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Coriolis force question

CurtC said:
The exact method would have to calculate the shape of the ellipse that the dropped item will follow, and then computer where that ellipse will intersect the circle of the Earth's surface, right?
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Hmm, my thought was that the distance the coin follows horizontally is just a function of the length of time that it is traveling and the speed that it is traveling at horizontally. This assumes an infinitely large circumference for the earth so that the gravitational force is parallel (and perpendicular to the original horizontal direction of the coin) throughout the drop path of the coin, but I think the Earth is reasonably close to infinitely large compared to a few inches so that no significant error is introduced. But maybe I'm missing something?
 
OK, thanks for sharing that you two. Your calculation method was the same that I was using to arrive at what I thought was the upper bound, which I then halved. I'll have to think a while to believe that it's valid to take the top speed minus the bottom speed, and multiply that by the elapsed time. That's not obvious to me.
 
Well, I took a shot at the plumb bob error issue.

Amazing, I come up with an angular difference from vertical of .1291 degrees. which seems small but it produces a pretty big displacement from vertical over a large distance.

I got 34 inches for 381 meters. That means the ESB is displaced to the north 34 inches assuming it is aligned with a plumb bob.

That seems too large. I need to do more checking.

But do you need to take the plumb bob error into account for this issue? The coin will just follow the path of the plumb bob on the way down?

No, I guess it won't. Once the coin is released it is no longer moving in a circle. It is just going in a straight line. So it only sees the earth's gravity and not the centrifugal force from the rotation of the earth.
 
If we're considering the inertial reference frame fixed at the Earth's center, the coin follows an ellipse, only very slightly eccentric, and that ellipse is slanted with respect to the equator at an angle of 40.75 degrees, the same as the latitude of where it's dropped.

So what's this with the plumb bob? There are two things going on here. One is that a plumb bob will not point at the center of the Earth, but at a point south of it. This is simply because the Earth is an ellipsoid, not a sphere. But "vertical" is not defined as being toward the center, it's defined as normal to the surface, so a plumb bob will be vertical.

The other thing is that ceptimus noted that a dropped item would hit slightly south, as well as west, of where it was dropped. This is because of the ellipse path I mentioned above. But ceptimus went on to postulate that a plumb bob would point slightly south of vertical for this same reason, which I don't think is true. Since it's static in that rotating reference frame, I think it would see no coriolis deflections.
 
I didn't take into account that the earth is an ellipsoid with my plumb bob error calculation.

The picture I had in my mind was the carnival ride where people sit in chairs suspended below a rotating circular support. The rotation of the circle causes the people to move away from the axis of the circular support.

OK, so the direction a plumb bob hangs down is a fundtion of two vectors. One the gravitational vector pointing to the center of the earth and the other a centrifugal force vector pointing away from the axis of rotation.

I calculated the resultant vector by breaking the gravitational vector into two components one pointing south and one pointing to the axis of earth's rotation. I subtracted the centrifugal acceleration from the vector pointing to the axis of the earth and calculated the angle of the resultant vector. This was about .075 degrees away from vertical.

BICBFOS.
 
davefoc wrote:
I calculated the resultant vector by breaking the gravitational vector into two components one pointing south and one pointing to the axis of earth's rotation. I subtracted the centrifugal acceleration from the vector pointing to the axis of the earth and calculated the angle of the resultant vector. This was about .075 degrees away from vertical.
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I have no problems with this, until the last sentence. However, you didn't calculate the angle between the composite vector and "vertical," because "vertical" is defined as that composite vector - their angular difference is zero, by definition.

What you calculated was the difference between that composite vector and another pointing to the center of the Earth, but the vector pointing at the center of the Earth is not vertical. If you wanted to do more figuring like this, you could show that your composite vector is normal to the surface of the ellipsoid at all points.
 
I made a point of saying that the plumb bob won't point at the centre of the earth, rather than 'down', as, of course, the way the bob points defines 'down'. The ground is (on average) at right angles to the plum bob, for the same reason that the plumb bob doesn't point at the centre - the reason being the earth's spin. You can take a normal to the oblate spheroid shape, or work out davefoc's carnival ride math and come to (approximately) the same answer. The line of plumb bob misses the earth's centre by a surpisingly large amount.
 
CurtC said:
OK, thanks for sharing that you two. Your calculation method was the same that I was using to arrive at what I thought was the upper bound, which I then halved. I'll have to think a while to believe that it's valid to take the top speed minus the bottom speed, and multiply that by the elapsed time. That's not obvious to me.
Click to expand...
I suspect your right to be suspicious of the validity of top speed - botton speed /2. It seems to me the con spends more time high up than low down, as it accelerates in our bell jar.

Walt
 
The horizontal speed is independant of the vertical speed (assuming a flat earth, which is a fair approximation over a few inches). Once the weight is in free fall, there is nothing to affect the horizontal component of the speed - this will remain constant all the way down.
 
OK, I've checked my numbers again. I get a plumb bob error of 25.7 inches for a plumb bob suspended from 381 meters at 40.5 degrees. The plumb bob hangs .098 degrees away from vertical as a result of the rotation of the earth at a latitude of 40.5 degrees.

This means that our coin will hit the side of the building if we drop it from the south side of the building even if that side is vertical.
 
ceptimus said:
OK here's mine (blush)

Distance of bottom of tower from earth's axis.

