Why does a detached tire roll faster than the car?
October 28, 2006 9:59 PM Subscribe
[PhysicsFilter] In movies, when a tire comes off of a speeding car, it begins to move faster than the car. Why does the velocity increase?
This is a topic of debate among some of the post-doc Physicists in my lab. There were a few theories proposed that I don't claim to understand:
* The lack of constraints on the tire causes a shift in energy (by conservation of momentum), which simply translates from rotational to kinetic velocity. I don't really understand their argument here, since there is a removal of active forces, and possibly a change in mass of the system (from rotating tire-wheel-axel combination to just tire).
* The diameter of the tire changes, since there's no more weight on it, and so this causes the acceleration. This is counterintuitive simply because the diameter should increase, causes a reduction in angular velocity.
* The process is akin to alpha decay, where the mass of the car is much greater than the tire.
Also, in the movies, the tire seems to jump first, then land and speed ahead of the car.
This is a topic of debate among some of the post-doc Physicists in my lab. There were a few theories proposed that I don't claim to understand:
* The lack of constraints on the tire causes a shift in energy (by conservation of momentum), which simply translates from rotational to kinetic velocity. I don't really understand their argument here, since there is a removal of active forces, and possibly a change in mass of the system (from rotating tire-wheel-axel combination to just tire).
* The diameter of the tire changes, since there's no more weight on it, and so this causes the acceleration. This is counterintuitive simply because the diameter should increase, causes a reduction in angular velocity.
* The process is akin to alpha decay, where the mass of the car is much greater than the tire.
Also, in the movies, the tire seems to jump first, then land and speed ahead of the car.
A car missing a wheel probably isn't going to be moving optimally. Perhaps it slows down and the wheel rolls on ahead.
posted by Good Brain at 10:07 PM on October 28, 2006
posted by Good Brain at 10:07 PM on October 28, 2006
Uhmm... Because the director said "hey, the tire should roll ahead of the car after it comes off."
Why do real-life physics enter into this at all?
posted by blenderfish at 10:09 PM on October 28, 2006
Why do real-life physics enter into this at all?
posted by blenderfish at 10:09 PM on October 28, 2006
I'm guessing the tire only moves foward relative to the car... the car is slowing down.
Which movies are we talking about?
posted by phrontist at 10:12 PM on October 28, 2006
Which movies are we talking about?
posted by phrontist at 10:12 PM on October 28, 2006
These must be theoretical physicists since they are offering explanations for something which has not been experimentally verified to even occur.
As others have mentioned, its the car which is slowing down, since it is now missing a tire. Unless of course you have a specific movie scene you are talking about where this is not the case.
posted by vacapinta at 10:18 PM on October 28, 2006
As others have mentioned, its the car which is slowing down, since it is now missing a tire. Unless of course you have a specific movie scene you are talking about where this is not the case.
posted by vacapinta at 10:18 PM on October 28, 2006
I always thought it was because the story required the characters inside the car to see the rolling tire.
posted by winston at 10:21 PM on October 28, 2006
posted by winston at 10:21 PM on October 28, 2006
Yea I agree with blenderfish that this makes a better movie. Also in my opinion the tire should just move relative to the speed it fell off the car.
posted by pwally at 10:21 PM on October 28, 2006
posted by pwally at 10:21 PM on October 28, 2006
The tire has both forward momentum and rotational momentum, and the ability to convert the rotational momentum to forward momentum. Assuming that both the car and the loose tire are experiencing similar (lack of) acceleration and similar air resistance and road friction, the tire has more total energy relative to its mass to keep it moving forward.
That said, I really have to go with the "once the tire falls off, the car is slowing down right quick" explanation.
posted by trevyn at 10:25 PM on October 28, 2006 [1 favorite]
That said, I really have to go with the "once the tire falls off, the car is slowing down right quick" explanation.
posted by trevyn at 10:25 PM on October 28, 2006 [1 favorite]
My favorite tire striking-out-on-its-own video (youtube).
This is a fairly famous photo of Raymond Mays losing a tire on a hillclimb.
posted by maxwelton at 10:31 PM on October 28, 2006 [3 favorites]
This is a fairly famous photo of Raymond Mays losing a tire on a hillclimb.
posted by maxwelton at 10:31 PM on October 28, 2006 [3 favorites]
When your tire falls off, your first instinct is to hit the brakes. It's a big jolt inside the car. Also, the car falls down on that end and scrapes the road, slowing it down more. (At least that's how it was the time I was in a car that lost a wheel. And the wheel went merrily speeding down the hill and smashed into another car. But nobody was hurt.)
posted by Addlepated at 10:37 PM on October 28, 2006
posted by Addlepated at 10:37 PM on October 28, 2006
A couple of months ago the back wheel of a car in front of me fell off. The wheel went slower than the car. So there's a real life data point to add to the movie data points.
posted by markr at 10:42 PM on October 28, 2006
posted by markr at 10:42 PM on October 28, 2006
This sounds like it should be simple first year momentum, but, seeing as a group of post-doc physicists are unable to agree, it would have to be the wonders of movie magic.
posted by cholly at 11:00 PM on October 28, 2006
posted by cholly at 11:00 PM on October 28, 2006
This question feels like there's a lot of information missing. I have no idea what "akin to alpha decay" means. Both atoms and cars obey the same laws of momentum.
posted by vacapinta at 11:02 PM on October 28, 2006
posted by vacapinta at 11:02 PM on October 28, 2006
I agree with the "car slowing down" theory that seems likely. I would have thought that once the tire is free from driving forward that heavy car, all its momentum can go into rolling itself forward, much lighter than wheel + car. Howeve,r it has now lost the power from the car. So I would expect it to slow down quite quickly.
posted by Joh at 11:28 PM on October 28, 2006
posted by Joh at 11:28 PM on October 28, 2006
Seems simple enough to me with what physics knowledge I have. Assuming that this is a perfect disconnection between the car and the wheel (that is to say, the tire doesn't loose any energy to friction or the like), the tire would go faster than than the car because the rotational monmentum that used to have to contribute to "pulling" the car is now freely available to simply turn the wheel and move it forward on its own. Same energy, less mas, more speed.
posted by dantekgeek at 12:00 AM on October 29, 2006
posted by dantekgeek at 12:00 AM on October 29, 2006
Has anyone here had to take two bikes somewhere? So, you ride one and you carefully hold the other one beside you.
Does anyone believe that if you let go of the second bike, it would suddenly zoom forward?
