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

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That's not my claim. It's Einstein's principle of equivalence. The uncertainty can be made arbitrarily small. Since you haven't got as far as understanding Newton, I don't have any hope of you understanding general relativity, so I'll leave it there.

Those experiments are Einstein's way of letting us know what he thinks. They are not 'real'
At first glance, there is at least one mistake in your citation. Astronauts upon the ISS are not "weightless" but in microgravity. The hull of the ISS, has its own gravity, too.

Amongst other reasons, such as determinism...
http://en.wikipedia.org/wiki/Principle_of_locality
 
We are keeping a list of where humber is incorrect regarding classical physics and striking out each item as humber learns it.


Where humber is wrong

  1. "You need constant force and energy to maintain riverspeed." >2

  2. Zen: "When there is no wind, the belt has no meaning." #1089

  3. A rare moment of doubt: "The problem of the orange bothers me. When I move it either with the belt, or against the belt, it seems to gain KE w.r.t. the belt and ground by equal amounts." #1105

  4. Struggling to understand: "The cart's frame is tenuous, because it would seem that it is both in the belt and windspeed frames." #1214

  5. Assertion = Evidence: "Windspeed is the same "frame" as being still on the ground in still air. This too I have claimed. This is evidence." #2738[/QUOTE]

  6. "The boat will reach a finite speed, lower than the water. This speed is the solution to the simultaneous equations of the forces driving the boat, and that of drag. The air will reduce that speed, but even in a vacuum, waterspeed will not be reached." #136

  7. "Motion relative to the supermarket belt is also relative to the ground. Both directions yield the same KE, if viewed from belt or ground." #3053

  8. "The force is approximately linearly proportional to the relative velocity of chute and wind." #2951

  9. "For a such harmonic motion dv/dt is greatest at zero crossing." #



Some more of humbers wrong doings

  • "Again, sorry. I am working on it!" #27

  • "Balloons move relative to the ground, unless they in the state we Groundians call "stationary"." #3066



Nominations (waiting for a second)
  • "The KE of two bodies may remain the same (relative KE = 0) but gain 1000 fold wrt another." #?

  • "A treadmill belt can't be a "frame of reference." #18

  • "Objects that have the same "velocity" have the same "frame". #26

  • "The KE goes with the moving body." #77

  • "A balloon can raise its altitude, and gain potential energy. When descending, that energy is converted to lateral velocity. " #147 #447
 
That is the static force in both cases. If you simplfy the situation as I think you are suggesting, then the answer is 200lbs


How could it fall, Mender? With 100lb's each way?
No. The parachute is moving, so the static force is no longer applicable. The drag, and therefore the velocity, can be calculated using the formula you know, but with gravity as the input force.
There is no 'drive' from the chute, only drag that opposes motion due to gravity.

Wow. I thought I made it simple enough that you wouldn't misunderstand. I don't want you to respond as you think I am suggesting, because that doesn't tell me what you are suggesting. Just answer the questions honestly according to what you think is the correct answer.

1) Same question as before, no changes. What do you think the correct answer is?

2) I'll restate the question. A parachutist weighing 100 lbs jumps out of a plane that is 3000 feet above the ground. He/she deploys the parachute and a steady state speed of 30 ft/s falling towards the ground is achieved a short time later. The next person to use that parachute weighs 200 lbs. He/she jumps out of the plane at 3000 feet and deploys the parachute. What is the steady state falling speed for the 200 lb parachutist? What do you think the correct answer is?
 
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We are keeping a list of where humber is incorrect regarding classical physics and striking out each item as humber learns it.


Where humber is wrong

  1. "You need constant force and energy to maintain riverspeed." >2

  2. Zen: "When there is no wind, the belt has no meaning." #1089

  3. A rare moment of doubt: "The problem of the orange bothers me. When I move it either with the belt, or against the belt, it seems to gain KE w.r.t. the belt and ground by equal amounts." #1105

  4. Struggling to understand: "The cart's frame is tenuous, because it would seem that it is both in the belt and windspeed frames." #1214

  5. Assertion = Evidence: "Windspeed is the same "frame" as being still on the ground in still air. This too I have claimed. This is evidence." #2738

  6. "The boat will reach a finite speed, lower than the water. This speed is the solution to the simultaneous equations of the forces driving the boat, and that of drag. The air will reduce that speed, but even in a vacuum, waterspeed will not be reached." #136

  7. "Motion relative to the supermarket belt is also relative to the ground. Both directions yield the same KE, if viewed from belt or ground." #3053

  8. "The force is approximately linearly proportional to the relative velocity of chute and wind." #2951

  9. "For a such harmonic motion dv/dt is greatest at zero crossing." #

All answered by me. But not by you. I await you rebuttals.

