Coriolis force question

[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]

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]

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]

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]

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]

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? So let me just despoil the rarefied air with the obvious question relating to this:

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!