Why not? After all the tires are going the same speed when they are let go and yet "pulling" less mass (one bike instead of two) Thats the argument people are making in this thread.
posted by vacapinta at 12:11 AM on October 29, 2006
Does anyone believe that if you let go of the second bike, it would suddenly zoom forward?
Why not? After all the tires are going the same speed when they are let go and yet "pulling" less mass (one bike instead of two) Thats the argument people are making in this thread.
posted by vacapinta at 12:11 AM on October 29, 2006
I think the problem here is with assumptions about the car being under engine power or not.
If the car is not maintaining speed under engine power, and the wheel seamlessly detaches, the wheel will lose velocity at a much lower rate than the car, thus the appearance of the wheel moving faster.
If the car is maintaining speed under engine power, the wheel will, at best, keep up with the car.
posted by trevyn at 12:45 AM on October 29, 2006
If the car is not maintaining speed under engine power, and the wheel seamlessly detaches, the wheel will lose velocity at a much lower rate than the car, thus the appearance of the wheel moving faster.
If the car is maintaining speed under engine power, the wheel will, at best, keep up with the car.
posted by trevyn at 12:45 AM on October 29, 2006
I've got a point to add to markr's. This happened about 14 years ago when I was in summer camp.
On the way back from a trip to somewhere in CT, driving on the highway in the lane next to our van, was a car pulling a small U-Haul trailer. Right as I happened to look out the window at the trailer, the left wheel popped off. The wheel slowed down but continued to roll down the highway behind our van as the car and trailer maintained their speed for at least half a minute before the driver noticed what had happened. The wheel gradually slowed down until we couldn't see it anymore (either because other cars were blocking our view or because it found its way to the side of the highway).
posted by Venadium at 12:54 AM on October 29, 2006
On the way back from a trip to somewhere in CT, driving on the highway in the lane next to our van, was a car pulling a small U-Haul trailer. Right as I happened to look out the window at the trailer, the left wheel popped off. The wheel slowed down but continued to roll down the highway behind our van as the car and trailer maintained their speed for at least half a minute before the driver noticed what had happened. The wheel gradually slowed down until we couldn't see it anymore (either because other cars were blocking our view or because it found its way to the side of the highway).
posted by Venadium at 12:54 AM on October 29, 2006
In family stories of the early days of motoring it seemed to be considered funny but not totally remarkable to be overtaken by one of your car's wheels when going downhill.
posted by Idcoytco at 1:13 AM on October 29, 2006
posted by Idcoytco at 1:13 AM on October 29, 2006
This is a topic of debate among some of the post-doc Physicists in my lab.
This is scaring the hell out of me. This is basic mechanics.
The tire has a given kinetic energy, and it will continue to roll until frictional forces slow it down. It will never go faster than the car unless the car is slowed down faster (which is likely, cars don't work too well after losing a wheel).
If the tire went faster, it was either a) movie magic, or b) an outside force acting on the tire. For example, perhaps the person driving was gunning the throttle when the car hit a bump. The tire was momentarily airborne, and picked up some extra rotational speed, and then it magically fell off. The extra rotational speed translated into forward motion when the tire hit the ground. But, more likely it's just movie fictional physics.
posted by knave at 1:46 AM on October 29, 2006
This is scaring the hell out of me. This is basic mechanics.
The tire has a given kinetic energy, and it will continue to roll until frictional forces slow it down. It will never go faster than the car unless the car is slowed down faster (which is likely, cars don't work too well after losing a wheel).
If the tire went faster, it was either a) movie magic, or b) an outside force acting on the tire. For example, perhaps the person driving was gunning the throttle when the car hit a bump. The tire was momentarily airborne, and picked up some extra rotational speed, and then it magically fell off. The extra rotational speed translated into forward motion when the tire hit the ground. But, more likely it's just movie fictional physics.
posted by knave at 1:46 AM on October 29, 2006
The diameter of the tire changes, since there's no more weight on it, and so this causes the acceleration. This is counterintuitive simply because the diameter should increase, causes a reduction in angular velocity.
Just to do a little extra debunking -- the change in diameter of a properly inflated tire is going to be negligible. Properly inflated, a tire is barely deformed at all by the weight of the car.
posted by knave at 1:49 AM on October 29, 2006
Just to do a little extra debunking -- the change in diameter of a properly inflated tire is going to be negligible. Properly inflated, a tire is barely deformed at all by the weight of the car.
posted by knave at 1:49 AM on October 29, 2006
I looked for this on Insultingly Stupid Movie Physics, but didn't find it.
Perhaps you could email them then question with an example movie?
There is a lot of good analysis & debunking (as well as movie criticism!) going on there, and they might take on your cause.
posted by Mutant at 1:16 AM on October 29, 2006
Perhaps you could email them then question with an example movie?
There is a lot of good analysis & debunking (as well as movie criticism!) going on there, and they might take on your cause.
posted by Mutant at 1:16 AM on October 29, 2006
another datapoint for you.
I can't explain the physics but I can vouch that the tyre can go faster than the car. I'd guess at conservation of momentum and change in mass though.
I overtook a truck one night and as I moved back into the correct lane a truck tyre overtook me. It was only visible for a few moments as it moved through the lume of the headlights and then it was gone. I was travelling significantly faster than the truck and I guess that the tyre was going 20 - 30 km/h faster than me.
It took a while to register what had happened and how lucky we had just been. Fortunately there was no traffic in the other direction at the time either.
posted by w.fugawe at 1:35 AM on October 29, 2006
I can't explain the physics but I can vouch that the tyre can go faster than the car. I'd guess at conservation of momentum and change in mass though.
I overtook a truck one night and as I moved back into the correct lane a truck tyre overtook me. It was only visible for a few moments as it moved through the lume of the headlights and then it was gone. I was travelling significantly faster than the truck and I guess that the tyre was going 20 - 30 km/h faster than me.
It took a while to register what had happened and how lucky we had just been. Fortunately there was no traffic in the other direction at the time either.
posted by w.fugawe at 1:35 AM on October 29, 2006
There's also the very real possibility that the tire is 'burning rubber' so that after it pops off it is able to convert more of its rotation to forward movement.
vacaptina's bike notwithstanding, I think we can all imagine a car in the middle of peeling out. The car + tire system is stationary, but if we remove the car instantly the tires will shoot forward.
posted by fleacircus at 2:02 AM on October 29, 2006
vacaptina's bike notwithstanding, I think we can all imagine a car in the middle of peeling out. The car + tire system is stationary, but if we remove the car instantly the tires will shoot forward.
posted by fleacircus at 2:02 AM on October 29, 2006
If the car is under heavy acceleration
AND
the wheel that comes off is one of the driving wheels
AND
the wheel was off the ground for a little bit before coming off, maybe due to the car having bounced over a heavy bump really really hard
THEN
the errant wheel MIGHT have been given enough extra spin, by virtue of the car's differential, to outpace the car temporarily once the rubber hits the road again.
