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

[humber]

Wrong again humber. In physics the event horizon is a definite distance from the center of mass of a black hole. It has no meaning outside of that usage. You do not enter into a planets gravitational gradient, you are always in it. If a planet is sufficiently far away you can ignore its gravitational field, but it is always there.

Yes SZ. I did say 'can be considered to have and event horizon', but I was not suggesting a black hole. So "event horizon", then
Of course there will no definite threshold, but that does not matter. It is impossible to introduce such an object instantaneously into the field, so the effect would be still be there. If the object were two masses coupled by a long spring, then the difference would be obvious. The point was to show that a gravitational gradient is not necessary for the stress in the object to be present due to acceleration as a result of that field.

 

[humber]

Sure - why not? I'm sure humber will correct me if (when) I get it wrong.



400 lbs.




42.4 ft/s

And?

 

[humber]

It looks like we can add "event horizon" to the long list of terms Humber has used without having any idea what they mean.

A scatter gun that always misses its target.

 

[Michael C]

Humber, what about your YouTube video of a cart spinning on a belt? You said you were going to post it last Monday.

 

[humber]

The entry continues: "For Ockham, the only truly necessary entity is God; everything else is contingent." I wonder how many times Occam's Razor is given as a reason not to posit a God!

Quite often, I think.

Newton's laws are "Ockham's Razor" at work, even though Newton was a believer.

 

[humber]

Humber, what about your YouTube video of a cart spinning on a belt? You said you were going to post it last Monday.

Sorrry, Micheal_C. I did say it was delayed. Only on weekends is it possible to work on it.

 

[Michael C]

Would it help if we instead used Ockham's reference frame which is defined as where the simplest laws of motion apply?

I'm happy to do that. I don't mind if anyone wants to use a more complex one than necessary. Sometimes it's useful to state the context.

Now comes the question: in which reference frame are the laws of motion simpler?

a) the surface of the earth
b) the elevator in free fall

 

[humber]

Now comes the question: in which reference frame are the laws of motion simpler?

a) the surface of the earth
b) the elevator in free fall

Things should be a simple as possible, but no simpler.

 

[RossFW]

The point was to show that a gravitational gradient is not necessary for the stress in the object to be present due to acceleration as a result of that field.

So you still believe an accelerometer will show a value whilst in uniform acceleration due to gravity?

Let me try and explain it this way- An accelerometer is usually accelerated by an external force that acts on the case of the instrument. The inertial member inside senses this, as it's own inertial resists the acceleration.

If you got into the inards of your accelerometer, and applied an equal force to the entire lenght of the inertial member, there would BE no strain on it as the force would be evenly distributed along it's lenght, and the whole accelerometer would accelerated, but not register it. That's what gravity does. It accelerates ALL parts of the accelerometer evenly, so they maintain an identical velocity relative to each other, and thus cannot sense they are being accelerated.

 

[John Freestone]

Now comes the question: in which reference frame are the laws of motion simpler?

a) the surface of the earth
b) the elevator in free fall

And that raises another question - what do we mean by simple? Your question - or your having posed it - seems to imply that the elevator in freefall is the answer, and perhaps in some strict sense that's true. I don't know. Is it simpler to consider the motion of a projectile inside a falling lift if it travels in a straight line? Maybe. It's not too much trouble to calculate the trajectory with a gravitational force involved either, even if you have to strain hard and pretend it's real.

OTOH, there is a version of "simple" that means "easy to understand" or "intuitive", and it was in that sense that I criticised the idea of a person accelerating up to meet a floating ball, or a non-accelerating lift falling in relation to an accelerating frame of reference, or however it's supposed to be expressed, which apparently you weren't arguing for anyway...or were you...or are you now...?

I suppose the question depends on what the problem is, and how much error is acceptable in the answer.

 

[RossFW]

Sorrry, Micheal_C. I did say it was delayed. Only on weekends is it possible to work on it.

And the delay in answering Reclusives question about Negative drag and Cavitation in Gasses?

 

[mender]

Thank you, spork! I didn't even have to add any extra data for you to answer (correctly of course), unlike humber who was mystified and had to guess (incorrectly of course).

I peeked at his response - apparently he doesn't understand the significance. Too bad.

 

[Clive]

Now comes the question: in which reference frame are the laws of motion simpler?

a) the surface of the earth
b) the elevator in free fall

The laws of motion that we can use are equally simple in both cases. Both of your examples are reasonably close approximations to inertial frames of reference for suitably "small scale" scenarios, and so Newton's Laws should be adequate. After all, one of the main points of using inertial reference frames is that the same (and also simplest) set of laws apply in all such frames.

 

[recursive prophet]

Things should be a simple as possible, but no simpler.

humber: Perhaps the Federal Patent Office in Berne has an opening for a third class clerk? I have faith, unlike Ross, that you will build on Lilienthal’s negative drag theory and become the ‘Batman’ of this millennium.

@Ross: Good one. Yes, I am also a reclusive prophet, and in time you’ll understand why. I’m working on it. Been busy lately triangulating humbert’s position in our universe, and I believe I now know who he is. humbert is really.….Kaiser Söze!

Anybody else having trouble navigating this site today? I keep thinking I hear Hal singing Daaaiiiissssyyyy in the background.

