1. What if a hole was drilled through the the earth from one side to the other, plated with a super strong metal so the earths center wont effect it, and a rock was dropped in it?

2.

3. Do you want the answer calculated with or without air in the tube?

4. I don't understand what you're getting at

5. Originally Posted by billiards
I don't understand what you're getting at

i thinks hes asking what would happen to the rock, would it go through or sit in the middle cause of gravity

6. If there was no air to resist, it would pass directly through the middle and halt at the opposite end of the tube and, unless arrested there, fall back to the other side again.

7. Originally Posted by SkinWalker
If there was no air to resist, it would pass directly through the middle and halt at the opposite end of the tube and, unless arrested there, fall back to the other side again.
Well it would 'oscillate' and eventually settle in the middle.
If there was air, the air may be so compressed that it would slow the rock down to come to a rest somewhere near the centre.

8. stay in the middle

9. thats what i thought. Your from South, Carolina, SealOtter?

10. Originally Posted by pendragon526
thats what i thought. Your from South, Carolina, SealOtter?
charleston, south carolina

11. Without air (read any forces of friction) the rock will oscillate infinitely from the point A to the point A' (wich is the symetric of A) because the sum of all the accelerating forces reserve from A to the point Zero (earth center) will be equal to the sum of decceleranting forces from zero to A'.

12. Originally Posted by Powerdoc
Without air (read any forces of friction) the rock will oscillate infinitely from the point A to the point A' (wich is the symetric of A) because the sum of all the accelerating forces reserve from A to the point Zero (earth center) will be equal to the sum of decceleranting forces from zero to A'.
only if your hole is through the axis of rotation. Any other path will result in collisions as the earth spins.
If for example your hole is started at the equator, the rock will be moving 'sideways' at about 1000mph. It will therefore 'drift' to the side. I think?

13. Originally Posted by billco
Originally Posted by Powerdoc
Without air (read any forces of friction) the rock will oscillate infinitely from the point A to the point A' (wich is the symetric of A) because the sum of all the accelerating forces reserve from A to the point Zero (earth center) will be equal to the sum of decceleranting forces from zero to A'.
only if your hole is through the axis of rotation. Any other path will result in collisions as the earth spins.
If for example your hole is started at the equator, the rock will be moving 'sideways' at about 1000mph. It will therefore 'drift' to the side. I think?
I was thinking about that too. I was thinking along the lines of Coriolis, but that inertial acceleration is caused by movement orthogonal to the "z-axis" (line pointing to the centre of the Earth). I'm not sure you would get such a deflection if you were travelling along the z-axis. Will have to check my lecture notes for the formulae.

14. I would think coriolis would be too weak a force to be applied to an object that small moving that fast. In addition, of course, to the axis question. Planes are slightly affected by coriolis, but only when heading in certain bearings and on the surface of the planet.

15. Originally Posted by SkinWalker
I would think coriolis would be too weak a force to be applied to an object that small moving that fast. In addition, of course, to the axis question. Planes are slightly affected by coriolis, but only when heading in certain bearings and on the surface of the planet.
It's gonna bash into the sides chaps..... Imagine if you were in the centre of this tube and you fired a projectile out - whichever direction you shoot, the earth is rotating, the stars will go by, the projectile will hit the side. Unless it was travelling very very fast. It's a pole to pole hole or a failed experiment!

EDit:

If it helps, remember if the hole is through the equator, then the rock at the surface has an angular velocity 2 pi radians/day, about 1000 mph.
This is in the direction of rotation of the earth, at the centre the angular velocity will be Zero. The rock by the time it gets to the centre must have lost this velocity.

16. Originally Posted by billco
Originally Posted by SkinWalker
I would think coriolis would be too weak a force to be applied to an object that small moving that fast. In addition, of course, to the axis question. Planes are slightly affected by coriolis, but only when heading in certain bearings and on the surface of the planet.
It's gonna bash into the sides chaps..... Imagine if you were in the centre of this tube and you fired a projectile out - whichever direction you shoot, the earth is rotating, the stars will go by, the projectile will hit the side. Unless it was travelling very very fast. It's a pole to pole hole or a failed experiment!

EDit:

If it helps, remember if the hole is through the equator, then the rock at the surface has an angular velocity 2 pi radians/day, about 1000 mph.
This is in the direction of rotation of the earth, at the centre the angular velocity will be Zero. The rock by the time it gets to the centre must have lost this velocity.

Oh Danny Boy...
I guess I'll "pipe" in...

This is certain;
If the hole is concertric with the earth's axis, then the rock will undergo simple harmonic motion if there is no friction of any kind - and if there is friction, then it will undergo damped harmonic motion. I think we all agree on this.

The question has become;
What if the hole is not along the axis of rotation?

This is what has given me pause. I haven't convinced myself either way yet, but I suspect the result will surprise some of you....

Anywhere on (or in) the earth, we are in a co-rotating reference frame. Where I stand now (about 41 deg latitude), I cannot feel the earth rotating.

While the tangential velocity may change, the angular velocity is the same anywhere on (or in) the earth. I still want to play with some equations before I make a verdict, but I think the rock will oscillate whether the hole is through the earth's axis of rotation or not.

