Coriolis force question

[QuarkChild]

This is a really dumb question, but it came up in the lab I teach today and I couldn't think of the answer.

If you let a ball drop from some height above the surface of the Earth, it's deflected easterly if you're in the Northern hemisphere, right? And if you throw a ball up in the air and let it come down again, you get a net westerly displacement. (At least that's how it came out when I did the calculation.) Conceptually, though, I can't figure out why the deflection as it travels upward doesn't cancel the deflection as it travels downward.

Tez? Anyone?

(I tried googling it but my google skills failed me.)

[Tez]

Originally posted by QuarkChild
This is a really dumb question, but it came up in the lab I teach today and I couldn't think of the answer.

If you let a ball drop from some height above the surface of the Earth, it's deflected easterly if you're in the Northern hemisphere, right? And if you throw a ball up in the air and let it come down again, you get a net westerly displacement. (At least that's how it came out when I did the calculation.) Conceptually, though, I can't figure out why the deflection as it travels upward doesn't cancel the deflection as it travels downward.

Tez? Anyone?

(I tried googling it but my google skills failed me.)

Hmm - I think I only understand coriolis force for firing projectiles tangentially (horizontally) not for vertical dropping etc. i.e. if a fire a missile northwards it'll be deflected east, if I fire it eastward it'll go south and so on. If I just fire a bullet straight up and down does it get deflected? If so, I'll have to think about it...

[Hamish]

It sounds crazy but it's true.

To solve this, just return to the tried and trusted method of solving all these types of problems: conservation of angular momentum.

In the case where you drop the ball from a height, the ball starts off with more angular momentum than the ball which starts on the ground. I'm assuming that you're dropping this from a tall building and not from an aircraft with motion relative to the earth. As you drop the ball, it's angular velocity increases to conserve angular momentum, so it speeds up relative to the surface of the Earth. Result: easterly deflection. If you throw the ball upwards, the angular velocity decreases so it slows down relative to the earth. It returns with the same angular velocity with which it left, but the avarage angular velocity over the throw is less than the angular velocity of the Earth's surface. Result: westerly deflection.

The fundamental difference, therefore, is that a drop is not the same as half a throw since in preparing for the drop, you have given the ball more angular momentum by raising it.

Hope this answers your question.

[Hamish]

Just a quick note to add that this is not, strictly speaking, a coriolis effect question. As I understand it, the coriolis effect deals with the difference in tangential velocitites of the earth's surface at different latitiudes. There is no change in latitude in this question. If you were throwing the ball north or south, you'd get a coriolis effect but since the ball falls to the same lattitude each time, there is no coriolis effect.

Somebody please correct me if I'm wrong.

[Tez]

Both your posts sound reasonable to me Hamish... Thanks.

[xouper]

Hamish: In the case where you drop the ball from a height, the ball starts off with more angular momentum than the ball which starts on the ground. I'm assuming that you're dropping this from a tall building and not from an aircraft with motion relative to the earth. As you drop the ball, it's angular velocity increases to conserve angular momentum, so it speeds up relative to the surface of the Earth. Result: easterly deflection.So, if I drop a U.S. penny from the Empire State Building (in New York) how much westerly offset should I use if I'm trying to hit a specific target on the ground (in addition to any wind drift corrections that might be needed)?

[Abdul Alhazred]

Originally posted by xouper
So, if I drop a U.S. penny from the Empire State Building (in New York) how much westerly offset should I use if I'm trying to hit a specific target on the ground (in addition to any wind drift corrections that might be needed)?

The ESB has setbacks rather than being straight up the whole way. Also the top floor is an enclosed observatory.

So lots of luck hitting the ground at all, unless you're on the 85th floor terrace (or perched on top like King Kong) with a strong wind behind you.

IIRC the first setback is somewhere around the twentieth floor.

http://memory.loc.gov/ammem/today/images/0501empire.jpg

:p

[ceptimus]

Originally posted by xouper
So, if I drop a U.S. penny from the Empire State Building (in New York) how much westerly offset should I use if I'm trying to hit a specific target on the ground (in addition to any wind drift corrections that might be needed)? I could work it out for a straight sided building in a vacuum. I'll post a guess though first to see how bad my guessing is.

I guess 7.33 inches. :)

[xouper]

Abdul Alhazred: The ESB has setbacks rather than being straight up the whole way.It's always someone. :D

OK, how about Sears Tower in Chicago? One of the sides doesn't have any setbacks, but I don't recall which one.

[CurtC]

Hamish wrote:
Just a quick note to add that this is not, strictly speaking, a coriolis effect question.Yes, it is the coriolis effect. The coriolis effect, in its simplest terms, describes the bahavior of the forces you feel on a merry-go-round. If you're on the outside and try to walk to the inside, you'll perceive a force pushing you in the direction of rotation. If the carousel is turning anticlockwise (viewed from above), the force will always appear to be pushing you to the right, whether you're headed in or out.

The reason the question about dropping objects is about coriolis is because what makes for the slightly strange behavior is that the top of the building is closer to the outside of the merry-go-round, going faster, than the bottom.

And my guess is much much less than 7.33 inches.

[CurtC]

Well, my own back-of-the-envelope calculations give something like three or four inches. Much more than I guessed.

[QuarkChild]

Originally posted by Hamish
It sounds crazy but it's true.

To solve this, just return to the tried and trusted method of solving all these types of problems: conservation of angular momentum.

In the case where you drop the ball from a height, the ball starts off with more angular momentum than the ball which starts on the ground. I'm assuming that you're dropping this from a tall building and not from an aircraft with motion relative to the earth. As you drop the ball, it's angular velocity increases to conserve angular momentum, so it speeds up relative to the surface of the Earth. Result: easterly deflection. If you throw the ball upwards, the angular velocity decreases so it slows down relative to the earth. It returns with the same angular velocity with which it left, but the avarage angular velocity over the throw is less than the angular velocity of the Earth's surface. Result: westerly deflection.
Wow. That makes sense.

Thank you Hamish.

[Hamish]

Curt C wrote:

Yes, it is the coriolis effect

You are, of course, correct. I could only remember the applications where latitude lines were crossed but this is just s special case. Thanks for pointing that out.

My calculations give a deflection of 20.3 cm (about 8 inches) for a penny dropped from the Sears Tower and 16.7 (about 6.6 inches) for the Empire State Building.

[davefoc]

Results of my calculations (dimensions are in meters)


height at equator
10 0.001
100 0.033
442.8 0.306

height at NYC latitude
10 0.001
100 0.023
442.8 0.215

Assumptions
gravitational acceleration - 9.8 meters/per sec./per sec
circumference of earth - 40075.0 meters
NYC latitude - 44.7 degrees
empire state bldg height - 442.8 meters



The predicted displacement of a coin dropped in a vacuum from height of the top of the antenna on the empire state building at the latitude of New York City is about 8.5 inches from point on the ground pointed to by a plumb bob suspended from the location the coin was dropped from. The contact point will be east of the point marked by the end of the plumb bob.

[davefoc]

I just noticed that Hamish and I came up with somewhat different answers.

When I subtracted the height of the antenna from the height of the overall height of the Empire State Building I got 379.7 meters. Using this height I got 6.7 inches of displacement. Perhaps the Observation deck is a little below this in which case my calculations would agree with Hamish's exactly. Which of course means that we're both right or that we both screwed up in similar ways.

New question: How much is the penny going to be displaced because of the gravitational pull of the empire state building?

[Hamish]

I just noticed that Hamish and I came up with somewhat different answers.

Yep, and the Empire State Building is 381m according to my sources. The Sears Tower is 442m. I did just assume that I was dropping this from the top of the antenna. I can redo the calculation really easily with whatever parameters you'd like.

I will admit to having forgotten about latitude. It sorts itself out in the earlier stage of my working since the angular deflection (longitudinal) works out the same wherever you are on the earth. Of course, this doesn't equate to the same displacement.

I've done some revised and ever so slightly more accurate calculations. I'll take 381 metres as the ESB height and assume 44.7 degrees latitude.

I get 0.123m (5.1 inches) as the displacement. So now we completely disagree again.

Shall we compare methods?

[ceptimus]

I just googled on NYC latitude and got 40.5 degrees :confused:

You also need to take into account how far the top of the tower is from the earth's axis. Remember the tower is leaning at an angle of approximately (90 - latitude) to the axis.

I will keep adding quibbles till you get nearer to my guess. :D

[Hamish]

I just googled on NYC latitude and got 40.5 degrees

That's what you get for trusting people.;) Yes, NYC is at about 40.5 latitude. However, due to a mistake I've just spotted in my calculation, my new figure happens to be 0.124m, so not much change.

You also need to take into account how far the top of the tower is from the earth's axis. Remember the tower is leaning at an angle of approximately (90 - latitude) to the axis.

No, this was implicit in my calculation.

Any more quibbles?

[Abdul Alhazred]

Originally posted by ceptimus
I just googled on NYC latitude and got 40.5 degrees :confused:

You also need to take into account how far the top of the tower is from the earth's axis. Remember the tower is leaning at an angle of approximately (90 - latitude) to the axis.

