A guy in a not well lit room.

Originally posted by sonhouse
In this case you need to use something that doesn't bounce so you get a true reading. My calc's put the drop at around 63 meters away from the drop point, a significant distance. In order to get one G at one Km radius, you need to be going about 0.95 RPM which means you are going pretty fast on your interplanetary carousel. So it takes about 63 seconds per ...[text shortened]... point to get an accurate distance reading and then backtrack the equations to get your radius.

You are way out but according to my calcs there is a measurable distance.

The object lands approximately 3 metres from you.

Originally posted by wolfgang59
You are way out but according to my calcs there is a measurable distance.

The object lands approximately 3 metres from you.

Checked my calculation. I now reckon its about 12cm. 🙂

Originally posted by wolfgang59
Checked my calculation. I now reckon its about 12cm. 🙂

This is my thinking:

The floor the man is standing on has a speed of 2 * pi * 1,000 / 60 metres/sec
The object 2m above the floor has a speed of 2 * pi * 998 / 60 metres/sec
(pi= approx 3)
The difference in speeds is therefore about 0.2 metres/sec
The object hits the floor in about 0.6 secs

Now IF we approximate the trajectories of the object and the floor initially below the object we can say that in 0.6 seconds the floor will have travelled 0.12 metres more than the object.

Hence the object lands 12cm 'behind' the point it was dropped above.

Not accurate but of the correct magnitude I think!

Originally posted by mtthw
You're on the "equator", there won't be a Coriolis effect.

For info.

http://en.wikipedia.org/wiki/Accelerated_reference_frame

"Neglecting air resistance, an object dropped from a 50 m high tower at the equator will fall 7.7 mm eastward of the spot below where it was dropped because of the Coriolis force."

Originally posted by wolfgang59
For info.

http://en.wikipedia.org/wiki/Accelerated_reference_frame

[b]"Neglecting air resistance, an object dropped from a 50 m high tower at the equator will fall 7.7 mm eastward of the spot below where it was dropped because of the Coriolis force."[/b]

If you used your method above for this example, would you get close to 7.7mm?

Originally posted by wolfgang59
For info.

http://en.wikipedia.org/wiki/Accelerated_reference_frame

[b]"Neglecting air resistance, an object dropped from a 50 m high tower at the equator will fall 7.7 mm eastward of the spot below where it was dropped because of the Coriolis force."[/b]

Yes, I've accepted that. I usually think of the Coriolis effect as being when you're travelling parallel to the surface - e.g. the Coriolis effect that affects the direction of rotation of hurricanes in the northern and southern hemispheres. This doesn't happen at the equator.

But you're quite right - the Coriolis effect exists whenever you have motion in a plane perpendicular to the axis of rotation. Which means on the equator you do have an effect for vertical motions.

Originally posted by Palynka
If you used your method above for this example, would you get close to 7.7mm?

mmmmmmm yes using that method a rough calculation gives me 1cm!

The ball at the top of the 50m tower travels further than the base of the tower by 100 * pi = approx 300m

300m in 24 hours = 300/(24 * 60 * 60) = 1/288 =approx 1/300

A 50m fall is approx 3 secs

3/300 = 1/100 metres = 1 cm

Originally posted by wolfgang59
mmmmmmm yes using that method a rough calculation gives me 1cm!

The ball at the top of the 50m tower travels further than the base of the tower by 100 * pi = approx 300m

300m in 24 hours = 300/(24 * 60 * 60) = 1/288 =approx 1/300

A 50m fall is approx 3 secs

3/300 = 1/100 metres = 1 cm

I forgot the ball would still have a velocity 2 meters up. So I get 2*PI*1000 meters=
6,283,185.307 mm circumference at the inner surface of the outer rim and 2 meters from that floor, it would travel 12,566.37 mm less. So divide that by 63 seconds (one revolution) and you get 199.46 mm/second difference and it takes 0.63 seconds to fall 2 meters at one G which puts the ball traveling 125 mm from the dropping point. A bit different than my original 70 meters for sure. The principle is still the same.
Anyone else get those #'s, 125 mm? About 5 inches. Ah, I see Wolfy did it to a couple of decimal places.

The guitar string was not a red herring, it can also be used to measure the circumference because you can slide the string under the 2 meter ruler if the 'floor' is sufficiently smooth, and you can make a decent measurement of the chord and figure the circumference that way and work out the radius. You can't just use the ruler by itself because you don't know the distance from the center of the line to the top of the curve. The guitar string gives you two pieces of information:
If the circumference is circular, the distance from where the string intersects the curve to the ends will be symmetrical, the same distance to the end of the ruler from both sides but if it is an ellipse, that distance will be non-symmetrical, say a 10 percent difference between the readings at one end V the other, to pick a # out of a hat.

All in all instructive at least to me! Good Catch!

It is clear there would be a coriolis effect almost all the way to the north and south poles, I wonder what the graph of the coriolis effect would look like plotted from pole to pole? It is not as obvious as it looks because the distance around the earth changes at each latitude so the velocity changes too.

Can anyone figure the distance a ball would move dropped from that same 50 meter high tower at a latitude of 45 degrees? If you got sticky enough, the top of the tower would not be at the same latitude, but slightly less, it sounds like a great problem for someone to solve!

Another angle: At what latitude would the ball drop 1 mm from the tower?