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

[humber]

An accelerometer will read 1g vertically upwards when stationary at the earth's surface. This is not a calibration issue: it's the way the thing works. Have a look at this instruction booklet for an accelerometer, in particular at the "Frequent Questions" section. Or the Wikipedia article. Look also at this page from Einstein-Online, which gives a good explanation of the equivalence between elevator in free fall and spaceship without power.

On of hundreds of types and accelerometer design. Your ears and stomach work also.

The question of and elevator falling down a tunnel through the earth is an interesting one. Assuming that the tunnel has no air, so that the elevator is falling in a vacuum, the elevator will indeed oscillate indefinitely back and forth along the tunnel. Although the gravitational acceleration will be constantly changing (it'll be zero at the centre of the earth), the "free fall" condition still applies. The accelerometer will always read zero, and a person inside the elevator will always be in a state of weightlessness, just like the people in the video of a zero-g flight.

For a such harmonic motion dv/dt is greatest at zero crossing, Micheal_C.
What phase relationship do you think that velocity and acceleration must have to oscillate indefinitely?

 

[John Freestone]

Acceleration is not relative.

Thank you sol, I had a little niggly thing knawing at the back of my brain, and you just scratched it! It might take me a while to work out exactly why that makes so much sense, but it does.

An accelerometer will read 1g vertically upwards when stationary at the earth's surface. This is not a calibration issue: it's the way the thing works. Have a look at this instruction booklet for an accelerometer, in particular at the "Frequent Questions" section. Or the Wikipedia article. Look also at this page from Einstein-Online, which gives a good explanation of the equivalence between elevator in free fall and spaceship without power.

I wonder if we're at cross purposes. Maybe I'm using something called kinematic acceleration, and you are using some measure that includes non-accelerating (kinematically) bodies in a gravitational field as having positive acceleration. Is that it? I just read some of that first reference and found:

Here we are being very careful to not call something an acceleration when it is not a kinematic acceleration. For example, an “acceleration” of 9.8 m/s2 for an object that remains at rest is clearly a problematic interpretation, yet that’s what the accelerometer reads.

You can correct the Accelerometer reading to get a true acceleration by adding the component of the gravitational acceleration field along the direction of the sensor arrow. For example, if the axis of the accelerometer is pointing upward, then the gravitational component is –9.8 m/s2. The Accelerometer reads 9.8 m/s2 when the arrow is upward and the device is at rest. By adding –9.8 m/s2, we get zero, which is the correct acceleration.

​

You can correct the accelerometer reading to get a true acceleration by adding -9.8 m/s/s. That's what I'm saying. In level flight your accelerometer gives a 1g reading, I thought, but you're not accelerating vertically. In a falling lift, you are accelerating, at 9.8 m/s/s...which is 1g. in the sense I mean. Accelerometers are calibrated to read 0 in freefall, not at zero acceleration.

Also, see http://www.lunar.org/docs/LUNARclips/v5/v5n1/Accelerometers.html​

The question of and elevator falling down a tunnel through the earth is an interesting one. Assuming that the tunnel has no air, so that the elevator is falling in a vacuum, the elevator will indeed oscillate indefinitely back and forth along the tunnel. Although the gravitational acceleration will be constantly changing (it'll be zero at the centre of the earth), the "free fall" condition still applies. The accelerometer will always read zero, and a person inside the elevator will always be in a state of weightlessness, just like the people in the video of a zero-g flight.

Weightlessness is not the same as zero acceleration, though, is it? You are weightless at the centre of the earth, with zero acceleration[ETA:see next post]. All the rest of the time you're weightless with acceleration. In a vomit comet, you're weightless during the fall because weight is that normal force against the force of gravity (or vice versa, if you like). Give into it and accelerate at 9.8 m/s/s and you don't have any "weight". In orbit, you're weightless, but your velocity is constantly changing, as your direction changes (and speed in an eliptical orbit). In a zero gravity field in deep space you're weightless with no acceleration because there's no gravity. If there's gravity, you have to fall to be weightless.​

Again, all this seems to be backed up by some sites, but maybe in a more relativistic view there's a way that your position is right (I haven't looked at the Einstein-Online page, but it didn't seem to link).​

School-level, admittedly, but http://www.physicsclassroom.com/Class/1Dkin/u1l5b.cfm​

Do you consider yourself to be accelerating when you're on the ground and your accelerometer measures 1g?​

How do you go from being in a supported elevator to one falling increasingly fast, and not be accelerating? How does it slow down, stop and come back again without accelerating?​

How do you go in orbit and not be accelerating?​

​

 

[John Freestone]

Thanks humber for that correction - your acceleration in the bottomless lift shaft is greatest as you pass the centre of the earth, not zero, I think. Newton's Law of Universal Gravitation:

F = G m₁ m₂ / d²
d is the distance between the two masses - the lift and the earth's c of g. F is the force on each. m 1&2 the masses. and acceleration will be proportional to the force:

a = F/m

Not so?