I will keep adding quibbles till you get nearer to my guess. :D

There's a Foucault's pendulum at the United Nations HQ, in but not of NYC.

It's been a stationary hanging weight every time I've seen it. The last time was August of 2001, the first time was sometime in the 1960s. Any updates on that? :p

[ceptimus]

Originally posted by Hamish
Any more quibbles? Err, Err... Ah!

You do realise, I take it, that the earth spins in a sidereal day, not a mean solar one? About 23 hours 56 minutes per rev.

Type 'radius of earth in inches' into google, and it will say: 251 106 299 inches

What's the latest figure? Are we getting near to 7.33 inches? I'm running out of ideas. :D

[davefoc]

I used the wrong latitude. Sorry. However, the source I just read lists the latitude as 40'45" which 40.75 degrees.

source for latitude
http://www.fourmilab.ch/cgi-bin/uncgi/Earth?imgsize=320&opt=-z&lat=40.7517&ns=North&lon=73.9942&ew=West&alt=15&img=learth.evif

Assuming this, I get 6.25 inches of deflection.

[davefoc]

You do realise, I take it, that the earth spins in a sidereal day, not a mean solar one? About 23 hours 56 minutes per rev.


Yes, I was aware of this. I rounded off to 24 hours. I'll redo calcs based on 23 hour, 56 minute day.

For those of you not familiar with this the other four minutes comes from our rotation around the sun.

[davefoc]

My calculations (modified to eliminate a stupid mistake and to change to 23 hours 56 minutes for the length of day.


earth circumference 40075.000 40075.000 40075.000
time for one earth revolution 23.930 23.930 23.930
speed at surface 1674.676 1674.676 1674.676
height ball dropped at 10.000 100.000 381.000
circumference @ drop height 40137.832 40703.319 42468.894
speed @ drop height 1677.302 1700.933 1774.713
gravitational acceleration 9.800 9.800 9.800
time for ball to drop (secs.) 1.429 4.518 8.818
time for ball to drop (hours) 0.000 0.001 0.002
latitude 40.750 40.750 40.750
latitude (radians) 0.711 0.711 0.711
displacement @ equator 0.001 0.033 0.245
displacement @ equator(inches) 0.041 1.298 9.654
displacement @ latitude 0.001 0.025 0.186
displacement @ latitude (inches) 0.031 0.983 7.314


So now Ceptimus and I are off by about ~.02. perhaps the difference is in latitude I used.

[davefoc]

OK, I used 40.5 for latitude (I think 40.7 might be more correct), improved the precision on the metric to inch conversion and used the Ceptimus suggested value for radius. There is now only a rounding error between the davefoc number and the Ceptimus number. My number was 7.336 (rounds to 7.34. Ceptimus listed his number as 7.33.

Neither of us has taken into account the fact that the earth is not quite a sphere, the exact latitude of the Empire state building, any gravitational effect from the building as the coin drops, etc. Maybe somebody would like to suggest some other things we haven't taken into account. I didn't find a better figure than 9.8 meters/sec/sec for gravitational acceleration. Maybe there's some error there.

My numbers:

earth circumference 40074.784
time for one earth revolution 23.930
speed at surface 1674.667
height ball dropped at 381.000
circumference @ drop height 42468.678
speed @ drop height 1774.704
gravitational acceleration 9.800
time for ball to drop (secs.) 8.818
time for ball to drop (hours) 0.002
latitude 40.500
latitude (radians) 0.707
displacement @ equator 0.245
displacement @ equator(inches) 9.647
displacement @ latitude 0.186
displacement @ latitude (inches) 7.336
ceptimus value 7.330
ceptimus - davefoc 0.006


Edited to add:
Nitpick away. It is obviously very important that we know this number down to at least 5 decimal places of precision.

[davefoc]

This site lists the latitude of the Empire State Building:
http://www.geosnapper.com/view.php?coll_id=65&wyp_id=624

The listed value is N 40° 44.9127 or 40.748545 degrees.

I don't think we've taken into account the tidal forces of the sun and moon and their effect on gravitational acceleration either.

[Abdul Alhazred]

Originally posted by davefoc
This site lists the latitude of the Empire State Building:
http://www.geosnapper.com/view.php?coll_id=65&wyp_id=624

The listed value is N 40° 44.9127 or 40.748545 degrees.

I don't think we've taken into account the tidal forces of the sun and moon and their effect on gravitational acceleration either.

Wind blowing around and up the ESB will far outweigh any of this. And is in principle not calculable.

Calculating from a non-set-back ESB in a vacuum is like calculating from a spherical cow.

That was my point before, even though I was also kidding around.

It's a nonsense question, not one with an answer.

http://www.laurasnyctales.com/temping/empire.jpg

[davefoc]

Abdul Alhazred,
It amazes me that you can't see how important this question is.

Clearly, we who have been calculating plan to encase the empire state building in a large bell jar and evacuate it for this experiment so that wind resistance won't be an issue. We also plan to build a nice precision release device that drops the coin cleanly to the sidewalk positioned precisely 381 meters above the sidewalk. We will then measure the point that the coin drops to five decimal places of accuracy.

If encasing the ESB in a bell jar and evacuating it turns out to be impractical I think we might consider doing the experiment in an elevator shaft. It will be possible to take the effect of air resistance into our calculations. Assuming a hundred meter elevator shaft we would expect to see about an inch of displacement.

Nice picture of the ESB by the way.

[ceptimus]

Type: 'g in m/s/s' into google and you'll get: 1 g = 9.80665 (meters / second) / second. And so you arrive at my guess. :P

However, I didn't calculate the distance from the plumb line, as that wasn't what xouper asked for. The plumb line hangs deflected to the south by the earth's rotation. There is a slight north-south discrepancy on where the dropped weight would land, as well as the calculated 7.33 inches of eastward offset.

Of course the plumb bob would hang parallel to a straight sided building, as that is what the building was lined up against when it was built! However, the plumb bob does not point down to the centre of the earth, missing it by a wide margin to the south. Anyone care to calculate by how much?

[Abdul Alhazred]

Originally posted by davefoc
Abdul Alhazred,
It amazes me that you can't see how important this question is...

Cute.

The original plan was to moor zeppelins to the tower and have a boarding area on the eighty-fifth floor terrace. Really.

If you go to the top floor (one hundred and first) and put your head against the window and look straight down you can see the great big mooring hooks.

There was one experimental zeppelin mooring.

http://www.poster.net/anonymous/anonymous-empire-state-building-with-graf-zeppelin-1931-2804912.jpg

Updrafts quickly made the zeppelin go tail up.

The idea was shelved, and shortly thereafter came the Hindenberg disaster. So no more zeppelins, don't bother about figuring out how to fix the problem.

It is entirely possible that a penny dropped from the top of the Empire State Building would never reach the ground at all.

[davefoc]

Using the Ceptimus suggested value for gravitational acceleration I now get 7.333 inches. exactly in agreement with Ceptimus's 7.33 value.

I get 7.306 when I use what I believe is the correct value for the ESB latitude.

As to the plumb bob issue, I'm going to think about that. I don't understand it right now.

It appears that there is still about a 2.2 inch discrepancy with respect to the Hamish estimate.

[Abdul Alhazred]

Originally posted by davefoc
As to the plumb bob issue, I'm going to think about that. I don't understand it right now.


Plumb Bob Squarepants?

http://www.webundies.com/images/mk1126.jpg

[Abdul Alhazred]

Originally posted by ceptimus
Err, Err... Ah!

You do realise, I take it, that the earth spins in a sidereal day, not a mean solar one? About 23 hours 56 minutes per rev.

Type 'radius of earth in inches' into google, and it will say: 251 106 299 inches

What's the latest figure? Are we getting near to 7.33 inches? I'm running out of ideas. :D

Inches? Let's not insult the rest of the world with our silly "traditional English measure".

That's 6378099994.6 millimeters.

It's nice knowing the radius of our own sweet planet to the nearest tenth of a millimeter.

A more useful figure is the circumference of the Earth: 40074784173.8 mm

That's 31705.340 furlongs for you "traditional English" (American) types.

BTW "traditional English" is not identical to "Imperial", a scheme that the British tried to foist on the world during Victoria's reign.

In a free country people measure things however they feel like it! :p :D

Obligatory ESB picture:
http://www.lileks.com/postcards/ny/emp.jpg

I was wrong. As can be seen from this picture, the first setback is well below the twentieth floor. I was thinking of the second setback.

You could do a better penny out the window experiment from the public housing project where I grew up (and my mother still lives).

Twenty one floors straight down. Sorry no picture.

[CurtC]

davefoc wrote:
earth circumference 40074.784
speed at surface 1674.667
height ball dropped at 381.000
circumference @ drop height 42468.678
speed @ drop height 1774.704
There's something off by three orders of magnitude here. The circumference is in kilometers, the building height in meters. It appears as though you've treated them both as one or the other. If you're calculating the circumference of a circle that the top of a building 381 m tall, sitting on the equator, would trace, that should be 40077.177 km, not 42468.678.