Some more of humbers wrong doings
  • "Again, sorry. I am working on it!" #27

  • He's a wrong 'un, that humber. Perhaps he is a busy cat, that also bides his time in pursuit.

    Not are like a cat, on a couch, licking itself.

    "Balloons move relative to the ground, unless they in the state we Groundians call "stationary".

    I do move about a lot. 'Where ever I lay my hat, that's my home', perhaps.
    Before becoming a Groundian, I was a Bonovian, referring my location to his hat. That did not work well.

    Nominations (waiting for a second)

    Pausing between licks, to rest a weary tongue.

    "The KE of two bodies may remain the same (relative KE = 0) but gain 1000 fold wrt another."
    Yes. It does seem an odd idea, but a consequence of relative KE. If would seem that a third object couls gain KE, without influencing the other two.

    "A treadmill belt can't be a "frame of reference."

    It can be a "frame of reference", you should they feel the need, for someone moving on the belt, but that is all. A Groundian can equally argue, that same person is moving w.r.t the ground, and the belt is but the means of propulsion.

    "Objects that have the same "velocity" have the same "frame". "

    Yes. All objects are in the same frame, but moving at different velocities. I believe that point has been made.

    "The KE goes with the moving body."

    Yes. It has been noted that moving objects hurt. However, I will definitely change my mind if you can demonstrate how the KE of a cart said to be traveling as 10m/s, can be transferred to the belt. It is not seen from ground, so it must be there.

    "A balloon can raise its altitude, and gain potential energy. When descending, that energy is converted to lateral velocity. "

    Yes. Even changing the attitude of the balloon can redirect some of that energy in a lateral direction. The lateral speed may approach that of the vertical speed, due to falling within a gravitational field.
 
That is the static force in both cases. If you simplfy the situation as I think you are suggesting, then the answer is 200lbs

WOW!!!! humber answered a question! I don't think I've seen that before. Perhaps I need to slow my fast scroll just a tad. Of course he got it horribly wrong - but answering a question is one small step in the right direction.

How could it fall, Mender? With 100lb's each way?

Balanced forces don't preclude velocity - they preclude acceleration. With 100 lbs each way the parachute will descend at it's terminal velocity.

Does this need to be added to Dan's list?
 
OK, that's enough playing with Humber for me for today.

John, I started composing this reply this afternoon. I'm trying to find ways to make things clearer, for myself as well! I have some sort of mad delusion that if I really understand the concepts of general relativity, I will one day be able to explain them using toy animals on a tabletop. I didn't think my little question about accelerometer readings would get us into such murky waters!

It began because Michael C made a very clear and "obvious" statement that of course an accelerometer in a falling lift would read zero, because it is not accelerating.

Well, that's not exactly what I said. Maybe I wasn't clear enough: what I actually said was "in free fall with no air drag you will experience no acceleration forces". The problem is here: in the Newtonian world if you are in free fall you will indeed be accelerating, but you won't feel any force. None of your internal organs will be telling you "you're accelerating". You'll only feel the force of gravity when you are being held up by the ground, or the floor, or whatever. I agree that all this can get very confusing, and we can get muddled up in the terminology. I prefer, as you do, to think of the immediate world around me in a "Newtonian" manner, but it's worth remembering from time to time that we see and describe things in a certain way only because we live on the surface of an almost spherical planet with an almost constant gravitational field. We'd describe things differently if we lived inside a hollow planet (no gravity), or in some place where the gravitational field varied greatly over short distances.

However, I feel that from the classical perspective, as wikipedia agrees, there is no acceleration for a body stationary on the earth, but when it falls due to gravity it accelerates with g, which is even called "the acceleration due to gravity". From a classical perspective and a chosen (Earth) frame, the elevator can be seen and measured moving faster over time by that amount.

