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

[Michael C]

Anyway, you have said quite adamantly that weight must always be measured with respect to some frame of reference. But even the site you've give to support that view allows that there are different ways you could derive the "g" when we calculate weight using W = mg. The first option he lists is based on Newton's universal law of gravitation. That formula doesn't depend on any frame of reference and it gives you a force. If you divide the force by the mass of the object in question we get one version of "g" (acceleration due to gravity). You might not like that approach but it certainly seems to at least one possible definition and a naive reading of wikipedia and many other pages on the net (and many textbooks also I would suggest) doesn't seem to disagree or even suggest anything further needs to be considered.

The first option for g on that page reads thus:

"I-gravity is denoted gI and can be calculated using Newton’s law of universal gravitation, as follows: It has magnitude

|gI| = GM/r²
and is directed toward the center of the planet."

The key phrase is "directed toward the center of the planet": as long as we remain Earth-centred, we choose Earth as "the planet". But which planet do we choose when we are away from Earth? In particular, can you answer my previous question: what is the "real", "true" or "actual" weight of the Earth?

In other words (and this is copied almost directly from another forum I was reading), your view is that weight should be defined as the net force required to make an object accelerate at a rate equal to the local free fall acceleration. Is this right?

By definition there is no net force required to make an object accelerate at the rate of free fall.

I would have said that was what wikipedia, etc., call "apparent weight^WP".

I am not saying that "weight" and "apparent weight" are the same thing. We can always choose a frame where our defined weight is the same as our apparent weight, but we don't have to. If we take another frame, accelerated with respect to the first frame, our weight will be something else. That is the essential point. For instance, if we are on the surface of the Earth, we usually take the local frame defined by the surface we're standing on. That means that our apparent weight, as measured by a simple spring balance, is the same as the weight we define with respect to the frame of reference of the Earth's surface. We could stay standing on the Earth and define our weight by the local free fall frame (accelerating with respect to the frame of the surface of the earth), in which case our weight is zero. Of course, usually we don't, and the "standard" frame of reference is the surface of the Earth. This is just like measuring the speed of a car running on the surface of the Earth. If we say that a car is travelling at 100 km/h, usually we don't need to specify "with respect to the surface of the Earth": it's understood.

For the person in a space station in orbit around the earth, it's no different: we may either choose to measure their weight in the freely-falling reference frame, or in one at a fixed distance from the surface of the earth. In the first case we conclude that they are "weightless", in the second case we conclude that they have weight. Both answers are right, as long as the frame of reference is made clear. It's like measuring the velocity of that same space station: it won't be the same if we measure it from a frame of reference rotating at the same speed as the Earth and from a frame of reference not rotating at the same speed as the Earth.

To sum up, I'm not really particularly concerned one way or the other because as I said earlier, it seems to me that "weight" is often likely to cause confusion rather than clarity, and so it's probably easier to just avoid it altogether whenever possible and talk in terms of mass and force and so forth instead.

I think you're right.

ETA: What's your opinion of this blog post: http://blog.dotphys.net/2008/09/gravity-weightlessness-and-apparent-weight/ ?

It's OK, but he doesn't give a clear definition of weight. It would be better if he simply insisted on the fact that your mass remains constant in all these situations. In any case, since this is the same guy who confidently predicts that a DDWFTTW vehicle can't work, calling it "free energy" and "magic" (see here and here), I'm not inclined to spend too much time reading his articles.

 

[Dan O.]

Humber's "test" for free-fall in a gravity field vs zero acceleration outside of a gravity field qualifies as another statement where he is wrong about the physics involved. If the instruments are sensitive enough and are far enough apart, they can detect the gravity field's gradient if held in position across that gradient.

Spinning them makes absolutely no sense, unless humber is referring to a slow reorientation of the paired accelerometers to figure out the direction of the gravity field's source. Which he isn't.

Another humberism to add to the list, Dan. Looks like it may be part of the subset that you just posted.

Humber has been revisiting this test in several posts. It looks like he is trying to replicate someone else's test for measuring micro-g gradients but doesn't understand the concept himself. Spinning the apparatus allows you to remove the DC bias of the sensors by filtering the output for just the component that matches the spin frequency's first harmonic.