[davefoc]

CurtC,
I noticed that I had moved from km's to meters without noting that in the data I posted. I added dimensions to the data this time. You were also correct that I made a mistake on the speed calculation because I didn't handle KM's and meters correctly. Wasn't I smart to make a mistake that canceled itself out.

One thing I've learned from this is that I shouldn't do anything math related at 3am in the morning when I can't sleep.


earth circumference (meters) 40074784.2000
earth radius (inches) 251106299.1642
time for one earth revolution (hrs) 23.9300
speed at surface (meters/hr) 1674667.1208
height ball dropped at (meters) 381.0000
earth circumference @ drop height 40077178.0936
speed @ drop height (meters/hr) 1674767.1581
gravitational acc. (meters/sec/sec) 9.8067
time for ball to drop (secs.) 8.8149
time for ball to drop (hours) 0.0024
latitude (degrees) 40.5000
latitude (radians) 0.7069
displacement @ equator (meters) 0.2449
displacement @ equator(inches) 9.6437
displacement @ latitude (meters) 0.1863
displacement @ latitude (inches) 7.3331
ceptimus value 7.3300
ceptimus - davefoc 0.0031

[Hamish]

Originally posted by davefoc

It appears that there is still about a 2.2 inch discrepancy with respect to the Hamish estimate.

Yes, and I can't easily reconcile that. I can just go on finding the discrepency to higher and higher orders of magnitude.

So, if both you and ceptimus are getting very similar answers then you're either both making the same mistake or there is a serious flaw in my approach to this problem. It's more likely that I'm wrong. How are you calculating the figure? Can you describe your method?

I'll show you mine if you show me yours.;)

[diddidit]

NASA Lewis (er, Glenn, I guess) outside Cleveland has a 500 foot deep vaccuum drop tank. Somebody write up a grant request...

did

[davefoc]

Your thought is that this is a more practical approach than putting a bell jar around the ESB?

[CurtC]

Yeah, davefoc, can you share your method of calculation? I used a simple method to show an upper bound, and cut that in half for my estimate of three to four inches. 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? That seems like a complicated algebraic excercise which I don't get any hint of in your tabular data.

[davefoc]

displacement =
(speed@top - speed@surface) * T

circumferenceOfEarth = COE
heightOfDrop = H
gravitationalAcceleration = g
earthRotationPeriod = P
timeForDrop = T
latitudeOfDropLocation = L


speed@top = ((COE + 2 * pi * H) / P ) * cos (L)

speed@surface = (COE / P) * cos(L)

T = SQR( 2 * H / g)

displacement =
((COE + 2 * pi * H ) / P - COE / P) * cos (L) * sqr ( 2 * H / g)

displacement =
(2 * pi * H / P) * cos (L) * sqr (2 * H / g)

displacement =
(2 * 3.1415 * 381 /( 23.93 * 3600)) * cos (40.5) * sqr (2 * 381 / 9.8)

displacement = .186 meters

displacement = 7.3 inches

Note: the precision of some of the numbers was truncated here for clarity.

Note 2: The latitude of the ESB appears to be about 40.75 degrees. 40.5 was used here so as to produce a result that better correlated with the Ceptimus result

[ceptimus]

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.

[davefoc]

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?

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?

[CurtC]

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.

[davefoc]

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.

[davefoc]

Fixed one error. Now I have a plumb bob error of about 19 inches. Still a lot.

[CurtC]

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.

[davefoc]

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.

[CurtC]

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.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.

[ceptimus]

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.

[Walter Wayne]

Originally posted by CurtC
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. 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

[ceptimus]

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.

[davefoc]

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.

[Hamish]

Originally posted by ceptimus
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.

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.

[Hamish]

Originally posted by ceptimus
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.

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.

[ceptimus]

Originally posted by Hamish


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. 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.

[davefoc]

Ceptimus, How is it that you knew about the plumb bob error? I have no recolletion of hearing about it.

[ceptimus]

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. :)

[davefoc]

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).

[Hamish]

Originally posted by ceptimus
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.

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.

[davefoc]

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.

[CurtC]

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.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.

[davefoc]

OK,
I'm back from raquetball, but way too tired to think about this now.

I think that CurtC is wrong, but I'm not sure. I think we both agree that the plumb bob doesn't point to the center of the earth because of the centrifugal force produced by the earth's rotation.

I don't think this effect has anything to with the fact that the earth is not a perfect sphere.

If the earth was a perfect sphere the plumb bob would still not point to the center of the earth because of the earth's rotation and there would be a small difference in the angle of a wall which was built using geometric techniques to set a 90 degree angle from a wall which was built using levels or plumb bob's.

Does the fact that the earth is a geoide (not quite an ellipsesoid I think) make it so what we think of as level is tilted slightly so that it ends up being exactly perpendicular to the plumb bob? I don't know. I think this might be what CurtC is saying and I'm not sure he's wrong about that. I need to think more about this to understand what CurtC is saying and what the facts are.

As to what Hamish was saying about the Ceptimus/davefoc algorythm violating conservation of angular momentum, I thought about this some and don't think it does but I can't focus enough right now to get my thoughts out.

[ceptimus]

The dropped coin actually falls in the trajectory of an elipse, with one focus at the earth's centre. Over the height of the building this elipse is indistinguishable from the parabola that davefoc and I are calculating.

Dave - the buliding doesn't 'lean' relative to the local inertial reference frame. It is correct that it should line up with a plumb bob, as the same forces act on the building as on the line.

Once the coin is in free fall it will follow an elipse and the plane of the elipse will intersect the centre of the earth, so the coin will land north of the plumb bob, and can't be dropped down the south side of the building, as davefoc already stated.

[Hamish]

Ok, I don't think the difference is insignificant, I think it accounts for the 2.2 inch discrepency.


As to what Hamish was saying about the Ceptimus/davefoc algorythm violating conservation of angular momentum, I thought about this some and don't think it does but I can't focus enough right now to get my thoughts out.

Yes it does, and it's quite easy to show.

Angular momentum, L = mr²w where m is the mass, r is the distance from the centre of rotation and w is angular velocity.

The tangential velocity,v, is rw


If you assume that v is constant over the fall then, at the top, where the velocity is v₁

[/i]L=m(r₀+h)²w=m(r₀+h)v₁
[/i]

Where h is the height of the building and r₀ is the distance to the centre of roatation at the ground. At the bottom, where the tangential velocity of the penny is still v₁

L=mr₀v₁

Which isn't the same as it was at the top, therefore conservation of angular momentum is violated.

I'm prepared to be proven wrong. If someone can show that the effect I'm describing really is negligible then I'll accept the 7.33 inch figure. I'll repeat that it is entirely possible that I've made a blunder but I don't think so at the moment.

[ceptimus]

The angular momentum is v_t² / r.

Where v_t is the tangential component of the velocity, and r is the distance of the tangent from the point of rotation. I think you are confusing the radial and tangential components of the velocity.

[CurtC]

davefoc wrote:
Does the fact that the earth is a geoide (not quite an ellipsesoid I think) make it so what we think of as level is tilted slightly so that it ends up being exactly perpendicular to the plumb bob?Yes.

Think of it this way - what we consider "level" is the plane that the surface of a body of water makes at that point. Think about an Earth with no land mass, just all ocean, and try to imagine it being perfectly spherical. At that point at 40.75 degrees latitude, the plumb bob would not be exactly perpindicular to the surface of the water. But since it's water and free-flowing, the water that's more "uphill" according to the plumb bob (North) would flow to the downhill side (South), eventually causing the Earth to bulge out at the Equator. Just like it does now. That shape will be an ellipsoid. I think the "geoid" term you use means "the shape of the Earth," which is very close to a mathematical ellipsoid.

So our definitions of "plumb" and "level" are always exactly perpindicular to each other, and are a combination of the effects of the gravitational pull toward the center of the Earth, and the centrifugal force outward.

[Hamish]

Originally posted by ceptimus
The angular momentum is v_t² / r.

Where v_t is the tangential component of the velocity, and r is the distance of the tangent from the point of rotation. I think you are confusing the radial and tangential components of the velocity.

Go and check in a textbook. I think you're getting confused with angular momentum and centripetal acceleration.

[ceptimus]

Originally posted by Hamish


Go and check in a textbook. I think you're getting confused with angular momentum and centripetal acceleration. Yes. I was wrong here.

[Skeptoid]

CurtC wrote:
That shape will be an ellipsoid. I think the "geoid" term you use means "the shape of the Earth," which is very close to a mathematical ellipsoid.

I usually hear of the shape of the Earth described as an oblate spheroid. Is there a mathematical difference between an ellipsoid and an oblate spheroid?

[davefoc]

Hamish,
What you said about the angular momentum is what I thought you were going to say.

My thought was that you needed to take into account the angular momentum of the whole coin/earth system. I now think this was wrong.

I am not agreeing that you are right here, only that I am not sure. But I need to go, so I can't think about this right now.

