Test to prove Earths spins:

Originally posted by joe shmo
There is no way this answer you provided is correct.

"[b]1) Ω=0 ...The Earth is not rotating (relative to an object at the equator)."

The Earth having a rotation is independent of where any body is placed on it.

The Coriolis Force doesn't disapear. It just becomes a radial force at the equator.

Lets align the standard coordinate system so ...[text shortened]... ce goes to zero please elaborate on what I'm donig incorrectly...I honestly can't figure it out?[/b]

That is because you are using the rotation vector in the inertial frame for a calculation in the rotating frame. In a local frame at the equator with a vector i pointing eastwards, j pointing north and k pointing straight upwards then at the equator Ω = Ωj. Assuming no upwards movement our velocity vector has components (in the rotating frame) v = v_xi + v_yj. Now the cross product comes to Ωxv = Ωv_xjxi + Ωv_yjxj = -Ωv_xk and the acceleration is directly upwards. So in the rotating frame in the restriction to the two dimensional plane tangent to the Earth's surface the Coriolis force is zero at the equator. In the full three dimensional setting there is a component upwards, but that's irrelevant to meteorology.

What one could do is measure the period of a pendulum on a train going round the equator at high(ish) speed. The Coriolis acceleration upwards is 4πv/T, where v is the speed of the train. There are 86400 seconds in a day which gives us an angular speed of Ω = 2π/T = 7.3e-5 radians per second. A realistic speed for a train is 125 mph or 200 kmh which comes to 55m/s. So the acceleration is 0.008 m/s/s for an effect of a little under 1 part in a thousand. This is a fairly small effect, but I don't see any reason that the experiment wouldn't work. For a long enough railway line at the equator one could just use a 25 cm pendulum which has a period of 1 second and run the experiment for an hour or so. The pendulum should swing around seven more times going west than it will going east. I think this really would demonstrate rotation of the Earth as it seems less vulnerable to local variations in the gravitational field and doesn't require one to work out the difference between downwards and the direction to the centre of the Earth.

Originally posted by sonhouse
With GPS systems I imagine the bell jar could be made to be 100% vertical I would think anyway, then the bob would either be dead center as you all say or it would show some deflection away from dead center. I think it would deflect because the bob is on a long string and would not feel the same circular motion the same extent as ground motion.

I don't ...[text shortened]... ction of motion, or at least be able to be shown a force against the magnetic confinement field?

The problems with GPS for this are that first it probably isn't precise enough, even if one could establish one's position to the nearest millimetre over the length of a metre long pendulum that is an error of 1 part in 1000 which is around the size of the effect we are looking at. The other problem is that GPS calculations take general relativity into account and the state of motion of the atomic clocks is relevant to the calculation, so you are establishing what the direction to the centre of the earth is using a method that assumes that the earth is rotating. Since the objective is to show that the Earth is rotating your method is in danger of begging the question.

Originally posted by DeepThought
The problems with GPS for this are that first it probably isn't precise enough, even if one could establish one's position to the nearest millimetre over the length of a metre long pendulum that is an error of 1 part in 1000 which is around the size of the effect we are looking at. The other problem is that GPS calculations take general relativity into ...[text shortened]... bjective is to show that the Earth is rotating your method is in danger of begging the question.

Well then, couldn't you do it geographically by having equal length ropes from the peak to ground and build a teepee like structure say 4 or more ropes each one touching the outer circle around the bell jar such that the top and bottom of the bell jar are shown to be vertical?
So the top and the bottom would be shown to be at the center of the larger circles. With the whole assembly on a flat plain like salt flats or some such.

It would seem to me if you had one at a pole, north or south and compared that to one on the equator, there should be no deflection at the pole and some at the equator.

That would be a bit more organic than using technology already depending on spinning Earth.

There are extensions of GPS that allows geologists to establish Earth movement on the order of centimeters accurately though.

Originally posted by sonhouse
Well then, couldn't you do it geographically by having equal length ropes from the peak to ground and build a teepee like structure say 4 or more ropes each one touching the outer circle around the bell jar such that the top and bottom of the bell jar are shown to be vertical?
So the top and the bottom would be shown to be at the center of the larger circ ...[text shortened]... hat allows geologists to establish Earth movement on the order of centimeters accurately though.

