Mythbusters punk'd whole internet

[smack Down]

Originally posted by: mobobuff

Originally posted by: smack Down

Originally posted by: mobobuff

Originally posted by: randay

The bottom line is that the treadmill speed = wheel speed interpretation is wrong, and the only correct one is treadmill speed = wind speed. or if you still want to entertain the crazy infinite theory, the only way its true is when wheel speed is zero.

Exactly. The original question was very blunt and didn't specify, but it makes much more sense to assume the treadmill was meant to match ground speed.

This question wasn't thought up by a physicist.

No it doesn't the whole idea behind a treadmill is that it matches the speed of the object on it so that the object on it stays in one position.

I'm just going to assume you're trolling now.

I don't go to the gym very often but when I do I have yet to see a treadmill longer then 5 feet? Maybe because it is used to keep moving object stationary that they don't need to buy 5 mile long treadmills? But hey what do I know.

 

[Number1]

No point in keeping this thread open if the Mythbusters are not doing it.


Lock her up. This is only going to turn into a flame fest.

 

[Jeff7]

Originally posted by: smack Down
What did you smoke to come to that conclusion? Did you mean to say that if the speed of the ground matches the speed of the car with respect to ground but in the opisite direction their is zero movement? If so I agree now again explain how the plane is special.

You're missing a big big part of the equation.
The car's forward force, and therefore velocity, is mathematically dependent on the speed of the tire surface relative to the ground.

The plane's forward force, and therefore velocity, is mathematically dependent on the speed of the air rushing through the engines. Since the ground, or the tires, are not even in the equation, the force is unaffected by the speed of the ground or the tires relative to each other.
Equations to determine thrust of a rocket or jet engine will factor in air velocity, not ground velocity.

Math doesn't lie.

 

[jagec]

Originally posted by: Jeff7
The ground moves beneath them, imparting a force which is converted into rotational momentum, not linear momentum. Since it is rotational only, it does not serve to push the plane forward, nor backward, thus it does not counteract the forward force of the jet engines, which derive their impulse from the air flowing through them. That part is completely independent of the ground's motion.
AND, since the wheels' only momentum is rotational, not linear, they generate no force to oppose that created by the engines.

Take a wheel out in space. Attach a string to the outside of the wheel. Pull the string. You will notice that the wheel spins faster (you added angular momentum), but the wheel also starts to move towards you (you added linear momentum). The relative amount of each depends on the weight distribution and total weight of the wheel, and how far from the center you tied the string. Even with the same force exerted in each case, you can change the relative amount of linear vs. angular momentum by adjusting these parameters. You can even make the angular momentum zero by attaching the string in the middle. You cannot make the linear term zero.

Why is a toy car with a flywheel harder to push across the ground than a Matchbox car? If rotational and linear motion are decoupled, such a thing should be impossible.

 

[waggy]

Originally posted by: Number1
No point in keeping this thread open if the Mythbusters are not doing it.


Lock her up. This is only going to turn into a flame fest.

yeap.

damn OP!

 

[Jeff7]

Originally posted by: jagec

Originally posted by: Jeff7
The ground moves beneath them, imparting a force which is converted into rotational momentum, not linear momentum. Since it is rotational only, it does not serve to push the plane forward, nor backward, thus it does not counteract the forward force of the jet engines, which derive their impulse from the air flowing through them. That part is completely independent of the ground's motion.
AND, since the wheels' only momentum is rotational, not linear, they generate no force to oppose that created by the engines.

Take a wheel out in space. Attach a string to the outside of the wheel. Pull the string. You will notice that the wheel spins faster (you added angular momentum), but the wheel also starts to move towards you (you added linear momentum). The relative amount of each depends on the weight distribution and total weight of the wheel, and how far from the center you tied the string. Even with the same force exerted in each case, you can change the relative amount of linear vs. angular momentum by adjusting these parameters. You can even make the angular momentum zero by attaching the string in the middle. You cannot make the linear term zero.

Why is a toy car with a flywheel harder to push across the ground than a Matchbox car? If rotational and linear motion are decoupled, such a thing should be impossible.

A valid point. I guess the question is then, how much is converted to rotation, and how much is converted to linear motion? For that, I think I should actually know, because I have had a dynamics course, and I'm fairly sure I should know how to do this.
I still don't think that the linear momentum imparted would be sufficient to significantly impair the plane's forward velocity. The mass of the wheels would be a factor. The wheel's mass is definitely very small in comparison to that of the plane.

