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

[humber]

I assume "s" is the displacement vector between the cars? If so, s goes through zero (or rather the component of it parallel to the motion does), but ds/dt does not - that's the relative velocity, and it's constant.

Yes, yes, no. The velocities are constant w.r.t. ground. The KE's also constant w.r.t ground. The difference is not when the instantaneous velocities are used. The infinitesimal may then be zero, so zero KE for how long?

Since KE depends on ds/dt (but not on s), KE is constant too.

Your are right. To gainor lose KE w.r.t the ground, ds/dt must be non-zero, meaning the object must accelerate or decelerate to exchange it. This is why the cart has no KE on the treadmill.

This will be hard for you because it's the difference between position and velocity and acceleration, which you've repeatedly demonstrated you have no clue about at all.
If you would like I will write the vectors down explicitly for you, but I very much doubt it would help.

I think acceleration, velocity and displacement inextricably linked together.

When was the last time you saw a stationary parked car pass another stationary parked car?

Normally they don't, because they have zero KE, but I did see the very thing you describe when a car transporter passed by. Stationary? What do you mean?

 

[jjcote]

Actually, running in a treadmill in a gym, the way most people do it, is different (and easier) than running on a road, because of two things: no need to change your velocity, and no air resistance. To make it the same, you'd need to do two things. The first is to use a long treadmill, stand still on it, and let it crank up to your desired running speed before you start moving your feet, so that you can accelerate (to zero relative to the room). The second is that you need a fan that is trying to blow off the back of the treadmill.

That said, these differences are of limited significance. The first correction would really only matter if you were trying to do sprint training, since in a normal running workout, there's little acceleration after the first few seconds. The second is of increasing significance the faster you go, and at jogging speed it arguably doesn't matter that much. Treadmills are typically on an incline (sometimes significantly so), and this properly corresponds with running up a hill.

What running on a treadmill does match well with is running with a tailwind equal to your ground speed, where you step off a moving sidewalk to get started. (This doesn't cover the practice of standing on the treadmill belt and letting it get up to speed as you start to jog, which doesn't really have an analogue on the road, but which also ceases to matter once the treadmill is up to speed.)

 

[Vorpal]

Yes, yes, no. The velocities are constant w.r.t. ground. The KE's also constant w.r.t ground. The difference is not when the instantaneous velocities are used. The infinitesimal may then be zero, so zero KE for how long?

What.

Instantaneous velocity at time t is simply the limit of the average velocity on a time interval around t as the duration of that interval goes to zero: lim{ [s(t+h)-s(t)]/h }. That's the calculus 101 definition that makes no reference to infinitesimals and never involves the case when h is zero. If you want to explicitly use infinitesimals, then the instantaneous velocity is the coefficient of ε in [s(t+ε)-s(t)], where s(t) is the position at time t--you don't "set" the infinitesimal to zero, because then it's a formal symbol is the appropriate system. There are other contraptions to model infinitesimals, and in exactly none of them does one set them to zero in order to find the instantaneous velocity (or any other kind of derivative).

Suppose a car's position is given by s(t) = 20t along a straight line from some particular reference point. Do you agree that the instantaneous velocity is always 20 along that line? [In whatever appropriate units you like best: say, 20 is in m/s and t in seconds.] If not, what exactly are you doing to calculate the instantaneous velocity?

 

[humber]

So all this time I've been avoiding other cars when I passed them on the road when I didn't have to! Zero KE as I pass - so a collision wouldn't transfer any KE between the vehicles and thus no damage! Wow!

Mender, let me send you an application for up to my new Xero KE insurance solution.

ETA: nuts, almost forgot, I don't drive a humber.

But you are driving humber, nuts.

But make sure it is a really long piece of paper and have someone pull it at a steady speed. Hold the car over the paper so the wheels just touch, let the wheels get up to speed, then set the car on the paper. See how long the car "balances" for.

A sheet of A4 is enough to get the gist. A small toy will fit in the pocket for supermarket checkout-belt tests.

In real life, the propeller cart not only would stay in place when the paper is pulled at a high enough speed, it would start moving the other way. The propeller not only "hooks" into the air, its wheels "hook" to the ground and the gearing system allows the cart to move forward between those two "hooked" substances.