As to what CurtC is saying about the shape of the earth making it so that the plumb bob hangs exactly perpendicular to the plane of the local ground, I think that's wrong. I think that plumb bob might hang perpendicular to the local ground though because the level we use to determine the local ground is level is caused to be tilted by the same mechanism that tilted the plumb bob. I think this would be true if the earth was a perfect sphere also though. I am going to have to think about this some more though. Maybe what CurtC is saying that the shape of the earth was formed because of centrifugal force from the rotation and per force it turns out to be just the right shape to make the local ground level. Interesting. Has the shape of the earth changed over time as the earth's rotation has slowed down? Maybe too small an effect to be able to detect.

It is interesting that something dropped will not follow the path of the plumb bob, but instead drops toward the center of the earth. I think CurtC agrees with this.

[ceptimus]

I think the way to get a pretty exact answer is:

Use a non-rotating intertial reference frame with origin at earth centre, and one axis (z) along the earth's spin axis.

1. Calculate the elipse the coin traces out.

2. Find where it intersects with the earth surface

3. Allow for the rotation of the earth during the fall.

I'm not sure my math is up to the task, but I'll maybe give it a go later. I expect the eastward offset (relative to a plumb line) to be about 7.33 inches, and a northward offset of about the same - I'm not sure - but I do expect it to be north rather than south.

[CurtC]

davefoc wrote:
Maybe what CurtC is saying that the shape of the earth was formed because of centrifugal force from the rotation and per force it turns out to be just the right shape to make the local ground level. Interesting. Has the shape of the earth changed over time as the earth's rotation has slowed down?Yes, that's what I'm saying. The Earth is the shape it is, exactly because of the centrifugal force. It's how we use "sea level" as the definition of level, and the seas are free to flow as a result of the centrifugal force.

And I don't think a dropped object will fall towards the center of the Earth - it will hit just East of the plumb bob, and very slightly South. Ceptimus, since the ellipse that the dropped object will follow is at its Northernmost peak at the moment it's dropped (imagine an ellipse slanted at 40.75 degrees, with the "pointy" end and the northernmost point at the same location), it can go only South from there. After only nine seconds, it won't have gone very far South, but only a slight amount. There's no way it will hit North of the plumb bob.

Edited to add: wait, I retract that. The ellipse will have the center of the Earth as one of its foci, so maybe will be slightly North. Hmmm. Good luck with that ellipse calculation.

[BillyJoe]

What's all this talk about centrifugal force?
I thought there was no such thing!

BillyJoe

[Hamish]

Is there a good reason why my approach was wrong? I don't see why you have to go on calculating elipses using linear dynamics when the problem can be much simplified using rotational dynamics without making any simplifying assumptions (except for all the ones about air resistance and the like). Ok, you have a nasty integral to do but I was able to do that without too much trouble. If I have made a serious mistake, can anybody determine where I made it?

To recap, I'm standing firm on my 12.4 cm (or 4.88 inches).

[davefoc]

Hamish,
I may be missing something here, but it looks to me if one takes into account the angular momentum issue the effect is to increase the eastward displacement slightly.

LT = m * (R+H)*VT
LS = m * R * VS

where
R = radius of earth at ESB latitude
H = height of drop
VT = velocity at top
VS = velocity at surface

assuming angular momentum conserved

m * (R+H)*VT = m * R * VS

VS/VT = 1.000060

This says that the coin is going slightly faster horizontally at the surface than it was at the top. This looks like the skater issue to me except that instead of pulling in her arms to spin faster the earth is pulling in the coin, but the earth doesn't pull it in very far compared to its radius so the increase in speed is small.

[ceptimus]

I don't think that's right davefoc. The path we calculated is for a parabola, whereas the real path, with angular momentum conserved, is an elipse. The math for the two curves is very much the same (if the one elipse focus is moved an infinite distance away the two are the same) but my feeling is that the 'arms' of the elipse will be inside those of the parabola.

I've not actually done the math yet, so I reserve the right to be completely wrong on this. :) My feeling is that the coin will fall slightly short of our calculated distance (7.33) but only by an infinitesimal margin. Hamish's calculation is way different, so at least one of us must be wrong.

[davefoc]

CurtC,
I think we agree that the coin is displaced eastward. The only issue on that score is how much.

But I think the coin is also going to drop significantly north of a locally vertical line and I think you don't, but it sounds like you are equivocating a bit there.

A thought experiment:
At the instant the coin is released gravity is turned off. Which way is it going. I think it is going exactly tangent to the circle that it has been following around the earth. That is it is going exactly east with no downward deflection.

Now gravity is turned back on. What happens? Gravity sucks it towards the center of the earth. Are there any other forces acting on it, excluding much smaller lunar and solar gravitational forces, no. So the coin is just pulled toward the center of the earth which at the ESB latitude is somewhat north of a direction pointed to by a plumb bob.

[davefoc]

The path we calculated is for a parabola, whereas the real path, with angular momentum conserved, is an elipse. The math for the two curves is very much the same (if the one elipse focus is moved an infinite distance away the two are the same) but my feeling is that the 'arms' of the elipse will be inside those of the parabola.


My mind is reeling. I need to go. I don't quite get what you're saying, but I'm going to think about it. It seems like the coin isn't going to follow a parabola because the earth is curved and gravity is not completely parallel through its path. I'll take your word for it that as a result the coin is now following the path of an ellipse instead. That seems like a very small effect here, though. But it seems like the nature of that effect is to increase slightly the velocity of the coin parallel to the surface. I don't see how it could slow it down.

[davefoc]

Just got the ellipse issue. I guess I can take old Isaac's word on that one. The coin is essentially orbiting the earth around the center of gravity. It just it runs into the surface before it completes a revolution.

[Hamish]

Originally posted by davefoc

assuming angular momentum conserved

m * (R+H)*VT = m * R * VS

VS/VT = 1.000060

This says that the coin is going slightly faster horizontally at the surface than it was at the top. This looks like the skater issue to me except that instead of pulling in her arms to spin faster the earth is pulling in the coin, but the earth doesn't pull it in very far compared to its radius so the increase in speed is small.

This is rather misleading, since the figures that matter are the tangential velocities relative to the ground. If you work out the relative velocities and take the same ratio, you get

VS/VT = 2.00006

The 0.00006 isn't that significant, but the fact that relative to the Earth's surface the tangential velocity of the coin at the surface is twice that at the top is quite relevant.

I think.:confused:

Trying to translate all the rotational dynamics into linear dynamics is a very tough prospect at the moment. I find it all so much simpler to start with spherical polar co-ordinates and do the whole problem in a co-ordinate set where the acceleration and velocity really are perpendicular.

I'll try another argument. Think about the motion of the penny relative to various points on the tower. If you use your original model you get very weird results since points on the tower must move relative to each other and the tower ends up slightly skewed (7.33 inches over it's height). Any vertical taken in this co-ordinate frame cannot remain a vertical over the course of the experiment.

It's very hard to convince myself, let alone others, that the errors in using a cartesian frame really do ammount to a significant difference but I'm fairly confident of my result.

As you may have guessed from my incoherent rambling, I'm quite tired. I'll try and think this through some more tomorrow.

[ceptimus]

After some research, I've decided that my math isn't good enough (or perhaps I'm just too lazy) to work out ellipses in three dimensions and their intersections with oblate spheroids. I might try a computer simulation of the problem though - the math is simpler there.

[Walter Wayne]

I cranked out an ugly matlab program that does a piece-wise simulation of a coin falling. Using:

earth radius = 40 000 km/2pi (spherical planet)
height = 381 m
latitude = 40.5 degrees
time step = 10 microseconds

I got that the coin landed 6.18" from the base of the tower, and took 15.3 milliseconds longer than a "straight" drop. I have found an error in the program, so ignore the two numbers given above.


I'll post my basic algorithm below so someone else can try if they want. I may have screwed up with the switching between spherical and cartesian co-ordinates.

Walt



Basic Piece-Wise Algorithm
while height is greater than one earth radius
- time=time + time_step
- end_velocity=begin_velocity - g*time_step*unit_radius_vector
- position=position + begin_velcocity*time_step + (end_velocity-begin_velocity)/2*time_step²
- begin_velocity=end_velocity
The third line is in error. Since (end_velocity-begin_velocity) = acceleration*time, and not acceleration, my thirld line should have been:
position=position + begin_velcocity*time_step + (end_velocity-begin_velocity)/2*time_step

Edited many, many times.

[davefoc]

Hamish said:
This is rather misleading, since the figures that matter are the tangential velocities relative to the ground. If you work out the relative velocities and take the same ratio, you get

VS/VT = 2.00006


Hamish,
I am not following your arguments here.

1. If the tangential velocity is faster at the bottom than the top wouldn't the horizontal displacement be larger than if you ignored it?

2. I agree with Ceptimus that the way to calculate this more accurately is to assume the coin drops in an ellipse and not a parabola and use the the initial height and velocity of the coin to establish the ellipse. Then you can calculate the speed of the coin at the surface.

