can imagine no plausible mechanism whereby the can and gas could separate totally and maintain independent momenta.)

And the question was about what happens afterwards. This is a classic sort of physics textbook question; if you do things the hard way (i.e. try and work out the dynamics) your brain will melt, but the answer they're after is actually trivial.
posted by Luddite at 4:35 PM on January 9, 2007

I'm pretty sure the mythbusters have their hands full with trying to find a jet airliner and a treadmill that can move at infinite speed.
posted by tehloki at 4:35 PM on January 9, 2007

Back and to the left. Back and to the left . . . .
posted by landis at 4:43 PM on January 9, 2007

And the question was about what happens afterwards. This is a classic sort of physics textbook question; if you do things the hard way (i.e. try and work out the dynamics) your brain will melt, but the answer they're after is actually trivial.
posted by Luddite at 4:35 PM PST on January 9 [+]
[!]

The question asks what happens after the can is filled. Yes, obviously if you wait long enough, the can will not be moving because of the drag imparted by the air. But the interesting non-trivial question is what is happening at the moment the can is filled (and since this can mean at the moment of puncture or at the moment the can's pressure equalizes with the outside air, people have been pretty specific with their answers).

The specific problem with your explanation is that it applies just as well to the answer to what happens to the first can (the overpressured can that's punctured). And it's clear -- he says it straight out -- that Epstein means for the answer for that one to be "the can moves to the right".
posted by UrineSoakedRube at 5:33 PM on January 9, 2007

UrineSoakedRube: The air.

The air can't move to the left, because it is inside the can which is moving to the right.

So start with the "rocket" can:

T0 (can closed)
MV_can = 0
MV_air = 0
Total MV = 0

T1 (can open)
MV_can = -1 (wrt the x axis)
MV_air = +1
Total MV = 0

Now look at the vacuum can.

T0 (can closed)
MV_can = 0
MV_air = 0
Total MV = 0

The problem with pushing or sucking the can to the right is that you not only are moving the can, but the air inside the can. So you have a state in which:
MV_can &gt: 0
MV_air &gt: 0
Total MV > 0

We know it is impossible to change the MV of a system without an external force. Where is this external force?
posted by KirkJobSluder at 6:50 PM on January 9, 2007

The air can't move to the left, because it is inside the can which is moving to the right.

Take a train that's moving to the west at a speed of 1 mile/hour, and suppose that I am inside the train running towards the back of the train. I pass someone sitting inside the train with a radar gun who clocks me at 10 miles/hour.

What is my velocity (how fast am I running and in what direction) with respect to the ground outside the train?
posted by UrineSoakedRube at 7:20 PM on January 9, 2007

Did the conductor have chili for lunch? Is the floor of the train a frictionless surface? What about the passengers, do they approach the median weight of a large hippo? Is the train going forwards or backwards?

9mph. Bu this has nothing to do with the can. Until you slam into the back of the train and break your neck, like an idiot. Also: any motion you impart to the train (infinitesimal, I know) while accelerating towards the back will be canceled by the force of you breaking your neck.

Problem is, your(air) velocity has no net effect on the train (can).
posted by IronLizard at 8:14 PM on January 9, 2007

You can try this more simply with a couple of skate boards and three large planks:

[___________]
0 0 0 0

Run to the left, the thing will shift to the right under you until you slam into the far left board an, in a perfect world, ending up right where you started. This still doesn't illustrate the can perfectly.
posted by IronLizard at 8:19 PM on January 9, 2007

Damn. Forgot about the whitespace thing.
posted by IronLizard at 8:20 PM on January 9, 2007

Did the conductor have chili for lunch? Is the floor of the train a frictionless surface? What about the passengers, do they approach the median weight of a large hippo? Is the train going forwards or backwards?

Oh my, that is just fucking hilarious.

9mph.

And you didn't even answer the question correctly!

Bu this has nothing to do with the can.

And here we see why: because you have problems with basic reading comprehension.

