Split Thread The validity of classical physics (split from: DWFTTW)

[humber]

No, I assure you I'm not bluffing. If it's that easy you should name the amount, and we'll put up the money.

I don't care.
You cannot do it by dint of your knowledge. nor refute my other claims. The cart cannot make progress up the belt. Your treadmill is false.

ETA:
Here they are:
The belt rim and belt move in opposition, so the velocity w.r.t the ground is zero. It does not accelerate, so its KE is zero.
The cart can only advance by increasing its angular velocity w.r.t. the belt, but it cannot do so, because that is determined by the belt. It cannot move.
Bluff or bet your way out of that.

 

[humber]

Oh yes, goody, more humber pictures! Too bad he put anibus in the last one instead of Charlie Brown but hey, if we haven't learned to ignore humber's goof-ups by now ...

You failed to deny Anubis his rightful claim.

Any bets on who is going to star in this one? Fred Flintstone maybe?

I think he writes your material.

3body, no problem, apology accepted, kudos for admitting it!

 

[humber]

Is this the long sought after Energy Energy?

You have no answer, That much has been found.

 

[mender]

Simple.
(1) Car parked on level ground. Contact patch directly under the axle
(2) Upend the same car, and put a support under the rear wheels. The contact patch will now move 90 degrees, but still to the rear.
Conclusion: On an incline, the contact patch will lie between those limits.
That is the case for force propelling the car forward.

No, humber, "modeling" the location of the tire patch isn't needed. Look at a picture of a car on a hill. See the tire patch cling tenaciously to the ground as the weight of the car causes the tire to flex. See the sidewalls in tension, not compression. Grok the significance of that.

To more easily see the movement of the contact patch due to force, check what happens to the tire patch as a car corners. Understand the need to have either static or dynamic negative camber to counter that effect.

Conclusion: humber loses. On the track as well as here; maybe you should try different tires next time, obviously your design isn't working.

 

[mender]

You failed to deny Anubis his rightful claim.

True, but I did deny your claim full of wrong.

Why are you dissing Fred? He uses the same propulsion method that you do - probably the same physics too.

 

[jjcote]

Anyway, if you park your car on an incline, but facing up it, where do you think the contact patch is? Yes, behind the axle.

Not my car. I always avoid parking it in the humberverse, where that sort of thing is apparently commonplace. Where I live, a car parked on an incline looks like this:

 

[humber]

No, humber, "modeling" the location of the tire patch isn't needed. Look at a picture of a car on a hill. See the tire patch cling tenaciously to the ground as the weight of the car causes the tire to flex. See the sidewalls in tension, not compression. Grok the significance of that.

The tyre opposes motion down the hill due to gravity, That opposing force comes from behind the axle, which is why you would not put chocks on the uphill side.
The sdewalls bulge because they are yielding to a compressive the load. The excursion is limited by the steel banding, which is of course under tension to some degree, but the tyre is pressurized, so the walls are pre-stressed.
It is hopeless to argue from appearances. The fact is, the load is met by a patch to the rear of the axle.

To more easily see the movement of the contact patch due to force, check what happens to the tire patch as a car corners. Understand the need to have either static or dynamic negative camber to counter that effect.

The patch moves to the opposite side of the cars direction. That is how a bicycle steers.

Conclusion: humber loses. On the track as well as here; maybe you should try different tires next time, obviously your design isn't working.

[/QUOTE]
Bluster.

 

[humber]

Not my car. I always avoid parking it in the humberverse, where that sort of thing is apparently commonplace. Where I live, a car parked on an incline looks like this:
[qimg]http://i.ehow.com/images/GlobalPhoto/Articles/2198837/16-main_Full.jpg[/qimg]

Good, it is behind the axle. Too bad you can't see 'force' isn't it. Any more random self-defeating answers. Perhaps you can photoshop one.

ETA: As I said, I need not bother with that, because it is part of the general case, but specifically the synchronization of belt and wheel means motion is impossible.

 

[humber]

True, but I did deny your claim full of wrong.

Why are you dissing Fred? He uses the same propulsion method that you do - probably the same physics too.

Perhaps it is Barney who writes your material.

 

[Dan O.]

I refuse to believe humber knows better about this one now. Not until he confirms that. I don't think we can strike an item from the list just because humber's random thoughts occassionally suggest that he might no longer be suffering a given delusion. If he confirms it affirmatively, I think it should get struck.

