Rotation Of A Flat Earth - Bishop Owned

1. FE rotates once every 24 hours, as Foucault's pendulum appears to indicate (even you must agree here Tom, in the Northern hemisphere the experiment demonstrates as much).

2. The North Pole is the centre of this rotation.

3. Melbourne is latitude 37.5 degrees South, putting it approximately 14,000 km from the North Pole.

4. The circle around the North Pole described by Melbourne is then approximately 88,000 km in circumference.

5. Melbourne travels 88,000 km in 24 hours, which is around 1 km/s or 3600 km/h.

6. On a flat Earth this would mean a 75 kg person in Melbourne leaning against a wall facing north would feel a force of around 5360 N pressing him to the wall, as in any rotating system there is constant acceleration toward the centre and it is natural in any reference frame for a body to resist acceleration (it is the basis of FE gravitation). If he was not leaning against anything, he would fly off the side of the Earth.

7. This does not occur.

8. Either:
    a. The Earth is not flat.
    b. The Earth does not rotate.

9. We have shown (b) to be false in part 1.

10. The Earth is not flat.

On a round Earth this is not a problem, but it has a visible effect. The 'bulge' around the equation is a demonstration of the effect, but those who live there, rotating fastest, experience this only as reduced gravity as it acts in the opposite direction to Earth's gravity. It should also be noted that in the RE the people at the equator aren't nearly as far as the people in an FE Australia so the effect is much less. If you dispute Australia's distance from the North Pole, I suggest you explain how you can fit the land that verifiably exists in between the two places into a space too small for it. If you dispute the Earth's rotation (listed in your own FAQ), go find a Foucault pendulum (a northern hemisphere one will do just fine).

Responses welcome.

Edit: A force of this size would launch the man towards the horizon at around 270 km/h....

1.) Realize that the Round Earth model is rotating at a speed of more than 1000 miles per hour at its equator. Why are we not thrown off into space by this tremendous rotation?

2.) We are not thrown off the earth in either model because the earth rotates at a constant velocity without accelerating or slowing down. Moving at a constant velocity in a rotating frame of reference is comparable to moving at a constant velocity in a moving frame of reference.

3.) Technically, a rotating frame of reference is a special case of a non-inertial reference frame in which the coordinate system is rotating relative to an inertial reference frame.

2) We are not thrown off because of something called gravity.

Your explanation is incorrect. The reason why we do not feel rotational acceleration is because the Earth is moving at a near-constant velocity. However, Gin is referring to the reactive centrifugal force inherent in any rotating system. With RE, gravity counteracts this force, but there is no explanation in the FE case.

As an analogy, imagine a tire with mud on it. Even if the tire accelerates veeery slowly so as to have negligible acceleration, it will eventually start flinging the mud off due to the reactive centrifugal force. Why aren't we, the mud of the tire that is the Earth, flung off into space?

With RE, gravity counteracts this force, but there is no explanation in the FE case.

Gravitation exists in FE. Haven't you seen all of the Acceleration = Gravitation threads?

As an analogy, imagine a tire with mud on it. Even if the tire accelerates veeery slowly so as to have negligible acceleration, it will eventually start flinging the mud off due to the reactive centrifugal force. Why aren't we, the mud of the tire that is the Earth, flung off into space?

There is a centrifugal force on a rotating FE, but it is extremely negligible. In past threads it has been calculated that the force felt along the southern pole is so slight that the difference between the Northern latitudes would be imperceptible. I'll try to find some of these calculations for you.

Okay, so gravitation holds us to the Earth. But the key here is that in RE, gravity directly opposes the centrifugal force whereas in FE gravitation works primarily perpendicular to the centrifugal force and should thus have very little effect.

I'll try to find some of these calculations for you.

I already did, Gin has them in his original post.

One thing to remember: the Earth must be spinning fast enough to stop gravitation from pulling it into a sphere.

The perpendicular action is indeed the key. In a RE, not only are you never as far as FE Melbourne is from the centre of rotation, but when you are furthest you are being flung outwards while gravity is pulling you inwards, the resultant of these forces is a slightly reduced gravity. In a rotating system there is ALWAYS acceleration. In fact TheEngineer would no doubt back me up here that, when an object is rotating in a circle around another, there is constant acceleration towards the centre of rotation by both objects. In RE case the centre of rotation is in the centre of the Earth, so gravity counteracts the force, in FE it is at the north pole, and gravity's (UAs) action is perpendicular to it.

