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Coriolis force question

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.
Click to expand...

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

Code: 
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.
 
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:
Code: 
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 said:
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.
Click to expand...

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.

empire.jpg
 
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.
 
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?
 
davefoc said:
Abdul Alhazred,
It amazes me that you can't see how important this question is...
Click to expand...

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.

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.
 
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.
 
ceptimus said:
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
Click to expand...

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:
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.
 
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
Click to expand...
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.
 
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.

Code: 
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
 
davefoc said:

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

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

did
 
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.
 
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
 
OK here's mine (blush)

Distance of bottom of tower from earth's axis.

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

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

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

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

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

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

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

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

0.832 ips * 8.813 s = 7.33 inches.
 
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