Maybe.
I'm voting movie magic.
posted by flabdablet at 3:31 AM on October 29, 2006
AND
the wheel that comes off is one of the driving wheels
AND
the wheel was off the ground for a little bit before coming off, maybe due to the car having bounced over a heavy bump really really hard
THEN
the errant wheel MIGHT have been given enough extra spin, by virtue of the car's differential, to outpace the car temporarily once the rubber hits the road again.
Maybe.
I'm voting movie magic.
posted by flabdablet at 3:31 AM on October 29, 2006
I guess the car slowing down is probably the more important factor, usually.
But there is also the fact that the tires are elastic. If it's a drive wheel, and there's torque on it, the tire will get twisted in such a way that when it's released it will spring forward, momentarily faster than the car. By how much, I have no idea.
posted by sfenders at 5:10 AM on October 29, 2006
But there is also the fact that the tires are elastic. If it's a drive wheel, and there's torque on it, the tire will get twisted in such a way that when it's released it will spring forward, momentarily faster than the car. By how much, I have no idea.
posted by sfenders at 5:10 AM on October 29, 2006
If . . . the wheel that comes off is one of the driving wheels
Assuming a front-wheel-drive vehicle with four wheels and (weight of the) the engine in front... If the wheel is one of the drive (front) wheels, the car is going to dive into the pavement and stop as though it hit a brick wall. (This happened to my Dad)
posted by winston at 5:22 AM on October 29, 2006
Assuming a front-wheel-drive vehicle with four wheels and (weight of the) the engine in front... If the wheel is one of the drive (front) wheels, the car is going to dive into the pavement and stop as though it hit a brick wall. (This happened to my Dad)
posted by winston at 5:22 AM on October 29, 2006
I can't believe there are people who haven't seen this in news footage. I don't believe the OP is referring to fictional movies. It appears to happen every single time a tire comes off a race car that I've ever seen.
I always assumed it was a combo of a) no weight to pull and b) slowing down car (either due to breaks or that it now has 3 wheels pulling the same weight 4 were).
posted by dobbs at 6:09 AM on October 29, 2006
I always assumed it was a combo of a) no weight to pull and b) slowing down car (either due to breaks or that it now has 3 wheels pulling the same weight 4 were).
posted by dobbs at 6:09 AM on October 29, 2006
Does anyone believe that if you let go of the second bike, it would suddenly zoom forward?
No, but the analogy doesn't hold because the bike would flop over pretty quickly without someone to balance it, while the car tire would be able to continue balancing on its own (as demonstrated by the linked videos)- though please don't take this as an attempted defense of hollywood.
posted by drezdn at 6:59 AM on October 29, 2006
Under normal circumstances, a gently-detached wheel should gradually slow down. The car may slow down more quickly due to braking or increased friction. The "conservation of momentum" idea is crazy: that just means that the wheel will continue at it's previous velocity (and somehow "combining" this with rotational momentum won't change that result).
HOWEVER, wheels don't usually come off of vehicles. I would imagine that a significant number of wheel detachments accompany violent events (serious bumps in the road, collisions, etc). Collisions with big, complicated shapes can produce odd trajectories for debris, so a wheel could get launched that way. It's also easy to imagine a wheel coming loose, and running into the wheel well of the car, where it would likely be pinched. Pinching or smacking into a rapidly spinning, elastic object is also something that could seemingly send the wheel flying at high speed.
posted by Humanzee at 9:30 AM on October 29, 2006
HOWEVER, wheels don't usually come off of vehicles. I would imagine that a significant number of wheel detachments accompany violent events (serious bumps in the road, collisions, etc). Collisions with big, complicated shapes can produce odd trajectories for debris, so a wheel could get launched that way. It's also easy to imagine a wheel coming loose, and running into the wheel well of the car, where it would likely be pinched. Pinching or smacking into a rapidly spinning, elastic object is also something that could seemingly send the wheel flying at high speed.
posted by Humanzee at 9:30 AM on October 29, 2006
Having experienced the dubious pleasure of having a wheel fly off a car I was driving I can state that the tire can overtake you, even on flat ground. Probably because the urge to let off the gas or brake is hard to resist.
posted by Mitheral at 9:50 AM on October 29, 2006
posted by Mitheral at 9:50 AM on October 29, 2006
maxwelton writes "This is a fairly famous photo of Raymond Mays losing a tire on a hillclimb."
That is an amazing picture. Perfectly composed, awesome expression and perfectly timed considering what must have been the photographic state of the art at the time.
posted by Mitheral at 10:00 AM on October 29, 2006
That is an amazing picture. Perfectly composed, awesome expression and perfectly timed considering what must have been the photographic state of the art at the time.
posted by Mitheral at 10:00 AM on October 29, 2006
HOWEVER, wheels don't usually come off of vehicles.
It happened to my mom. She had her tires rotated and they forgot to put the bolts back onto one of the wheels. Hers accelerated past the car because it rolled down a big hill. Luckily some guys were sitting out on their porch near where it happened. They retrieved the wheel for her and let her use the phone to call the tire place.
posted by caddis at 10:33 AM on October 29, 2006
It happened to my mom. She had her tires rotated and they forgot to put the bolts back onto one of the wheels. Hers accelerated past the car because it rolled down a big hill. Luckily some guys were sitting out on their porch near where it happened. They retrieved the wheel for her and let her use the phone to call the tire place.
posted by caddis at 10:33 AM on October 29, 2006
This is the reversal of an inelastic collision of two rotating bodies.
Let Ia be the moment of inertia of the tire, Ib the moment of the wheel, w their shared angular speed before, wa the angular speed of the tire after, and wb the angular speed of the wheel after. Then simple conservation of angular momentum gives:
(Ia+Ib)w=Ia*wa+Ib*wb
Or:
Ia/Ib=-(Δwb)/(Δwa)
where Δwa=(w-wa), and similarly for Δwb. The negative sign enters because one slows down and one speeds up (the one with a smaller moment of inertia speeds up).