1. FE rotates once every 24 hours, as Foucault's pendulum appears to indicate (even you must agree here Tom, in the Northern hemisphere the experiment demonstrates as much).

2. The North Pole is the centre of this rotation.

3. Melbourne is latitude 37.5 degrees South, putting it approximately 14,000 km from the North Pole.

4. The circle around the North Pole described by Melbourne is then approximately 88,000 km in circumference.

5. Melbourne travels 88,000 km in 24 hours, which is around 1 km/s or 3600 km/h.

6. On a flat Earth this would mean a 75 kg person in Melbourne leaning against a wall facing north would feel a force of around 5360 N pressing him to the wall, as in any rotating system there is constant acceleration toward the centre and it is natural in any reference frame for a body to resist acceleration (it is the basis of FE gravitation). If he was not leaning against anything, he would fly off the side of the Earth.

7. This does not occur.

8. Either:
    a. The Earth is not flat.
    b. The Earth does not rotate.

9. We have shown (b) to be false in part 1.

10. The Earth is not flat.

On a round Earth this is not a problem, but it has a visible effect. The 'bulge' around the equation is a demonstration of the effect, but those who live there, rotating fastest, experience this only as reduced gravity as it acts in the opposite direction to Earth's gravity. It should also be noted that in the RE the people at the equator aren't nearly as far as the people in an FE Australia so the effect is much less. If you dispute Australia's distance from the North Pole, I suggest you explain how you can fit the land that verifiably exists in between the two places into a space too small for it. If you dispute the Earth's rotation (listed in your own FAQ), go find a Foucault pendulum (a northern hemisphere one will do just fine).

Responses welcome.

Edit: A force of this size would launch the man towards the horizon at around 270 km/h....

This is an Excellent post.

And Tom's response proves that he is either a troll, or a moron.

One thing to remember: the Earth must be spinning fast enough to stop gravitation from pulling it into a sphere.

Gravitation of the FE wants to flatten it out, not pull it into a sphere.

Only if Earth is made of magical mass that has no gravitation of its own.

Still no answer on the problem of 'flying men in Melbourne'.....

Well done, Gin.  Well done.

Still no answer on the problem of 'flying men in Melbourne'.....

TheEngineer is correct. If the centripetal force of a rotating object moving at a constant velocity was really as significant as you claim, then a round earth should have a vast desire to flatten like the spinning of pizza dough.

One thing I believe some people are missing is that the Earth does not have the same rate of rotation of a rapidly spinning tire with mud on it. As an analogy, lets imagine a metallic merry-go-round which rotates at 1/24 of a revolution per hour. A bacterium located towards the rim of the carousel would feel hardly any centripetal action at all.

Now make this carousel as big as your FE and see what happens....poor bacterium.

EDIT:  The reason the Earth wouldn't flatten out in the RE model is gravitation caused by mass bending space-time.  This effect acts against the rotation of the Earth at its most intense place, being the equator.  The fact that the RE model indicates that the equator is wider than other places and that weights are less only prove this point more.  Also, the equatorial distance from the center of rotation is HALF of that in the FE model to the South Poles....if you're calling the ice wall the south poles now.

Now make this carousel as big as your FE and see what happens....poor bacterium.

As the carousel increases in radius, so does its coordinate system.

Quote

Now make this carousel as big as your FE and see what happens....poor bacterium.

As the carousel increases in radius, so does its coordinate system.

OK, and so does the force on the passengers...

Centripetal force is directly correlated to mass, so your bacteria would indeed feel very little. However, if you multiply the dimensions of the whole situation by a scale factor k, the area of the disc increases by k squared, and the volume (and hence, mass) of the bacteria/people increases by k cubed. So in a larger version of the situation the effect is greatly amplified. Simply changing co-ordinate systems is a fallacy, and it is this cubic/quadratic relation which prevents spiders from reaching beyond a certain size amongst other things (if you want me to elaborate on this please ask). Because the centripetal force is a direct result of mass' tendency to stay at constant velocity, the force felt is directly proportional to the mass of the object.

Edit: the significance of the linear/quadratic/cubic relation is that for a scaled-up system the speed of rotation at the edge increases linearly (the distance covered, the circumference, is directly proportional to the radius of the circle) but the mass increases cubically so the centripetal force is far more significant.

2.) We are not thrown off the earth in either model because the earth rotates at a constant velocity without accelerating or slowing down. Moving at a constant velocity in a rotating frame of reference is comparable to moving at a constant velocity in a moving frame of reference.