Flat is not the same as level, you need to show that the plain you are setting your pendulum up on is level to a precision of 1 part in 1,000. There is no effect at the equator that will cause the bob, assuming it is stationary with respect to the Earth's surface, to be hanging in any direction but vertically down. For that to happen you need the rate of rotation of the Earth to be changing. You can measure the strength of gravity locally by setting the bob swinging, but you haven't got a way of filtering out local gravitational variation from the rotation of the Earth when you try to compare it with the theoretical value. This isn't a problem for my 'pendulum on a train' set up because you can compare the number of swings of the pendulum for travel east and west.

I wonder if there are any gyroscopic systems sensitive enough to measure the earths rotation.

Originally posted by DeepThought
Flat is not the same as level, you need to show that the plain you are setting your pendulum up on is level to a precision of 1 part in 1,000. There is no effect at the equator that will cause the bob, assuming it is stationary with respect to the Earth's surface, to be hanging in any direction but vertically down. For that to happen you need th ...[text shortened]... n' set up because you can compare the number of swings of the pendulum for travel east and west.

But if you look at my other thought experiment, a horizontal wheel spinning at some rate and a balanced affair with a bell jar on the perifery spinning with the wheel and the bob held in place weakly, with a magnet, how could the bob not move in opposite direction as the wheel?

Originally posted by sonhouse
But if you look at my other thought experiment, a horizontal wheel spinning at some rate and a balanced affair with a bell jar on the perifery spinning with the wheel and the bob held in place weakly, with a magnet, how could the bob not move in opposite direction as the wheel?

No, provided the wheel isn't accelerating. Imagine a bar resting on a carousel across a diameter and that there is no friction. By conservation of angular momentum the bar should just rotate with the carousel. If what you are saying were true then there would be a torque on it and its angular velocity would be falling in inertial frames and increasing in the frame co-rotating with the carousel, but there isn't a physical mechanism to generate this torque so there isn't a mechanism to transfer the rotational energy away from the bar.

Originally posted by DeepThought
No, provided the wheel isn't accelerating. Imagine a bar resting on a carousel across a diameter and that there is no friction. By conservation of angular momentum the bar should just rotate with the carousel. If what you are saying were true then there would be a torque on it and its angular velocity would be falling in inertial frames and increasing ...[text shortened]... rate this torque so there isn't a mechanism to transfer the rotational energy away from the bar.

I thought a spinning disk or wheel is in a state of constant acceleration, since it is changing the direction of motion at all times, not needing energy to do that of course, it could spin in outer space in a perfect vacuum, forever. But it seems the bob would not 'like' to be spun around like that and would bang up against the bell jar. I must be missing something fundamental if that is not true.

Originally posted by sonhouse
I thought a spinning disk or wheel is in a state of constant acceleration, since it is changing the direction of motion at all times, not needing energy to do that of course, it could spin in outer space in a perfect vacuum, forever. But it seems the bob would not 'like' to be spun around like that and would bang up against the bell jar. I must be missing something fundamental if that is not true.

It depends on the reference frame. From a reference frame rotating with the wheel, there is no acceleration. In a stationary reference frame, the acceleration is towards the centre of the wheel. The overall result is centrifugal force which is directly away from the centre and must be countered by a force directly towards the centre in order to keep the object in place. If this were not the case, then a weight on a string being spun would either start to go faster or slow down even without friction. This doesn't happen.

Originally posted by sonhouse
I thought a spinning disk or wheel is in a state of constant acceleration, since it is changing the direction of motion at all times, not needing energy to do that of course, it could spin in outer space in a perfect vacuum, forever. But it seems the bob would not 'like' to be spun around like that and would bang up against the bell jar. I must be missing something fundamental if that is not true.

By accelerating I mean that the rate of rotation of the carousel isn't increasing. In your carousel set-up if you set up the pendulum when the carousel is stationary, and then start the carousel spinning the pendulum weight will lag behind the rotation of the carousel. But once the carousel is up to speed and stops accelerating the only force other than gravity on the weight is outwards.