3. I believe the result of step 2 will be an iniinitesimal change in the displacement estimate. I see two errors from assuming the earth is flat as was done in the Ceptimus/davefoc estimate.
a. The earth is curving away from the coin so that the actual distance traveled is slightly more than the calculated horizontal distance. I did a calculation to estimate this error and it is close enough to zero that I can't get a non-zero number for it with excel.
b. Gravity is slightly accelerating the coin tangent to the surface of the earth throughout the flight of the coin except at the release point where gravity is exactly perpendicular to the flight of the coin. I did an upper bound calculation of this effect by taking the maximum tangential force on the coin which occurs at the instant before the end of its flight and applied it to the whole flight. My number was 3 millionth of a meter/second additional speed. Compared to the roughly 353 meters/second horizontal speed that the coin started with the tangential acceleration looks insignificant to me.

As an aside, the tangential force acting on the coin is the reason why when a mass is pulled towards the center in a rotating object the mass accelerates. So the spinning object not only spins faster because the mass has less distance to travel around the axis because of the reduced radius it spins faster still because the mass was accelerated by the pull towards the center. I never quite understood this before this thread and your comments about angular momentum.

[davefoc]

Walter Wayne,
It seems like vbulleting is not going to be happy unless you put a space some place in the line. I put a space between the second plus and the parenthesis in the line in question and it displayed it correctly.

As to your algorithm, what is the unit_radius_vector?

Thanks, Dave

[Hamish]

Originally posted by davefoc
Hamish said:

Hamish,
I am not following your arguments here.

I'm not surprised. I don't really follow them myself. I was quite tired when I wrote them and was stuggling to find ways to get my ideas across. I don't think I succeeded.


1. If the tangential velocity is faster at the bottom than the top wouldn't the horizontal displacement be larger than if you ignored it?

No, because you are calculating the cumulative displacement from a vertical reference line which is moving with different tangential velocity at different heights. My problem is with the simplification of assuming a flat Earth. The deviation is small but so is the effect we're calculating.


2. I agree with Ceptimus that the way to calculate this more accurately is to assume the coin drops in an ellipse and not a parabola and use the the initial height and velocity of the coin to establish the ellipse. Then you can calculate the speed of the coin at the surface.

I agree with this. In fact, I believe this approach just reduces down to my original method. You will run into serious difficulties when trying to do it in cartesian co-ordinates.


3. I believe the result of step 2 will be an iniinitesimal change in the displacement estimate. I see two errors from assuming the earth is flat as was done in the Ceptimus/davefoc estimate.
a. The earth is curving away from the coin so that the actual distance traveled is slightly more than the calculated horizontal distance. I did a calculation to estimate this error and it is close enough to zero that I can't get a non-zero number for it with excel.
b. Gravity is slightly accelerating the coin tangent to the surface of the earth throughout the flight of the coin except at the release point where gravity is exactly perpendicular to the flight of the coin. I did an upper bound calculation of this effect by taking the maximum tangential force on the coin which occurs at the instant before the end of its flight and applied it to the whole flight. My number was 3 millionth of a meter/second additional speed. Compared to the roughly 353 meters/second horizontal speed that the coin started with the tangential acceleration looks insignificant to me.

Which is about the ammount you reckoned that the tangential velocities will differ between top and bottom as outlined in your previous post. I agree, the change is fairly inconsequential. I do now believe that it is the difference in the geometry frames that causes the discrepency between our answers. It's the flat Earth assumption which I currently believe is the main source of discrepency.

I'm still willing to be proved wrong about this. The thing is that my model didn't make any simplifying assumptions, whereas yours did. It is possible that I've made an error in calculation and I'll check again.

[Walter Wayne]

Originally posted by davefoc
Walter Wayne,
It seems like vbulleting is not going to be happy unless you put a space some place in the line. I put a space between the second plus and the parenthesis in the line in question and it displayed it correctly.

As to your algorithm, what is the unit_radius_vector?

Thanks, Dave Thanks for the help.

The unit radius vector is a vector from the origin to the location of the coin of length one.

So if the coin is at position (x,y,z)
Then the unit radius vector is <x,y,z>/sqrt(x²+y²+z²)

Thus (-g * unit_radius_vector) is a vector of magnitude g, pointing at the origin (i.e. the acceleration vector).

Walt

[DrMatt]

Quite exciting.
Of course, as you know by the minute amounts of these outcomes, the Coriolis effect has no significant impact on water going down drains in the kitchen or bathroom. The initial conditions--intial mean angular momentum--determine these smaller situations conclusively. A slight clockwise twist translates into stupendous clockwise angular speed when the radius of rotation is compressed from the length of your bathtub to the radius of your drainpipe. Likewise if the prevailing angular momentum is counterclockwise.
If your bathtub starts out Really Really Becalmed and it's as tall as the Empire State Building, well, you might see a prevailing result in that case. Make it a few miles tall, and we can just about guarantee that you'll see such an effect.

[Walter Wayne]

I've discovered a discrepancy in my program. So ignore the 6.18" I mentioned above until I figure out what is wrong.

Walt

[ceptimus]

My lashed up C program, which tries to calculate a trajectory from first principles, using Newton's Law of gravitation, and google's figure for the mass of the Earth, has come up with 4.898 inches, which agrees very closely with Hamish's figure.

Of course there may be a bug here. It's only a hack.// tower.c : Drop a weight from the top of a tower in a vacuum.
// Where will it land?
// assumes spherical earth, Tower vertical to surface.
// assume mass of weight = 1.0 to simplify math

#include <stdio.h>
#include <math.h>

#define PI 3.1415926535897932384626433832795
#define G 6.67300E-11

#define EARTH_RADIUS 6378100.0
#define TOWER_HEIGHT 381
#define TOWER_LATITUDE 40.5
#define ROTATION_PERIOD 86164.0905
#define EARTH_MASS 5.9742E24
#define TIME_INCREMENT 0.00001

/* coordinate system: We are looking along the Earth's axis down on the north pole
* origin is at Earth centre with z axis is pointing straight at us. North = +z
* The simulation starts with the tower parallel to the x-axis, so y = 0 at t = 0
*/

int main(int argc, char* argv[])
{
double x, y, z, xv, yv, zv, t; // weight's position and speed, time

double d, lat, lon, f; // distance from Earth centre, latitude, longitude, force

double xb, yb; // position of tower base after time t

y = xv = zv = t = 0.0;

// inital distance of weight from Earth's axis
x = (EARTH_RADIUS + TOWER_HEIGHT) * cos(TOWER_LATITUDE * PI / 180.0);

// inital distance of weight from equatorial plane
z = (EARTH_RADIUS + TOWER_HEIGHT) * sin(TOWER_LATITUDE * PI / 180.0);

// Inital speed of weight due to rotation
yv = x * 2.0 * PI / ROTATION_PERIOD;

// loop till weight collides with surface. d = distance from Earth centre
while ((d = sqrt(x * x + y * y + z * z)) > EARTH_RADIUS)
{
f = (EARTH_MASS * G) / (d * d); // force on mass
lon = atan(y / x); // longitude of weight (relative to t=0 earth position)
lat = asin(z / d); // latitude of weight

xv -= f * cos(lat) * cos(lon) * TIME_INCREMENT;
yv -= f * cos(lat) * sin(lon) * TIME_INCREMENT;
zv -= f * sin(lat) * TIME_INCREMENT;

x += xv * TIME_INCREMENT;
y += yv * TIME_INCREMENT;
z += zv * TIME_INCREMENT;
t += TIME_INCREMENT;
}

// Calculate position for base of tower after time t.
xb = cos(2 * PI * t / ROTATION_PERIOD) * EARTH_RADIUS * cos(TOWER_LATITUDE * PI / 180.0);
yb = sin(2 * PI * t / ROTATION_PERIOD) * EARTH_RADIUS * cos(TOWER_LATITUDE * PI / 180.0);

// Subtract tower base position from weight position
x -= xb;
y -= yb;
z -= EARTH_RADIUS * sin(TOWER_LATITUDE * PI / 180.0);

printf("Impact at t=%f x=%f, y=%f, z=%f\n", t, x, y, z);

// make d = distance travelled east relative to tower base
d = y / cos(2 * PI * t / ROTATION_PERIOD);
printf("Distance east =%8.5f metres =%8.3f inches\n", d, d * 1000.0 / 25.4);

// make d = distance travelled south relative to tower base
d = -z / cos(TOWER_LATITUDE * PI / 180.0);
printf("Distance south =%8.5f metres =%8.3f inches\n", d, d * 1000.0 / 25.4);

return 0;
}

resulting output:

Impact at t=8.827240 x=0.423558, y=0.124414, z=-0.496396
Distance east = 0.12441 metres = 4.898 inches
Distance south = 0.65280 metres = 25.701 inches

[edit: updated version with more sensible constants - same answer though]

[CurtC]

ceptimus, you're assuming a spherical Earth, right? There's that one little annoying thing that I can't figure out for an ellipsoid Earth, which I'm wondering whether it's significant.

It would seem, intuitively, that the dropped item would follow a plumb line until it starts to deviate to the East. However, since the plumb line points way South of the Earth's center, does the object really do that? The object is now following an ellipse, which has the Earth's center in its plane, right? And the bottom of the plumb line is not in that plane. So what does it really do?

Or does the idea that gravity attracts to the center of a sphere, not apply to an ellipsoid? Anyone remember how to do triple integrals? It's been 24 years since I did that problem in freshman physics.