The moment of the tire is obviously much smaller than the moment of the "wheel," where we should really be considering the "wheel" to include the other tires, the axles, etc. This implies that the change in the angular speed of the tire is much greater than the change for the rest of the wheel system.
The situation is analagous to a gun going off, where a small explosion results in the light bullet going away at a much higher speed than the much heavier gun.
posted by dsword at 10:45 AM on October 29, 2006
Let Ia be the moment of inertia of the tire, Ib the moment of the wheel, w their shared angular speed before, wa the angular speed of the tire after, and wb the angular speed of the wheel after. Then simple conservation of angular momentum gives:
(Ia+Ib)w=Ia*wa+Ib*wb
Or:
Ia/Ib=-(Δwb)/(Δwa)
where Δwa=(w-wa), and similarly for Δwb. The negative sign enters because one slows down and one speeds up (the one with a smaller moment of inertia speeds up).
The moment of the tire is obviously much smaller than the moment of the "wheel," where we should really be considering the "wheel" to include the other tires, the axles, etc. This implies that the change in the angular speed of the tire is much greater than the change for the rest of the wheel system.
The situation is analagous to a gun going off, where a small explosion results in the light bullet going away at a much higher speed than the much heavier gun.
posted by dsword at 10:45 AM on October 29, 2006
dsword: That doesn't make sense. For starters, the instant after the tire seperates, every rotating part will continue rotating at the same rate. Your equation actually has an infinite number of solutions. The one that maintains conservation of energy is the one with wa = wb = w (that is, rotation is unchanged). Secondly, your equations seem to be solving for differing rotation rates of the tire and the wheel. If you're talking about a freely-rotating wheel where the tire flies off somehow, that might make sense. From what I can gather, we're discussing the case where the wheel seperates from the car, and the tire remains attached to the wheel.
posted by Humanzee at 11:30 AM on October 29, 2006
posted by Humanzee at 11:30 AM on October 29, 2006
There are obviously an infinite number of solutions, as there are an infinite number of ways the wheel can be forced off. The solution that satisfies energy conservation is not the right one to use. Energy is not conserved. There's a small amount of energy supplied to the system. That's why the tire comes off in the first place. The specifics of the process are irrelevant, though. Because the wheel itself has a much lower moment of inertia, the energy supplied to the system results in a much larger change in its rotational speed relative to the car's.
Consider the inelastic collision of two bodies rotating at different speeds. If one has a much higher moment, its rate of rotation changes very little compared to the smaller moment object. Energy is lost because the collision is inelastic. Now, to reverse the process, take two bodies rotating together, and push them apart while supplying a small amount of energy. Again, it's exactly the same as a bullet firing from a gun.
posted by dsword at 12:03 PM on October 29, 2006
Consider the inelastic collision of two bodies rotating at different speeds. If one has a much higher moment, its rate of rotation changes very little compared to the smaller moment object. Energy is lost because the collision is inelastic. Now, to reverse the process, take two bodies rotating together, and push them apart while supplying a small amount of energy. Again, it's exactly the same as a bullet firing from a gun.
posted by dsword at 12:03 PM on October 29, 2006
What collision are you talking about?
Conservation of momentum says that that wheel, with the tire attached, will continue to rotate at the same speed absent some force being supplied. Friction is a force which will slow it down. Unless something occurs at the moment of separation which applies a force there is no force to speed up the rotation. Of course, in the case of my mother's incident, gravity supplied a force, but I think for this thought experiment we are assuming level ground.
posted by caddis at 12:21 PM on October 29, 2006
Conservation of momentum says that that wheel, with the tire attached, will continue to rotate at the same speed absent some force being supplied. Friction is a force which will slow it down. Unless something occurs at the moment of separation which applies a force there is no force to speed up the rotation. Of course, in the case of my mother's incident, gravity supplied a force, but I think for this thought experiment we are assuming level ground.
posted by caddis at 12:21 PM on October 29, 2006
bike would flop over pretty quickly without someone to balance it
Depends on the bike but, with a bit of speed, most bikes are self-balancing. That's why riding no-hands is possible.
posted by randomstriker at 12:53 PM on October 29, 2006
Depends on the bike but, with a bit of speed, most bikes are self-balancing. That's why riding no-hands is possible.
posted by randomstriker at 12:53 PM on October 29, 2006
The tire isn't moving faster, the car is moving slower. If you take a wheel off the car, there is a LOT more friction at that corner of the car; from inside the car it probably feels like slamming on the brakes.
The tire, on the other, isn't slowing down near as much because it is rolling, not sliding, down the road.
posted by Doohickie at 1:27 PM on October 29, 2006
The tire, on the other, isn't slowing down near as much because it is rolling, not sliding, down the road.
posted by Doohickie at 1:27 PM on October 29, 2006
This sounds like it should be simple first year momentum, but, seeing as a group of post-doc physicists are unable to agree
Nope. Simple first year stuff. When people get into that post-doc world, they sometimes lose touch with common sense. I think that is the case here.
I worked with a guy that had a doctorate in statistics and probability. When California started their lottery, he started tracking every Lotto draw, in an attempt to predict future draws. I asked if he were trying to determine whether certain balls were more likely to be drawn, and therefore he would pick those numbers and he replied that, no, he was looking for the ones that *hadn't* come up and pick those because after a while the law of averages would have to take over.
The flaw in his reasoning is that if you assume a totally random system (which is what he insisted the lottery was), past events do not affect future events- if you flip a coin 20 times and it comes up heads, but it is a truly random coin, the coin still has a 50-50 chance of coming up heads or tails. If you want to predict with any greater accuracy after that point, the only reasonable assumption worth betting on is that it is a trick coin that somehow comes up heads more frequently.
posted by Doohickie at 1:38 PM on October 29, 2006
Nope. Simple first year stuff. When people get into that post-doc world, they sometimes lose touch with common sense. I think that is the case here.
I worked with a guy that had a doctorate in statistics and probability. When California started their lottery, he started tracking every Lotto draw, in an attempt to predict future draws. I asked if he were trying to determine whether certain balls were more likely to be drawn, and therefore he would pick those numbers and he replied that, no, he was looking for the ones that *hadn't* come up and pick those because after a while the law of averages would have to take over.