 

[humber]

So you still believe an accelerometer will show a value whilst in uniform acceleration due to gravity?

Yes.

Let me try and explain it this way- An accelerometer is usually accelerated by an external force that acts on the case of the instrument. The inertial member inside senses this, as it's own inertial resists the acceleration.

A mass suspended by a compliance, yes.

If you got into the inards of your accelerometer, and applied an equal force to the entire lenght of the inertial member, there would BE no strain on it as the force would be evenly distributed along it's lenght, and the whole accelerometer would accelerated, but not register it.

The mass is coupled to the body by the compliance. Accelerating the body will mean that the "inertia" of the sensing mass bends the compliance and so the strain gauge. This is not what I am suggesting.

That's what gravity does. It accelerates ALL parts of the accelerometer evenly, so they maintain an identical velocity relative to each other, and thus cannot sense they are being accelerated.

That is so for a simple mass/compliance accelerometer. In a uniform gravitational field, it may register zero. (However, it does depend upon how it is introduced into the field.)

The Earth has a gravitational gradient, so it is always possible to detect that. I know of a small, simple and commercially available force-balance device that can detect a difference in the gravitational field over a difference in height as little as 1 meter.

To argue the case for a uniform gravitational field is pointless, because then the claim becomes the obvious statement that "absolutely uniform acceleration, cannot be distinguished from absolutely uniform acceleration."
You all seem to be making a lot of fuss about a simple fact. Objects fall at the same rate, regardless of mass. (But the force upon the more massive object is greater.)

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.
A few more simple tests should allow you to conclude that you are in free-fall in an aircraft, and not in zero G. Spinning an accelerometer, will allow a pair of three-axis devices to provide conclusive proof.

As far as terminal velocity goes, it is possible to separate the forces of acceleration and drag.
http://www.sciencedirect.com/scienc...serid=10&md5=2558c0b3fe39390d1e4c9f5bff513e97

ETA:
If you turn on a tap/faucet, the water will from a stream at the outlet, but break into drops as time progresses. The water that first leaves the tap/faucet is subjected to a longer period of acceleration than the following water, and so has a higher velocity. The source is stationary, but the water moves.
In a falling elevator, you would expect the water poured from a glass not to flow, but if you move the glass upwards, what do you think will happen?

 

[semper]

Load lines for a chute or balloon or other airborne object, and the same for the driven object. A skater and parachute have been used, but you can chose another combination. You should show how you derive the terminal velocity.

So at least we can now understand what you meant when you seemed to be saying that the drag on the parachute is proportional to the velocity, or whatever you said, at the moment I don't have time to trawl back through the posts. Certainly what happened is that many of us could only interpret your statements that way, when what you were trying to describe was the velocity of a sled and parachute coupled by a rope as a function of windspeed, I think. Or the tension in the rope, or then the drag of the parachute wrt the wind under these system conditions. Anyway, so what are you trying to say by the result, and why would we need load curve analysis? I get that V_{systemwrtground}= V_wind- SQRT(KW_sled)
More or less with a few assumptions (please correct if thats wrong), so what is the point?

 

[humber]

And the delay in answering Reclusives question about Negative drag and Cavitation in Gasses?

When it suits me. I am not compelled.

 

[humber]

So at least we can now understand what you meant when you seemed to be saying that the drag on the parachute is proportional to the velocity, or whatever you said, at the moment I don't have time to trawl back through the posts. Certainly what happened is that many of us could only interpret your statements that way, when what you were trying to describe was the velocity of a sled and parachute coupled by a rope as a function of windspeed, I think. Or the tension in the rope, or then the drag of the parachute wrt the wind under these system conditions. Anyway, so what are you trying to say by the result, and why would we need load curve analysis? I get that V_{systemwrtground}= V_wind- SQRT(KW_sled)
More or less with a few assumptions (please correct if thats wrong), so what is the point?

Well, Semper, it is difficult for me to say what others have misinterpreted. I would not have expected such a fundamental idea to be questioned. That is the problem. It is not what I say, but what is not understood of first principles.

It's not the tension that is of concern, as that is the result of the force in the rope. The parachute/balloon generates a force that falls with velocity, whereas the load, the skater, has a retarding force that increases with perhaps the square of the velocity.
If you plot these on a graph with say x =velocity y =force, where they intersect, is the maximum velocity for that case.
Somewhere between zero and windspeed, these forces will balance, and that will be the maximum velocity for that case. It will always be less than windspeed. If the same is done for the power delivered and consumed, that too will give the same result.
MPT says that the maximum power that can be transmitted to the load is 50% of that available, but that does not mean that condition is always met, nor that it is necessary to achieving maximum velocity. ( A big chute and small load, for example)

 

[humber]

Thank you, spork! I didn't even have to add any extra data for you to answer (correctly of course), unlike humber who was mystified and had to guess (incorrectly of course).

I peeked at his response - apparently he doesn't understand the significance. Too bad.

Power = force * velocity
Simple drag (force) is proportional to V^2.

Power = (V^2) * V = V^3. Familiar?

 

[sol invictus]

The drag force is generally proportional to velocity squared, for the simple reason that both the number of molecules encountered and the momentum transferred to each is proportional to velocity.

Simple drag (force) is proportional to V^2.

humber learned something - duck, pigs overhead!