 

[RossFW]

Well, jeeze guys, welcome to the party!

It seems obvious to me that I am the least qualified person here wrt physics, yet you left me all by my lonesome for two days countering Humbers rubbish about freefall!

 

[Furcifer]

I think you might have missed where humber said "Now, she is a cast-iron idiot and a feminist."

Incidentally, I say you don't have to be a feminist to be an idiot, but it helps.

Indeed, I got "whooshed".

Too busy bolting things down and installing K filters as the harmonics in this place are causing things to float away.

Everything is all Fouriered up around here.

 

[Belz...]

I'm having trouble following this thread.

What's it about, now ?

 

[fredriks]

This is very of topic, but why are you using the very large spaces in the equations?

[latex] \begin {eqnarray} E & = & mc^2 ( 1 - v^2/ c^2 ) ^ {- 1/2} [/latex]

instead of

[latex] \[ E = mc^2 ( 1 - v^2/ c^2 ) ^ {- 1/2}\] [/latex]

 

[humber]

The Unbearable Lightness of Being Dan_O;

• Humber has been revisiting this test in several posts. It looks like he is trying to replicate someone else's test for measuring micro-g gradients but doesn't understand the concept himself. [/COLOR]

The Unbearable Lightness of Dan_O;

• Spinning the apparatus allows you to remove the DC bias of the sensors by filtering the output for just the component that matches the spin frequency's first harmonic.

The test is of my own. The two accelerometers know nothing of DC bias.
If thrown into the air, they will register the suggested periodic motion, according to angular velocity.
In a free-falling elevator, that will also be the case, but superposition of forces means that the constant force of g, will become part of the "AC Component".
In addition, gradients or other non-uniform gravitational forces, will produce distortion, and therefore analysis. This is why I first suggested a 3-axis arrangement. Free-fall is not zero-g.

 

[humber]

What utter nonsense.

Pish-posh and tiddlewaddle.

Treadmill belts are very smooth and stable - far more so than any surface you're likely to find outside. Even if they were there, "resonance effects" are going to reduce the friction between the wheels and the belt and decrease the cart's performance, not increase it.

As the friction to the road increases the cart's performance improves ?
The value of friction on the belt is arbitrary, as long as it is within the bounds of the cart's balance mechanism, so you can indeed make that claim

And why can't there be a resonance effect with the cart running on the street? Just make it bumpy with a certain period.

I think a few blobs of chewing gum and the occasional flat-cat may help.
Still, I can't help thinking that if the belt were a roller, it would drive the cart in the same way, but not at all lend the impression of motion. A wise choice by Walmart.

 

[humber]

Indeed, I got "whooshed".

Twice.

 

[humber]

Well, jeeze guys, welcome to the party!

One more drink at "The Flight Deck" and we're off!.

It seems obvious to me that I am the least qualified person here wrt physics, yet you left me all by my lonesome for two days countering Humbers rubbish about freefall!

You will do no better, Captain.

 

[RossFW]

[QUOTE Free-fall is not zero-g.[/QUOTE]

So, as you are under the influence of gravity everywhere in the universe, what IS zero G?

 

[ThinAirDesigns]

As the friction to the road increases the cart's performance improves ?

Yes.

JB

 

[humber]

It might if it were just a tad over 40075 km (but we should ask TAD)

Do you think twee should?

Humber has been prolific over the last few pages but the random stream of words has little coherence most of the time. However, here is one concept that was even repeated and since it had already been questioned I don't need to wait for another second to add it to the list of...

Rush, rush..

Where humber is wrong

"If in free-fall inside an aircraft, the simple act of lifting your arm and measuring the force, will tell you that you are in a gravitational field." #3255

Now, this one's a real brain-teaser. Apparently, even in free-fall, work can be done against gravity. In fact, if you do enough work, you can stop falling Free-fall is not zero-g. Perhaps we can make a new class for you; xero-KE, xero-g.

"If you are stationary in a gravitational field, then you will do work against that field should you raise your arm. That does not change if you jump from a table, or fall in an enclosed elevator." #3327

Yes, this is also correct. At one time you were all telling me that it is "all the same" as long as "ya can'na tell", but now free-fall has you all in a tizz trying to claim the contrary.