[Walter Wayne]

Well I think the biggest problem is that we are all using the radius of the earth ar equator as far as I can tell and then assume a spherical earth. We should find the distance from the base of the building to the center of the earth and use that to determine the radius of the spherical chicken in a void, upon which we have placed the ESB.

Using ceptimus' numbers for mass, earth radius and G, I also believe we get 9.7992 m/s². This slight error is due I think to the non-uniform density of the earth and the oblate spheroid we are on.

Walt

[69dodge]

Originally posted by CurtC
It would seem, intuitively, that the dropped item would follow a plumb line until it starts to deviate to the East. However, since the plumb line points way South of the Earth's center, does the object really do that? The object is now following an ellipse, which has the Earth's center in its plane, right? And the bottom of the plumb line is not in that plane. So what does it really do?Ignoring the eastward deflection due to the Coriolis force, the object follows both the plumb line and the ellipse. Here's how that's possible:

The ellipse and its plane are fixed (in an inertial reference frame). The plumb line rotates with the earth.

At the moment the object is dropped, the bottom of the plumb line is not in the ellipse's plane. But that's not important, because the object doesn't hit the earth at the bottom of that plumb line. It hits the earth at the bottom of a later, rotated, plumb line, the bottom of which is in the plane of the ellipse. The ellipse's plane is flat while the earth's surface is curved, so the edges of the ellipse hit the earth closer to the equator than its center-line does.

[ceptimus]

I spotted one mistake in my code, but the answer is hardly affected. At the end, where I convert the y offset between the tower base and the weight into an eastward offset, I should have divided by the cosine of the Earth's rotation, not multiplied.

However, the Earth rotates through such a small angle in 8.8 seconds, that it makes hardly any difference. The revised figure is 4.898175 inches.

I will tweak the program later today to calculate the northward offset relative to the tower base, and allow for the Earth's slightly flattened shape.

[ceptimus]

Originally posted by CurtC
ceptimus, you're assuming a spherical Earth, right? There's that one little annoying thing that I can't figure out for an ellipsoid Earth, which I'm wondering whether it's significant.

It would seem, intuitively, that the dropped item would follow a plumb line until it starts to deviate to the East. However, since the plumb line points way South of the Earth's center, does the object really do that? The object is now following an ellipse, which has the Earth's center in its plane, right? And the bottom of the plumb line is not in that plane. So what does it really do?

Or does the idea that gravity attracts to the center of a sphere, not apply to an ellipsoid? Anyone remember how to do triple integrals? It's been 24 years since I did that problem in freshman physics. I'm not sure about the local direction of gravity near to an oblate spheroid. I'll have a go at doing the ingegrals for a flat thin disc. If the gravity field for that always points at the centre of the disc, I'll be more confident that Earth's does too.

If the field doesn't always point at the Earth's centre, it would mean that a satellite in a low Earth orbit, inclined to the equatorial plane, would not orbit in a flat plane. I guess there must be some stuff on the web about this. I'll google. :)

[ceptimus]

Originally posted by 69dodge
Ignoring the eastward deflection due to the Coriolis force, the object follows both the plumb line and the ellipse. Here's how that's possible:

The ellipse and its plane are fixed (in an inertial reference frame). The plumb line rotates with the earth.

At the moment the object is dropped, the bottom of the plumb line is not in the ellipse's plane. But that's not important, because the object doesn't hit the earth at the bottom of that plumb line. It hits the earth at the bottom of a later, rotated, plumb line, the bottom of which is in the plane of the ellipse. The ellipse's plane is flat while the earth's surface is curved, so the edges of the ellipse hit the earth closer to the equator than its center-line does. I don't think that's quite right 69dodge. I think we all agree that the orbit of the weight (if the Earth didn't get in the way) would take it around the Earth's centre, and as it starts 40.5 degrees north of the equatorial plane, at some point it would dip south of that plane. So there is no way that the plumb line can do that.

The 3D shape traced out by the plumb line as the Earth rotates is a cone, with the apex of the cone on the Earth's axis, but some way south of the centre of the earth, due to the centrifugal reaction that the rotation imparts on the plumb bob.

There is no way that the plane of the elliptical orbit can remain superimposed on the cone for long. Of course, it may do so for the roughly nine seconds that the coin takes to drop - I don't think any of us are completely sure about this yet. :)

[Hamish]

Originally posted by Hamish

I agree with this. In fact, I believe this approach just reduces down to my original method. You will run into serious difficulties when trying to do it in cartesian co-ordinates.


I'll take that back. Nice job, ceptimus. I think your answer is probably more accurate than mine.

Wow, you even took the change in gravity over the drop into account. No way was I going to try that, it just made the maths a bit too hairy. I guess it didn't make that much difference to the final answer though.

[69dodge]

Originally posted by ceptimus
There is no way that the plane of the elliptical orbit can remain superimposed on the cone for long. Of course, it may do so for the roughly nine seconds that the coin takes to drop - I don't think any of us are completely sure about this yet. :)The direction of the plumb line is determined by gravity and centrifugal force. The path of the falling object is affected by the Coriolis force too. But the gravitational force and centrifugal force are identical on the falling object as on the plumb line. That's what I was trying to say. So, basically, as long as the Coriolis force is negligable, we can, um . . . , neglect it. :)

The Coriolis force is proportional to the speed of the falling object towards the earth's axis of rotation, and is always perpendicular to the object's velocity as well as to the earth's axis. If the object is falling quickly, approximately towards the center of the earth, its velocity has a significant component towards the earth's axis. So the Coriolis force will deflect it east (in the northern hemisphere). But this eastward velocity, due to the Coriolis force, is much smaller than the velocity toward the axis, due to gravity. So any further Coriolis force due to this eastward velocity, which force would be directed away from the axis (i.e., southward and upward), is negligable, I'm pretty sure, compared to the centrifugal force in the same direction.

[Walter Wayne]

BTW, correcting my original program to yield proper results (I had missed a cos(latitude) at one point), I find that:

t=8.82368 sec
east deflection = 4.88549"
south deflection = 25.63639"

Still based on a spherical planet with a 40 000 km circumfrance and a constant g of 9.8067 m/s²

Walt

[Floyt]

You are aware that you can acquire merit by giving to paupers, right? :D So let me just despoil the rarefied air with the obvious question relating to this:

Originally posted by davefoc


Yes, I was aware of this. I rounded off to 24 hours. I'll redo calcs based on 23 hour, 56 minute day.

For those of you not familiar with this the other four minutes comes from our rotation around the sun.

Why is that? How does Earth's rotation around the sun add to the duration of the planet's own rotation?

cheers
floyt
(used to write upside-down messages on the pocket calculator in math class, in case you're wondering)

[ceptimus]

The Earth rotates in a sidereal day (aproximately 23 hours 56 minutes). This is the time it takes for a very distant object, like a star or galaxy to appear to make one rotation.

The Earth will have moved part of the way around its orbit of the sun during the day, and to get the sun back into the same position in the sky as yesterday (as near as possible) the earth has to rotate another 4 minutes. The earth actualy spins about 366.25 times per year, but as the earth also moves round the sun once in a year, the sun appears to go around the Earth one turn less, or approximately 365.25 times.

The diffierence between the two kinds of 'day' is approximately 1/366.25 of a sidereal day which is about 4 minutes.

[Floyt]

Got it. Thanks!

[davefoc]

Walter Wayne's south displacement estimate: 25.63639 inches
Davefoc's south displacement estimate: 25.6512 inches

Using
40,000 km circumference
9.8067 gravitational acceleration
40.5 degrees latitude
381 meter drop
23 hours 56 minute sidereal day

Hmm, the last time I had a result this close to somebody else's we were both wrong. Still it is nice to see a result that corresponds roughly with one of mine since my 7.33 inch eastward displacement seems to be getting battered by a raft of results that don't agree with it.

I'm going to spend some time looking at what's going on here with respect to the eastward displacement. I don't understand why the 7.33 inch estimate is so wrong. But now there have been two simulations and a theoretical calculation by Hamish that suggest its about 4.8 inches.

[ceptimus]

Did you notice that my program came up with a z displacement (relative to the release point) of 247.936664 metres? My model worked with a tower 381 metres high ata latitude of 40.5, perpendicular to a spherical Earth, so that accounts for 247.4397 metres. The difference is 0.4969 metres or 19.564 inches, and indicates that the weight would land that far to the south of such a tower.

I've thoroughly confused myself reading about oblate spheroids, geoids and so on. Apparently, local gravity to an oblate spheroid is not perpendicular to the surface, unless you count centrifugal force as part of the gravity. Neither does local gravity point through the centre of the Earth when the rotation effect is ignored - the bulge of material around the equator causes a 'towards the equator' component, that produces a torque on the orbits of satellites in low inclined orbits, causing the orbits to precess. I don't understand how all these effects combine, and they are not easy to model.

I confess I have no idea now whether the weight would land north or south of the tower, or dead in line with it.

[69dodge]

Originally posted by ceptimus
Apparently, local gravity to an oblate spheroid is not perpendicular to the surface, unless you count centrifugal force as part of the gravity.Right. So count it.