The flaw in his reasoning is that if you assume a totally random system (which is what he insisted the lottery was), past events do not affect future events- if you flip a coin 20 times and it comes up heads, but it is a truly random coin, the coin still has a 50-50 chance of coming up heads or tails. If you want to predict with any greater accuracy after that point, the only reasonable assumption worth betting on is that it is a trick coin that somehow comes up heads more frequently.
posted by Doohickie at 1:38 PM on October 29, 2006
Right, so is anyone with some knowledge of physics going to consider the intuitive reason why I think the wheel should speed up? The tire is flexible rubber, makes a pretty good spring. Imagine you were pushing the car along with a compressed linear spring at the proper angle with a tiny little wheel at the end, and both the spring and the wheel came loose together; then it's obvious that the spring would bounce up and forward. Same thing for a flexible tire, except with some angular momentum complicating things.
Anyway, that may be what your physicists are talking about when they say "The lack of constraints on the tire causes a shift in energy, which simply translates from rotational to kinetic velocity." The total angular momentum of the wheel wouldn't change if it were released when not in contact with the road, just the edge would be momentarily rotating faster than the centre. Just like the linear spring above wouldn't change its total momentum if both road and car disappeared. But since it is in contact with the road when that happens, some of its rotational energy is translated to kinetic velocity, making it go faster than the car, at least for the first bounce.
That's for a drive wheel; for a free wheel with the car at constant speed, I guess you'd get the opposite effect because of the friction of the axle etc., but it'd be relatively a very small change in velocity. For braking too, the wheel would bounce slower than the car, but then it'd quickly catch up.
posted by sfenders at 2:39 PM on October 29, 2006
Anyway, that may be what your physicists are talking about when they say "The lack of constraints on the tire causes a shift in energy, which simply translates from rotational to kinetic velocity." The total angular momentum of the wheel wouldn't change if it were released when not in contact with the road, just the edge would be momentarily rotating faster than the centre. Just like the linear spring above wouldn't change its total momentum if both road and car disappeared. But since it is in contact with the road when that happens, some of its rotational energy is translated to kinetic velocity, making it go faster than the car, at least for the first bounce.
That's for a drive wheel; for a free wheel with the car at constant speed, I guess you'd get the opposite effect because of the friction of the axle etc., but it'd be relatively a very small change in velocity. For braking too, the wheel would bounce slower than the car, but then it'd quickly catch up.
posted by sfenders at 2:39 PM on October 29, 2006
sfenders: It's difficult to discuss this stuff without diagrams. I'll give it a shot.
If I'm reading you correctly, the idea is that the drive wheel is torqued by the engine from the inside, and torqued by friction from the outside. The two torques balance at constant speed, but since the wheel is made of real material instead of adamantium, there is some shear: the inner portion of the wheel leads the outer portion of the wheel. Now the wheel comes loose and we want to imagine what this shear does.
For a moment, set aside the magnitude of this effect. Because of conservation of angular momentum, the overall rotation rate can't change. But the elastic force that decreases the shear will cause the rotation of the inner portion of the wheel to decrease, and the outer part to increase, speeding up the wheel .... momentarily. Eventually the shear decrease to zero, then due to the inertia (angular momentum of the different components of the wheel) it will reverse sign. It won't shear quite as much due to friction, but on the return trip, the wheel will tend to slow down. This cycle will repeat: overall wheel speed will oscillate, with decreasing amplitude.
Now for the magnitude. I don't have any numbers, but I'm guessing that it's negligible. If tires flexed a lot, driving would be incredibly difficult. With a (linear) spring,
Force = k * displacement (k is stiffness)
Energy = .5 * k * displacement ^ 2.
Solving for displacement,
Energy = .5 * (Force^2) / k
In other words, with a given applied force, the stiffer the spring, the less energy is stored. The story is similar with circular springs. I would expect tires to be very stiff circular springs.
Finally, when the postdocs were talking about "The lack of constraints on the tire causes a shift in energy, which simply translates from rotational to kinetic velocity" I suspect that they were thinking in a sloppy fashion. If you roll a tire up a hill, it will travel farther than a puck sliding up a frictionless incline. This is because the rotational energy must be used up before the tire stops rotating. Having seen problems like this in the past, the postdocs probably just had the vague idea that sometimes rotational kinetic energy matters for translational motion.
posted by Humanzee at 3:21 PM on October 29, 2006
If I'm reading you correctly, the idea is that the drive wheel is torqued by the engine from the inside, and torqued by friction from the outside. The two torques balance at constant speed, but since the wheel is made of real material instead of adamantium, there is some shear: the inner portion of the wheel leads the outer portion of the wheel. Now the wheel comes loose and we want to imagine what this shear does.
For a moment, set aside the magnitude of this effect. Because of conservation of angular momentum, the overall rotation rate can't change. But the elastic force that decreases the shear will cause the rotation of the inner portion of the wheel to decrease, and the outer part to increase, speeding up the wheel .... momentarily. Eventually the shear decrease to zero, then due to the inertia (angular momentum of the different components of the wheel) it will reverse sign. It won't shear quite as much due to friction, but on the return trip, the wheel will tend to slow down. This cycle will repeat: overall wheel speed will oscillate, with decreasing amplitude.
Now for the magnitude. I don't have any numbers, but I'm guessing that it's negligible. If tires flexed a lot, driving would be incredibly difficult. With a (linear) spring,
Force = k * displacement (k is stiffness)
Energy = .5 * k * displacement ^ 2.
Solving for displacement,
Energy = .5 * (Force^2) / k
In other words, with a given applied force, the stiffer the spring, the less energy is stored. The story is similar with circular springs. I would expect tires to be very stiff circular springs.
Finally, when the postdocs were talking about "The lack of constraints on the tire causes a shift in energy, which simply translates from rotational to kinetic velocity" I suspect that they were thinking in a sloppy fashion. If you roll a tire up a hill, it will travel farther than a puck sliding up a frictionless incline. This is because the rotational energy must be used up before the tire stops rotating. Having seen problems like this in the past, the postdocs probably just had the vague idea that sometimes rotational kinetic energy matters for translational motion.
posted by Humanzee at 3:21 PM on October 29, 2006
What collision are you talking about?
No collision. I'm saying the process can be thought of in the same way, only backwards.
Conservation of momentum says that that wheel, with the tire attached, will continue to rotate at the same speed absent some force being supplied.
That's not true. Conservation of angular momentum says that the combined momentum of the wheel/tire system remains the same absent external torques. That's all. It is completely separate from rotational kinetic energy.
