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

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It must come from behind the axle. The axle cannot lead itself.

Who believes in fairy tales?

Your claim that the contact patch must be behind the tire in order for the tire to move forward is the fairy tale.

Then again, your explanation of how the cart works is a fairy tale as well.
 
No. The photo (which I see you do understand) shows cornering force.
That is due to applying a force to keep the car from going straight ahead. Like the force in a string of an object rotated above the head.

No what?

1) No, the car isn't cornering to the left.
2) No, the contact patch isn't left of the centreline of the rim.
3) No, cornering to the left is not accelerating to the left.
4) No, the forces on the tire when accelerating don't displace the contact patch in the direction of the acceleration.
5) No, the forces on the tire when accelerating a car around a corner aren't comparable in reaction to the forces on a tire that is accelerating a car forward.
6) No, the contact patch on a tire when accelerating a car forward is not displaced in the direction of the acceleration.

Where do you disagree?
 
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Can anyone here do a Matlab Simulink job for the Cart and Treadmill? Also A Cart on Road with wind for comparison would be useful. I have to admit, I can't work that software well enough, the guy next to me here can though, if I could get him interested. I know that no matter what resulted, Humber would never believe it, and in any case, nearly all the input blocks would be in dispute.
Why not at least get a block level system diagram going? Can you help here Humber?
BTW, congrats to all for reaching the ton.
 
The torques are in opposition. This is the basis of
(a) The differential in your car
(b) The way reverse gear (th' towin' one to you) works.

The shafts have opposing torques. You had better hope that is so, or the cart would not work in wind.

Wrong again. Goodman's cart uses a belt with a 90 degree twist. No gears. But I guess "opposing torque" has something to do with your delusion about how the cart works.
 
A simple example of load matching.

The Balloon:
(a)
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(1) The balloon is a meteorological balloon, behaving in accordance with (a) and simplified below.
(2) The force from the balloon is simplified to F = k(V - Vb)^2, where k=1
(3) Power = force * Velocity, so that makes P = Vb(v- Vb)^3

The Skater:
(b)
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(1) The total drag. Aerodynamic and other frictional losses are lumped as (b)
(2) The drag force is simplified to Fd= k(Vb)^2
(3) The power demanded by the skater is therefore Pd = k(Vb)^3


The plots are those of the forces and power involved. It can be seen that maximum velocity is where power and force balance, and less than windspeed.


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The x axis is the velocity of the balloon (Vb)
The wind is therefore equated to one. Vwind =1.
Windspeed is therefore 1.0 on the x axis.
 

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No opposite is not the same as not-backwardness, Captain.

If the two shafts are at 180 degrees, you get the simple straight-kut gears that everyone agrees counter-rotate. (the green gears of the second set of photos)


Go fetch a stick...

The first picture is good enough to meet your peculiar demands, but as the text says, "independently of the angle the gears axes must intersect (at point 0). This is true for all cases.
Including you ETA.

Once again raising irrelevant crap in a futile effort to hide the fact that he doesn't know what "counter rotate" means.
 
Wrong again. Goodman's cart uses a belt with a 90 degree twist. No gears. But I guess "opposing torque" has something to do with your delusion about how the cart works.

Twisting the belt does not matter, nor does Goodman's cart, ot the operation of a cart in wind. The treadmill cart is not like at all like a cart in wind.

The cart has opposing torques of propeller and wheel. This means that without a load, no real work can be done.
The cart does no travel, so does not meet the real-world velocity dependent drags of the air and road surface.
It does no work, and so falls to a state of minimum power consumption.
 
I'm bursting with pride. With all these votes for me over humber I feel like I've just come in 2nd to last in the special olympics. :D

Yes, they are your "target audience" as you put it. You are failing in front of a larger one, though.
 
Can anyone here do a Matlab Simulink job for the Cart and Treadmill? Also A Cart on Road with wind for comparison would be useful. I have to admit, I can't work that software well enough, the guy next to me here can though, if I could get him interested. I know that no matter what resulted, Humber would never believe it, and in any case, nearly all the input blocks would be in dispute.

Why not at least get a block level system diagram going? Can you help here Humber?
BTW, congrats to all for reaching the ton.

Not really, Sumper. If you get "the guy next to you" to write the code or Simulink blocks, we can run it and take it from there.
 
That zero-g remark is a quote from you. I think that it is easy to see how you would make such blunders when thinking about gears.
Interestingly, as usual, Spork has provided evidence to his disadvantage. (only Dan_O tops him in this regard)
The skate wheels are in parallel, so quite obviously go the same way. That means that when geared, as in the cart, they go the opposite way, unless you thing that all gear combinations are the same.
It does make you wonder why so many posters are unclear regarding the relative direction of the prop and wheels. They see what you don't.

There is only one poster who is unclear (or more correctly, flat wrong) about the relative directon of prop and wheels. Everybody but that one poster knows that the axes are at a 90 degree angle to each other. The poster who is wrong has a delusion that because they are connnected by gears, that they are counter-rotating, and the even more bizarre delustion that that "fact" is somehow relevant to the functioning of the cart.
 
No, to your assertion that cornering or axle drive, is as you describe.
1) No, the car isn't cornering to the left.
The car is being impeded from moving to the left as from the view of the camera

2) No, the contact patch isn't left of the centreline of the rim.
There is no need for it to be related to the centre of the rim

3) No, cornering to the left is not accelerating to the left.
Correct. There is a force causing it to accelerate in a 'circular' motion. Like the force of the string of an object rotated above the head.

4) No, the forces on the tire when accelerating don't displace the contact patch in the direction of the acceleration.
The road force is a reaction. It comes from the rear, because the car is in motion in the opposite direction.