Let's take a look at what KirkJobSluder wrote:

The air can't move to the left, because it is inside the can which is moving to the right.

Now, if anyone else -- including KirkJobSluder -- would like me to explain what the relevance of my train question was to KJS's comment, I'd be glad to explain. IronLizard can go fuck himself.
posted by UrineSoakedRube at 8:33 PM on January 9, 2007

Iron Lizard nailed it. You have a system with zero momentum. What happens inside the system is not going to change the total momentum of the system.
posted by KirkJobSluder at 8:27 PM PST on January 9 [+]
[!]

Okay, look, the system taken as a whole has zero total momentum. You are absolutely right about that.

The problem with your statement was that you assume that the momentum of the air inside the can cannot be directed to the left if the momentum of the can itself is directed to the right. This is just not true, and that's the point of my example. If you mean that once the system reaches equilibrium the air taken as a whole can't have a momentum that's directed opposite to that of the can's, you're right.

But the non-equilibrium state after the puncture and before the pressure equalizes is the crucial stage of the thought experiment. The can may or may not move, depending on what you mean by "when the can is filled", but you can't argue that the can is unable to move because nothing can have an opposing momentum in the system.
posted by UrineSoakedRube at 8:40 PM on January 9, 2007

Oh my, that is just fucking hilarious.

Yes, it's exactly the kind of crap this thread has been full of from the start.

Oh, look! You don't even know how to answer your own question. I think you need to re-read it.
posted by IronLizard at 9:11 PM on January 9, 2007

UrineSoakedRube: The problem with your statement was that you assume that the momentum of the air inside the can cannot be directed to the left if the momentum of the can itself is directed to the right. This is just not true, and that's the point of my example. If you mean that once the system reaches equilibrium the air taken as a whole can't have a momentum that's directed opposite to that of the can's, you're right.

KirkJobSluder: No, I did not make that assumption. And yes, the question is worded in such a way that the equilibrium state is the critical condition we need to worry about. The key words are, "After the vacuum is filled..."

Even if we assume that "after the vacuum is filled" is the point at which the net inflow of air equals the net outflow or air, that isn't necessarily the moment at which the system is in equilibrium in a thermodynamic sense. It's certainly not clear to me that the rest frame of the air has yet become the same as the rest frame of the can.

And that's why I asked the question that I did. I didn't make the same assumption you did as to when we're supposed to be seeing if the can is moving or not -- actually, I specifically stated that I didn't know when the moment of observation was.

If the can has a net positive momentum along the x-axis in the equilibrium state, then something else in that equilibrium state must have a net negative momentum along the y-axis. What is it?
posted by KirkJobSluder at 8:57 PM PST on January 9 [+]
[!]

Okay, so let's take a look at the equilibrium state. You're right, if the can is moving then, it's the air outside the can has to have a net momentum in the opposite direction. You're arguing that this is impossible (if I'm wrong, etc.), so the can isn't moving at all at the moment the air in the can reaches equilibrium.

I don't buy it -- not that you're wrong, but I don't buy that it's obvious. I spent the day at work at a place where about about a third of the employees are Ph.D.s in physics or engineering, and the ones I spoke to debated this to a standstill. There wasn't any agreement on what "after the vacuum is filled" meant, on whether equalization of air density inside and outside the can was equivalent to thermodynamic equilibrium, and whether or not there is any weird fluid dynamic effect that can give the outside air net momentum, and whether this depends on can shape and the system boundaries (if any).

On a personal level, you (by which I mean you, KirkJobSluder) may be pissed at me because you read my thought experiment as condescending and missing the point -- and if that's the case, I am truly sorry about that, and I can tell you that I genuinely didn't mean it to come off like that. Before this post gets way too long, I just want to add that I don't think the answer is trivially obvious -- judging from the arguments I had, I don't even think the question is trivially obvious.
posted by UrineSoakedRube at 10:42 PM on January 9, 2007

USR> Take a train that's moving to the west at a speed of 1 mile/hour, and suppose that I am inside the train running towards the back of the train. I pass someone sitting inside the train with a radar gun who clocks me at 10 miles/hour.