Cheers,
william

17. Originally Posted by billco
Originally Posted by SkinWalker
I would think coriolis would be too weak a force to be applied to an object that small moving that fast. In addition, of course, to the axis question. Planes are slightly affected by coriolis, but only when heading in certain bearings and on the surface of the planet.
It's gonna bash into the sides chaps..... Imagine if you were in the centre of this tube and you fired a projectile out - whichever direction you shoot, the earth is rotating, the stars will go by, the projectile will hit the side. Unless it was travelling very very fast. It's a pole to pole hole or a failed experiment!

EDit:

If it helps, remember if the hole is through the equator, then the rock at the surface has an angular velocity 2 pi radians/day, about 1000 mph.
This is in the direction of rotation of the earth, at the centre the angular velocity will be Zero. The rock by the time it gets to the centre must have lost this velocity.
I would disagree with that because the angular velocity of the hole is the same as that of the rock at all depths. Also the coriolis acceleration shouldn't factor because neither latitude or longitude vary with depth. I don't think the stone would hit the side. Prove me wrong......

EDIT: If you fired from the centre then the rock would hit the side, but that isn't what we were asked to consider.

18. Okay, here are the equations I promised;

In cylindrical coordinates, the accelerations are

a(r) = (d^2r/dt^2) - r(d(theta)/dt)^2
a(theta) = r(d^2theta/dt^2) + 2(dr/dt)(d(theta)/dt)
a(z) = d^2z/dt^2

which you can either look up or derive from variational principles.
In the first equation, the first term is just the acceleration in the r-direction and the second term is the centripetal acceleration. In the second equation, the first term is just the angular acceleration multiplied by r (the tangential acceleration) and the second term is the coriolis acceleration. The third equation is just the acceleration in the z-direction.

We can easily approach this from two perspectives - either viewed from an outside observer, or from within the co-rotating reference frame. I'll do both.

Assuming the hole is drilled on the equator through the center of the earth...

Co-rotating reference frame (standing on the earth):
In this case, the angular velocity and acceleration are zero and we get

a(r) = (d^2r/dt^2) - r(0)^2 = (d^2r/dt^2)
a(theta) = r(0) + 2(dr/dt)(0) = 0
a(z) = d^2z/dt^2 = 0

and the only remaining acceleration is in the r-direction.
=> the rock makes it through without hitting the side and it will oscillate the same way as if the hole were through the earth's axis of rotation.

Outside observer:
An outside observer sees the earth (and rock) rotating with the same constant angular velocity and the accelerations to her become

a(r) = (d^2r/dt^2) - r(d(theta)/dt)^2
a(theta) = r(0) + 2(dr/dt)(d(theta)/dt) = 2(dr/dt)(d(theta)/dt)
a(z) = d^2z/dt^2 = 0

and since the angular velocity is constant (and more importantly, the angular acceleration is zero), the rock again will make it through without bumping the side of the hole.

The thing to notice, is that for the rock to hit the side, it must be given an initial angular acceleration (or tangential acceleration). That is, the first term in this equation

a(theta) = r(d^2theta/dt^2) + 2(dr/dt)(d(theta)/dt)

must not be zero in either reference frame.

The upshot is that the rock will not hit the side no matter where the hole is drilled so long as the hole is through the earth's center.

Cheers,
william

19. I maintain it will hit the sides provided the earth is spinning. Just like swinging a bucket around your head on a rope then pulling the rope in. the speed of the bucket will remain the same but as the rope gets shorter the time taken to complete a revolution will decrease. Imagine therefore the bucket is the rock, the rope is the tube.

If you are still unhappy, explain where the initial kinetic energy[due to the rock being on the equator and therefore moving at 1000mph] goes, when the rock is at the centre of the earth and has none of this motion. and then magically gets it back when at the other side.

Sorry Will, I'll leave you find the flaw in your maths!

20. Originally Posted by billco
I maintain it will hit the sides provided the earth is spinning.
...

If you are still unhappy, explain where the initial kinetic energy[due to the rock being on the equator and therefore moving at 1000mph] goes, when the rock is at the centre of the earth and has none of this motion. and then magically gets it back when at the other side.

Sorry Will, I'll leave you find the flaw in your maths!
Hi billco,
If you are in the co-rotating reference frame, the rock doesn't have that "1000 mph" that you see if you're in the non-rotating "outside observer" reference frame.

Sorry billco, I'll leave it to you to remedy your confusion between reference frames!

21. Next time you are at a fairground, jump on a merry-go-round stand on the outside and roll a marble or something towards the centre, keep your eye fixed on the centrepoint of the merry-go-round and watch the marble roll. Then come back and tell me what path the marble took.

Have a look at this, then tell my why question (1) specifically states the hole drilled from the south pole!

http://physics.pdx.edu/~egertonr/ph425/P425mt00.htm

22. Originally Posted by billco
Next time you are at a fairground, jump on a merry-go-round stand on the outside and roll a marble or something towards the centre, keep your eye fixed on the centrepoint of the merry-go-round and watch the marble roll. Then come back and tell me what path the marble took
Ahhh...
Prom previous posts of yours, I now your fondness for air resistance. That would make the marble's path curve.

The next time you are on an elevator, let me know if you are slammed into the side of it.

cheers

23. Originally Posted by william
Originally Posted by billco
Next time you are at a fairground, jump on a merry-go-round stand on the outside and roll a marble or something towards the centre, keep your eye fixed on the centrepoint of the merry-go-round and watch the marble roll. Then come back and tell me what path the marble took
Ahhh...
Prom previous posts of yours, I now your fondness for air resistance. That would make the marble's path curve.

