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

[humber]

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

I doubt that you can speak for Newton, if in in jest, Captain.
He would ignore your math, and roll his eyes at your logic.

 

[humber]

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.

A brief resume of some of the differences of treadmill and road. Not at all like a fish in a tank.

http://www.pponline.co.uk/encyc/0969.htm

 

[RossFW]

So, all motion IS relative....

Except that a car moving backward RELATIVE to you isn't.................

Gotcha

Humber, you have made statements implying or outright saying that you do NOT believe motion is relative several times.

You SAID:-

The cart has zero KE in all frames

As the only factors in KE are Mass and velocity, and I think we can agree it has Mass, you could only think this if you believed the cart had no velocity IN ALL FRAMES.

As (which you now admit) velocity is RELATIVE, the cart has zero KE in ONE frame (that in which it has no velocity) and positive KE In the INFINATE number of others that exist. This was implicitly told to you at the time, but you tried to brush it off with the usual Humber blather.

Another case of you finding out your wrong, then pretending you knoew something you didn't.

So, tellme, does the cart have relative motin and therefore KE to things moving RELATIVE to it, or does it have "Zero KE in all frames". Both cannot be true.

Also liked your retort to H'eath that his "Support was weak". Kindly list yours.....

 

[sol invictus]

Don't you recognize this as a derivative of hyperbola

Hyperbole, maybe....

 

[GregLondon]

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.

KE has nothing to do with inertial frames. Inertial frames are nothing more than a frame of reference moving at a fixed velocity (fixed speed in a fixed direction). If you're moving at a fixed velocity, you're in an inertial frame and that's all there is to it.

The motor is tied to the ground and moves the treadmill, certainly, but that has nothing to do with whether standing on the treadmill puts you in an inertial frame or not. ALl that is needed to be in an inertial frame is to be moving at a fixed velocity.

As for the kinetic energy being isolated from the ground or not, I assume someone has pointed out that neither the treadmill nor being outside in the wind is a closed system, right?

The treadmill motor injects energy into the system by moving the treadmill belt in an enclosed room with no wind. This energy is equivalent to being outside with a wind at your back moving at whatever speed the treadmill was moving indoors.

in other words, if you have a windmill on your farm that generates a kilowatt of power in a 10 mph wind, if you mounted that windmill on a giant indoor treadmill and made the treadmill move at 10 mph, the windmill would still generate a kilowatt.

It would take a lot of power to move that treadmill, but the windmill doesn't know and doesn't care, either way it generates a kilowatt.

The cart is similar to the windmill. Whether you have it outside in a 10 mph wind, or you have in inside on a treadmill moving at 10 mph, the cart will behave teh same. It extracts power using the difference between the velocity of air versus velocity of inertial frame to keep the cart moving.

That you have to put a motor under the treadmill to make it move doesn't change the fact that as far as the windmill is concerned it gnerates a kilowatt or as far as the propeller cart is concerned it pushes itself forward.

 

[Tunny]

So many ridiculous statements, so little time...

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.

Here, yet again, you are mixing up frames. [PYTHON]The bleedin' ground don't enter into it![/PYTHON] Items moving relative to the treadmill surface have KE relative to the treadmill surface. Glue a block to the treadmill surface, and it will smack the cart with the same KE that a cart moving at windspeed w.r.t. the ground would smack a wall that is stationary w.r.t. the ground.

Tunny

 

[RossFW]

BEAUTIFULLY put Greg!

That was the clearest explanation I've read.

When Humber doesn't understand it, it will be pretty to watch.....

 

[D'rok]

in other words, if you have a windmill on your farm that generates a kilowatt of power in a 10 mph wind, if you mounted that windmill on a giant indoor treadmill and made the treadmill move at 10 mph, the windmill would still generate a kilowatt.

It would take a lot of power to move that treadmill, but the windmill doesn't know and doesn't care, either way it generates a kilowatt.

The cart is similar to the windmill. Whether you have it outside in a 10 mph wind, or you have in inside on a treadmill moving at 10 mph, the cart will behave teh same. It extracts power using the difference between the velocity of air versus velocity of inertial frame to keep the cart moving.