 

[mender]

John, before accepting humber's "correction" you may want to consider again what the acceleration would be if you were stationary at the exact centre of the earth. Also, consider several positions as you are approaching the centre of the earth.

The key is that you are now inside the mass rather than approaching it.

 

[mender]

For a such harmonic motion dv/dt is greatest at zero crossing, Micheal_C.

And this one, Dan. Not that it has any bearing on the problem at hand, but another example of humber "correcting" a correct statement.

 

[sol invictus]

Weightlessness is not the same as zero acceleration, though, is it?

You're teetering on the edge of something profound, called the equivalence principle. Acceleration and gravitation are very closely related - to the point that unless you can do an experiment sensitive enough to detect the gradients in the gravitational field they cannot be distinguished.

So it's up to you whether you want to consider as "real" the acceleration due to gravity an accelerometer on earth will measure, but if you don't, you're drawing a distinction that does not really exist.

 

[Subduction Zone]

In the universe you are always accelerating towards something. Wegiht is the sensation that you get when a surface opposes acceleration. An accelerometer measures the acceleration caused by the body being accelerated itself. In other words if it was a car you would first zero out the force of gravity and then measure any forces created by the car. Of course humber will claim that he can tell the difference between "real" weight and artificially generated weight.

 

[mender]

Equation 3, I think. (There are other approximations.)
http://www.cee1.org/ind/mot-rep/mot-rep-fanlaws.pdfi]k

Thank you, humber, equation 3 shows that the power absorbed by the fan is the difference in rpm cubed, which basically restates what I stated earlier.

The fan laws are based on the fact when something moves twice as fast through air, it encounters twice as much air and that air is moving twice as fast relative to whatever is moving through the air. That means that the fan must overcome four times as much KE to move the air. Taken over a specific time to show power, that increases to a cube function.

Are you now saying that air is somehow exempt from the normal rate of KE change with speed (velocity squared) when encountering a parachute instead of a fan?

 

[John Freestone]

John, before accepting humber's "correction" you may want to consider again what the acceleration would be if you were stationary at the exact centre of the earth. Also, consider several positions as you are approaching the centre of the earth.

Zero. F=m1*m2/d*d, where d=0

...but actually, you are right also that this formula does not apply within the Earth. Assuming it were spherical and of the same density, the gravitational force would vary linearly, becoming zero at the centre. Sorry, humber, I have to defect again! On the other hand, this was entirely a footnote of mine, and for an elevator at great height, the formula above would seem to be correct.

You seem to be ignoring the main issue, so maybe I should just agree to differ and move on? [ETA:You said that of course if there is no drag you feel no acceleration forces. What happened to the acceleration due to gravity? Are you accelerating when you're stationary in a gravitational field, is that what this is about? Why ignore my questions and focus on the footnote about travelling through the earth?]

The key is that you are now inside the mass rather than approaching it.

I beg your pardon, that seems to be so. However, outside the sphere of the earth:

http://en.wikipedia.org/wiki/Kepler_orbit

By symmetry, the net gravitational force attracting a mass point towards a homogeneous sphere must be directed towards the centre of the sphere. The shell theorem (also proven by Isaac Newton) says that the magnitude of this force is the same as if all mass was concentrated in the middle of the sphere, even if the density of the sphere varies with depth (as it does for most celestial bodies). From this immediately follows that the attraction between two homogeneous spheres is as if both had its mass concentrated to its center.

The shell theorem linked to in that passage gives the linear reduction of the force towards the centre.

I guess maybe your conception of acceleration is including the version sol commented on, where the 1g at the surface is real acceleration. And that must be upward, yeah, so that as we "accelerate" (as we used to call it) down towards the surface of the earth, that negates the "not unreal" acceleration we had upwards away from the mass of the Earth.

Isn't this getting a little far from classical mechanics guys? Why was Newton right every other time there's something to decide, and now he's wrong? What is this, Einstein? Doesn't classical mechanics include the old version of space as 3 non-warping dimensions and another of time?

 

[John Freestone]

You're teetering on the edge of something profound, called the equivalence principle. Acceleration and gravitation are very closely related - to the point that unless you can do an experiment sensitive enough to detect the gradients in the gravitational field they cannot be distinguished.

So it's up to you whether you want to consider as "real" the acceleration due to gravity an accelerometer on earth will measure, but if you don't, you're drawing a distinction that does not really exist.

Ah yes, having gone to wikipedia on that, I see that my answer is right according to Newtonian mechanics, but not General Relativity, it seems.