It is impossible to rotate at a constant velocity because velocity is a vector quantity. If you are on the edge of the earth, you rotate at a constant speed, but your direction is changing, and therefore you are accelerating.

I think static friction should also be taken into account.

I'd just like to chime in and say that the Earth doesn't actually rotate. Focault's pendulum is a sham as I have pointed out in another post.

I see... and the fact that cyclones rotate ccw in the northern hemisphere and cw in the southern hemisphere is all just a matter of coincidence, right?

I'd just like to chime in and say that the Earth doesn't actually rotate. Focault's pendulum is a sham as I have pointed out in another post.

How is it a sham?  I don't remember reading how it's flawed exactly in the other posts.

OK, and so does the force on the passengers...

The passengers would feel a centripetal force relative to the rate of rotation. The speeds don't add up like that in a rotating frame of reference.

See: http://www.kwon3d.com/theory/rotsys/rotfrm.html#rot

Take note of the part which says

"Since the inertia tensor of the body changes as the body rotates in the inertial frame's perspective. Computing the time-derivative of the inertia tensor is not a simple job to do. Now, applying [9] instead, one can obtainsince the inertia tensor observed in the rotating frame (I_CM) does not change as the body rotates. a_r shown in [13] is the angular acceleration of the body relative to the rotating frame."

How is it a sham?  I don't remember reading how it's flawed exactly in the other posts.

Samuel Birley Rowbotham points out the inconsistencies of the Foucault Pendulum in Chapter 14, Section 21 of Earth Not a Globe. He also provides analysis of the variability of pendulum vibrations at different latitudes.

Later authors have found that the direction the pendulum rotation is not always consistent, and that the direction of rotation has statistical significance to the right or left handedness of the person putting it into motion.

So what about when they tried it at the equator and it didn't rotate at all?

You can't use a 'rotating frame of reference' and expect the forces to disappear. If you are accelerating, your inertia will cause you to resist acceleration. In rotational motion you are constantly accelerating (your velocity is changing). That's all there is to it.

The passengers would feel a centripetal force relative to the rate of rotation. The speeds don't add up like that in a rotating frame of reference.

I still don't follow you.  I have experienced the forces getting larger on one of those little kids spin around thingys.  You know, where they spin it really fast and try to hold on until someone pukes?  If you stand in the center you can stay on better, if you stand on the edge you probably will fall off.

I still don't follow you.  I have experienced the forces getting larger on one of those little kids spin around thingys.  You know, where they spin it really fast and try to hold on until someone pukes?  If you stand in the center you can stay on better, if you stand on the edge you probably will fall off.

Again, you are directly comparing a slowly rotating reference frame which moves at one revolution per every 24 hours with a quickly rotating reference frame spinning on an axis at dozens of revolutions per minute. There is no direct comparison. When comparing two rotating reference frames we must compare their rates of revolutions.

Lets go to a marble sitting on a carousel example. The carousel is moving at the extraordinary slow pace of 1/24^ths of a revolution per hour. Its clear that a marble at the edge of this snail paced metallic carousel wouldn't roll off through centripetal force alone. The centripetal action at that rate of rotation is very negligible.

Now what happens if we increase the radius of the carousel inch by inch and place the marble at the edge each time? The carousel is still moving at its extremely slow constant pace, exerting a negligible constant outward proportional to the shallowness of ts radius. The outward force does not add up exponentially by increasing the radius of the carousel.

You can't use a 'rotating frame of reference' and expect the forces to disappear.

The forces don't disappear. The same outward force is projected onto the marble at the expanding edges of the carousel despite its radius.

Doesn't the marble experience more force because it's going a further distance in the same amount of time?  And no, I was comparing the center of the turning to the outside of the turning.  In case you haven't read before start at the top left, go to the top right.  Now move down a line and do the same thing.  Spaces ( ) indicate the end a word and periods (.) indicate the end of a sentence.

Doesn't the marble experience more force because it's going a further distance in the same amount of time?

As far as the marble is concerned it is traveling at a single constant velocity, sitting at the edges of the carousel. The centripetal action is very negligible since it needs to move less as the radius increases. Why should it feel a need to shift? Only a significant acceleration or deceleration of its velocity would cause it to move.

But it IS moving...according to your logic if I got on the spinny kids thing and was on it while it spun at a constant speed, I wouldn't feel anything.  Guess what, Tommy boy, I have done this and I DID feel the force.