What is my velocity (how fast am I running and in what direction [Please note: emphasis was not added]) with respect to the ground outside the train?

IronLizard> 9mph.

USR>And you didn't even answer the question correctly!

IronLizard> Oh, look! You don't even know how to answer your own question. I think you need to re-read it.

Back when I was in graduate school in physics, I was a TA in 2 courses: the first-year physics class for engineering and physics and the "physics for poets" class for non-science majors. The future engineers and physicists were a bit callow, to be sure, but were fundamentally quiet and serious students. If anything, they were too reticent to question or correct me, so I started to make mistakes on purpose and ask "Does everyone understand how I get this result?" in order to get them into the habit of speaking up when they thought I was wrong or mistaken.

Many of the physics for poets students were also serious, hardworking students who were going into the social sciences or humanities and didn't necessarily need to take a hardcore physics class. Which struck me as being perfectly okay.

But a good number of them were colossal assholes. My fellow TAs traded stories of the 3 students who sat together and all wrote "36%" as an answer to the question "What are the factors which determine the efficiency of a solar cell?" We marvelled at the students who didn't know the answer to a question on a test and instead of leaving the space blank would launch into diatribes about how the topic had never been covered in class (completely untrue, of course). There was the guy who cheated on the final and couldn't be expelled for it, as he had never signed the copy of the school honor code which was on the front of every test booklet. So instead he just got an F on the final and squeaked by with a passing grade in the class.

My favorite one wasn't nearly as egregious as that -- I was explaining the relation between the energy of a planet's orbit and its radius (assuming circular orbits -- this was physics for poets), and pointed out that for a given mass, planets orbiting at larger radii had more energy than those orbiting at smaller radii. At this point, one of the students raised his hand and told me flat out that I was wrong, and that energy decreased as radius increased.

I said, okay, let's derive the formula. I went to the chalkboard and derived the kinetic and potential energies of the orbit and showed that the total energy did, in fact, increase with an increase in radius. I then asked him why he said I was wrong. I stood in front of him for a good minute. He said nothing, which wasn't really all that surprising, as he had struck me as the kind of dick who would let the entire hour go by rather than apologize. Anyway, I blinked first -- I gave up and returned to my lecture.

Now I no longer have to teach obnoxious pricks in order to pay my stipend at graduate school. So if someone else wants to explain to IronLizard why he didn't answer the question correctly, go right ahead. I feel obligated to warn you that it'll be like "slam[ming] into the back of [a] train and break[ing] your neck, like [you just tried to knock some sense into] an idiot". Also, you can expect some tasteless rejoinders from, well, a colossal asshole.
posted by UrineSoakedRube at 11:35 PM on January 9, 2007 [1 favorite]

"you can be the saintly, endlessly patient professor or the obnoxious mefi asshole, but not both at the same time."

actually, quantum theory recognizes something called a "superposition of states" where he _can_ be both at the same time, so long as nobody reads his comments to determine what state he's in presently. in this context, this phenomenon would be called "schrodinger's asshole".
posted by bruce at 7:26 AM on January 10, 2007 [1 favorite]

Cool! I knew I should have stuck with physics. (I'd never gotten grades like 47 on tests until I unwisely took the regular physics course in college instead of Physics for Math Majors, which is what I was. Hubris will getcha every time.)
posted by languagehat at 7:44 AM on January 10, 2007

Invective-laden physics disagreements. It's like college, without the beer run. I love this place.
posted by cortex at 7:50 AM on January 10, 2007

You don't make beer runs?
posted by languagehat at 8:35 AM on January 10, 2007

No, I mean, I can't yell at someone else in the thread and expect them to.
posted by cortex at 9:18 AM on January 10, 2007

USR, you'd have a lot more standing to complain about pricks and assholes if you hadn't gone from zero to "Oh my, that is just fucking hilarious... you have problems with basic reading comprehension... IronLizard can go fuck himself" from one comment to the next. You can be the saintly, endlessly patient professor or the obnoxious MeFi asshole, but not both at the same time.
posted by languagehat at 5:08 AM PST on January 10

Okay, what is the proper response when someone responds to a post by including lines like "Until you slam into the back of the train and break your neck, like an idiot. Also: any motion you impart to the train (infinitesimal, I know) while accelerating towards the back will be canceled by the force of you breaking your neck"? That's when I went from "zero" to pissed off.