Agreed. I, as an Earth surface dweller, consider that an object accelerates towards the ground when I drop it. However, the changing acceleration of the elevator dropping through the tunnel in the earth can only be seen and measured from outside it, from the perspective of somebody on Earth. The accelerometer inside cannot measure the changing acceleration.

Yet the link to instructions on using one discusses adding -9.8 m/s/s to get the "true acceleration".

That's a confusing term. In fact they are talking about the acceleration with respect to the frame of reference of the Earth. If we were using the accelerometer on the Moon, we'd need to add -1.6 m/ s2 to get the "true" acceleration at the moons' surface. If we were using it on Jupiter, we'd need to add -25.9 m/s2. We need to make these different adjustments because the frame of reference of the surface of a planet is not an inertial frame.

But wait a minute: didn't we say that a reasonably small section of the surface of the earth could be considered to be an inertial frame? Well yes, we did: it was in fact a cheat, a crafty manoeuvre to serve our purposes (I wonder what Humber will say to that...). We can only consider the place where we're standing on Earth to define an inertial frame if we include gravity as a "fictitious force". Oh dear! I'm not going to go into a big explanation of fictitious forces here: click on the link if you wish. In fact the surface of the earth qualifies as a constantly accelerating reference frame, whereas an elevator in free fall, whether it is near the surface of the earth, in orbit around the earth, in a tunnel through the earth or somewhere in deep space, defines an inertial frame of reference: throw a ball in it and it will continue in the same direction at the same speed until it hits something (Newton's 1st law). If the elevators are identical and the ball is thrown in the same way in each elevator, its trajectory will be identical in each elevator. Throw a ball in an elevator parked at the surface of some planet and it will describe a particular parabola. Depending on the gravitational field of the planet in question, an identical ball thrown with the same direction and force will not describe the same parabola: you have a way of distinguishing between different non-inertial frames. There is no way to distinguish between one inertial frame and another (principle of Relativity).

So the straight dope is here: when we're talking about the "frame of the ground", the "frame of the treadmill" or the "frame of the wind" we are not talking about real inertial frames in the strict sense. We can say that they are inertial frames that all have the same fictitious force of 1 g, or we can say that they are non-inertial frames that are all accelerating upwards at the same rate. For the Newtonian discussion, it's best to stick with the idea that they are inertial frames with an identical fictitious force in each frame. Since the force is the same in each frame, we can consider all these frames to be equivalent just as we can consider all real inertial frames to be equivalent.
 
Wow. I thought I made it simple enough that you wouldn't misunderstand. I don't want you to respond as you think I am suggesting, because that doesn't tell me what you are suggesting. Just answer the questions honestly according to what you think is the correct answer.

1) Same question as before, no changes. What do you think the correct answer is?
For that you need specific data. It seems that is not easy find. There is such information available balloons, and the support of the meteorological book, so in that case, the difference may be calculated, otherwise the "correct" answer would be the on that is assumed by you. No?

2) I'll restate the question. A parachutist weighing 100 lbs jumps out of a plane that is 3000 feet above the ground. He/she deploys the parachute and a steady state speed of 30 ft/s falling towards the ground is achieved a short time later. The next person to use that parachute weighs 200 lbs. He/she jumps out of the plane at 3000 feet and deploys the parachute. What is the steady state falling speed for the 200 lb parachutist? What do you think the correct answer is?

It depends on the design of the parachute. What if the parachute were enormous? The increase of 100lbs would make little difference. A parachute that is 'just so' for 100lbs may simply collapse under 200lbs.

There is simply the general case, Mender. Terminal velocity will be achieved when drag and drive are in balance. When the force of gravity is no longer adequate to accelerate the object against the forces of drag against it.
 
For that you need specific data...

Well it was a brief but shining moment in the history of this forum. Having gotten these two questions miserably wrong, we're back to the total non-answer approach.
 
No problem to use nice looking math notation on the forum if you want.