The next time you are on an elevator, let me know if you are slammed into the side of it.

cheers
Air resistance has nothing to do with it. Since you have decided to take the piss out of me, I'll withdraw from the thread, but ask you to keep this question in mind, until you are able to work it out properly.

http://mathforum.org/library/drmath/view/56370.html

Date: 12/31/2001 at 18:25:31
From: Doctor Tom
Subject: Re: Falling through the Earth

If the earth is spinning, then it depends on where the hole is
drilled. If it's from equator to equator, you'd bounce against the
walls as you went down, since your velocity on the surface is
1000 mi/hr (approximately), and halfway down the tunnel would only be
turning at 500 mi/hr. If it were from pole to pole, or if the earth
weren't spinning, and the earth were a perfect sphere, and there were
no wind resistance, you would keep gaining speed until you passed
through the center, and then would lose speed as you came up the other
side, and would stop going up just as you got to the surface at the
opposite end. Then you would drop through again, and cycle that way
forever, just rising to the surface at each end.

Now of course with air resistance, you would slow down on each cycle,
and eventually would stop right in the center because your path would
be constantly opposed by friction with the air.

Here's an amazing additional fact (assuming that the earth is not
spinning and there is no air resistance). If you drill a hole at an
angle, and slide along a frictionless track back and forth, exactly
the same thing would happen. You'd cycle forever, coming just to the
surface at each end each time. But the time for a complete cycle is
exactly the same for every straight-line path - approximately a
90-minute round-trip. This is also the time that a satellite in orbit
would take to do one loop around the earth right at the surface (or a
fraction of an inch above it) with no air resistance and a perfectly
spherical earth and circular orbit.

- Doctor Tom, The Math Forum

24. Billco wrote:
Since you have decided to take the piss out of me, I'll withdraw from the thread, but ask you to keep this question in mind, until you are able to work it out properly.
Hi billco,
I surely did not intend to take the piss out of you. I was just having fun. Please don't take offense.

As it stands, I think "Dr. Tom" is incorrect. If, however, it turns out that I'm the one incorrect, it wouldn't be the first time. But I think I'm correct.

Cheers,
william

25. Originally Posted by william
Billco wrote:
Since you have decided to take the piss out of me, I'll withdraw from the thread, but ask you to keep this question in mind, until you are able to work it out properly.
Hi billco,
I surely did not intend to take the piss out of you. I was just having fun. Please don't take offense.

As it stands, I think "Dr. Tom" is incorrect. If, however, it turns out that I'm the one incorrect, it wouldn't be the first time. But I think I'm correct.

Cheers,
william
hmmmm I note you were very quick to accept a journalist's confirmation of the helium balloon when it agreed with your [and mine] conclusion.....

If you check this link, you will see in a similar question that the origin of the hole is precisely mentioned.
http://physics.pdx.edu/~egertonr/ph425/P425mt00.htm

So that's three physicists all clearly indicating there is a difference between polar and equatorial orientation. There ought to be a message in there for you. You must agree that there is a diff between dropping pole or equator, there's one extra force involved, just tell me what happens?
I tell you it is dissipated by bumping into the sides. :wink:

As a matter of Interest look into the pendulum of St'Pauls....

26. is there any way to measure dropping something off a high building to see how much lateral movement before it hits? or would the drop be too short to measure a lateral shift(from rotation)? Thinking about it, it probably would

I remember the pendulem and sure it was replicated on tv, probably an adam hart davies program

27. Originally Posted by captaincaveman
is there any way to measure dropping something off a high building to see how much lateral movement before it hits? or would the drop be too short to measure a lateral shift(from rotation)? Thinking about it, it probably would

I remember the pendulem and sure it was replicated on tv, probably an adam hart davies program

It's Foucault's pendulum - if you want to look up, I think the sideways motion is about 0.078 mm/Metre fall, if the hole is from 51.2 latitutde [london] - but I've lost the piece of paper. It is irrespective of [vertical] velocity/acceleration.

28. Originally Posted by billco
Originally Posted by william
Billco wrote:
Since you have decided to take the piss out of me, I'll withdraw from the thread, but ask you to keep this question in mind, until you are able to work it out properly.
Hi billco,
I surely did not intend to take the piss out of you. I was just having fun. Please don't take offense.

As it stands, I think "Dr. Tom" is incorrect. If, however, it turns out that I'm the one incorrect, it wouldn't be the first time. But I think I'm correct.

Cheers,
william
hmmmm I note you were very quick to accept a journalist's confirmation of the helium balloon when it agreed with your [and mine] conclusion.....

If you check this link, you will see in a similar question that the origin of the hole is precisely mentioned.
http://physics.pdx.edu/~egertonr/ph425/P425mt00.htm

So that's three physicists all clearly indicating there is a difference between polar and equatorial orientation. There ought to be a message in there for you. You must agree that there is a diff between dropping pole or equator, there's one extra force involved, just tell me what happens?
I tell you it is dissipated by bumping into the sides. :wink:

As a matter of Interest look into the pendulum of St'Pauls....
Billco, at the moment I think I'm correct but I mentioned that if indeed I am wrong, it wouldn't be the first time.

hmmmm I note you were very quick to accept a journalist's confirmation of the helium balloon when it agreed with your [and mine] conclusion.....
It is one thing to be "quick to accept" experimental evidence and yet another to accept theoretical conclusions. You can't argue with experimental results!