5) No, the forces on the tire when accelerating a car around a corner aren't comparable in reaction to the forces on a tire that is accelerating a car forward.
No, but they have the same relationship. You introduced cornering.

6) No, the contact patch on a tire when accelerating a car forward is not in the direction of the acceleration.
It must be behind the axle, for the axle to move forward. You insist on tyres, where there are none.

Where do you disagree?
The cart's wheels are disks, not pneumatic tyres under cornering.
No matter how you try, and how many strawmen you post, the belt cannot impose forward drive upon the wheel.
 
My vote, by the way, is for spork, in case anyone is wondering.

Mender, I feel I should apologise for suggesting that we take humber's ideas of how wheels work and discuss them with a vehicle dynamics expert by email. I was forgetting that we have you here on the thread already. These are very interesting posts:

The tires do rotate without slip at low driven rates. It's only when the mechanical intermeshing of the rubber tread with the road surface is inadequate to prevent slippage that slip actually occurs. In my line of work that is essentially all the time but certainly not that case of the average car driving down the highway.

By the way, the more common term is percentage of slip when discussing longitudinal tire to surface speed differences and slip angle for lateral differences. Note: for higher level discussions, shear angle should be used instead of slip angle to more clearly distinguish between steering angle (direction of the rim), slip angle (direction of the tread), and shear angle (direction of the tread vs the actual direction of travel).

If you had bothered to answer my tire quiz, you might have found out that tension in the sidewalls of the tire transmits the torque to the contact patch on the tread face, and that is what produces motion. The amount of contact patch displacement (in the opposite direction of what you keep insisting on) will depend on the magnitude of the torque and the construction of the tire. None of this has to involve breaking traction or slip.

Direct observation of a stressed tire through a heavy glass plate and across a pressure sensitive surface shows the centre of force on the displaced contact patch to be located off to the side of the centreline of the tire in the direction of the induced force (centripetal in cornering), showing that the tire is reacting to the force. The tire is elastic and bends, resulting in the attached car trying to continue along its path while the tire tries to pull the car in the desired (by the driver) direction. The contact patch is in "front" of the tire, not behind.

That's why your description is exactly like trying to push a rope; the contact patch of the tire is pulling, not pushing, on the rim of the tire which is attached to the rest of the car.

And then this:

The contact patch is to the "left" of the centreline of the rim, because the car is turning to the "left" (right rear tire). If you consider the change in direction of the car as acceleration and compare that to a forward acceleration of a car, you'll see that the contact patch is in the direction of the acceleration - in "front" of the tire. The contact patch is again "pulling" on the rim. Just like all the pictures have shown so far. See the tension in the sidewall trying to unseat the bead? If the tire comes off the rim, the tire ends up piled up on the inside of the rim, not the outside.

Does anyone else claim that the contact patch is "pushing" or piling up behind the tire?
Not me. I'm very happy to stick with my original assessment. Thanks.
 
Wow. I didn't include that easily accessed picture because it doesn't show the centre of force in the contact patch, only that the contact patch is shifted under the tire as it is deflected by the cornering force. I'm shocked that you would misinterpret the picture so badly. Well, maybe not so shocked.

The contact patch is to the "left" of the centreline of the rim, because the car is turning to the "left" (right rear tire). If you consider the change in direction of the car as acceleration and compare that to a forward acceleration of a car, you'll see that the contact patch is in the direction of the acceleration - in "front" of the tire. The contact patch is again "pulling" on the rim. Just like all the pictures have shown so far. See the tension in the sidewall trying to unseat the bead? If the tire comes off the rim, the tire ends up piled up on the inside of the rim, not the outside.

Does anyone else claim that the contact patch is "pushing" or piling up behind the tire?

Of course not. given the direction of the forces, it couldn't be otherwise. Not that this has any relevance whatsoever to the working of the cart, or the validity of the treadmill as a model.
 
There is only one poster who is unclear (or more correctly, flat wrong) about the relative directon of prop and wheels. Everybody but that one poster knows that the axes are at a 90 degree angle to each other. The poster who is wrong has a delusion that because they are connnected by gears, that they are counter-rotating, and the even more bizarre delustion that that "fact" is somehow relevant to the functioning of the cart.
attachment.php


I situation (4) A drives B drives A. No work. Just like a cart on a treadmill.
 
Not really, Sumper. If you get "the guy next to you" to write the code or Simulink blocks, we can run it and take it from there.

Oh thankyou Mr Humber, thankyou so much! I'm so glad you will be able to run it and take it from there! Only after the hard work and thinking parts are done!

And, you do realize that a propeller works differently from a parachute don't you? And that wheels being driven from a surface and transferring energy into torque on a rotating shaft is different from simple skull-dragging friction. You see these are some of the subtleties which may come up in the modeling.
 
There is only one poster who is unclear (or more correctly, flat wrong) about the relative directon of prop and wheels. Everybody but that one poster knows that the axes are at a 90 degree angle to each other. The poster who is wrong has a delusion that because they are connnected by gears, that they are counter-rotating, and the even more bizarre delustion that that "fact" is somehow relevant to the functioning of the cart.
I wonder if ynot's wheel and prop are counter-rotating in humberville.

100 Today. Do I get a telegram from the Queen?
 
A simple example of load matching...

I think this is just darling! humber has found some graphs and equations to try and impress us with.

But I do have to admit - I'm starting to doubt my vote that humber is truly as confused as he presents. It's just hard to imagine anyone can be this consistently wrong on every single nuance of every single sub-problem that arises. Seriously - is that possible?
 
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