That you have to put a motor under the treadmill to make it move doesn't change the fact that as far as the windmill is concerned it gnerates a kilowatt or as far as the propeller cart is concerned it pushes itself forward.

This lurking physics ignoramus says thanks for the clear explanation.

 

[humber]

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.

By that reasoning, the speed of sound with a series of them. There is no power gain. That is a fact.
I did not look too closely at your ratio calculations, but the bicycle analogy is not correct.
The sprockets of wheel and pedal may rotate at different angular velocities and torques. The yo-yo's wheels are concentric, so that independence is broken. You may want to recheck your calculations. Force and velocity are directly linked, but not like the block and tackle. Remember, that in those arrangements ,there are also independent axles, and linear motion of those axles. Not so for the yo-yo. It makes no difference to the terminal speed, but to acceleration.

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?

It's like the trick called 'Walking the dog'. So, yes.

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.

Deja vu, again. I need to say it every time, many times. Yes, it goes faster.
Faster, yes. It's not magic. It is a device old as time.

There were/are toy car like this. Internally geared, but with a flywheel added to give it some momentum. So what if it's faster? If you take one of those whistles from Christmas cracker and blow in it, the end uncurls very fast, and there is real linear motion, with almost static air !!. Free air travel here we come.

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?

What do you think,Greg? The last post said no.

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.

Power. That is the critical difference, otherwise you could not only do as you say, and for the same reasoning, but even faster.

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).

OK. But you are simply describing the simple arithmetic, and supporting metaphors, posited on the assumption that it will certainly happen.

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.

Can you pedal a bike in top gear from standstill? Yes? Go higher. You could go to say 60kph in half a turn?. Why doesn't everything travel at windspeed?
You assume that the chute force starts at a maximum. Not quite right, but close enough, and then falls to zero at winspeed, aon the basis that the load( the dragged item) will also need no force. No, the process stops before then. It is limited by the losses of motion of both cart and chute.

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.

A parachute will not get to windspeed. They are slower than balloons, and even almost ideal versions of those( lighter-than-air meteorological balloons) only come close, but even then because the work is very low, and 100% is
applied to the balloon. External load will certainly slow even them.

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?

Segways are an abomination.
This is yet another "relative velocity" argument at the expense of all other relative quantities, or the conservation of energy.
I don't even need to read it too closely, because I am sure you can do the appropriate arithmetic.

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?

Humans are not designed to monitor the world in that way. It would be false to think that a vacuum is just a very still wind. But that's no the point.
To go back to your home-built paper plate yo-yo. Try it at say a 1:1 ratio, but pull it with a spring balance. Change to say 2:1 and repeat.
If you don't have a spring balance, you can perhaps try rubber bands, even better, drive it with a series of weights. You should see that you need an ever increasing level of power.
Parachutes have a load line, that gives the force/velocity ratio for any given velocity. The load has similar, and where these load lines intersect, is the maximum velocity of that situation.

 

[mender]

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?

I find it hard to believe that you can't even remember the actual claim. "The cart uses the relative speed difference between the air and the ground to power the cart."

Did you get it this time? Two things are needed, and they have to be moving relative to each other. One is not enough, and the only person ever to say "one is enough" is you because you don't understand how the cart works.

Your comments are contrary to the claim. How do you reconcile your inability to even state the claim correctly after all this time?

 

[Clive]

KE has nothing to do with inertial frames. Inertial frames are nothing more than a frame of reference moving at a fixed velocity (fixed speed in a fixed direction). If you're moving at a fixed velocity, you're in an inertial frame and that's all there is to it.

This stuff is not really correct Greg. If an inertial frame is "nothing more than a frame of reference moving at a fixed velocity", then what frame are you using to measure that velocity against. Can it be any old frame? I think not! And "if you're moving at a fixed velocity, you're in an inertial frame and that's all there is to it" is also incorrect. Let's say I'm fixed to a rotating surface and my frame of reference is rotating with that plane. My velocity is zero, but I'm certainly not in an inertial frame.