So that's three physicists all clearly indicating there is a difference between polar and equatorial orientation. There ought to be a message in there for you.
Sure, three physicists comments seem to suggest I'm wrong. I may be. From the "get-go," I stated that the rotation of earth and an off-polar hole gave me some pause. The "message" was in my head before seeing these other points of view.

Does it make you feel any better to know that I still have reservations? Certain aspects of the problem still bug me. After all, I'm only human. The important thing (and the sign of a good physicist) is that I still am open to the possibility of me being wrong.

Cheers,
william

29. Originally Posted by captaincaveman
is there any way to measure dropping something off a high building to see how much lateral movement before it hits? or would the drop be too short to measure a lateral shift(from rotation)? Thinking about it, it probably would
This is a good point. I'd be interested in seeing the results. If I'm indeed wrong, I want to know where I messed up.

Edit: Coriolis still bugs me. I think I may have messed up....

Cheers,
william

30. Originally Posted by william
Originally Posted by captaincaveman
is there any way to measure dropping something off a high building to see how much lateral movement before it hits? or would the drop be too short to measure a lateral shift(from rotation)? Thinking about it, it probably would
This is a good point. I'd be interested in seeing the results. If I'm indeed wrong, I want to know where I messed up.

Edit: Coriolis still bugs me. I think I may have messed up....

Cheers,
william
If you hung a pendulum from a tall building and marked it's position at the base, then dropped a pellet [after moving the string] down the string line, the deviation [in london] would be 0.078mm / Metre fall. - that was a very quick calculation, so I'll not swear to it.

William

"The upshot is that the rock will not hit the side no matter where the hole is drilled so long as the hole is through the earth's center".

That is a quote from an earlier post of yours, NEVER write it like that, you have stated this as a fact.

May I suggest you take out a little insurance in future and ALWAYS write

"Therefore my conclusions are, blah blah blah"

It allows for your conclusions to be wrong, for which you will not be punished by your peers.

if you say something like "It is a fact that a photon is a particle" then be prepared to be ridiculed.

Assuring you of my best intent, Billco.

31. Originally Posted by billco
Originally Posted by william
Originally Posted by captaincaveman
is there any way to measure dropping something off a high building to see how much lateral movement before it hits? or would the drop be too short to measure a lateral shift(from rotation)? Thinking about it, it probably would
This is a good point. I'd be interested in seeing the results. If I'm indeed wrong, I want to know where I messed up.

Edit: Coriolis still bugs me. I think I may have messed up....

Cheers,
william
If you hung a pendulum from a tall building and marked it's position at the base, then dropped a pellet [after moving the string] down the string line, the deviation [in london] would be 0.078mm / Metre fall. - that was a very quick calculation, so I'll not swear to it.

ah, ok, so its do-able given a tall enough building to see the results with a naked eye i suppose a vacuum would be better to take out deflections from wind and air resistance

32. Originally Posted by captaincaveman
Originally Posted by billco
Originally Posted by william
Originally Posted by captaincaveman
is there any way to measure dropping something off a high building to see how much lateral movement before it hits? or would the drop be too short to measure a lateral shift(from rotation)? Thinking about it, it probably would
This is a good point. I'd be interested in seeing the results. If I'm indeed wrong, I want to know where I messed up.

Edit: Coriolis still bugs me. I think I may have messed up....

Cheers,
william
If you hung a pendulum from a tall building and marked it's position at the base, then dropped a pellet [after moving the string] down the string line, the deviation [in london] would be 0.078mm / Metre fall. - that was a very quick calculation, so I'll not swear to it.

ah, ok, so its do-able given a tall enough building to see the results with a naked eye i suppose a vacuum would be better to take out deflections from wind and air resistance
100 Metre building, 7.8 millimetres?, you'd have to drop thousands of them and record each one to see a 'shift' in the data.

33. Originally Posted by billco
Originally Posted by captaincaveman
Originally Posted by billco
Originally Posted by william
Originally Posted by captaincaveman
is there any way to measure dropping something off a high building to see how much lateral movement before it hits? or would the drop be too short to measure a lateral shift(from rotation)? Thinking about it, it probably would
This is a good point. I'd be interested in seeing the results. If I'm indeed wrong, I want to know where I messed up.

Edit: Coriolis still bugs me. I think I may have messed up....

Cheers,
william
If you hung a pendulum from a tall building and marked it's position at the base, then dropped a pellet [after moving the string] down the string line, the deviation [in london] would be 0.078mm / Metre fall. - that was a very quick calculation, so I'll not swear to it.

ah, ok, so its do-able given a tall enough building to see the results with a naked eye i suppose a vacuum would be better to take out deflections from wind and air resistance
100 Metre building, 7.8 millimetres?, you'd have to drop thousands of them and record each one to see a 'shift' in the data.

thats quite a bit more than i thought

34. Billco wrote:
William

"The upshot is that the rock will not hit the side no matter where the hole is drilled so long as the hole is through the earth's center".

That is a quote from an earlier post of yours, NEVER write it like that, you have stated this as a fact.

May I suggest you take out a little insurance in future and ALWAYS write

"Therefore my conclusions are, blah blah blah"

It allows for your conclusions to be wrong, for which you will not be punished by your peers.

if you say something like "It is a fact that a photon is a particle" then be prepared to be ridiculed.