Originally posted by DeepThought
No, provided the wheel isn't accelerating. Imagine a bar resting on a carousel across a diameter and that there is no friction. By conservation of angular momentum the bar should just rotate with the carousel. If what you are saying were true then there would be a torque on it and its angular velocity would be falling in inertial frames and increasing ...[text shortened]... rate this torque so there isn't a mechanism to transfer the rotational energy away from the bar.

So if we modify my thought experiment replacing my magnet with just exending the string to center line fixed to the outer wheel at 5 meters radius and now the bob is just a mass say one meter further up the string, I can see there would be centrifugal/centrifical force that would try to fling the mass up the string, it would go out if it was not held tight to the string but say a hole in the center of the mass and the string in the hold, then the mass would slide up to the end of the bell jar.

But would there also be a torque sideways too?

I also thought about the original posit, seeing if there was some experiment you could do that would prove the spinning Earth and I see now there doesn't need to be a measurment of the mass sideways, there would always be a centripital force trying to fling the mass away from center, so in the case of a mass on the ground at the equator, there would be that centrifugal/centripital force on the mass making it weigh slightly less than a non spinning Earth.

I guess that would still open up lines of flatass defense, saying stuff like you don't know the density or the free space mass, etc. But a known mass on the north pole taken to the equator will show a slight lessening of apparent weight and that should be evidence of the spinning of Earth.

But back to the original posit, lets change things where now there is a nice railway on the equator, where it is perfectly round, that is to say it follows the curvature of the Earth at the equator and further, there is a car on the railway and we have super science that renders the rails 100% friction free. Under those conditions, considering a spinning Earth, wouldn't there be a force causing the car to move, lets assume we are in a big tunnel evacuated of air so wind effects are out. Wouldn't the car move in the opposite direction of the circular motion of Earth?

Originally posted by sonhouse
I guess that would still open up lines of flatass defense, saying stuff like you don't know the density or the free space mass, etc. But a known mass on the north pole taken to the equator will show a slight lessening of apparent weight and that should be evidence of the spinning of Earth.

There are several problems there. Firstly, that is not an experiment that the average person can actually do, nor is it an experiment that I am aware of having been done, so no references to provide. So although you can claim that that would produce a given result, you have no actual experimental evidence so you've got nothing.
Next is the issue that altitude matters. So how are you going to determine altitude at the different locations where you weight the mass?
What about local variations in gravity?
What if the there is stronger gravity near the equator? (easily explained by means other than rotation).

Also we are not discussing flat earth beliefs, but stationary earth. There is a difference.

Wouldn't the car move in the opposite direction of the circular motion of Earth?
Take a small weight, tie it to a string then swirl it around you. Next keep your hand still. If there were no friction, would you expect the weight to keep going indefinitely, slow down or speed up? Whatever your answer it applies identically to your train.

Originally posted by twhitehead
There are several problems there. Firstly, that is not an experiment that the average person can actually do, nor is it an experiment that I am aware of having been done, so no references to provide. So although you can claim that that would produce a given result, you have no actual experimental evidence so you've got nothing.
Next is the issue that alt ...[text shortened]... indefinitely, slow down or speed up? Whatever your answer it applies identically to your train.

In my thought experiment, the car is carefully placed on the track so it starts out with no motion. The question is would the movement of Earth be enough to impart a motion to the car in the opposite direction of Earth's rotation. In an evacuated tunnel, good vacuum.

I am assuming the track to be going east to west not north to south, which if there was a force like I posit, a north south track would lead to a situation where the car tries to tilt off the track. Assuming there such a force.

Originally posted by sonhouse
In my thought experiment, the car is carefully placed on the track so it starts out with no motion.

That is no motion relative to the earth, so in actuality, it is going around the earths centre every 24 hours. That is equivalent to my weight on a string going around at a constant angular velocity.

The question is would the movement of Earth be enough to impart a motion to the car in the opposite direction of Earth's rotation. In an evacuated tunnel, good vacuum.
No.

I am assuming the track to be going east to west not north to south, which if there was a force like I posit, a north south track would lead to a situation where the car tries to tilt off the track. Assuming there such a force.
A north-south track would experience a Coriolis force everywhere except the equator and the poles.

I could be mistaken, but a mass being raised or lowered should experience a sideways force.