The second statement would technically be true under a consideration of the separate forces. But the simple experiment of the first statement precludes such an interpretation.

No, there are many ways. There really is no need though, as the gravitational gradient alone is enough to tell zero-g from free-fall. There are all sorts of ways that are a consequence of acceleration and it relationship to time. Rather like you can tell a stationary cart from a moving one.​

 

[humber]

Yes.

JB

Let the air out of your tires, JB. Flat tires improve fuel consumption?

Then you also won't mind demonstrating what happens if you increase the friction of the cart to the belt. Let's say, a 1kg mass over the wheel. If friction increases performance, then why not glue it to the belt?

 

[fredriks]

Lets see, drive a car with summer tires on a icy steep road vs driving the car on the same road in the summer without ice. What is going to work best?

Why do we have rubber tires and pigs in the winter tires? I have no idea because friction between the road and tire is clearly bad.

 

[humber]

Free-fall is not zero-g.

So, as you are under the influence of gravity everywhere in the universe, what IS zero G?

But not "under the influence".

There is no true zero-g; gravity is everywhere, but the claim has been made (by you, for example), that "free-fall is the same as being in zero-g".
Even if the background gravity were to be zero or ignored, then free-fall in a gravitational field would still be distinguishable from that.
When on the 'vomit comet' you are in free-fall, but not zero-g. They are not the same, even if undetectable, but differences are detectable.

 

[humber]

Lets see, drive a car with summer tires on a icy steep road vs driving the car on the same road in the summer without ice. What is going to work best?

That is not enough friction.

Why do we have rubber tires and pigs in the winter tires? I have no idea because friction between the road and tire is clearly bad.

You will go slower in that case, Fredriks. You will consume more fuel to make up for the losses of the soft tires, etc.

Friction is a loss. You need only the enough friction to meet the force required for motion. All else is a waste. So, it cannot be true that increasing friction always produces improved performance.
Tires flex with rotation as the 'contact patch' moves from contact with the ground. That generates heat, that is a loss.
The level of friction on the belt is arbitrary. It has not been demonstrated that it is like that of the road, or that the force normal to the belt is enough to sustain motion in the real world, at the equivalent beltspeed.

 

[fredriks]

I got a strange thought. There might be different frictions involved with a rolling tire. It might not even be the same tire that make it possibly to accelerate very fast and the one that make it possibly to roll the longest distance.

Another strange thought. What if we can talk about the friction needed for the tire to not slip against the road and another force/friction/energy that is about the deformation of the wheels and implies how easy the wheel turns, lets call that the rolling resistance.

No I guess this cant be correct because a wheel is clearly false and a wheel is also not a reference frame because most frames are square.

 

[humber]

I got a strange thought. There might be different frictions involved with a rolling tire. It might not even be the same tire that make it possibly to accelerate very fast and the one that make it possibly to roll the longest distance.

Depends upon the application. Tour de France riders are after maximum efficiency, so they gave tire pumped up to very high pressurs. Reduces the loss due to flexure. F1 cars leave a lot of rubber on the track. Acceleration and grip are important to them.

Another strange thought.

'Thought' is a stretch.

What if we can talk about the friction needed for the tire to not slip against the road and another force/friction/energy that is about the deformation of the wheels and implies how easy the wheel turns, lets call that the rolling resistance.

Not quite. See tire/wheel 'slip'. Now, I know that will confuse you, it does not mean the wheel is completely slipping, but is the standard term for the way that a tire will lfex, so that the contact patch move w.r.t the axle.

No I guess this cant be correct because a wheel is clearly false and a wheel is also not a reference frame because most frames are square.

A wheel is not a frame of reference. It is a simple machine.

 

[fredriks]

A wheel is not a frame of reference. It is a simple machine.

You are of course correct that it isn't a reference frame but your reason are not correct. All my frames are square and a wheel is round, only a fool can't see this. The same is of course true for the treadmill, the treadmill are not square and I have never seen a frame that looks like a treadmill. Where are you supposed to put the picture on a treadmill? Doh. It is also very inconvenient to put the treadmill on the wall above the sofa. It is clearly not a good frame.