Assuring you of my best intent, Billco.
Thanks billco.
If I am wrong (and I'm starting to think that I am...) I'll be the first to admit it. And from there, I would try to find out what lead me to the wrong conclusion.

If one does enough of these type of problems, one's bound to get it wrong every once in a while. And believe me, I'm wrong way more than I'm right.

Anything I say, whether I state it as a fact or as a conjecture, if I'm wrong, I would like someone to show me my error - to convince me that I'm wrong. It is not anyone's duty to do this for me, but it is what I would like. Also, it is my duty to convince others that I'm right when I think I am. Somewhere in this process hopefully the correct answer will come out.

I am the type of person that likes to see the math. That's just me. The statement I made that you quoted was my conclusion based on my math. From the context of where that statement was made, it should be clear that it was my conclusion - and it goes without saying that it may be wrong, even if it reads as if I stated it as fact.

As I said, if I'm wrong, I would hope my peers would show me why and not "punish" and "ridicule" me. And once I'm shown the error, when I finally agree that I made a mistake, I would further hope that any "ridicule" would cease.

At the onset of this problem, I was of the mindset that your conclusions were wrong. That's why it took me a few days before I posted here. And when I did post, I did my best to show you why I felt you were wrong. I hope no one would interpret what I did as "ridicule" or as me "punishing" you. And with all due respect, I don't feel that you left open the option that you could also be wrong. C'est la vie.

Now, about this problem. I am becoming more and more convinced that the rock will hit the side of the hole and continue to bounce. Or, in other words, I am becoming more and more convinced that you are right billco and that my original conclusion was wrong.

Again, I want to play with some equations and see if I can convince myself. This newfound doubt has more to do with coriolis and conservation of angular momentum than the three other physicists you found to contradict me.

Regards,
william

35. Ahem...
I have changed my conclusion and I now believe the rock will indeed strike the side of the hole and continue to bounce the rest of its journey if the hole in not coaxial.

Before I explain why I have come to this conclusion, it would be appropriate to begin with an explanation of why my previous attempt gave me the wrong answer.

For a refresher, here is my previous attempt:

Originally Posted by william
Okay, here are the equations I promised;

In cylindrical coordinates, the accelerations are

a(r) = (d^2r/dt^2) - r(d(theta)/dt)^2
a(theta) = r(d^2theta/dt^2) + 2(dr/dt)(d(theta)/dt)
a(z) = d^2z/dt^2

which you can either look up or derive from variational principles.
In the first equation, the first term is just the acceleration in the r-direction and the second term is the centripetal acceleration. In the second equation, the first term is just the angular acceleration multiplied by r (the tangential acceleration) and the second term is the coriolis acceleration. The third equation is just the acceleration in the z-direction.

We can easily approach this from two perspectives - either viewed from an outside observer, or from within the co-rotating reference frame. I'll do both.

Assuming the hole is drilled on the equator through the center of the earth...

Co-rotating reference frame (standing on the earth):
In this case, the angular velocity and acceleration are zero and we get

a(r) = (d^2r/dt^2) - r(0)^2 = (d^2r/dt^2)
a(theta) = r(0) + 2(dr/dt)(0) = 0
a(z) = d^2z/dt^2 = 0

and the only remaining acceleration is in the r-direction.
=> the rock makes it through without hitting the side and it will oscillate the same way as if the hole were through the earth's axis of rotation.

Outside observer:
An outside observer sees the earth (and rock) rotating with the same constant angular velocity and the accelerations to her become

a(r) = (d^2r/dt^2) - r(d(theta)/dt)^2
a(theta) = r(0) + 2(dr/dt)(d(theta)/dt) = 2(dr/dt)(d(theta)/dt)
a(z) = d^2z/dt^2 = 0

and since the angular velocity is constant (and more importantly, the angular acceleration is zero), the rock again will make it through without bumping the side of the hole.

The thing to notice, is that for the rock to hit the side, it must be given an initial angular acceleration (or tangential acceleration). That is, the first term in this equation

a(theta) = r(d^2theta/dt^2) + 2(dr/dt)(d(theta)/dt)

must not be zero in either reference frame.

The upshot is that the rock will not hit the side no matter where the hole is drilled so long as the hole is through the earth's center.

Cheers,
william
Okay, for any physics students; feel free to use the equations for the accelerations I provided. They are correct. What I did with them however, was not.

Where I messed up in the co-rotating frame of reference:

Let's begin this with the definition of an inertial system. An inertial system is one in which Newton's laws of motion are valid. For many purposes, a system of coordinates fixed in the rotating Earth is a good approximation to an inertial system, however, the rotating Earth is not an ideal inertial system. I hope this bugs all of you with the knowledge that all your calculations on a rotating Earth will be plagued with the effects that the earth is not an inertial system. (But don't loose sleep over it, for the most part your calculations will almost agree with reality.)

As an example, in the co-rotating frame of reference, assuming a perfectly spherical Earth with uniform density, my weight as recorded on a scale would be more if I was at the pole than if I was at the equator. From an outside observer, they would simply see this effect was the centripetal acceleration but to me it would remain a mystery because I would say that I was not rotating (since I'm in co-moving coordinates). The Coriolis effect would also seem like a mystery to me but not to the outside observer.

Bottom line: The earth is not an inertial system.

