Dropped into the Deep

To calculate this you would need to consider some rubbing friction on the walls depending on the shape, surface finish, and mass of the object, but the drag due to air in the hole will dwarf that. If you drew a vacuum and evacuated the hole (could be very hard since thehole is ice which would sublimate at low pressure), you could have essentially free fall and create conditions of microgravity (or almost) on the falling body.

O. K. I'll ask the question: Can you accurately calculate the travel time of a lead ball dropped out of an airplane flying at an altitude of 3.2 kilometers and verify it with actual measurements?

If so, can one be sure that the lead ball will fall in a spiral path after contacting the wall of a vertical(?) pipe due to Coriolis effects?

The terminal velocity for lead balls is reached in a few seconds, and let's say for the sake of arguement that it's about 320 km/hr. At that rate, the time of flight for a distance of 3.2 km is 1/100 of an hour or 36 seconds, ignoring the acceleration distance.

The earth rotates at surface speed of about 0.45 km/sec at the equator and very much slower at 15 degrees latitude, about 0.06 km/s. In 36 seconds, and ignoring air drag at 0.06 km/s, the ball will have tried to move about 2 km, or plenty enough for it to have contacted the surface of a bore hole.

Will centrifugal force overwhelm the Coriolis effect and make the ball travel in more or less a straight line?

Yes. You can calculate that time

S = 0.5 * g * t ^ 2

where S is the distance in cm
g is acceleration due to gravity ~981 m / cm ^ 2
t is time in seconds
t ^ 2 is quare of time = t * t

This formula is from Neuton

Shyam

Sorry Shyam, but you are missing it completely! Did you think about terminal velocity, amongst others? Look at Bill's post #8097 above.

Does anyone know how to calculate the terminal velocity of a round lead ball? I believe the 'most unknown' factor is the drag coefficient. Friction with the walls will introduce further uncertainty, but probably to a lesser extent.

Nicely summarized Bill! One other uncertainty is: what is the average 'g' inside that vertical hole? We know that the gravitational acceleration is slightly higher at the poles than at the equator. Then, the average height of the pipe is at least 1.5 km above sea level. Finally, do we need Newton's 'inside matter gravity', or 'outside matter gravity' equation? Or are the other uncertainties much bigger than this one?

You are right.

Perhaps you are looking for 100% perfact answer.

"g" will reduce as ball moves in the hole and is a continuous function of depth.

Air friction comes from surface contact which will be almost constant if you assume air density contact, which is not true.

Temperature difference will make the ball to expand, and drag calculation will change so you need that parameter also.

There will be humidity difference, gas mixture difference and they affect density more than "g" changes.

That is more to fluid dynamics and will be similat to JET moving in air.

Are you collecting information for your class lectures or trying to teach your son or daughter at home? Parents and teachers do more home work these days.

I think you can get some data from Oilwell drilling companies as they drill that deep and also drop neutron detectors in the hole to collect data on water and oil.

Shyam
India

Quote: Shyam wrote: "Are you collecting information for your class lectures or trying to teach your son or daughter at home?"

Thanks Shyam, interesting and useful reply! Your question: no, thank goodness, I'm not teaching - just trying to 'tweak' the occupants of Conference Room 4 into discussion. I think that's more or less the purpose of the "Challenge Questions" category.

Thanks for the compliment and same to you, not to mention your posting of a seemingly simple problem for us to ponder and discuss, which would have only been done by someone with enough insight to know that it's either extremely difficult or impossible to model.

There are so many other uncertainties with greater impact on the time of flight that I didn't mention the ones with much smaller influence, such as the earth's fairly linear gravity gradient of about 0.05%.

For example, air density should show up as a significant factor (my gut tells me it's a non-linear drag thing) which would affect time of flight in a similarly positioned pipe above the earth's surface versus one below the earth's surface, all other conditions being held the same.

I have an idea.

I just thought if I really have to do this then what exactly I will be doing.

I will sure find a way to get the result and one way that comes to my mind is

I will attach a battery and laser light flaser and impact switch to the heavt lead ball at the bottom and laser pointing upward. I believe a light flash will be see even after 3km depth if this hole is streight line.

I can also attach a small wireless transmitter to get the information.

Time taken by these devices will be nearly zero.

These taguchi Methods work for me. Calculation requires no investment other than time and brain so comes much cheaper.