 

[humber]

KE has nothing to do with inertial frames. Inertial frames are nothing more than a frame of reference moving at a fixed velocity (fixed speed in a fixed direction). If you're moving at a fixed velocity, you're in an inertial frame and that's all there is to it.

So?

The motor is tied to the ground and moves the treadmill, certainly, but that has nothing to do with whether standing on the treadmill puts you in an inertial frame or not. ALl that is needed to be in an inertial frame is to be moving at a fixed velocity.

Conservation of Energy holds across "frames". Motor proves that the energy source is in the ground frame, so that can be the datum. It is the source of the Ke, and so must flow from it and be measurable from the ground or any other frame.

As for the kinetic energy being isolated from the ground or not, I assume someone has pointed out that neither the treadmill nor being outside in the wind is a closed system, right?

It is. Got another planet in mind?

The treadmill motor injects energy into the system by moving the treadmill belt in an enclosed room with no wind. This energy is equivalent to being outside with a wind at your back moving at whatever speed the treadmill was moving indoors.

Velocity is not energy. Only the demand governs the consumption.

in other words, if you have a windmill on your farm that generates a kilowatt of power in a 10 mph wind, if you mounted that windmill on a giant indoor treadmill and made the treadmill move at 10 mph, the windmill would still generate a kilowatt.

Then you need 1kW wind power. That would come from the motor, but the overall efficiency would be poor ( See Carnot cycle). Better to connect the motor the the generator, or take the 1KW from the supply that runs the motor.

It would take a lot of power to move that treadmill, but the windmill doesn't know and doesn't care, either way it generates a kilowatt.

No, but there is no need to be careless.

The cart is similar to the windmill. Whether you have it outside in a 10 mph wind, or you have in inside on a treadmill moving at 10 mph, the cart will behave teh same. It extracts power using the difference between the velocity of air versus velocity of inertial frame to keep the cart moving.

Prove that.

That you have to put a motor under the treadmill to make it move doesn't change the fact that as far as the windmill is concerned it gnerates a kilowatt or as far as the propeller cart is concerned it pushes itself forward.

Prove that, too.

 

[humber]

So many ridiculous statements, so little time...


Here, yet again, you are mixing up frames. [PYTHON]The bleedin' ground don't enter into it![/PYTHON] Items moving relative to the treadmill surface have KE relative to the treadmill surface. Glue a block to the treadmill surface, and it will smack the cart with the same KE that a cart moving at windspeed w.r.t. the ground would smack a wall that is stationary w.r.t. the ground.

Tunny

Well Tunny. You do have this world weary tone. OK, I say that I have dealt with that matter, months ago.

(1) Nobody has posted a plausible explanation of how the belt implies motion or how the KE is transferred.

(2) Your test will not give the correct result (this should be the KE of the carts approximate 150g at the appropriate beltspeed or close enough)

(3) Conservation of energy says I can measure from any frame and get the same result, or do you believe in perpetual motion?

I can knock down any arguments that you put forward.

Ready?

 

[mender]

I can knock down any arguments that you put forward.

Ready?

How Monty Python-ish!

"Come back here you coward! I'll bite your ankles!" - or some such, I can't remember the exact line that the dismembered knight yells at King Arthur.

Your confidence is remarkable, humbre, considering you can't even do simple physics. Oh, here's a couple of questions that I asked quite a while ago, maybe you've had time to phone a friend:

"I will ask you to consider the following scenario that makes use of normal physics. Remember that the ground is considered to be flat and smooth.

A fishing reel with lots of line is attached by a scale to the ground. The other end of the fishing line is attached to a sail cart. The sail cart with a fishing line attached to it is released in a 10 mph wind. It reaches 9 mph, losing 1 mph because of rolling resistance.

You apply the brake on the fishing reel to slow the cart down to 8 mph and read the amount of pull (force) on the scale. Let's say it reads 100 grams. Based on these numbers, what is the rolling resistance of the cart when it is traveling at 9 mph in a 10 mph wind?