If you like, yes, but the power must come from the propeller not the ground.

If you think of the belt velocity-plane/displacement argument, the two quantities are related. The ratio is 1:1. This is due to the forced synchronization of wheel and belt, and so the propeller. Seen this way, it becomes clear that the propeller must turn faster than can be obtained from the wheels, if it is to make progress up the belt.
That can't happen, so it does not move any more than the small 'drift' that you see it do.

Another humber gem: the drag of the chute against the air only increases at the same rate as the difference in the speed between the air and the chute, but the drag of the cart is the square of the difference in speed.

You are mixing two different matters, Mender. As I said earlier, (re: the met. balloon) the force to the chute from the wind is related to (V_Wind - V_Chute), but the drag that opposes that motion, is related to the chute's velocity through the wind, and that is at least proportional to V^2. When these two forces balance, the chute itself reaches its maximum velocity. An a approximation.

A cart pulled by the chute, will have both frictional and aerodynamic drag to contend with, and I approximated that to V^2, but that will likely limit the velocity before the chute itself does. If not, then you have a powerful chute that can take you to some higher terminal velocity, but still not windspeed.

Does anyone see something wrong with this? Or are we all supposed to consider this inverted view to be another normal occurance in the humberverse?

If it's wrong, it's wrong.

 

[H'ethetheth]

Actually, running in a treadmill in a gym, the way most people do it, is different (and easier) than running on a road, because of two things: no need to change your velocity, and no air resistance. To make it the same, you'd need to do two things. The first is to use a long treadmill, stand still on it, and let it crank up to your desired running speed before you start moving your feet, so that you can accelerate (to zero relative to the room). The second is that you need a fan that is trying to blow off the back of the treadmill.

That said, these differences are of limited significance. The first correction would really only matter if you were trying to do sprint training, since in a normal running workout, there's little acceleration after the first few seconds. The second is of increasing significance the faster you go, and at jogging speed it arguably doesn't matter that much. Treadmills are typically on an incline (sometimes significantly so), and this properly corresponds with running up a hill.

What running on a treadmill does match well with is running with a tailwind equal to your ground speed, where you step off a moving sidewalk to get started. (This doesn't cover the practice of standing on the treadmill belt and letting it get up to speed as you start to jog, which doesn't really have an analogue on the road, but which also ceases to matter once the treadmill is up to speed.)

What I meant is that the difference between running outside and running on a treadmill is certainly not nearly as large as humber appears to think. Wind speeds are unsteady, variable in direction and usually larger than run-of-the-mill jogging speed. So your running speed is almost like noise on your airspeed, or a bias at most. I thought I'd keep it simple to avoid injecting confusing tangents and details into the discussion. More importantly, for the fish in my fish wheel there is no difference at all.

 

[mender]

Yes, yes, no. The velocities are constant w.r.t. ground. The KE's also constant w.r.t ground. The difference is not when the instantaneous velocities are used. The infinitesimal may then be zero, so zero KE for how long?

Now we have a new concept from humber, his interpretation of "instantaneous velocity".

But what about the object's history, humbre? You can't just throw that out with the bathwater! You have to know what came before to figure out anything! The instantaneous velocity for anything is zero (according to you), so nothing can have any KE wrt to anything else. But I guess that's okay:KE doesn't exist, therefore any arguement using KE as "proof" for or against is invalid. As has been said about humbre's KE arguements all along.

Instantaneous velocity of zero no matter what. Zero KE as a result. Another aspect of humberphysics.

Who was keeping track of all these? Chris?

Normally they don't, because they have zero KE, but I did see the very thing you describe when a car transporter passed by. Stationary? What do you mean?

It's been 'splained to you.

 

[humber]

Uote

Hey humber, did you know that if you have a hang glider that can fly at 25 mph, and there's a 25mph wind blowing, you can fly straight into the wind, and it works just fine?

How many times jjcote? Hang gliders are aircraft. I have flown them, but I am not a big fan. I see more expert people head in directions where they can expect a thermal or rising currents from a bluff. I am sure it works for them.

Groundspeed will be zero, and I guess you have no HKE,..

With potential energy that will be KE should you hit the ground...