 

[jagec]

Originally posted by: Jeff7
A valid point. I guess the question is then, how much is converted to rotation, and how much is converted to linear motion? For that, I think I should actually know, because I have had a dynamics course, and I'm fairly sure I should know how to do this.
I still don't think that the linear momentum imparted would be sufficient to significantly impair the plane's forward velocity. The mass of the wheels would be a factor. The wheel's mass is definitely very small in comparison to that of the plane.

Ah, but even if it's a small amount, you can always accelerate the treadmill faster to make that term comparable to the force of the engines.

I agree that it's a tiny factor in any real-world situation, but if you have a ridiculously oversized treadmill power device, you can spin those wheels so fast that you can apply as much linear force as you like.

I did some rough calculations in the last thread. By the time the plane runs out of fuel, there's twice as much energy locked up in the angular momentum of the wheels as was in the fuel to begin with (after engine efficiency is taken into account).

The wheels were spinning at a ridiculous, but non-relativistic, speed. Certainly fast enough to grenade any wheels that currently exist on planes.

 

[spidey07]

Originally posted by: Jeff7

Originally posted by: jagec

Originally posted by: Jeff7
The ground moves beneath them, imparting a force which is converted into rotational momentum, not linear momentum. Since it is rotational only, it does not serve to push the plane forward, nor backward, thus it does not counteract the forward force of the jet engines, which derive their impulse from the air flowing through them. That part is completely independent of the ground's motion.
AND, since the wheels' only momentum is rotational, not linear, they generate no force to oppose that created by the engines.

Take a wheel out in space. Attach a string to the outside of the wheel. Pull the string. You will notice that the wheel spins faster (you added angular momentum), but the wheel also starts to move towards you (you added linear momentum). The relative amount of each depends on the weight distribution and total weight of the wheel, and how far from the center you tied the string. Even with the same force exerted in each case, you can change the relative amount of linear vs. angular momentum by adjusting these parameters. You can even make the angular momentum zero by attaching the string in the middle. You cannot make the linear term zero.

Why is a toy car with a flywheel harder to push across the ground than a Matchbox car? If rotational and linear motion are decoupled, such a thing should be impossible.

A valid point. I guess the question is then, how much is converted to rotation, and how much is converted to linear motion? For that, I think I should actually know, because I have had a dynamics course, and I'm fairly sure I should know how to do this.
I still don't think that the linear momentum imparted would be sufficient to significantly impair the plane's forward velocity. The mass of the wheels would be a factor. The wheel's mass is definitely very small in comparison to that of the plane.

You're forgetting the primary restraint of the question.

Those speaking of the plane and treadmill not taking off seem to have a VERY firm grasp of statics and dynamics. It's not that hard. You're making assumptions. Work within the initial question.

By very definition of the problem I have a treadmill with near infinite acceleration, if not infinite. Actually the problem suggests (and actually contains you to) that you can have infinite acceleration.

Do the math and work within the bounds of the original problem. No cheating allowed. If I could divide by zero I could meet the bounds of the original problem.

 

[Jeff7]

Originally posted by: jagec
Ah, but even if it's a small amount, you can always accelerate the treadmill faster to make that term comparable to the force of the engines.

I agree that it's a tiny factor in any real-world situation, but if you have a ridiculously oversized treadmill power device, you can spin those wheels so fast that you can apply as much linear force as you like.

I did some rough calculations in the last thread. By the time the plane runs out of fuel, there's twice as much energy locked up in the angular momentum of the wheels as was in the fuel to begin with (after engine efficiency is taken into account).

The wheels were spinning at a ridiculous, but non-relativistic, speed. Certainly fast enough to grenade any wheels that currently exist on planes.

So then, back to the original test conditions. On what is the treadmill's speed made dependent? Rotational momentum? Linear momentum? Air speed through the engines?
By your example of the "grenading" wheels, then it becomes a materials question, not physics.

 

[mobobuff]

Originally posted by: jagec

Originally posted by: Jeff7
A valid point. I guess the question is then, how much is converted to rotation, and how much is converted to linear motion? For that, I think I should actually know, because I have had a dynamics course, and I'm fairly sure I should know how to do this.
I still don't think that the linear momentum imparted would be sufficient to significantly impair the plane's forward velocity. The mass of the wheels would be a factor. The wheel's mass is definitely very small in comparison to that of the plane.

Ah, but even if it's a small amount, you can always accelerate the treadmill faster to make that term comparable to the force of the engines.