I'm not saying every time the wheel will shoot off incredibly fast. The point is that a small addition of rotational energy to the system--however it happens, it doesn't really matter--may increase the angular momentum of the tire a lot by decreasing the angular momentum of the much "heavier" wheel a small amount. I've assumed there are no external torques, so angular momentum is conserved, but rotational kinetic energy is not.
posted by dsword at 3:31 PM on October 29, 2006
No collision. I'm saying the process can be thought of in the same way, only backwards.
Conservation of momentum says that that wheel, with the tire attached, will continue to rotate at the same speed absent some force being supplied.
That's not true. Conservation of angular momentum says that the combined momentum of the wheel/tire system remains the same absent external torques. That's all. It is completely separate from rotational kinetic energy.
I'm not saying every time the wheel will shoot off incredibly fast. The point is that a small addition of rotational energy to the system--however it happens, it doesn't really matter--may increase the angular momentum of the tire a lot by decreasing the angular momentum of the much "heavier" wheel a small amount. I've assumed there are no external torques, so angular momentum is conserved, but rotational kinetic energy is not.
posted by dsword at 3:31 PM on October 29, 2006
According to this the tire actually has a greater moment of inertia (and mass) than the rest of the wheel, at least for the FWD wheels that they considered. It's not obvious to me that wheels are ever much more massive than tires, and I still fail to see what your "gunshot" mechanism is.
posted by Humanzee at 4:22 PM on October 29, 2006
posted by Humanzee at 4:22 PM on October 29, 2006
Humanzee,
The gunshot analogy was meant to provide a parallel between translational and rotational mechanics. The basic idea is that whenever there's some internal process that supplies kinetic energy to the system, it will be divided among the constituents so as to conserve momentum.
Here's an experiment to try. Get a tape measure and an office chair. Extend the tape measure, sit in the chair, and let go of the tape. Now repeat, but throw the thing in the air right after you let it go and watch it spin like crazy. There are no external torques in either case, so the angular momentum of the system is conserved.
Now, in this case, you know what's causing the change in the rotational speed of the tape measure: it's the spring inside it and the metal piece at the end of the tape smacking into the body of the tape measure, and it clearly happens after you throw the thing. That's fine. Take a wheel, which is fairly complicated. I don't know what goes on when the wheel comes off. I don't even know if it actually speeds up. That's fine. I don't need to. I can't think of any external torques, so I assume that angular momentum is conserved. But I see no reason to believe that rotational energy is conserved, and I know how to conserve the combined angular momentum of two bodies given a change in the total rotational energy. That's all I need to know. I don't need to know any of the specifics of the forces or torques involved. That's the whole reason physicists like using conservation laws.
As for the link you provided, that does surprise me. However, I was considering the axles and other wheels along with the tire, because the system as a whole is constrained.
posted by dsword at 4:55 PM on October 29, 2006
The gunshot analogy was meant to provide a parallel between translational and rotational mechanics. The basic idea is that whenever there's some internal process that supplies kinetic energy to the system, it will be divided among the constituents so as to conserve momentum.
Here's an experiment to try. Get a tape measure and an office chair. Extend the tape measure, sit in the chair, and let go of the tape. Now repeat, but throw the thing in the air right after you let it go and watch it spin like crazy. There are no external torques in either case, so the angular momentum of the system is conserved.
Now, in this case, you know what's causing the change in the rotational speed of the tape measure: it's the spring inside it and the metal piece at the end of the tape smacking into the body of the tape measure, and it clearly happens after you throw the thing. That's fine. Take a wheel, which is fairly complicated. I don't know what goes on when the wheel comes off. I don't even know if it actually speeds up. That's fine. I don't need to. I can't think of any external torques, so I assume that angular momentum is conserved. But I see no reason to believe that rotational energy is conserved, and I know how to conserve the combined angular momentum of two bodies given a change in the total rotational energy. That's all I need to know. I don't need to know any of the specifics of the forces or torques involved. That's the whole reason physicists like using conservation laws.
As for the link you provided, that does surprise me. However, I was considering the axles and other wheels along with the tire, because the system as a whole is constrained.
posted by dsword at 4:55 PM on October 29, 2006
Simple observation: a car with 3 wheels is going to slow down. The wheel alone has no such limits and will continue on its journey, free of the burden of its former master.
oooommmmm... ooommmm... ooooommmmm...
posted by chairface at 4:56 PM on October 29, 2006
oooommmmm... ooommmm... ooooommmmm...
posted by chairface at 4:56 PM on October 29, 2006
the elastic force that decreases the shear will cause the rotation of the inner portion of the wheel to decrease, and the outer part to increase, speeding up the wheel .... momentarily. Eventually the shear decrease to zero, then due to the inertia it will reverse sign.
Yes, that's what I had in mind. It's during the first part of that cycle that the rubber is in contact with the road, so that's why some angular momentum is immediately converted to forward motion. The oscillation briefly continues while it's in the air, and stuff happens when the tire hits the ground again, but the wheel will initially be thrown forward.
As for the magnitude of that effect with normal road car tires, I've no idea. But I seem to remember that you can actually see the tires deform on some of those crazy drag racing cars when they launch. Normal tires may be hard to compress this way, but then the force on them is very large compared to the mass of the wheel, so it's hard to guess.
posted by sfenders at 5:44 PM on October 29, 2006
Yes, that's what I had in mind. It's during the first part of that cycle that the rubber is in contact with the road, so that's why some angular momentum is immediately converted to forward motion. The oscillation briefly continues while it's in the air, and stuff happens when the tire hits the ground again, but the wheel will initially be thrown forward.
As for the magnitude of that effect with normal road car tires, I've no idea. But I seem to remember that you can actually see the tires deform on some of those crazy drag racing cars when they launch. Normal tires may be hard to compress this way, but then the force on them is very large compared to the mass of the wheel, so it's hard to guess.
posted by sfenders at 5:44 PM on October 29, 2006
stuff happens when the tire hits the ground again
... which is a big problem with that theory, given that the tire is supposed to land and keep on rolling faster than the car was going.
So, on further thought: The tire is compressed by the weight of the car pushing down on it. It comes free from the car while rolling forward. Now, as it rolls forward, the bit that was squashed against the road is free to resume its normal form, but by the time it can do so it is already centred slightly behind the point where the it makes contact with the ground. So, the rebounding tire not only pushes itself forward, it maybe even contributes to its rotation at the same time.
posted by sfenders at 6:08 PM on October 29, 2006
... which is a big problem with that theory, given that the tire is supposed to land and keep on rolling faster than the car was going.