Did I "yell" at anyone before that comment? No, I didn't. I responded in kind; I'm not sorry about it; and if you think that comments like that deserve the same tone of response as comments which argue that the system is an adiabatic expansion into a vacuum, you're in no position to determine anyone's "standing".
posted by UrineSoakedRube at 12:32 PM on January 10, 2007

Also, I'm out of here -- I have to get back to work and I don't have time to think about this, let alone write another post on this topic. You can view this as a retreat or surrender; I don't care.
posted by UrineSoakedRube at 12:37 PM on January 10, 2007

Did I "yell" at anyone before that comment?

What? I was talking about actually yelling at actual people to get me actual beer. That had not a goddam thing to do with you, man.
posted by cortex at 12:43 PM on January 10, 2007

Yeah, quit butting in, pal. But I'll respond anyway, just 'cause I'm that kind of guy:

I responded in kind; I'm not sorry about it

And why should you be? I'm not saying it's wrong to yell at people, I'm just saying if you do that, you kind of lose your standing as saintly, endlessly patient professor. One or the other, capeesh? Now, could you bring cortex and me some brewskis? We're working up a powerful thirst.
posted by languagehat at 12:53 PM on January 10, 2007

Beer's over here.
posted by flabdablet at 1:07 PM on January 10, 2007

Iron Lizard nailed it. You have a system with zero momentum. What happens inside the system is not going to change the total momentum of the system.

No, the air molecules have momentum. What moves to the left isn't the air inside it's the air outside the can. Specifically the air molecules that bounce against the left side of the can that push it right. The molecules inside the can do cancel that effect, as you said, but the point is there are not as many air molecules inside the can as there are outside of it.

You're only considering molecules coming in from the right. Those push the can to the left. There are more molecules hitting the can from the left. That's why it moves, and momentum is conserved, because the air on the left side of the can bounces off to the right after it hits the can.
posted by delmoi at 7:30 AM on January 11, 2007

Also talking about other scenarios (trains, skateboards, etc) is a waste of time and a distraction.
posted by delmoi at 7:31 AM on January 11, 2007

Here is another point you guys seem confused about. When an air molecule moves into the hole from the right it does not move the can at all in fact, it doesn't interact with the can in any way. It's only when the can strikes the inside wall that it has any effect. And what effect is that? To push the can in the left direction, not the right. it's not canceling anything, it's just pushing to the left.

Here is a diagram



⇓ ⇓ ⇓ ⇓ ⇓ ⇓ ⇓

⇓ ⇓ ⇓ ⇓ ⇓ ⇓ ⇓

⇓ ⇓ ⇓ ⇓ ⇓ ⇓ ⇓

⇓ ⇓ ⇓ ⇓ ⇓ ⇓ ⇓

⇓ ⇓ ⇓ ⇓ ⇓ ⇓ ⇓

⇓⇑⇓⇑⇓⇑⇓⇑⇓⇑⇓⇑

⇓⇑⇓⇑⇓⇑⇓⇑⇓⇑⇓⇑

| Left Side   |

|             |

|             |

|             |

|             |

|     ⇑ ⇑     |

|     ⇑ ⇑     | Right Side.