 

[humber]

Thank you, humber, equation 3 shows that the power absorbed by the fan is the difference in rpm cubed, which basically restates what I stated earlier.

That does not apply to a parachute, Mender. If you recall, I told you that is was a fan law. Also, equation 3 that is the for power and not force.

The fan laws are based on the fact when something moves twice as fast through air, it encounters twice as much air and that air is moving twice as fast relative to whatever is moving through the air. That means that the fan must overcome four times as much KE to move the air. Taken over a specific time to show power, that increases to a cube function.

Nice try.

Are you now saying that air is somehow exempt from the normal rate of KE change with speed (velocity squared) when encountering a parachute instead of a fan?

A fan is stationary, so the velocity of the wind is constant. This is not so of a moving fan or other object, in which the available force falls with velocity.

 

[John Freestone]

Zero. F=m1*m2/d*d, where d=0

Whoops. Spot the 'deliberate' mistake!

 

[mender]

That does not apply to a parachute, Mender. If you recall, I told you that is was a fan law. Also, equation 3 that is the for power and not force.

Maybe you could go back and reread what I said in response to what you said so you can respond to what was actually said and not what you think was said.

Nice try.

Ah. I see. Well reasoned responses as always, humber.

A fan is stationary, so the velocity of the wind is constant. This is not so of a moving fan or other object, in which the available force falls with velocity.

You're so busy trying to deny the cart and/or the treadmill that you end up making some pretty wild claims. So the parachute (other object) experiences a loss of force with velocity?

Perhaps this link will help you - if you read it:

http://www.aeroconsystems.com/chutes/drag_calculator.htm

And here's one that saves me the trouble of doing my own real world testing:

http://www.nakka-rocketry.net/paratest.html

 

[humber]

Maybe you could go back and reread what I said in response to what you said so you can respond to what was actually said and not what you think was said.



Ah. I see. Well reasoned responses as always, humber.



You're so busy trying to deny the cart and/or the treadmill that you end up making some pretty wild claims. So the parachute (other object) experiences a loss of force with velocity?

Perhaps this link will help you - if you read it:

http://www.aeroconsystems.com/chutes/drag_calculator.htm

And here's one that saves me the trouble of doing my own real world testing:

http://www.nakka-rocketry.net/paratest.html

The link to the parachute is for the drag of a chute at a forced velocity. This is not the same as force developed by a chute in wind.

ETA: It's the general equation for drag, (D = 1/2 * p * V^2 * Cd ) proportional to the square of the velocity, which is assumed for the load of the skater.

 

[mender]

The link to the parachute is for the drag of a chute at a forced velocity. This is not the same as force developed by a chute in wind.

So a chute moving through air is not the same as a chute moving in wind. Interesting.

ETA: It's the general equation for drag, (D = 1/2 * p * V^2 * Cd ) proportional to the square of the velocity, which is assumed for the load of the skater.

Or did you change your mind? Are you agreeing that the general equation for drag applies to a parachute as it is moving through air or in wind?

 

[Michael C]

Accelerometers are calibrated to read 0 in freefall, not at zero acceleration.

Once again, this isn't a question of calibration: an accelerometer measures the force due to acceleration. This is an absolute, not a relative value (Humber has spread some confusion with the idea of "absolute accelerometers": in fact all accelerometers produce absolute measurements). The accelerometer cannot tell if the force is due to gravitational pull, rocket engine or some other source. An accelerometer will read zero when it experiences zero acceleration.

As Sol said, acceleration and gravity are equivalent: just by looking at the accelerometer, you can't tell if you're standing in a room at the surface of the earth or standing in a room that is being accelerated through space with a rocket engine.

If you want to make a simple accelerometer, just attach a known mass to the end of a spring scale. If the mass is 1 kg and the scale reads a weight of 1 kg (we need to be careful with "mass"' and 'weight", since we use the same units for both), the accelerometer is experiencing an acceleration of 1 g. If the scale is in a free fall situation (for instance if you hold it while jumping off a diving board) it will read zero weight, corresponding to zero acceleration.

How do you go from being in a supported elevator to one falling increasingly fast, and not be accelerating? How does it slow down, stop and come back again without accelerating?

Here again the equivalence of gravitational force and force due to acceleration comes into play. When the elevator is stationary with respect to the earth, it has to be supported by something that exerts an upward force on it exactly equivalent to its weight, to counteract the acceleration due to gravity. The accelerometer in the elevator reads 1 g in an upwards direction. As soon as this force is removed, the elevator is in a free fall situation. We see it start to accelerate at 1 g towards the centre of the earth: now the accelerometer inside reads zero.