...but there's is no difference at all in terms of how the glider handles, compared to flying in still air.

No difference? Similar in some respects. What do you expect of a simple machine?

Isn't it strange how, even though there is no motion, somehow the dynamics of motion at speed are reproduced perfectly? I've done it, and it seems like a really really accurate model of flying. Maybe that's because I was actually flying. Kind of like a treadmill for aircraft, don't you think? How does that work?

Nothing like a treadmill, but a lot like flying. There are real air currents and real air motion over the wings, even if only by motion through the air, by dint of the force of gravity.
Take the hang glider in to the same room as the treadmill and then compare the results, or try and launch from the belt.

 

[John Freestone]

John,

Did you see this exchange??

If ONLY he'd been around to set Newton straight......

Yeah, he was probably googling, or "wikkiing", saw "the equations are application dependent" and thought we'd be impressed.

Yes, yes, no. The velocities are constant w.r.t. ground. The KE's also constant w.r.t ground. The difference is not when the instantaneous velocities are used. The infinitesimal may then be zero, so zero KE for how long?

So the difference in two objects' velocities is non-zero, but not when we take the instantaneous velocity (wrong, BTW, for anyone new to humberphysics). So that would be all the time, then, during all instantaneous instants?...or just, as you said earlier, when the displacement is zero? You keep asking, "so for how long?", as though someone had suggested it ever was. Never is the answer we keep giving over and over again, and then you rephrase the same question incorrectly.

Look, speed is d/t. 10 m/s = 600 m/min = 1cm/.001s

An instant is just a theoretical time. Is it 0s? What distance could be covered by something going 10 m/s in 0s? If it's positive, it's velocity has suddenly jumped to infinite. For how long does it have infinite velocity? You seem to be suggesting that the closing velocity becomes 0/0 m/s if we take the instantaneous velocity, but in that kind of mad universe there's just no measurement at all. They could be at the speed of light, and in the instant, have no velocity? You're just messing with math like a kid playing with his food.

Also, like I said, you said it was only as things passed each other before. So then they could be going massive speeds and yet have no velocity suddenly as they pass, which suggests an infinite acceleration suddenly happens.

Your arguments are like someone saying that there is no place on their ruler that is exactly the 1 inch position, so 1 inch doesn't exist. If there was such a position, it would have to have width to exist, and then it would have a distance from one side of it to another. Then, you can't get from the 1 inch position to 2 inches, because to get there you have to go an infinite number of infintessimal stages, and you can't do infinity, and infintessimal steps don't get you anywhere.

Or in short, you're trolling. You really, really should team up with doronshadmi. Your understanding of maths is as banal and messed up as his, although he makes better squiggles. He could probably help you devise whole books of equations for humberversity courses.

Any thoughts yet about which way a bit of paper bends as you wave it through unreal wind?

 

[humber]

No Ross, distance is a scalar. Velocity is defined as the rate of change of position between an object and an arbitrary reference point. Position being a vector quantity and the reference point would be the origin of the reference frame.
ETA: If you were to use distance in defining velocity, your world would be almost as screwed up as humbers.

I hear 1st Class engineers say 'pressure', when they mean force, but we do know what they mean.

That's only because it is a real wind, which you would have known if you carried your handy dandy little paddle device (which we still don't really know what it does), it keeps your hang glider from getting defrangimenated.

You don't know how it works, but you will bellow anyway?

Humber, if you somehow came to realize that you have some fundamental misunderstanding about basic physics, and came to understand that error and repair it, would you admit it?

Have you ever said anything that may require that? I don't think I need lesson from you, SD.

Speed has no direction.

Yes, Brian_M, that's why I use signs for my flows, rather than "looks like it's going backwards"

Compare car going 100mph West with a car going 100mph East.
The difference in speed is 0mph.
The difference in velocity is 200mph.

Know that.

I'm fairly sure that the kinetic energy involved in a 200mph collision is somewhat larger than the kinetic energy involved in a 0mph collision.

That too

So no, "speed" alone is not enough.

It's a car on a road. Speed signs...
Strawman to evade the point.

ETA:
Since the example is only moving in one dimension this is sufficient.
In case you hadn't noticed, with the math the rest of us are using negative values to indicate the reverse direction, but I don't expect you to understand it fully.