I agree that it's a tiny factor in any real-world situation, but if you have a ridiculously oversized treadmill power device, you can spin those wheels so fast that you can apply as much linear force as you like.

I did some rough calculations in the last thread. By the time the plane runs out of fuel, there's twice as much energy locked up in the angular momentum of the wheels as was in the fuel to begin with (after engine efficiency is taken into account).

The wheels were spinning at a ridiculous, but non-relativistic, speed. Certainly fast enough to grenade any wheels that currently exist on planes.

In the original question, the treadmill moves in the opposite direction in equal speed to the forward movement of the plane. Neither moves faster than the other. In the original interpretation of the question that started this all, the plane takes off.

 

[FoBoT]

mods, please BAN adam jamie kari and grant and tory

:|

 

[Jeff7]

I wonder if HL2's physics engine is accurate enough to simulate this, courtesy of Garry's Mod?

Originally posted by: FoBoT
mods, please BAN adam jamie kari and grant and tory

:|

Sorry, Kari can do no wrong. And Jamie's just cool. Adam - he does his "voices" too much now, and it's getting irritating. He gets a vacation until he stops. Grant and Tory, I'm indifferent about.

 

[randay]

Originally posted by: jagec

Originally posted by: randay
You have not addressed the issue that the treadmill can NEVER match the speed of the wheels relative to itself. Which makes the whole infinite-super-treadmill theory bunk.

Also the problem with what you believe is that you assume that the treadmill will stop once net motion is zero, thats not true since the wheels are still spinning and therefore the treadmill is still accelerating and angular momentum is still building. The treadmill will overcome the thrust of the engines(through angular momentum achieved by infinite acceleration and velocity), and then the airplane will accelerate backwards until it takes off anyway. granted, it will not take off very long since there will be no way to control it backwards, however it will gain an infinite amount of speed and eventually and messily, catch air.

Yes I have. Take your initial setup: Plane speed with respect to stationary object = P, treadmill speed with respect to stationary object = T, wheel speed = (P + T). If we try to match the treadmill speed to the wheel speed (T = (P + T)), this looks like a paradox until we realize that as long as the treadmill is accelerating, it can interact with the plane and reduce its speed. So the equation becomes (T = (P-AX) + T), where AX is the acceleration of the treadmill (A) times a term (X) which encapsulates the dimensions of the wheels and converts AX into a velocity. The treadmill has to accelerate at a rate where AX = P. If it accelerates faster, the plane moves backwards on the runway. If it accelerates more slowly, the plane moves forwards on the runway. Once the plane runs out of fuel, the treadmill must stop accelerating and continue at a constant speed.

The bolded portion is untrue. I said that the treadmill will stop ACCELERATING when the plane returns to the control position (its initial position on the runway). It will continue moving at a constant speed, with the wheels turning at a constant speed and angular momentum, with no acceleration and no buildup of further forces.

Explain please, why the treadmill will stop accelerating. The setup is thus: The treadmill will match the speed of the airplane in reverse. speed must be either air speed(same as "speed with respect to a stationary object"), or ground speed(treadmill is the ground). It cant be both, and so the underlined formula seems to be where you went wrong.

 

[videogames101]

WTF!!!??? they didn't show it!

 

[mugs]

Originally posted by: FoBoT
mods, please BAN adam jamie kari and grant and tory

:|

assfuckers :|

 

[mobobuff]

Originally posted by: FoBoT
mods, please BAN adam jamie kari and grant and tory

:|

Folks are pissed.

 

[videogames101]

Originally posted by: mobobuff

Originally posted by: FoBoT
mods, please BAN adam jamie kari and grant and tory

:|

Folks are pissed.

good!

 

[skulkingghost]

DAMNIT MYTHBUSTERS I sat through your "show" that really should be called "commercial hour" and I get nothing. I'm pissed.

 

[NatePo717]

:|

 

[mobobuff]

Originally posted by: videogames101

Originally posted by: mobobuff

Originally posted by: FoBoT
mods, please BAN adam jamie kari and grant and tory

:|

Folks are pissed.

good!

"After scamming viewers to get ratings, Jamie and Adam need to test the myth that they can survive skydiving without any parachutes."

 

[BillGates]

We got screwed. Luckily, the plane still takes off.

 

[videogames101]

close this thread plz

 

[smack Down]

Originally posted by: randay

Originally posted by: jagec

Originally posted by: randay
You have not addressed the issue that the treadmill can NEVER match the speed of the wheels relative to itself. Which makes the whole infinite-super-treadmill theory bunk.