I think the easiest approach here is to work in a rotating coordinate system fixed to the earth; all we need to do is include in our calculations the Coriolis force and centrifugal force. (These are simply fictional forces that are due to the use of a rotating coordinate system; they have nothing to do with the shape of the earth, just its rotation rate.) So the three forces on our falling object are gravity, centrifugal force, and the Coriolis force. But the sum of gravity and centrifugal force has a known direction: it's the local vertical, based on which buildings are built, and relative to which we wish to measure the displacement of the falling object.

So I'm pretty sure that the object will not be deflected north or south of the building's base, except for an exceedingly small southward Coriolis force due to the object's small velocity eastward, which in turn is caused by the Coriolis force due to its significant velocity toward the earth's axis.

The only thing missing that we need to know is the angle of the local vertical relative to the plane of the equator. Then we can calculate the component of the object's velocity toward the axis, which, together with the earth's rotation rate, determines the eastward Coriolis force. On a spherical earth, this angle is just the latitude. On an ellipsoidal earth, a quick Google search yielded this page (http://www.colorado.edu/geography/gcraft/notes/datum/datum.html), which says, I think, that we want the "geodetic latitude" of the point. I don't know whether that is the same as the usual latitude. (This is just to calculate the size of the eastward deflection. It's not related to the question of the existence of a northward or southward deflection.)

[CurtC]

I think the difference between local gravity, and gravity towards the center, will be a significant difference at this level of precision. So we've probably gone about as far as we can without modelling that.

I'm still intrigued by the difference between the two methods. It does seem like the difference in speed times the fall time would give a pretty darn precise answer, but it's quite a bit off from the numerical approaches. I'd like to understand this better.

[ceptimus]

Although the net force, due to gravity and rotation is normal to the surface, at the surface this effect varies with height above the surface. This means that a very tall building, with each floor accurately lined up to plumb bob, would actually have a slight curve!

It is easy to calculate the centrifugal component of the force - this can be subtracted from the normal-to-the-surface total force to give the local gravitational component. This has a slight southward-pulling component in New York, due to the extra 'bulge' of material around the equator.

Once you are far enough away from the surface (say as far as the moon) the gravitational component does point through the centre of the Earth, and the normal to the point on the Earth that happens to be closest to the moon, even after the centrifugal component is subtracted, will point in a slightly different direction.

I don't know how fast the effect falls off with height, and as we are only considering small deflections, this may be important. A true simulation would need to cater for how the Earth's mass is distributed, taking into account not just the non-spherical shape, but also the differing density with depth.

I don't think this affects the eastward offset, as symmetry cancels out most of these awkward effects, but to get an accurate number for the north/south offset, maybe such complications need to be considered?

I'm going to abandon any further changes to my simulation unless someone comes up with an accurate model that is still simple enough to work with.

[Walter Wayne]

I did a calculation of the plumb bob displacement on a spherical planet, and interestingly I got 25.52" south displacement. Not far from the 25.64" I got for the dropped ball using a numerical approach.

The method I used to come up with the formula was fairly simply. I calculated the net force given that the plumb line must see given that it is following a latitude circle and subtract the force of gravity. The result must be the force applied by the string upon which we hang the plumb bob. Calculate the angle of that. Note that the result for the angle were identical to about several decimals wether I looked at the top or bottom of the tower, so I didn't need to integrate over the height of the plumb bob.

The equation below is my result, where:
g = acceleration due to gravity at the spheres surface (m/s²)
R = radius of sphere (m)
h = height of plumb bob above surface (m)
omega = radial velocity (radians/second)
phi = latitude (radians)
theta = deviation from vertical of plumb line (radians)

Walt

[davefoc]

Walter it is surprising how different your formula looks to the way I got my answer, but the results seem quite similar.

Could you try a few different latitudes and heights. I'd like to put them into my calculation and see if we continue to get the same results.

Dave

[Skeptoid]

Suppose you scrap the plumb bob entirely and replace it with a laser that is leveled is such a way as to point to the Earth's center of mass. Drop the penny from the point where the laser light is emitted. Where does the penny strike the sidewalk in relation to the red dot on the sidewalk?

[ceptimus]

Originally posted by Skeptoid
Suppose you scrap the plumb bob entirely and replace it with a laser that is leveled is such a way as to point to the Earth's center of mass. Drop the penny from the point where the laser light is emitted. Where does the penny strike the sidewalk in relation to the red dot on the sidewalk? According to my model, which simulated a spherical Earth, the penny strikes the ground 19.564 inches closer to the equatorial plane than the laser. Measured along the ground, this figure must be divided by cos(latitude) and as I used 40.5 for my latitude figure, this gives a southward offset of 25.728 inches from the laser.

With a spherical Earth, the density distribution doesn't matter, but as soon as you consider a non-sphere, you need to take into account the Earth's density variations with depth.

[davefoc]

What is the force that is pulling the weight south?

If the earth stopped rotating and all we were dealing with was earth gravity wouldn't the weight be drawn exactly towards the center in parallel with the laser beam?

Does the weight only have eastward motion when it is released? If so what force is making it deviate from its eastward path?

Do we agree that building are leaning to the north relative to the laser beam?

Sorry, several questions asking basically the same thing, but I just don't get why weight would end up south of drop point.

edited to add: I just noticed that I have been using the words south displacement for what we are talking about. I meant that the weight is displaced north of the building base in my thinking. Hmm, is Walter saying he thinks the weight is displaced south of the building base? We end up with nearly the same quantitative estimate but Walters is South and mine is north?

[ceptimus]

My model assumes a spherical Earth, with the building perpendicular to the surface. So the axis of the building points through the centre of the Earth.

The reason for the 25.7 inch southward deviation of the penny (from the base of the building, or the laser in Skeptoid's version) is that the penny starts out on an orbit that would eventually take it around the centre of the Earth (if the Earth didn't get in the way). The furthest north the penny would ever get on it's orbit is the release point. The plane of the orbit is at the same angle to the equatorial plane as the latitude of the building, 40.5 degrees in my version.

Now the axis of the building is in this same plane when the penny is released, but when the penny reaches the ground, some 8.8 seconds later, the Earth has rotated carrying the building out of (north of) this plane. The base of the building is now 25.7 inches north of the penny's impact point.

[edit to add: This only applies to the model. I've still no idea on the north/south offset of a real coin with real building, on a real Earth. I think the 4.898 inches eastward offset is about right though.]

[Walter Wayne]

Same goes for a plumb bob. Anything on a sphere moving in a straight line follows a great circle route (circle centered on center of sphere with radius r). If one walks along a line of latitude one must continually turn away from the equator. Thus New York turns to the north, due to lateral forces the spinning earth places upon it.


On an analytical solution:
I am getting close to an analytical solution for a spherical planet. I have found Keplers Laws somewhat anoying for finding east-west displacement. With them one can precisely define where the elipse intersects the planet. Thus I have an analytical solution for south displacement of the coin (25.63").

However, to find the east displacement is pretty difficult. Even finding the solution for where the coin lands doesn't help, if you don't know when it landed. Using Kepler's second law, I end up with a very difficult integral to t. If I approximate the integral to find t, I can bound the time of impact to within about 1.1 milliseconds. However, in 1.1 milliseconds, the earth surface travels about 14". Not very accurate.

The ironic thing is, my previous approximations were more accurate on this, as any error in time ended up effecting both the building movement and coin impact. However, my "exacting" solution is less accurate because I can determine the impact point precisely, but the amount the building has moved has a significant error due to approximations in impact time.

If I have time a might post a bit more about it later.

If anybody is really bored, you can try the integral below. :D
I'll feel real stupid if someone does this quickly.

[Hamish]

Ok, enough theorising. I say we submit a grant request for a 382 metre bell jar, a bloody huge vacuum pump, some lasers, a plumb bob, a cunning scaffold to fix to the roof and the exclusive use of the Empire Sate Building for a couple of days.

Oh, and a penny too.

That ought to settle the issue.

[davefoc]

ceptimus,
If the spherical earth you are assuming the buildings and the plumb bob will not point toward the center.

Spin the earth faster so the effect is more pronounced.

What happens? Centrifugal force pulls at the weight.in the plane of the rotation. So the angle the plumb bob hangs at is the sum of two vectors, one the gravity vector and the other the centrifugal force vector.

The above is true even if the earth is a perfect sphere.

Eventually as the earth spins faster and faster the weight at 381 meters won't fall at all. 381 meters will become high enough for a geosynchronous orbit.

Wayne,
I'd really like to see you try your equation out on a different latitudes so I can see how what it predicts compared to my calculations.

My numbers
displacement (inches)

latitude nortward tilt(degrees) displacement from base of 381 meter drop
0 0 0
10 .0340 8.9
20 .0639 16.74
30 .0861 22.55
40 .0978 25.63
50 .0977 25.61
60 .0859 22.51
70 .0637 16.7
80 .0339 8.824
90 0 0

[ceptimus]

davefoc,

Yes, I agree the buildings in New York would lean to the north on a spherical Earth. Maybe I was the first to mention that? The problem, as I see it, is that there are other effects to take into account:

1. The Earth isn't round. It's approximately an oblate spheroid (the shape generated by rotating an ellipse around its minor axis).