See ALL of my boundary flow remarks, Brian_M.

 

[humber]

This thread is more!!
It has ceased to be!!
It's curled up it's tootsies and joined the bleedin' choir Invisibule!!
If it hadn't run into Humbers brain, it would be puhing up the Daisies!!!

THIS IS AN EX-THREAD !!!!

How original. The Parrot Sketch.

 

[mender]

If you like, yes, but the power must come from the propeller not the ground.

As you have been told so many times, the propeller is powered by the movement of the cart over the ground. The cart is moving because of a difference in speed between the air and the ground. That's why the treadmill and the paper and the cart outside in the wind are all the same as far as the cart is concerned, if the speed difference is the same. The difference between those is that outside, we can't force the ground to move but fortunately the air does occasionally. Inside we can move the "ground" and therefore don't have to wait for the air to move.

It doesn't matter to the cart which of the two substances are moving, nor which substances is stationary, as long as one of them is moving and the other isn't. The air outside uses thermal energy from the sun to make it move over the ground. The treadmill uses energy that ultimately came from the sun to move the belt. The energy flow through the cart is the same regardless.

If you think of the belt velocity-plane/displacement argument, the two quantities are related. The ratio is 1:1. This is due to the forced synchronization of wheel and belt, and so the propeller. Seen this way, it becomes clear that the propeller must turn faster than can be obtained from the wheels, if it is to make progress up the belt.

Maybe there is; in fact that is how the thing works. Concentrate real hard, humber. If you can understand this, you might get it finally. The pitch of the prop is important; it determines how far the prop would move forward if there was no slippage against the air. The wheel diameter on the cart is important too; it determines how many times the prop turns as the cart moves a specific distance over the ground. Either one can be changed to alter the travel ratio; in other words, the ratio is only 1:1 in one specific case. By changing either the prop pitch or the wheel diameter or both that ratio can be altered to be below 1:1, in which case the prop acts like a wind turbine and the prop has more "leverage" than the wheels and the prop turns the wheels, or it can be over 1:1, in which case the wheels have more leverage than the prop and the wheels turn the prop. When the ratio is over 2:1, the cart will want to move backwards, into the wind. What we're interested in is when the ratio is over 1:1 and under 2:1 for the obvious reasons.

My cart has a ratio of around 1.8:1. When the cart is moving over the ground or on the treadmill at 10 mph, the theoretical progress of the prop is 18 mph.That changes the momentum of the air going through the prop, transferring energy from the treadmill belt to the surrounding air. The amount of energy transferred equals the total drag of the cart when the cart is in equilibrium. If the total drag of the cart is now reduced by using better bearings for instance, the cart will reach equilibrium at a higher speed or the speed difference between the moving substances can be reduced (less energy) to maintain the previous speed.

My cart doesn't need 10 mph of relative air or ground movement to stay stationary with the air. In its best configuration it only needs 5.5 mph. Using different combinations of prop pitch and wheel diameter to change the travel ratio results in different equilibrium speeds, exactly as predicted by physics (the rest of the world's, not yours). Yours predicts no change; your concept of what is happening is wrong.

I doubt that you bothered to try and follow this and doubt even more that you understand or accept this. It's too bad because this is what is happening. Feel free to ask questions about this though.

If it's wrong, it's wrong.

I really would like to believe that you could accept that you are wrong on anything, humber.

 

[GregLondon]

humber, there are block-and-tackles that have mechanical advantages of 2, 3, 4, and up as far as you want. I'm trying to figure out why you couldn't have a parachute at the end of a block and tackle, pick one with enough pulleys, and end up having the chute move less than the speed of the wind due to innefficiency and whatnot, but have the block and tackle make up for that by have a 3-to1 or 4-to-1 ratio.

Also, I have a yes/no question for you. have you ever experimented with moving a yoyo along the ground by pulling on teh string so that the string wraps up under the axle?

If not, could you try it? ANd when you do, here's the next question: can you see that at any time, no matter how fast or how slow you pull the string, the yoyo is ALWAYS moving faster than you are pulling the string? That the speed of the yoyo is equal to the speed of the string multiplied by some constant? The constant being the "gear ratio" you get between the diameter of the wheel versus the diameter fo teh axle the string wraps around?