I realized that I have to unstrike it anyway because I don't have the post reference where humber recants. This is about scientific documentation, we must be consistent.

I will also need to reorganize the list into a hierarchy so related concepts can be found together.

 

[mender]

The patch moves to the opposite side of the cars direction.

Sorry, still wrong, stumble. You just won't learn anything will you. Incredible.

Go ahead, post some silly retort rather than address the issue.

"That Barney, what an actor!"

 

[humber]

Sorry, still wrong, stumble. You just won't learn anything will you. Incredible.

Go ahead, post some silly retort rather than address the issue.

"That Barney, what an actor!"

The car reacts against the force of the patch. The car cannot roll over the patch it creates to turn. That is a fact.
ETA: When it does overcome the friction, the car slides opposite to the direction in which it is steered

 

[mender]

Humber, the tire isn't "active". It doesn't brace itself, it gets "bent" by the forces acting on it. Haven't you ever watched any car racing?

Look at the left rear tire. See how the whole carcass of the tire is shifted?

http://img233.imageshack.us/img233/8361/f2009mockupeq7.jpg

 

[humber]

Humber, the tire isn't "active". It doesn't brace itself, it gets "bent" by the forces acting on it. Haven't you ever watched any car racing?

They are all your words, not mine. That changes nothing.
If you look at the way bicycle wheels are spoked, that should tell you something about what a tyre faces.

 

[mender]

They are all your words, not mine. That changes nothing. If you look at the way bicycle wheels are spoked, that should tell you something about what a tyre faces.

Well, that about sums it up. Humber can't figure it out, goes to plan A: change the subject.

I guess you haven't seen any racing. F1 doesn't use bicycles.

 

[humber]

Well, that about sums it up. Humber can't figure it out, goes to plan A: change the subject.

I guess you haven't seen any racing. F1 doesn't use bicycles.

The spokes tell of expected forces FI cars are set up according to the 'direction' of the circuit.
The cart is not an F1 racer, nor does it steer. The point is one of logic, and if you can support your side.

 

[mender]

Humber's Plan B: deny that it is relevant but says that he wins anyway.

You're right this time, humber: your "argument" about the contact patch was another distraction (pun intended) despite your attempt to tie it in.

Next obfuscation?

 

[Dan O.]

Simple.
(1) Car parked on level ground. Contact patch directly under the axle
(2) Upend the same car, and put a support under the rear wheels. The contact patch will now move 90 degrees, but still to the rear.
Conclusion: On an incline, the contact patch will lie between those limits.
That is the case for force propelling the car forward.

No humber, that only shows that a when an external force accelerating the car is infinite relative to the normal force from the road, the contact patch needs to be 90º to the rear. Of course, this won't work if the car's center of mass is above the axle.


But since you showed me yours, I'll show you mine:

<http://books.google.com/books?id=3GnIfH6K5GEC>

p267 ¶5.4.5 Driving force ... As the angular velocity of the wheel increases, a driving force is generated that tends to move the contact patch froward relative to the wheel centre.

 

[humber]

Humber's Plan B: deny that it is relevant but says that he wins anyway.

If a take a domestic fan, say 50W, 1000rpm. I could make a duct to couple a driven fan to a second slave fan. Perhaps the fans would reach close to the same speed, but the slave fan would not producing 50W, because there is no load. It is spinning, but not producing any power.
That is the case for the cart. There is no load, because the cart does not move, so no power is consumed. That's why it hovers. It is simply doing nothing.
Equivalence does not allow sources of power to be casually transposed.

 

[humber]

No humber, that only shows that a when an external force accelerating the car is infinite relative to the normal force from the road, the contact patch needs to be 90º to the rear. Of course, this won't work if the car's center of mass is above the axle.

A simple recanting of my statement that the contact patch lies between those limits. The rest is flannel.

But since you showed me yours, I'll show you mine:
<http://books.google.com/books?id=3GnIfH6K5GEC>

I would put it away.
Not forward of the axle, but towards it. If angular velocity increases, the angle between the axle and trailing patch is reduced. That means that after acceleration, when the force is lower, the patch moves towards the position under the axle, so as to reduce rolling resistance .That is one of the advantages of tyres. They are levers, and work to keep friction to a minimum according to load.

The cart wheels are essentially rigid, so they are a disk and the force is to the rear of the axle. Tyres are complicated.