So, on further thought: The tire is compressed by the weight of the car pushing down on it. It comes free from the car while rolling forward. Now, as it rolls forward, the bit that was squashed against the road is free to resume its normal form, but by the time it can do so it is already centred slightly behind the point where the it makes contact with the ground. So, the rebounding tire not only pushes itself forward, it maybe even contributes to its rotation at the same time.
posted by sfenders at 6:08 PM on October 29, 2006
dsword, your profile says grad student. I surely hope it is not in physics or mechanical engineering. You need to learn how to draw the boxes around objects to study their physics. Draw your box around the wheel, forget about the car, and your problems will be solved.
posted by caddis at 6:20 PM on October 29, 2006
posted by caddis at 6:20 PM on October 29, 2006
How is it possible for a detached tire to accelerate? Acceleration requires an applied force. It doesn't matter if the car is accelerating--when the tire detaches, it cannot go faster than the velocity at that instant; in fact, it must start to decelerate due to friction. There may be some miniscule stored elastic energy from the car's weight, but even this must instantly leave the tire when the car's weight is off. Even a bullet starts to decelerate when it leaves the gun barrel. If you were to shoot a bullet under water, pointed towards the surface, it would not suddenly accelerate when it hit the air. So the only determining factor here is the acceleration or deceleration of the car relative to the certain deceleration of the tire.
posted by weapons-grade pandemonium at 7:00 PM on October 29, 2006
posted by weapons-grade pandemonium at 7:00 PM on October 29, 2006
There may be some miniscule stored elastic energy from the car's weight
From how high do you think you'd have to drop a pressurized tire to get it to compress as much as it does with the weight of the car on it? How high would it bounce? There's way more than enough energy there to get it going a few miles per hour faster than the car. The tire most certainly accelerates, the question is what direction it accelerates in, and how much is lost to the friction of the wheel/car interface in a typical failure of that connection.
posted by sfenders at 7:49 PM on October 29, 2006
From how high do you think you'd have to drop a pressurized tire to get it to compress as much as it does with the weight of the car on it? How high would it bounce? There's way more than enough energy there to get it going a few miles per hour faster than the car. The tire most certainly accelerates, the question is what direction it accelerates in, and how much is lost to the friction of the wheel/car interface in a typical failure of that connection.
posted by sfenders at 7:49 PM on October 29, 2006
Well, I think the previous explanations about the wheel going the same speed while the car slows down is the most likely explanation.
But here is one alternate possibility. If a wheel comes free due to a large bump and the wheel is not in contact with the road, then all of the torque of the engine will be transferred briefly to the airborne wheel. This is the same thing that happens when you are stuck in snow and one wheel spins because it has no traction. In fact, in the case of one fixed wheel and the other spinning, the unfixed wheel will spin at exactly twice the normal speed due to the rotation of the planetary gears in the differential.
In this scenario, the one wheel hits a big bump, loses contact with the ground, is nearly instantaneously accelerated by the differential, and then while still airborne becomes detached. When the detached wheel hits the ground, the faster rotation causes it to overtake the car. In fact, the instantaneous doubling of torque on the lifted wheel by the differential could contribute to its detachment.
Note that this would only work on the driven wheels of a car and would be reduced by a limited slip differential. Also, it is not possible to instantaneously double the speed of the free wheel due to its rotational inertia. This causes backward torque on the differential that causes the engine's torque to be split in some fractional amount between the free wheel and the traction wheel. So the free wheel would actually take a couple of seconds to truly get up to twice normal speed. But certainly the wheel could accelerate in a fraction of a second to substantially more than normal.
So if the stories are true about the wheel being accelerated, it would have to be due to the differential.
posted by JackFlash at 8:28 PM on October 29, 2006
But here is one alternate possibility. If a wheel comes free due to a large bump and the wheel is not in contact with the road, then all of the torque of the engine will be transferred briefly to the airborne wheel. This is the same thing that happens when you are stuck in snow and one wheel spins because it has no traction. In fact, in the case of one fixed wheel and the other spinning, the unfixed wheel will spin at exactly twice the normal speed due to the rotation of the planetary gears in the differential.
In this scenario, the one wheel hits a big bump, loses contact with the ground, is nearly instantaneously accelerated by the differential, and then while still airborne becomes detached. When the detached wheel hits the ground, the faster rotation causes it to overtake the car. In fact, the instantaneous doubling of torque on the lifted wheel by the differential could contribute to its detachment.
Note that this would only work on the driven wheels of a car and would be reduced by a limited slip differential. Also, it is not possible to instantaneously double the speed of the free wheel due to its rotational inertia. This causes backward torque on the differential that causes the engine's torque to be split in some fractional amount between the free wheel and the traction wheel. So the free wheel would actually take a couple of seconds to truly get up to twice normal speed. But certainly the wheel could accelerate in a fraction of a second to substantially more than normal.
So if the stories are true about the wheel being accelerated, it would have to be due to the differential.
posted by JackFlash at 8:28 PM on October 29, 2006
Actually, the torque on both wheels remains the same, but the airborne wheel spins twice as fast.
posted by JackFlash at 9:23 PM on October 29, 2006
posted by JackFlash at 9:23 PM on October 29, 2006
The tire most certainly accelerates, the question is what direction it accelerates in...
The forces that deform the tire when it is on the car are pushing the tire forward and down onto the road. Wouldn't you expect the releasing elastic energy to act in the opposite direction when the wheel is detached from the car? That would be up and backwards.
My son and I have been doing some testing with a simple four-wheeled Lego frame on the kitchen floor. The tires are almost solid (not very elastic), and the “car” is extremely light, so deformation of the tires is negligible. We put one of the wheels only slightly onto the axle, and pushed the car across the floor. When we got the wheel to release halfway across the floor, it went ahead of the car. It appears that the wheel on its own has very little friction compared to the friction of the remaining three wheels carrying the Lego “car”, which slows relatively quickly. There is no appearance of acceleration here, only relative deceleration.
As for the truck tire/tyre passing the car, it is possible, even likely, that it was hit by another vehicle, and bumped ahead. In that case, the elastic energy comes into play.
posted by weapons-grade pandemonium at 10:03 PM on October 29, 2006
The forces that deform the tire when it is on the car are pushing the tire forward and down onto the road. Wouldn't you expect the releasing elastic energy to act in the opposite direction when the wheel is detached from the car? That would be up and backwards.