It should only take you a minute or two to build a yoyo using paper plates, an empty soda can, string, and tape. Say the ratio is 4 to 1. If you pull the string 1 inch, the yoyo will move 4 inches. if you pull the string 20 inches, the yoyo moves 80 inches. Which then translates directly to speeds. If you pull hte string at .1 inch per second, the yoyo will move at .4 inch per second. If you pull the string at 10 inches per second, the yoyo moves at 40 inches per second.

at ANY TIME you move the string, the yoyo is ALWAYS moving faster than the string.

So, at any time the parachute is pulling on the string, the yoyo is always moving faster than the chute.

So, if in a 10 mph wind, the chute will only move at 6 mph, then all you need to do is make the diameter of the axle versus the diameter of the wheel give you enough of a ratio so that pulling the string at 6 mph will move the yoyo at 10 mph.

Right? Are we both in agreement to this point?

If you're with me so far, then the only issue is that we need to account for the yoyo moving forward, and how it affects the speed of the parachute.

So, imagine supersizing the yoyo until its 6 feet in diameter or so. Put a shaft through the center, put it on bearings so the shaft just floats. From that shaft, hang a seat so you can ride along as the yoyo is rolling forward. you will see the ground go by underneath you at 12 mph. and you will see the rope coming towards you and getting wrapped up around the axle at 6 mph (or 4 or 8 or whatever).

At ANY speed, you will see the ground move back faster than the rope moves towards you. slower than the wind or faster than the wind, it doesn't matter. THe ground is always moving faster than the rope.

When teh yoyo is moving at the speed of the wind, you won't feel any wind on your face sitting in your seat on the yoyo. But the parachute does. the ground is going by underneat you at teh speed of the wind, say 10 mph, and the chute is getting reeled in at some slower velocity, say 5 mph. You're sitting in your seat on the yoyo and you will see the chute coming towards you. So even though you don't feel teh wind because you're moving exactly at the speed of the wind, the parachute is getting reeled in by the axle, getting pulled towards you, and so the string pulls the yoyo forward.

In your seat, you don't feel any wind because the yoyo is rolling at 10 mph. But you see the parachute coming towards you because it is getting reeled in by the axle. Since you don't feel any wind, and since the parachute is moving relative to you, the parachute must feel wind.

That doesn't matter if youre on a yoyo and a chute is tied to a rope, or if you're on a seqway and a friend of your is driving towards you on another seqway. If you're driving your segway at 10 mph, and you don't feel any wind, but you see your friend ALice coming towards you, you know she MUST feel wind on her face. The only way she cannot feel wind is if she's moving in formation with you at exactly the same speed adn direction as you, so that neither of you feel wind.

So, first of the last questions: Do you agree with the segway analogy above? That if you're moving, but don't feel wind, but if Alice is coming towards you, she must be feeling some wind?

Second, can you see that translating to you compared to the parachute? If you don't feel wind hanging from the yoyo as its moving forward, but the parachute is coming towards you, then it must feel some wind?

 

[John Freestone]

That's only because it is a real wind, which you would have known if you carried your handy dandy little paddle device (which we still don't really know what it does), it keeps your hang glider from getting defrangimenated.

You don't know how it works, but you will bellow anyway?

Yeah, SZ. Besides, keep up, I told you how it works. It's quite simple. It's a device for measuring two adjacent wind velocities and subtracting one from the other, the result being expressed as a velocity, but with an arbitrary clock rotation attached to the end of the numbers instead of a preceding sign.

It's a very helpful way to think of different windspeeds, because if you held a paddle up in them, it would turn at that speed in the specified manner, or rather it isn't, because it wouldn't, but it might have been if it did.

Paddles are handy not just for defrangimenatizing hang-gliders, but also if you're lost up a certain creek in discussing basic physics, due to the operator-error feature. In fact, there's a whole suite of them built into the handle. You can reverse the sign of either wind, take either of them away from 10, or use the Scramble button, which just randomizes everything.