Where I messed up in the outside observer frame of reference:

Okay, the co-rotating reference frame error was not so bad because it is very subtle. But this mistake was simply an outright brain-fart. It is true that the tangential acceleration would be zero (unless you gave the rock a sideways push when dropping it). But the centripetal acceleration and Coriolis effect were staring me in the face the whole time.

The centripetal acceleration would account for why an object weighs less at the equator. But it is only directed radially (in the cylindrical coordinate sense of the word).

The real brain-fart was overlooking the Coriolis effect. This is directed in the theta direction and there is nothing in my equations that would cancel this effect (recall centrip. is directed radially).

Bottom line: You can't just sweep the Coriolis effect under the rug like I did before.

So what is the effect of Coriolis as applied to the equatorial hole...?

Another way to write the Coriolis effect is

m(d^2x/dt^2) = F(cor) = -2m(w x v)_x = -2mwv_z sin(theta)

where theta is co-latitude, w (omega) is angular velocity, m is mass, and v_z is the velocity in the z-direction (perpendicular to the surface).

In this coordinate system, y points North, z points up, and x points East.

The velocity in the z-direction (v_z) is itself effected by Coriolis, but it is small enough to approximate v_z as

v_z = -gt.

Integrating this with appropriate limits gives

t = sqrt(2z/g)

which we'll use in a moment.

Integrating the diff-eq for Coriolis gives

x = [(wgt^3)/3]sin(theta)

and replacing t with what we previously found gives

x = [(w/3)sqrt({(2z)^3}/g)]sin(theta).

This gives the deflection due to Coriolis.

For w (the angular velocity of Earth) we use

w = [2pi/(24*3600)](366.5/365.5).

The [...] term is the angular velocity relative to the sun, and the (...) term is the ratio of the number of sidereal days in a year to the number of solar days in a year. The (...) term is simply a correction factor to give us w relative to the "fixed" stars.

If you had the 100 meter tower billco mentioned and dropped a marble from it at the equator (theta = pi/2), the deflection is ~ 2.2 cm Eastward.

The only thing left for our particular problem is to address the gravitational acceleration g. Inside the earth it takes the form gr/R where R is Earth's radius.
So g --> gr/R inside the earth.

The moral of the story is;
a. "It is my most recent conclusion that" the rock will be deflected for non-polar holes.
b. If you're wrong, then you're wrong, and it's best to figure out why.

This was fun!

Cheers,
william

36. I calculated the sidways error due to coriolis to be 0.078mm for the first metre dropped, I used 12,700 KM as the radius of the earth at a latitude of 51.2 degrees, what would you make it?

37. But the coriolis force at the equator is 0 because the latitude there is 0, not pi/2!!!!!! sin0=0.

A quick google search says that the stone will hit the side, but I don't think it's due to Coriolis. Better get back to the books to try and find out why.

38. Originally Posted by billiards
But the coriolis force at the equator is 0 because the latitude there is 0, not pi/2!!!!!! sin0=0.

A quick google search says that the stone will hit the side, but I don't think it's due to Coriolis. Better get back to the books to try and find out why.
Edit, Erroneous post deleted.

39. Originally Posted by billco
Originally Posted by billiards
But the coriolis force at the equator is 0 because the latitude there is 0, not pi/2!!!!!! sin0=0.

A quick google search says that the stone will hit the side, but I don't think it's due to Coriolis. Better get back to the books to try and find out why.
Coriolis is maximum at the equator.
Want a bet? it's not! Coriolis is maximum at the poles.

40. Originally Posted by billco
I calculated the sidways error due to coriolis to be 0.078mm for the first metre dropped, I used 12,700 KM as the radius of the earth at a latitude of 51.2 degrees, what would you make it?
Without doing the calculation, I first notice that 12,700 km is Earth's diameter. 6,378 km is the radius. See what that gives you....

For "my" formula, latitude = 51.2 => co-latitude = 38.8. Give me a minute....

cheers

41. Originally Posted by billiards
But the coriolis force at the equator is 0 because the latitude there is 0, not pi/2!!!!!! sin0=0.

A quick google search says that the stone will hit the side, but I don't think it's due to Coriolis. Better get back to the books to try and find out why.
Co-latitude!

cheers

42. Originally Posted by billiards
Originally Posted by billco
Originally Posted by billiards
But the coriolis force at the equator is 0 because the latitude there is 0, not pi/2!!!!!! sin0=0.

A quick google search says that the stone will hit the side, but I don't think it's due to Coriolis. Better get back to the books to try and find out why.
Coriolis is maximum at the equator.
Want a bet? it's not! Coriolis is maximum at the poles.
I wouldn't advise you make that bet billiards unless you want to lose....

cheers

43. Thanks for the advice but I know what I'm talking about, and I'll be happy to prove you both wrong.

44. The coriolis effect is at a maximum at the poles, I had gotten mixed up what I trying to say, and pressed the wrong button.

The corioliis effect is usually associated with the weather, northern/southern hemisphere rotation of clouds - obviously more pronounced at the poles.

45. Originally Posted by billco
The coriolis effect is at a maximum at the poles, I had gotten mixed up what I trying to say, and pressed the wrong button.
8)

46. Originally Posted by william
Originally Posted by billiards
Originally Posted by billco
Originally Posted by billiards
But the coriolis force at the equator is 0 because the latitude there is 0, not pi/2!!!!!! sin0=0.