[latex]
\rho \ \Delta \ \int_{-\infty}^\alpa f(\theta) d\theta \ \sum_{k=1}^N \ , \forall \lambda \leq 0
[/latex]

[latex]
\begin{align}humber \triangleq \begin{bmatrix} \delta & \psi \\ I & \mathcal{H}_2 \end{bmatrix}
\label{m1}
\end{align}
[/latex]

[latex]
\begin{tabular}{|c|c|}
\hline
$\Lambda$ & $\xi$ \\
\hline
$1234$ & $-0.1$\\
\hline
\end{tabular}
[/latex]
 
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WOW!!!! humber answered a question! I don't think I've seen that before. Perhaps I need to slow my fast scroll just a tad.
Don't stumble over any rocks.

Of course he got it horribly wrong - but answering a question is one small step in the right direction.
Not like your elevator fiasco?

Balanced forces don't preclude velocity - they preclude acceleration. With 100 lbs each way the parachute will descend at it's terminal velocity.

Things fall under gravity, not only because gravity is kinda weird, but because there is a force applied to them.
However, during that process and as velocity increases, drag is created, and that opposes the gravitational force. That opposing force is not independent of gravity, but a product of the motion produced by gravity. Terminal velocity is reached when the force of gravity is insufficient to further accelerate the object against the forces of drag.

Does this need to be added to Dan's list?

No, I wouldn't. Dan_O already has a fur ball.
 
No problem to use nice looking math notation on the forum if you want.

[latex]
\rho \ \Delta \ \int_{-\infty}^\alpa f(\theta) d\theta \ \sum_{k=1}^N \ , \forall \lambda \leq 0
[/latex]

[latex]
\begin{align}humber \triangleq \begin{bmatrix} \delta & \psi \\ I & \mathcal{H}_2 \end{bmatrix}
\label{m1}
\end{align}
[/latex]

[latex]
\begin{tabular}{|c|c|}
\hline
$\Lambda$ & $\xi$ \\
\hline
$1234$ & $-0.1$\\
\hline
\end{tabular}
[/latex]

I am sure you understand no more than 'nice'
 
How could it fall, Mender? With 100lb's each way?

Then...

Things fall under gravity, not only because gravity is kinda weird, but because there is a force applied to them.
However, during that process and as velocity increases, drag is created, and that opposes the gravitational force. That opposing force is not independent of gravity, but a product of the motion produced by gravity. Terminal velocity is reached when the force of gravity is insufficient to further accelerate the object against the forces of drag.

So how about that! You got it both ways, but you weren't wrong either time. The parachute will both fall and not fall when forces are balanced.

Good for you.


So we're up to 80 pages almost entirely dedicated to humber's extreme wrongness. Can we make it to 100?
 
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So how about that! You got it both ways, but you weren't wrong either time. The parachute will both fall and not fall when forces are balanced.

Good for you.

So we're up to 80 pages almost entirely dedicated to humber's extreme wrongness. Can we make it to 100?

A Shrine of Extreme Wrongitude. However, you cheekily failed to mention that you were wrong.

I was trying pointing out to Mender, the difference between the static (tethered) force, and that generated by motion. If that distinction is not made, you get the paradoxical situation of a chute that cannot fall.
 
Balloons pass overhead quite often. This one was at night, and surprisingly fast. I see them, but I do not assume that the operators of a commercial enterprise are physicists, let alone knowledgeable.
It this why you seem to take your instruments at face value?

Many devices can gain lateral velocity by using energy gained from the gravitational field.
I submit for your approval; The Roller Coaster.
Balloons can do the same. If not why not?
You need more than money to deny that.

Right,

So if you asked a licensed, commercial Balloon pilot something like "How is it propelled laterally" and he said "It simply goes at wind speed", you'd chucle in your beard and think "Well I know more about balloons than HIM!!".

It really does speak volumes.

By the way, I agree it would be theoretically possible to have some form of angled plane on a balloon, and by using the relative airflow in the vertical as the balloon ascended and decsended, have some very limited form of propultion.

But the balloon that flew over your house didn't, and I've never seen one that does.
 
I was trying pointing out to Mender, the difference between the static (tethered) force, and that generated by motion...


That might be a swell explanation if there were such a difference - but there's not.

RossFW said:
But the balloon that flew over your house didn't, and I've never seen one that does.

Well, that's hard to say. I suspect no one else could see the balloon humber "saw" (like a lot of things humber sees I suspect).
 
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