 

[GregLondon]

Humber, page 38 of my doc shows a diagram of a yoyo moving at the speed of teh wind, with a parachute getting reeled in, putting a force on the yoyo to speed it up.

it's basically a drawing of what you wold see if you were sitting on that seat on the yoyo and you're moving at the speed fo the wind so you don't feel any wind on your face.


page 9 shows a timelapse fo the yoyo moving and the parachute getting reeled in.

http://www.greglondon.com/tumbleweed/tumbleweed.pdf

 

[Dan O.]

So there is a diminishing closing gap? That is ds/dt? Going through zero as it "passes" the car? So zero KE?

I assume "s" is the displacement vector between the cars? If so, s goes through zero (or rather the component of it parallel to the motion does), but ds/dt does not - that's the relative velocity, and it's constant. Since KE depends on ds/dt (but not on s), KE is constant too.

Don't you recognize this as a derivative of hyperbola

 

[Christian Klippel]

Who was keeping track of all these? Chris?

Hello mender,

i'll try to collect some "best of humberphysics" in a week or two. I'm quite busy at the moment, but already made some notes about some really interesting stuff. Thing is, the more humber posts, the more ridiculous his line of "argument" gets. Makes it pretty hard to decide

Greetings,

Chris

 

[Christian Klippel]

Or in short, you're trolling. You really, really should team up with doronshadmi. Your understanding of maths is as banal and messed up as his, although he makes better squiggles. He could probably help you devise whole books of equations for humberversity courses.

Hello John,

as i said earlier, i'd really like to see humber, GMB and Dutch sitting in one room and arguing about physics and stuff. But i wouldn't mind to have doronshadmi there as well.

That would truly make a really good physics daily soap.

Greetings,

Chris

Edit: Just make sure that there are no things in that room that could be used as weapons in any way, and bolt the chairs and table to the ground. Otherwise hell would break loose.

 

[humb]

Also, I have a yes/no question for you. have you ever experimented with moving a yoyo along the ground by pulling on teh string so that the string wraps up under the axle? If not, could you try it?

You're obviously new around here. We'll have to get you up to speed. But clearly not windspeed.

No, neither humber nor I has any need to perform your time-wasting experiments when it is obvious what the outcome will be. The yoyo is notional, and it is not a kite. The parachute may be a kite, but it is not a treadmill. The cart may be a treadmill, but it is not my tinfoil hat, and you have no more chance of getting that off of my head than you do of getting the stiletto heels off of the slut in that physics quiz.

In your seat, you don't feel any wind because the yoyo is rolling at 10 mph. But you see the parachute coming towards you because it is getting reeled in by the axle. Since you don't feel any wind, and since the parachute is moving relative to you, the parachute must feel wind.

The wind propels the parachute, but the parachute moves though the air, and must overcome drag. It is therefore a case of diminishing returns. From the perspective of the observer in the seat, the yoyo has zero KE, and therefore cannot move. If you put the yoyo on a treadmill, the parachute will have no KE, and the boundary layer will be flowing in the wrong direction. It will remain in place by a simple force balance. It cannot do so on the road, because the friction is insufficient for it to hover.

Do we have to explain this to everyone?

 

[sol invictus]

Yes, yes, no. The velocities are constant w.r.t. ground. The KE's also constant w.r.t ground. The difference is not when the instantaneous velocities are used. The infinitesimal may then be zero, so zero KE for how long?

Wrong. In this situation KE is exactly constant - it is never zero, since it never changes at all. I ask again - would you like to see the full mathematical solution (which is trivial, but I can write it down if you ask)?

To gainor lose KE w.r.t the ground, ds/dt must be non-zero, meaning the object must accelerate or decelerate to exchange it. This is why the cart has no KE on the treadmill.

Wrong yet again. Acceleration is d²s/dt², not ds/dt (ds/dt is called the "velocity" - see, you can learn some new terms). In this situation, where ds/dt is constant, d²s/dt² =0 and so there is zero acceleration.

I think acceleration, velocity and displacement inextricably linked together.

They are, but you fail miserably to understand how. This is the very first bit of high school (or even grade school physics). It doesn't get any more basic than this.

Normally they don't, because they have zero KE, but I did see the very thing you describe when a car transporter passed by. Stationary? What do you mean?

With respect to each other, obviously.