A quick google search says that the stone will hit the side, but I don't think it's due to Coriolis. Better get back to the books to try and find out why.
Coriolis is maximum at the equator.
Want a bet? it's not! Coriolis is maximum at the poles.
I wouldn't advise you make that bet billiards unless you want to lose....

cheers
William,

Please be aware that Billco CAN make mistakes, do NOT follow me onto the ice blindly..... :wink:

47. Originally Posted by billco
The coriolis effect is at a maximum at the poles, I had gotten mixed up what I trying to say, and pressed the wrong button.

The corioliis effect is usually associated with the weather, northern/southern hemisphere rotation of clouds - obviously more pronounced at the poles.
Billiards wrote:
Thanks for the advice but I know what I'm talking about, and I'll be happy to prove you both wrong.
Sorry guys, you are both right and you are both wrong. It depends on how you look at it and the direction of the particle's trajectory. For a downward (or upward) trajectory, the largest effect is at the equator.

Why do you think they want to launch space rockets from an equatorial location?

cheers

48. Originally Posted by william
Originally Posted by billco
The coriolis effect is at a maximum at the poles, I had gotten mixed up what I trying to say, and pressed the wrong button.

The corioliis effect is usually associated with the weather, northern/southern hemisphere rotation of clouds - obviously more pronounced at the poles.
Billiards wrote:
Thanks for the advice but I know what I'm talking about, and I'll be happy to prove you both wrong.
Sorry guys, you are both right and you are both wrong. It depends on how you look at it and the direction of the particles trajectory. For a downward (or upward) trajectory, the largest effect is at the equator.

Why do you think they want to launch space rockets from an equatorial location?

cheers
Strictly speaking the coriolis effect was to describe the weather, in terms of the weather it's effect is most prominent at the poles. You have said NO air in your tube, ergo NO weather, ergo NO coriolis. I think I indicated the rotation of the earth in my first or second post, that is the reason our rock gets bashed on the way down.

49. Originally Posted by billco
Originally Posted by william
Originally Posted by billco
The coriolis effect is at a maximum at the poles, I had gotten mixed up what I trying to say, and pressed the wrong button.

The corioliis effect is usually associated with the weather, northern/southern hemisphere rotation of clouds - obviously more pronounced at the poles.
Billiards wrote:
Thanks for the advice but I know what I'm talking about, and I'll be happy to prove you both wrong.
Sorry guys, you are both right and you are both wrong. It depends on how you look at it and the direction of the particles trajectory. For a downward (or upward) trajectory, the largest effect is at the equator.

Why do you think they want to launch space rockets from an equatorial location?

cheers
Strictly speaking the coriolis effect was to describe the weather, in terms of the weather it's effect is most prominent at the poles. You have said NO air in your tube, ergo NO weather, ergo NO coriolis. I think I indicated the rotation of the earth in my first or second post, that is the reason our rock gets bashed on the way down.

For crying out F'n loud!!!
Now this is getting rediculus.

For one thing, make sure you know the difference between latitude and co-latitude billiards.

The equation I provided can be found or derived. Don't take my word for it. Do it yourself. But is was derived via Coriolis! (Thought I made that clear...?)

If a f'n cannon was shot from the North pole Southward, the trajectory would appear to move West. Shot form the equator North, it would appear to move East. For crying out loud guys....

This reminds me of a scene in the (2004) remake of "Dawn of the Dead"... "If you want to argue then you can argue with this (as I wave my gun in your face...)!"

Yes... I'm getting frustrated.
Yes... I still like you guys.

Cheers

50. William!

I don't think I've ever seen you profane.

http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/crls.rxml

Scroll down to the bottom and run the movie..... :wink:

51. William wrote:
If a f'n cannon was shot from the North pole Southward, the trajectory would appear to move West. Shot form the equator North, it would appear to move East.
This direction ({Edit:} the cannon ball's that is {end edit}) is purely from Earth's rotation - not wind effects.

cheers

52. Originally Posted by william
William wrote:
If a f'n cannon was shot from the North pole Southward, the trajectory would appear to move West. Shot form the equator North, it would appear to move East.
This direction is purely from Earth's rotation - not wind effects.

cheers
Yes I know that, my point is that the coriolis effect was to describe the rotation of storms, and weather systems, the rock hitting the side of the tube on it's 45min 20.48462538 second journey to the centre of the earth is due to the rotation of the earth and NOT the weather system.

In other words the coriolis effect is simply the name given to the rotation of weather systems. - did I just repeat my self? :wink: 8)

53. Originally Posted by billco
Originally Posted by william
William wrote:
If a f'n cannon was shot from the North pole Southward, the trajectory would appear to move West. Shot form the equator North, it would appear to move East.
This direction is purely from Earth's rotation - not wind effects.

cheers
Yes I know that, my point is that the coriolis effect was to describe the rotation of storms, and weather systems, the rock hitting the side of the tube on it's 45min 20.48462538 second journey to the centre of the earth is due to the rotation of the earth and NOT the weather system.

In other words the coriolis effect is simply the name given to the rotation of weather systems. - did I just repeat my self? :wink: 8)

Sorry, I can't be bothered right now. I'm in the process of slitting my wrists.

54. just let me know if the blood falls vertically or at a slight angle..... :wink:

55. Originally Posted by billco
just let me know if the blood falls vertically or at a slight angle..... :wink:
It's falling at slight angles.