Now you slow the cart to 6 mph. What is the reading on the scale now?"

Any answers yet humbre, or are you still stumped? (sorry about the bad pun but ...)

 

[GregLondon]

It's like the trick called 'Walking the dog'. So, yes.

No, its not like walkign the dog. walking the dog has the yoyo spinning with inertia. this doesn't require inertia. You start with the yoyo on the ground, stationary, not spinning, and teh string coming off the bottom of the axle. Then you pull the string, and the string makes the yoyo roll, and the yoyo rolling reels in teh string. the moment you start pulling the string, no matter how slow, or how fast, or how much, or how little, the yoyo moves faster than the string, because the yoyo reels in the string.

A parachute will not get to windspeed. They are slower than balloons,

If you jump out of an airplane and open your parachute, sure, you'll never go faster than the wind. But if you're in a vehicle that has a long rope with a parachute at the end of it, and that vehicle is reeling the rope in at 50 mph, but the weather conditions report a 10 mph wind, the vehicle will go faster than the wind.

that's all this does.

THe parachute is NOT a simple tether. It wraps around the axle, so it acts like a block and tackle, not a balloon, not a sail, but a block and tackle. The wind pushes the parachute 5 feet and the yoyo ends up rolling 10 feet.

If you actually take a yoyo and roll it around on the floor so that the string wraps up around the axle, this should be immediately obvious.

If you let the string unwind and just DRAG the yoyo around, not have the string wrap around the axle, then it's just a tether and will only go as fast as you drag it.

If I had to guess, I'd guess you've never actually done this to see the difference.

 

[humber]

How Monty Python-ish!

"Come back here you coward! I'll bite your ankles!" - or some such, I can't remember the exact line that the dismembered knight yells at King Arthur.

Your confidence is remarkable, humbre, considering you can't even do simple physics. Oh, here's a couple of questions that I asked quite a while ago, maybe you've had time to phone a friend:

"I will ask you to consider the following scenario that makes use of normal physics. Remember that the ground is considered to be flat and smooth.


A fishing reel with lots of line is attached by a scale to the ground. The other end of the fishing line is attached to a sail cart. The sail cart with a fishing line attached to it is released in a 10 mph wind. It reaches 9 mph, losing 1 mph because of rolling resistance.

The cart is going 9mph in a 10mph wind? It how does the scale measure the force of the cart?

You apply the brake on the fishing reel to slow the cart down to 8 mph and read the amount of pull (force) on the scale. Let's say it reads 100 grams. Based on these numbers, what is the rolling resistance of the cart when it is traveling at 9 mph in a 10 mph wind?

This calculation is no possible without the actual data of the cart, otherwise all carts would be the same.
Your technique. Form a conclusion. Make an ad-hoc numerical example that supports that conclusion, and ask then that it be denied, while the only data is that which you have imagined. Real cart data please. ( or chute, kite..)

Now you slow the cart to 6 mph. What is the reading on the scale now?"

Show me your proof, and we can talk about that.

Any answers yet humbre, or are you still stumped? (sorry about the bad pun but ...)

My turn. Provide a method of measuring the cart's KE.

Still, the cart seems to work a lot like a toy car on a sheet of paper. I can get mine to stay still on a supermarket belt by adding the appropriate number of coins to weight it. Still, with your knowledge of physics, you should be able to tell me why.

ETA:
One is enough. The BMW team think so. One fewer results in no model at all.
The mistake is to think that a car; a powered car driving along a road, is like that same car, stationary on a backwards belt running at the same speed. It is not. Now, here we do not even have air to worry about, yet it is not only wrong, but impossible. Care to tell me that is not true? Assume no air drag if you like.

 

[GregLondon]

greg:

there are block-and-tackles that have mechanical advantages of 2, 3, 4, and up as far as you want

humber:

By that reasoning, the speed of sound with a series of them. There is no power gain. That is a fact.

The average human being can output about 0.1 horsepower continuous. 74 watts. Power = work done per second. Work = force * distance.

power = force * distance / time

Say you put that into linear motion. A human being walking on level ground, pulling with some force on a rope. say he walks at 1 meters per second.