 

[humber]

In the treadmill experiment not all vectors are zero, so there's no need to.

All quantities are near zero too. KE, velocity, acceleration, displacement, work.

Of course it does, just like you want your aicraft wind tunnel models to have zero KE w.r.t. the ground. That way you won't have to leave your comfy chair.

You have zero KE when in your chair, too.

This made me laugh out loud. I'm sorry, but math is now an obfuscation for physical modelling? That's just rich. Anyway, I assume you won't be talking sense then.

Math as obfuscation is not the same as math as education, H'ethetheth
OK, please post your analysis of the treadmill. I will address it then.
Math cannot repair a false model, but if you insist.

No no no. Sorry, my friend, pretty much everyone has told you how the car can be said to be in motion. Motion is relative. The question is whether or not the relative motions of all parts involved are preserved. The answer, of course, is: yes they are.

Please explain why it is said, and how it is like the fish tank you said was equivalent. You argued for the belt as evidence of motion. I say it is not needed, so you have now contradicted yourself. All relative quantities are not preserved on the treadmill.

Yes, but the gravitational field, which is vertical, is very nearly invariant under a transformation of horizontal velocity.

Please explain how the system can be said to be isolated from the ground ( eg treadmill side) when the system that provides the energy (the motor) is common to ground. The KE of the cart would be from this source, and so measurable from the same ground. To deny that would be to deny the conservation of energy between frames. The KE should therefore appear to all observers.

Our dead travel around the sun each year. They travel around the earth every day; motion is relative. There is no preferential inertial frame. How many times do people have to say this before you consider the possibility that you're wrong about this?

I say there is a palpable difference between those making motion over the ground, and those buried in it.
They perhaps both experience that same remote force, so that can be ignored as a common factor. How many times are you going to exhume pre-modern ideas? It seems to me that absolutism must be raised so that it can be shot down with what is obvious. Motion is relative. A big discovery apparently missed by all wind-tunnel engineers

Have you considered that you may be wrong, H'ethetheth? Your support and replies seem weak and selective.

What? What the hell does the word treadmill even mean to you? Have you ever seen someone running on a tread mill? They have no KE w.r.t. the ground. The fish need to exert exactly as much power to stay in one place as they would need to swim through still water.

Like the cart, the fish has no KE. You agree.
Yet another false analogy. The runner does the work of moving the limbs and generating muscle force. The object is cardio-muscular exercise, not to emulate motion. The belt passes under the runner's feet, to accommodate human ambulation. ( Essentially we walk by a sequence of free-falls and re-stabilization). This is a fact.

If that's the point, then I agree. You don't need a belt. You don't even need a car, really. The question is what you want to model. If you want to model the interactions of a car that is driven by the difference in velocity of two media, your going to need a car, and two media at different speeds. That simple.

You agree. The belt is not necessary, so there can be no frame for the said observer, but wind is, and some of the conditions of travel can be emulated by the motion of a medium. One is enough, and for F1 cars and wind carts, it is the wind. This is contrary to your earlier statement that the wind was not necessary because of the belt. This is essentially the claim of the treadmill. How do you reconcile the lack of wind on the treadmill?

Yes, and incidentally, that's also everything that is needed to be similar.

No, it is not. They are quite different. The force of the real wind is not self limiting in this way. That is demonstrable. The real wind is not limited by the object that is blown by it. This is not a small difference, but one that the treadmill depends upon. I see no refutation of the self-regulatory mechansim I described. This is a critical difference, that demands more of an answer than the repetition that all is relative.

In both cases the dynamics depend on gravity, and in both cases the dynamics deped on the difference in velocity between air and surface.

The force of the wind itself does not depend upon gravity, whereas on the treadmill it does. It is not possible for the force of the wind upon the cart to exceed the demand (load) presented by the cart. One is a reaction to the other. That is not like real wind.
The real cart's dynamics depend upon the friction to the road, and the dynamic response of the system to perturbation. Without the correct level of KE, or real wind to account for that influence, the dynamics are false. This is why F1 use wind, and you agree that is necessary.

At standstill, it is already at windspeed. When it climbs off, it is faster than wind speed.