56. Now bandage it quick, you are a valued member of the forum

57. Originally Posted by billco
Now bandage it quick, you are a valued member of the forum

Awwwe...

How much value? \$\$\$

58. Just enough to keep communications open... :-D

I know this is off topic, but it does say humour me...

A chap lives at the top of a very tall apartment building, on sunny days he uses the elevator to go down to the ground floor, upon his return he uses the elevator to go to the third floor but walks the rest of the way and uses the stairs. On rainy days he can use the lift all the way in both directions, why is this?

59. Originally Posted by billco
Originally Posted by william
William wrote:
If a f'n cannon was shot from the North pole Southward, the trajectory would appear to move West. Shot form the equator North, it would appear to move East.
This direction is purely from Earth's rotation - not wind effects.

cheers
Yes I know that, my point is that the coriolis effect was to describe the rotation of storms, and weather systems, the rock hitting the side of the tube on it's 45min 20.48462538 second journey to the centre of the earth is due to the rotation of the earth and NOT the weather system.

In other words the coriolis effect is simply the name given to the rotation of weather systems. - did I just repeat my self? :wink: 8)
I just calculated that it takes ~1300 seconds for the rock to get to the centre (on a non-rotating earth), corresponding to just over 21 minutes. How did you arrive at your ever so precise figure

60. Originally Posted by billiards
Originally Posted by billco
Originally Posted by william
William wrote:
If a f'n cannon was shot from the North pole Southward, the trajectory would appear to move West. Shot form the equator North, it would appear to move East.
This direction is purely from Earth's rotation - not wind effects.

cheers
Yes I know that, my point is that the coriolis effect was to describe the rotation of storms, and weather systems, the rock hitting the side of the tube on it's 45min 20.48462538 second journey to the centre of the earth is due to the rotation of the earth and NOT the weather system.

In other words the coriolis effect is simply the name given to the rotation of weather systems. - did I just repeat my self? :wink: 8)
I just calculated that it takes ~1300 seconds for the rock to get to the centre (on a non-rotating earth), corresponding to just over 21 minutes. How did you arrive at your ever so precise figure

Through what I now see as an erroneous route.......

as well as forgetting to divide by two.

61. Okay this is how I worked out that the rock would hit the side. Surprise surprise Coriolis appears to be the culprit, although the centripetal acceleration also factors.
As the older/more knowledgable members will know, there are three components to acceleration within a rotating frame: linear, coriolis and centripetal. Because the angular velocity is constant we needn't consider the linear acceleration.

I am considering a frame with an origin at the earth's centre and a spin axis w.
The coriolis acceleration is given by 2wx[dr/dt|rotating frame]. The only time this will be zero is when the spin axis is perfectly alligned with the projectile's trajectory. (Which is why this is felt most strongly at the poles where the ground is perpendicular to the spin axis.)

The centripetal acceleration is given by wx(wxr). Again this will be non-zero provided that the spin axis is inclined to the tunnel. However, this acceleration will be directed perpendicular and outwards from the spin axis such that if the tunnel meets the equator, it will not cause the rock to hit the tunnel wall. If the tunnel is inclined at some angle other than along or perpendicular to the spin axis, this force will act to cause a tunnel wall collision.
All in all, I have satisfied myself at least that the rock will hit the side if the tunnel is at an angle to the spin axis. I'm still in the process of working out how large the effect will be, but it is safe to say that it is there. Unless of course I've fudged up, criticism welcome...

62. Then there's 'after a few bounces' the rock will simply roll down the edge of the tube and friction will take its course, the eventual fate of the rock will be to 'oscillate' with a peak deviation from centre equal to the radius of the tube - I think..

63. Originally Posted by billiards
Okay this is how I worked out that the rock would hit the side. Surprise surprise Coriolis appears to be the culprit, although the centripetal acceleration also factors.
As the older/more knowledgable members will know, there are three components to acceleration within a rotating frame: linear, coriolis and centripetal. Because the angular velocity is constant we needn't consider the linear acceleration.

I am considering a frame with an origin at the earth's centre and a spin axis w.
The coriolis acceleration is given by 2wx[dr/dt|rotating frame]. The only time this will be zero is when the spin axis is perfectly alligned with the projectile's trajectory. (Which is why this is felt most strongly at the poles where the ground is perpendicular to the spin axis.)

The centripetal acceleration is given by wx(wxr). Again this will be non-zero provided that the spin axis is inclined to the tunnel. However, this acceleration will be directed perpendicular and outwards from the spin axis such that if the tunnel meets the equator, it will not cause the rock to hit the tunnel wall. If the tunnel is inclined at some angle other than along or perpendicular to the spin axis, this force will act to cause a tunnel wall collision.
All in all, I have satisfied myself at least that the rock will hit the side if the tunnel is at an angle to the spin axis. I'm still in the process of working out how large the effect will be, but it is safe to say that it is there. Unless of course I've fudged up, criticism welcome...

I concur that for non-polar and non-equatorial holes the centripetal acceleration also tends to "wall" the rock.

I can't express how happy I am that we all finally agree! Shall we share the Nobel prize? :-D

cheers

64. That'd be nice , I've never been awarded a prize before

65. You guys take the shiny thing, I'll grab the money, then I can go get that beer McGuyver owes me!

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