Also the problem with what you believe is that you assume that the treadmill will stop once net motion is zero, thats not true since the wheels are still spinning and therefore the treadmill is still accelerating and angular momentum is still building. The treadmill will overcome the thrust of the engines(through angular momentum achieved by infinite acceleration and velocity), and then the airplane will accelerate backwards until it takes off anyway. granted, it will not take off very long since there will be no way to control it backwards, however it will gain an infinite amount of speed and eventually and messily, catch air.

Yes I have. Take your initial setup: Plane speed with respect to stationary object = P, treadmill speed with respect to stationary object = T, wheel speed = (P + T). If we try to match the treadmill speed to the wheel speed (T = (P + T)), this looks like a paradox until we realize that as long as the treadmill is accelerating, it can interact with the plane and reduce its speed. So the equation becomes (T = (P-AX) + T), where AX is the acceleration of the treadmill (A) times a term (X) which encapsulates the dimensions of the wheels and converts AX into a velocity. The treadmill has to accelerate at a rate where AX = P. If it accelerates faster, the plane moves backwards on the runway. If it accelerates more slowly, the plane moves forwards on the runway. Once the plane runs out of fuel, the treadmill must stop accelerating and continue at a constant speed.

The bolded portion is untrue. I said that the treadmill will stop ACCELERATING when the plane returns to the control position (its initial position on the runway). It will continue moving at a constant speed, with the wheels turning at a constant speed and angular momentum, with no acceleration and no buildup of further forces.

Explain please, why the treadmill will stop accelerating. The setup is thus: The treadmill will match the speed of the airplane in reverse. speed must be either air speed(same as "speed with respect to a stationary object"), or ground speed(treadmill is the ground). It cant be both, and so the underlined formula seems to be where you went wrong.

If the plane is going backwards then it would mean that the treadmill is no longer matching the wheel speed of the plane. The treadmill would have to be going faster then the wheels. Stop think of the plane as something special just think of it like a car.

 

[Exterous]

I feel so....so....used. I got so excited but left hanging. And now nothing Bastards!

 

[jagec]

Originally posted by: Jeff7
So then, back to the original test conditions. On what is the treadmill's speed made dependent? Rotational momentum? Linear momentum? Air speed through the engines?
By your example of the "grenading" wheels, then it becomes a materials question, not physics.

I think the easiest way to set up a control system to effect the second interpretation of the original problem is to use a position sensor, and keep the plane stationary next to it via manipulation of the treadmill. This is mathematically the same as matching the "wheel speed", but less conceptually difficult since it resolves the apparent paradox of A = A + 1 which often comes up in these threads.

Originally posted by: mobobuff
In the original question, the treadmill moves in the opposite direction in equal speed to the forward movement of the plane. Neither moves faster than the other. In the original interpretation of the question that started this all, the plane takes off.

I totally agree. The first interpretation (which I believe is correct) will result in the plane taking off every time. The second interpretation, which is harder to conceptualize, has some interesting possibilities depending on the assumptions that go into it.

Originally posted by: randay
Explain please, why the treadmill will stop accelerating. The setup is thus: The treadmill will match the speed of the airplane in reverse. speed must be either air speed(same as "speed with respect to a stationary object"), or ground speed(treadmill is the ground). It cant be both, and so the underlined formula seems to be where you went wrong.

If it's matching airspeed, that's what I call the "first interpretation", which results in the plane taking off. If it's matching "wheel speed", that's the second interpretation. I believe I have resolved the apparent paradox that this seems to present at first glance.

Airspeed is "P", AKA "plane speed" or "speed with respect to a stationary object".
The treadmill will reduce the speed of the plane via the angular momentum term, albeit at a very small level. In order to do so, it must be accelerating...when the treadmill moves at a constant speed, it has no effect on the plane save through friction, which we are ignoring. When we reduce the speed of the plane to zero by applying an equal and opposite force via the angular momentum term (constantly accelerating the treadmill), the paradox disappears since treadmill speed T = P (plane speed) + W ("wheel" speed), P= 0 so T=W.

I really do prefer to think of it as "hold the aircraft in one spot by twiddling with the treadmill" rather than speed matching...simpler to conceptualize. Either way is the same...no control system is able to keep the error signal at zero at all times, so the treadmill speed will oscillate around the "wheel speed" as the controller changes the rate of acceleration of the treadmill to keep the plane in one spot.

Also, please keep my arguments separate from smack Down's, we're clearly talking about different things.