Meaning that if you don't pull the trolley it is at windspeed too? OK the prop is not turning. Spin it up, and put it on the ground, so that it runs on the momentum of the propeller (as in the video) Hard to argue that is not going faster than windspeed? Because of the 1:1 relation ob belt and wheel, it is not possible to separate increased velocity over displacement. The latter appears in all frames, meaning that it is displacement and not velocity increase. It has essentially no motion.

That's where you're wrong. When you do this experiment you will see that, if you stand on the trolley, the cart will be blown away from you, and will act just like it did on the road.

No, my analysis stands. The trolley model is equivalent. I can easily do more measurements than on the current treadmill. The cart will remain essentially stationary w.r.t the ground. The trolley will pass under the cart so to speak, as its wheels turn.
If an observer stands on the belt, he will move relative to the cart, but that is because he is being accelerated by the trolley, but the cart is not. Measurement will show that the expended energy is stored as KE in the observer. Furthermore, it does not change if the cart is not running, but stopped on the belt. No increase can be detected between those two conditions. This suggests that all KE is that of the observer, and there is none in the cart. No measurement can demonstrate KE for the cart, and that is flat.

What? Did you even see spork's movies?
You really need to look at those movies again.

Yes, hence the remark. I noticed things. The axle does not move w.r.t the belt. When in motion, the cart is not in motion by the means employed by a cart in real wind. The acceleration of the cart seen in the videos can be estimated, and so the number of turns of the propeller calculated. The velocity is not at all in accordance with that estimate.

Similarly, to make progress down the belt, the axle must also move w.r.t the belt. When pushed to be seen as going backwards, the wheels should rotate in the opposite direction, but clearly they do not. The wheels spin much as before, in direct opposition to the normal motion of the belt. Motion of a wheeled vehicle in either direction along the belt is NOT possible unless there is contact patch forward of, or to the rear of the axle. On the belt, this contact point is directly in line with the axle because of the fixed relation of belt and wheel. No motion as found on the road, is possible.

I can understand that you believe this if you think the ground is the absolute frame of reference, but unfortunately, it isn't.

I believe no such thing, and you know it, whereas you are denying that it can be. It is the equal of all others, and I need not be biased or absolutist to use it. This you also know.
You may believe reiteration to be refutation. In any case, I do not need to assume that the ground is absolute to find fault. I have my trolley equivalent, that allows me to make all claims from that, so the absolutist lament is no longer going to work.
There are enough errors without resort to that claim. anyway.
Also in the real world, there are no known examples of objects traveling w.r.t the ground that do not have KE, w.r.t the ground. The treadmill flatly contradicts nature and that is the reference.

You have obviously never run on a treadmill. It is not easier than running on the ground, believe me.

Put that first point to Steven Hawking. I imagine he can know nothing about anything to do with motion.

A runner's treadmill is not the treadmill, nor your fish tank. There is work done, but no linear motion.
(1) Human is self-powered.
(2) Cart is not human, and does not have the evolved balance to keep it stable in the required manner.
(3) See human ambulation.

I imagine it does to you, yes. Please go to a gym and run on a treadmill for me.

Low-hanging fruit taken. Not an answer.

Yes it does. It is only the relative motion that is relevant, because motion always is only relative!!I will repeat again: All motion is relative.

Of course it is, but that does not mean that the treadmill is a valid representation of that fact. The treadmill is not consistent with that claim, that very fact! "Velocity is relative" is only one aspect of the fact that "All motion is relative". The treadmill takes only that first over-simplified form, and it fails. Please do not offer again the trite observation that all motion is relative. It would be extraordinary if it were not the case. However, acceleration is absolute for our purposes. That can be wikkied.

Try to understand what this means, humber. It really means what it says. Everything can always said to be moving, and not in a metaphorical way: Cars can be said to move across the road with exactly equal validity as the earth can be said to move along the car. Deal with this, please, humber.

No, H'ethetheth. That is not the case. Is there anything that I have written in this post alone that does not fully indicate that I understand that? It is the most basic of ideas. What is there not to understand? I think that I can safely say that I understand that, and its consequences better than you.
You offer bipedal ambulation as equivalent to a cart? The treadmill is a very simple idea taken to the breaking point, and it is said that nothing matters, because it's all relative.