For us surface-dwellers, who are used to the acceleration of 1 g we always feel through our feet, the elevator was stationary and is now accelerating. For the accelerometer, which cannot differentiate between gravitational acceleration and other acceleration, it was accelerating but is now no longer doing so.

As the elevator falls through the earth, here's what we on earth see: its speed increases but its acceleration decreases. When it reaches the centre of the earth, its speed is at a maximum but its acceleration is zero: at the centre of the earth gravitational force is zero. As soon as it goes past the centre of the earth, it starts being accelerated in the opposite direction, which causes it to slow down: it will continue to slow down (while its acceleration, in the opposite direction to that in which it is travelling, increases!), until it reaches the other end of the tunnel, at which point its acceleration is once more at 1 g but its speed is zero. Now its starts going back the other way.

The weird thing is that we see all these changes in acceleration, but the accelerometer inside the elevator always shows a value of zero. If you're using Newtonian mechanics to analyse this, you might say that at any point along the voyage, the "actual" acceleration of the elevator is producing a force in a direction away from the centre of the earth, that exactly counteracts the force of gravity towards the centre. This isn't surprising, since the "actual" acceleration is being directly produced by the force of gravity experienced at that point. If you're using general relativity to analyse the movement of the elevator, you'll say that it is following the shortest path in a region of space-time that is curved due to the mass of the earth.

 

[spork]

To a first approximation, and over most of its range, the force generated by a parachute is proportional to the difference between its velocity and that of the wind.

You clearly make things up just for effect. I don't think humb would even say this to make fun of you.

Recursive Prophet asked that also. I will get back to it.

No you won't.

Thanks humber for that correction - your acceleration in the bottomless lift shaft is greatest as you pass the centre of the earth, not zero, I think...

John, if we model the Earth as a true sphere we can think of it as an infinite number of infinitely thin shells. When we are outside the Earth we can model the whole thing as being the total mass of the shells at the very center. When we are inside the Earth something whacky happens... When you are inside a homogeneous spherical shell you feel no gravitational pull from that shell - no matter where in the shell you are. So, whatever your depth beneath the ground, you have to ignore the contribution of all of the shells whose radius is greater than your distance from the center of the Earth. As you get closer to the center there is therefore less Earth mass effectively acting on you. And this reduction in mass happens more rapidly (cubed law) than the increase in gravity due to your closer proximity to the center (squared law).

Also, what Sol said about gravity and acceleration is on the money (of course). Basically, you can consider yourself to be in an inertial frame if:

A) You're free-falling in a gravitational field (and therefore accelerating!) - in which case you will ignore the effect of gravity entirely.

OR

B) if you are not accelerating - and then you must consider the force of gravity as an external force.

Gravity and acceleration are wierd that way. Of course as Sol points out, the only distinction that can be made between gravity and linear acceleration are the gradient effects (lines of gravity are not parallel, but converge toward a point - and gravity is stronger toward the source).

 

[humb]

To a first approximation, and over most of its range, the force generated by a parachute is proportional to the difference between its velocity and that of the wind.

You clearly make things up just for effect. I don't think humb would even say this to make fun of you.

Damn straight. Of course I would not say that -- or anything -- to make fun of humber! He's the only voice of reason around here, and you can take that to the bank.

Calculating the force from a parachute is a complex matter, and that's why the ones used for rockets are designed by rocket scientists. Determining the difference is not a simple matter, because you need to know the real velocity of the parachute, and the real wind, and these may not be in the same direction. The force is entirely load-dependent as well. Furthermore, there is a small glitch in the middle of the range, when the difference is equal to one half of the average between the two and the other, during which the force is directly proportional to the reciprocal of the inverse difference. This is why the balloon never reaches windspeed, because it can be viewed as two parachutes sewn together, and the drag on one can never equal the force on the other, because of inefficiencies and real-world imperfections. Besides, if as you assert the drag were as the square, then when it opposed the inverse square of gravity, they would cancel out, produce a constant force, and the skydiver would continue to accelerate and never reach terminal velocity. Which would certainly be terminal! Don't even ask me. It's redundant.

 

[spork]

Calculating the force from a parachute is a complex matter, and that's why the ones used for rockets are designed by rocket scientists....

Very interesting. I learned a great deal from that post. I am a rocket scientist as you may know, but I think I was absent the day they covered a lot of the stuff you've covered here - which of course makes perfect sense once you stop and think about it.

Incidentally, I have to admit I got slightly lost at one point. When calculating the inverse reciprocal, do you have to invert the matrix in the complex plane to compute the eigen-vector, or can you simply translate into frequency-space using Laplace and thus linearize the system?

 

[recursive prophet]

You have come up with a lot of great posts here humb, but I would vote this one as your very best. What clarity! You do humber proud!