2. The distribution of mass within the Earth isn't constant - it has a dense nickel-iron core, a mantel and a lighter crust etc.

3. Gravity direction close to an oblate spheroid (even a constant density one) varies. Close to the spheroid, there is a 'towards the equator' component of gravity. This component decreases with height.

4. Because of these effects, the direction of 'up' is not a straight line, so buildings with each floor accurately aligned to 'up' will be slightly curved.

I can't see the point of allowing just for the centrifugal component, while neglecting the others, until someone works out, or finds a reference on, the relative magnitude of each effect.

[davefoc]

Ceptimus, thank you for your response. I think I understand better what you are saying now.

My thoughts:
I think to a very high degree of accuracy one could assume that the earth is a sphere plus some material that causes it to take the shape it does. The sphere will pull towards the center. It doesn't matter that the sphere is made up of different density material. The material is arranged concentrically so when a mass is outside the sphere the gravitational field it experiences is just like all the mass was at center.

The material that is outside the sphere also contribute to the local gravity. Since there is more of it towards the equator there will be a net southward pull (in northern latitudes) from this material so buildings will lean somewhat more north to compensate for that. In addition there are local gravitational variations caused by variations in local densities and large masses near by that will effect the direction of gravity.

However, I think these variations in gravity will effect the plumb bob and the dropped weight equally whereas the centrifugal force on the weight is removed once it is let free.

So my thought here is that the shape of the earth will not cause a significant displacement between where the plumb bob points and where the weight lands. The shape of the earth causes a variation in the gravitational force. This is a minor effect and would only affect the calculation slightly in that the drop time is changed sligthly as a result of the gravitational variation.

[davefoc]

This is a link to some calculations done on the effect of rotation and the earth's shape on local gravity.

http://www.geo.cornell.edu/geology/classes/isacks/geoid.pdf

[69dodge]

Originally posted by davefoc
But now there have been two simulations and a theoretical calculation by Hamish that suggest its about 4.8 inches.Add the formula at the bottom of this page (http://scienceworld.wolfram.com/physics/CoriolisAcceleration.html) to the list.

[arcticpenguin]

Originally posted by 69dodge
Add the formula at the bottom of this page (http://scienceworld.wolfram.com/physics/CoriolisAcceleration.html) to the list.
Apologies for the intrusion, but Dodgemeister, please check your PM box.

[davefoc]

I plugged the numbers into equation referenced by 69dodge.

Using their value for sidereal rotation I got 4.8878 inches. Using value for sidereal rotation we have been using I got 4.8887 inches.

Either way the estimates are very close to the Hamish calculation and the Ceptimus and Walter Wayne simulations and very far from the davefoc number.

Walter Wayne 4.88549 inches
Ceptimus 4.8982 inches
Hamish 4.8819 inches (.124 meters)

So at this point what is davefoc to conclude? That he is wrong and the rest of the world is right or that he is right and the rest of the world just doesn't see this quite correctly?

Well, davefoc is going to think about this some more but I sense a lack of confidence in his original estimate.

[sorgoth]

Originally posted by davefoc


New question: How much is the penny going to be displaced because of the gravitational pull of the empire state building?


Very, VERY little.

[Skeptoid]

davefoc wrote:
So at this point what is davefoc to conclude? That he is wrong and the rest of the world is right or that he is right and the rest of the world just doesn't see this quite correctly?
This is where davefoc should go woo-woo and defend his result despite evidence to the contrary. Invent fudge factors justified by new and undetectable forces (use of the word quantum will come in handy here). Accuse the others of being muddled in scientific dogmatism. Tell them how Galileo was ridiculed when he dropped things from tall buildings and got the "wrong answers" and won't they be the stupids when you are proven correct.

[Walter Wayne]

Originally posted by davefoc
Wayne,
I'd really like to see you try your equation out on a different latitudes so I can see how what it predicts compared to my calculations. Virtually identical, but I have southward displacement.


lat tilt displacement
0 0 0
10 0.0337 8.8252
20 0.0634 16.5910
30 0.0854 22.3632
40 0.0972 25.4447
50 0.0972 25.4600
60 0.0856 22.4017
70 0.0635 16.6348
80 0.0338 8.8538
90 0 0

[Walter Wayne]

Does anyone have data on the approximate direction of gravity in New York on oblate earth. I think with the small change of height (relative to earth scale) and the fact the coin is prevented from doing a much of an ordit, we could approximate its fall as an elipse. Ideally, I like the value and angle of g at both the top and bottom of the building as a reality check.

So with two values of g one could find the center of gravity as seen from New York, at that point one could calculate the equation of the coin orbit around said center and then calculate it impact point. With one value of g, we would assume constant gravity center, and calculate from their.

My (almost) analytical solution for a spherical went:
1. Calculate equation of elipse in the plane of orbit.
2. Calculate at what point in orbit coin hit sphere.
3. Move co-ordinate system by (latitude) degrees, to bring elipse into "earth" co-ordinate system.
4. Calculate latitude and longitude displacement of coin impact
5. Calculate time it takes coin to hit ground (this is step where I had to resort to numerical methodology)
6. Calculate longitude displacement of building over that time.

I think for an oblate spheroid the changes to this methodology would be: first, a more complicated change in the co-ordinate system as the orbital center is no longer the center of the earth and second, calculate the impact point after changing co-ordinate systems. All lot more intensive, but unless the gravity over the height of the building is more varied than I believe, it is probably doable.

Walt

[davefoc]

69dodge said:
[QUOTE]So I'm pretty sure that the object will not be deflected north or south of the building's base, except for an exceedingly small southward Coriolis force due to the object's small velocity eastward, which in turn is caused by the Coriolis force due to its significant velocity toward the earth's axis.QUOTE]

And I think he was right.

The 25.701 inches that Ceptimus's simulation determined for the southward motion is almost exactly offset by the northward lean of the building. I got 25.699 inches for the offset of a plumb bob hanging 381 meters above the ground at a latitude of 40.5 degrees based on the vector addition of gravity and centrifugal force.

The key thing that I didn't get was that the coin continues to experience centrifugal force even after it is released.

[Walter Wayne]

Originally posted by davefoc
The key thing that I didn't get was that the coin continues to experience centrifugal force even after it is released. I'm not sure I understand this. The coin stops experience no centripetal force and thus orbits the middle of the spheroid earth. The building continues to experience a centripetal force, that keeps in orbiting the axis of rotation in the x-y plane.

If this doesn't help you, see if you have a globe that indicates the ecliptic. The ecliptic is a circle centered on the earth's center which goes from the tropic of cancer, down to the tropic of capricorn. Looking at the globe, a building on the intersection of the tropic of cancer and the ecliptic, would travel along the tropic (a latitude line) because the ground exerts a centripetal force on the building. A coin released from the top will try to follow the ecliptic because the only force on it is gravity.

Walt

[davefoc]

Walter,
I apologize, but I'm not quite sure I understood your comment.

Do we agree on the following?

1.The building points north relative to a line drawn from the center of the earth.

2. The reason the earth points north is that centrifugal force caused by its rotation around the earth changes the direction slightly of the downward force on the building.

3. If the earth did not rotate the coin would drop straight towards the center of the earth. (I set the rotation period to a very large number in the Ceptimus simulation and in fact this is what happened. There was no offset from the base).

4. The coin follows a path somewhat south of a line that points towards towards the center of the earth.

5. The coin, almost exactly, follows the line of the plumb bob as it falls (excluding the eastward motion caused by the slightly higher speed at the top of the building). My thought here is that there is a slight variation because the coin experiences a slightly different gravitational field as it falls whereas the plumb bob only experiences the gravity field near the earth's surface. Assuming my calculation for the angle of the plumb bob and Ceptimus's simulation are correct, this is a very slight variation.

As to the specifics of what you said about the ecliptic, I put 23.5 degrees in to Ceptimus's simulation and came up with a south displacement of 19.044 inches for the coin. So at least by the results of the Ceptimus simulation the coin does not follow the ecliptic but rather follows a path slightly south of it. I put 23.5 degrees into my vector addition formula. The formula predicted a northward lean of the building of 19.047 inches at a latitude of 23.5 degrees. Once again the northward lean of the building seems to almost exactly match the southward path of the coin as it drops.

I realize that the calculations above are all based on a rigid, spherical earth and the actual results would be somewhat different for the real earth.

[Walter Wayne]

Originally posted by davefoc
2. The reason the earth points north is that centrifugal force caused by its rotation around the earth changes the direction slightly of the downward force on the building.[pedantic mode]Centrifugal force is an "apparent" force, so when we talk about this, we should talk about centripetal force. When you do a loop on a roller coaster, you don't push down on the seat, the seat pushes up on you. It just so happens having your chair pushed into you, feels like you are being pushed into your chair.[/pedantic mode]

Take a non-rotating earth again. If you are standing a few feet from the north pole and drop a coin it lands directly at your feet.