My son and I have been doing some testing with a simple four-wheeled Lego frame on the kitchen floor. The tires are almost solid (not very elastic), and the “car” is extremely light, so deformation of the tires is negligible. We put one of the wheels only slightly onto the axle, and pushed the car across the floor. When we got the wheel to release halfway across the floor, it went ahead of the car. It appears that the wheel on its own has very little friction compared to the friction of the remaining three wheels carrying the Lego “car”, which slows relatively quickly. There is no appearance of acceleration here, only relative deceleration.
As for the truck tire/tyre passing the car, it is possible, even likely, that it was hit by another vehicle, and bumped ahead. In that case, the elastic energy comes into play.
posted by weapons-grade pandemonium at 10:03 PM on October 29, 2006
caddis,
I'm sorry you feel the need to question my ability to do physics. Why don't you go try your "box" method on the decay of a rho meson into a positive and negative pion? Once you've figured out how drawing boxes around pions explains how a spin-1 particle decays into two spinless particles, then you can come back and tell me how great drawing boxes is for understanding physics, and how it has made you an expert in the field of angular momentum conservation.
posted by dsword at 6:53 AM on October 30, 2006
I'm sorry you feel the need to question my ability to do physics. Why don't you go try your "box" method on the decay of a rho meson into a positive and negative pion? Once you've figured out how drawing boxes around pions explains how a spin-1 particle decays into two spinless particles, then you can come back and tell me how great drawing boxes is for understanding physics, and how it has made you an expert in the field of angular momentum conservation.
posted by dsword at 6:53 AM on October 30, 2006
The diameter of the tire changes, since there's no more weight on it, and so this causes the acceleration.
To explain that bit of the question, consider what would happen if you had a wheel rolling along, and it suddenly increased in diameter by extending an infinite number of very light spokes out from its rim. Since they have no appreciable mass, the thing would want to keep rotating at about the same speed, but its edge, being further out, would be going faster relative to the ground. Contact with the road surface would slow down the rotation, and make it go faster. Thus, some angular momentum is converted to forward speed.
posted by sfenders at 7:31 AM on October 30, 2006
To explain that bit of the question, consider what would happen if you had a wheel rolling along, and it suddenly increased in diameter by extending an infinite number of very light spokes out from its rim. Since they have no appreciable mass, the thing would want to keep rotating at about the same speed, but its edge, being further out, would be going faster relative to the ground. Contact with the road surface would slow down the rotation, and make it go faster. Thus, some angular momentum is converted to forward speed.
posted by sfenders at 7:31 AM on October 30, 2006
I just looked at maxwelton's famous photo, and noticed that not only does the wheel (which is indeed one of the driving wheels) appear to be overtaking the car, but that it's airborne. What I'm guessing happened there is something like this:
1. Rear end of car tries to slide out on a screaming left-hand turn, tearing the inside wheel off the hub. The wheel doesn't detach cleanly straight away; instead, it comes off all but one stud.
2. The torque applied to the hub is no longer balanced by an equal but opposite torque applied to the wheel via its point of contact with the road, and instead causes the hub's rotation to accelerate sharply. The differential also causes a loss of torque to the other driving wheel, causing the hill-climbing car to decelerate.
3. The last remaining stud hurls the detaching wheel up and forward, like a little ballista. At the same time, that corner of the car is falling onto the road (note the angle of the other rear wheel; that back axle is seriously tilted).
Result: the car slows down and the wheel is hurled up and forward.
posted by flabdablet at 2:54 PM on October 30, 2006
1. Rear end of car tries to slide out on a screaming left-hand turn, tearing the inside wheel off the hub. The wheel doesn't detach cleanly straight away; instead, it comes off all but one stud.
2. The torque applied to the hub is no longer balanced by an equal but opposite torque applied to the wheel via its point of contact with the road, and instead causes the hub's rotation to accelerate sharply. The differential also causes a loss of torque to the other driving wheel, causing the hill-climbing car to decelerate.
3. The last remaining stud hurls the detaching wheel up and forward, like a little ballista. At the same time, that corner of the car is falling onto the road (note the angle of the other rear wheel; that back axle is seriously tilted).
Result: the car slows down and the wheel is hurled up and forward.
posted by flabdablet at 2:54 PM on October 30, 2006
Why must everybody insist that the natural thing for the tire to do is to come off cleanly and smoothly, with the axles remaining perfectly rigid and everything staying perfectly symmetrical, so that the tire has the same angular speed it had before the car hit the bump? It's much more likely for the thing to be flung in some way, so that some angular momentum (either parallel or anti-parallel to its initial angular momentum) is transferred from the car to the tire.
If you drop a marble on a merry-go-round, the marble comes off spinning. You can't draw a box around the marble after the fact and say that it was spinning at the same speed before you put it on the merry-go-round.
The idea that a wheel can gain speed as it's flung off isn't strange at all. It doesn't even hint at a violation of a physical law, and it doesn't necessitate any complicated explanation. There's absolutely no need to consider the torque differentials or to assume that the car suddenly slows a great deal. Tape a compact disc to the end of a pencil and flick your wrist slightly, then watch the CD spin through the air.
posted by dsword at 6:56 AM on October 31, 2006
If you drop a marble on a merry-go-round, the marble comes off spinning. You can't draw a box around the marble after the fact and say that it was spinning at the same speed before you put it on the merry-go-round.
The idea that a wheel can gain speed as it's flung off isn't strange at all. It doesn't even hint at a violation of a physical law, and it doesn't necessitate any complicated explanation. There's absolutely no need to consider the torque differentials or to assume that the car suddenly slows a great deal. Tape a compact disc to the end of a pencil and flick your wrist slightly, then watch the CD spin through the air.
posted by dsword at 6:56 AM on October 31, 2006
Response by poster: I think sfenders is on the right track. It's more than just angular momentum at work -- there's a spring factor involved too. The reason the postdocs (who, incidentally, are incredibly intelligent people) were arguing about it is because they were debating the various ways acceleration could possibly occur. The fact that the wheel jumps has a lot to do with this, I imagine.
Anyway, there's no consensus, unfortunately. Thanks for the responses!
posted by spiderskull at 10:22 AM on November 6, 2006
Anyway, there's no consensus, unfortunately. Thanks for the responses!
posted by spiderskull at 10:22 AM on November 6, 2006
This thread is closed to new comments.
posted by ernie at 10:04 PM on October 28, 2006