Radius of earth is 251,106,299 inches, at a latitude of 40.5 degrees distance is:

251,106,299 cos(40.5) = 190,942,728 inches.

Distance travelled in 23 hours 56 mins = 2 π r, so speed in inches per second is:

2 * π * 190,942,728 / ((23 * 60 + 56) * 60) = 13,924.426 ips

Tower is 15,000 inches high. Add this to earth radius and repeat calculation to get speed of top of tower.

2 * π * (251,106,299 + 15,000) cos(40.5) / ((23 * 60 + 56) * 60) = 13,925.258 ips

Take difference to find speed of top of tower, relative to ground = 0.832 ips

Fall time = sqrt(2 * s / a) (in SI units) = sqrt (2 * 381 / 9.81) = 8.813 seconds

0.832 ips * 8.813 s = 7.33 inches.
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Ok, now I see where we differ.

I don't think you can just work out the difference between speed at top and speed at bottom. The speed changes as the penny drops. The speed does not vary linearly with time. By speed here I mean the tangential velocity.

I did this by working out the angular velocity with respect to the earth's axis based on conservation of initial angular momentum. I then worked out how angular velocity varies with height and how height varies with time.

So now you have angular velocity as a function of time. Integrate over the fall time to get the total angular displacement. Take away the angular displacement due to the earth rotating. Convert this into linear displacement.

The integration is not easy - you can't do it analytically. I had to do it numerically using simpsons rule and a computer.
 
ceptimus said:
The horizontal speed is independant of the vertical speed (assuming a flat earth, which is a fair approximation over a few inches). Once the weight is in free fall, there is nothing to affect the horizontal component of the speed - this will remain constant all the way down.
Click to expand...

I don't think you can assume a flat earth. The tangential velocity is not a constant. It varies as 1/r where r is the distance to the axis of rotation. "Horizontal speed" is meaningless in a rotational system where the body is being accelerated radially.
 
Hamish said:


I don't think you can assume a flat earth. The tangential velocity is not a constant. It varies as 1/r where r is the distance to the axis of rotation. "Horizontal speed" is meaningless in a rotational system where the body is being accelerated radially.
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I think you're wrong. Once the weight has been released, it is, by definition, in free-fall. The only force acting on it, is gravity.
 
I didn't know about it. I just reasoned it out after I read the start of this thread. Of course, I may still be wrong. :)
 
Well, if you're right, I wonder if it is a consideration in the construction of very tall structures?

Maybe it just happens because people use level and/or plumb bobs to establish vertical and the building tilts a small amount and who cares.

On the other hand if a geometric technique was used for establishing vertical the building wouldn't tilt slightly and therefore the building's mass would be adding a little torque to the structure trying to tear it down. And maybe that amount is still so small that nobody cares.

To Hamish, CurtC, et al, I would have responded but I tend to agree with what Ceptimus said and I don't have anything to add. I just don't get why once the coin is in free flight there are any significant forces acting on its velocity parallel to the earth's surface.

This sounds like the old after one second of flight how far has a 1000 ft/sec bullet dropped versus how far has a 2000 ft/sec bullet dropped issue to me. Gravity isn't doing anything with the horizontal velocity so the distance dropped by both bullets is the same (assuming a flat earth).
 
ceptimus said:
I think you're wrong. Once the weight has been released, it is, by definition, in free-fall. The only force acting on it, is gravity.
Click to expand...

Yes, gravity is the only force. But you assume that the penny retains it's linear tangential velocity in the same direction all the way to the ground. Due to rotation of the earth, this linear velocity will not remain perpendicular to the acceleration vector. As such, there will be a component of the acceleration along it's velocity vector. It's small, but so is the deflection we're calculating.

Under your argument, the angular momentum of the penny is not conserved.
 
Hamish,
I think I understand what you're saying. It seems to me the effect you are talking about is insignificant in this case.

I have to go. But if somebody hasn't calculated it I will try when I get back.
 
davefoc wrote:OK, I've checked my numbers again. I get a plumb bob error of 25.7 inches for a plumb bob suspended from 381 meters at 40.5 degrees. The plumb bob hangs .098 degrees away from vertical as a result of the rotation of the earth at a latitude of 40.5 degrees.

This means that our coin will hit the side of the building if we drop it from the south side of the building even if that side is vertical.
Click to expand...
My point earlier was that the effect you're calculating is not plumb bob error, but is the angle between the vertical (which a plumb bob indicates) and the vector that would point to the center of the Earth. This centrifugal component is already figured into what we define as vertical, and this is the direction a plumb bob hangs, and it is perpindicular to the ideal surface of the Earth (ellipsoid shape). What we consider "straight down" is not towards the exact center of the Earth, but is affected by our centrifugal force.

And although I agree with ceptimus that there will be a very slight southward drift to our dropped object (it will be following an ellipse that's inclined to the Earth's equator by 40.75 degrees), I maintain that the plumb bob will not be thrown off by this same effect, because it's not moving, and the coriolis effect is only for things that are moving in our rotating reference frame.

Also, Hamish, I can't think of any reason to object to the linear velocity model that they came up with. It's not exact, but any difference is a second-order effect and will be much smaller than the already small first-order effect that we're considering. I think.
 
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