⇑⇓⇑⇓⇑ ⇑ ⇑⇓⇑⇓⇑

⇑⇓⇑⇓⇑ ⇑ ⇑⇓⇑⇓⇑

⇑ ⇑ ⇑ ⇑ ⇑ ⇑ ⇑

⇑ ⇑ ⇑ ⇑ ⇑ ⇑ ⇑

⇑ ⇑ ⇑ ⇑ ⇑ ⇑ ⇑

The arrows represent air molecules moving in the right and left direction. We're only interested in the perpendicular projection of the air, not it's "up down" movement (which is right/left in this diagram).

The bold arrows represent air molecules that have just bounced up against the walls of the can.

As you can see, more air is "bouncing" off the left side of the can, then off the right side, so the can is pushed to the right. this accelerates the can. When the air pressure is normalized, the can stops accelerating, it doesn't stop moving though, it continues moving until friction slows it down.

USR, you'd have a lot more standing to complain about pricks and assholes if you hadn't gone from zero to "Oh my, that is just fucking hilarious... you have problems with basic reading comprehension... IronLizard can go fuck himself" from one comment to the next. You can be the saintly, endlessly patient professor or the obnoxious MeFi asshole, but not both at the same time.

IronLizard was both incredibly wrong and incredibly rude about it in this thread. Those two traits together are very irritating.
posted by delmoi at 8:21 AM on January 11, 2007



⇓ ⇓ ⇓ ⇓ ⇓ ⇓ ⇓

⇓ ⇓ ⇓ ⇓ ⇓ ⇓ ⇓

⇓ ⇓ ⇓ ⇓ ⇓ ⇓ ⇓

⇓ ⇓ ⇓ ⇓ ⇓ ⇓ ⇓

⇓ ⇓ ⇓ ⇓ ⇓ ⇓ ⇓

⇓⇑⇓⇑⇓⇑⇓⇑⇓⇑⇓⇑

⇓⇑⇓⇑⇓⇑⇓⇑⇓⇑⇓⇑

| Left Side   |

|             |

|             |

|             |

|             |

|     ⇑ ⇑     |

|     ⇑ ⇑     | Right Side.

⇑⇓⇑⇓⇑ ⇑ ⇑⇓⇑⇓⇑

⇑⇓⇑⇓⇑ ⇑ ⇑⇓⇑⇓⇑

⇑ ⇑ ⇑ ⇑ ⇑ ⇑ ⇑

⇑ ⇑ ⇑ ⇑ ⇑ ⇑ ⇑

⇑ ⇑ ⇑ ⇑ ⇑ ⇑ ⇑

posted by delmoi at 8:21 AM on January 11, 2007

There we go.
posted by delmoi at 8:22 AM on January 11, 2007

IronLizard was both incredibly wrong and incredibly rude about it in this thread. Those two traits together are very irritating.

How ironic. You're still wrong.

Also talking about other scenarios (trains, skateboards, etc) is a waste of time and a distraction.

Yes, that's exactly what I was saying. But you're still wrong. I realized some time ago that ansii drawings aren;t going to change anyone's mind. It will require YouTube. So, if anyone has a good, strong vacuum pump, they should really get to it.
posted by IronLizard at 4:47 PM on January 11, 2007

Who's going to buy the damn book already?
posted by caddis at 5:13 PM on January 11, 2007

IrontLizard, just out of curiosity, what credentials do you have to be making this argument? Have you ever taught a physics class like USR?

Do you even know anything about Newtonian physics? Here is a question for you:

Suppose a cylindrical can is sitting in outer space. It 20 centimeters long, and 10 centimeters in diameter, and its full of oxygen which is 90°F and compressed to one ATM. Now, suppose a circular door one inch in diameter is opened on one of the flat sides. How fast will the can be moving after one second?

Since you seem to think you're the thermodynamics king, you should have no trouble figuring that out. If you can't figure that out, you clearly have no idea what you're talking about, and everything you've said in this thread can be ignored.
posted by delmoi at 1:18 PM on January 12, 2007

This is the sound of somebody who took a 1st-year physics class in university frantically looking for his textbook.
posted by tehloki at 1:30 PM on January 13, 2007 [1 favorite]

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