74 watts = force * 1 meter / 1 second

force = 74 newtons.

So, if our guy has the rope going to a simple pulley and a weight hanging on the other end, he could move about 7 kg vertically at 1 meter per second.

If he uses a 2-to-1 block and tackle, he could move 15 kg at about .5 m/s.
3-to-1 => 21kg at .3 m/s
4-to-1 => 30 kg at .25 m/s.

If the ratio of the block and tackle is reversed and is instead 1-to-2, then he could move 3.5 kg at 2 m/s.
1-to-3 => 2.5 kg at 3 m/s
1-to-4 => 1.8 kg at 4 m/s

At every stage, regardless of the ratio, the power from the human is 74 watts and the power being applied to the object being lifted is 74 watts.

The yoyo takes the power from the wind, say its 74 watts, puts it through a transmission to gear it up, and allows the cart to go faster than the wind, as long as the force being exerted is reduced, so that the power remains the same.

Our human moved the rope 1 meter, and moved the object 2 meters. He did this by applying a 74 newton force on the rope, and this was able to move a 3.5 kg object at 2 m/s

The cart does exactly the same thing. THere is no magic increase of power. LIke a block and tackle the increase in speed is won by trading off and losing some of the force working on the cart.

greg:

That you have to put a motor under the treadmill to make it move doesn't change the fact that as far as the windmill is concerned it gnerates a kilowatt or as far as the propeller cart is concerned it pushes itself forward.

humber:

Prove that, too.

wait, what? If I have a windmill that generates a kilowatt of power in a 10 mph wind, I could put that same windmill indoors and mount it on a treadmill or moving track of some kind, and if I move that track at 10 mph, the windmill will generate a kilowatt of power.

You don't agree?

It'll take power to move the treadmill or track, but the windmill generates the same amount of power in either situation, either outside in a 10 mph wind, or indoors on a track movign at 10 mph.

You need me to prove that?

 

[GregLondon]

My turn. Provide a method of measuring the cart's KE.

The kinetic energy is irrelevant. the only thing that matters is the resistance from friction on teh cart, its wheels, axles, bearings, and from the wind. This is some force measured in newtons. You could measure it by simply dragging a cart around (take the prop off so it doesn't push the cart forward) and measuring the tension it takes to pull the cart at a certain speed.

Say its. a newton at 12 meters per second. That's probably too high, but its easy to work with. and if you want an exact number, you can measure soem specific cart.

So, the only thign that matters is that you need to exert a newton of force to keep the cart moving at 12 m/s.

Say the wind is moving at 10 m/s relative to the ground. The cart is moving at 12 m/s relative to the ground, so the cart has a 2 m/s relative wind that it can convert into 1 newton of force.

So, all you need to do is select a prop big enough that a 2 m/s wind can create 1 newton of force at 12 m/s.

1 newton at 12 m/s is 12 watts. So, you would need a prop big enough to extract 12 watts of power from a 2 mph wind. That's it.

The kinetic energy of the cart is irrelavent. ALl you need to do is figure out the force needed to overcome the friction in the cart at the speed you want, and that gives you the power the prop needs to pull out of the wind. That's it. KE is irrelvant.

KE is 0.5 * mass * velocity^2, and it doesn't matter how much teh cart weighs. If it weighs a thousand pounds, but somehow manages to only require 1 newton of force to move it at 12 m/s, then it needs the same size prop as above.

 

[GregLondon]

Let's say I'm fixed to a rotating surface and my frame of reference is rotating with that plane. My velocity is zero, but I'm certainly not in an inertial frame.

Clive, if you're point of reference is on the surface of a rotating cylinder, then your velocity is changing. Velocity is speed in a specific direction. Change the direction and you change the velocity.

 

[ThinAirDesigns]

I just love this typical Humber move:

Mender asks a question ... no assertion, just a question.

Humber responds with:

Show me your proof, and we can talk about that.

LOL .. "proof" of what humber ... it's a QUESTION.

JB