Shyam
India

Have you ever looked two miles down a railroad track? It's five feet wide and goes to a point two miles away as far as human vision can tell. If the bore hole is very straight, it would suprise me, and I'd bet a dollar it ain't very large in diameter. I'd bet two dollars nobody's gonna let you drop something into it.

And just out of curiosity, how does dropping a ball (instrumented with all sorts of stuff) in a bore hole (using Taguchi Method or not) tell you how to calculate the time of flight, which is the original question? Besides, a simpler way to measure the time of flight is to electronically time the flight of a lead ball -- start it (electronically) when it is dropped and end it (electronically) when it lands -- then read the register. No lasers (which wouldn't point straight up the bore hole anyway) needed!

Idea is not seeing the picture by eye. It may be possible to get few photons that are scattered through the borehole and reach a detector on top. We can detect just few photons to get the signal.

For tapered hole, perhaps wireless signal may reach if used in low frequency band, which follow the curvature.

These are only ideas and not tested success stories.

Shyam
India

Numerically integrating the full formula for free-fall velocity in a fluid from Wikipedia, I arrived at an interesting result for the lead ball falling down the 3.2 km shaft.

As a first approximation, I used an average air density of 1 kg/m^3, a drag coefficient of 0.1 for a smooth sphere, an average 'g' of 9.8 m/s^2 and a lead ball of 10 mm diameter (mass 5.76 grams, if I got that right, because it's important).

Ignoring contact from the sides and possible "ground effects" caused by the sides, the ball approaches the terminal velocity of about 430 km/h more or less as it hits the bottom. The total travel time is 35 sec., for an average velocity of about 330 km/h.

The interesting part is that the ball speed is heavily dependent on the ball size! Double the lead ball size (20mm diam, mass 46 grams) and the speed is significantly larger at the bottom: 575 km/h, with a fall time of about 30 sec. This ball did not reach terminal velocity, which is 610km/h.

In numerical integration, one can also easily model the changes in air density and gravity, if you care. How to bring in the friction effect of the walls of the hole is another animal - with teeth...

Jorrie, you have posed an interesting point!

At first glance, "hmmmm" came to mind when I read, "The interesting part is that the ball speed is heavily dependent on the ball size! Double the lead ball size (20mm diam, mass 46 grams) and the speed is significantly larger at the bottom: 575 km/h, with a fall time of about 30 sec. This ball did not reach terminal velocity, which is 610km/h," but I almost immediately thought, "but the surface area of a sphere does not increase linearly with the mass, so this may make sense afterall!"

I'll have to get back to you on this . . . If I don't get a cup of coffee soon, I'll go nuts.

Quoting Jorrie's "The hole was drilled roughly 15 degrees (1000 miles) east of the geographical South Pole, so one has to take into account Coriolis and many other effects." Coriolis acceleration being -2wxv is always perpendicular to the ball's velocity, so should not effect flight time.

Quote: cameo wrote "Coriolis acceleration being -2wxv is always perpendicular to the ball's velocity, so should not effect flight time."

Yea, except that Coriolis will cause the lead ball to hit the sides of the hole! The Coriolis deviation is about 1 meter at 15 degrees off the Pole and at 3.2 km depth. The hole is around 30 cm in diameter, so even if the hole is perfectly straight down, the lead ball is bound to hit the sidewalls.

I must confess, it is probably impossible to model the actual drag caused by the ball hitting the sidewalls (without a lot of experimentation).

Dear Jorrie,

I think the friction will be too high and temperature of the ball will change the shape and will be no longer a ball so it will move much faster like a bullet. This experiment is same as shooting the ball up vertically 3.2km and let it fall to the shooting point (2xt). I am not sure if bullet with that velocity will have same shape all the way. Lead melts at 327.46C but softens much earlier. Try U238 ball which is used in Tank hitter shots. It is high temperature stuff but Hazardous.

Hi Shyam,

I don't believe that 400 - 600 km/h speed is enough to heat a lead ball enough to be softened and deformed. There are (jet) cars that travel much faster than that without excessive aerodynamic heating, I think!

Dear Jorrie, Perhaps make the car of Lead then try. You may be right, but sounds difficult to me.

The canopy of an F-16 is made of plexiglas (I know, I have one). The F-16 is capable of exceeding Mach 2. Plexiglas has a much lower melting point than lead and Mach 2 friction does not distort the canopy.

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I'm sure it can be modeled without actually testing it and that through discussion like this eventually all facets of the problem will be brought to light. One of those facets is as the ball drops depending on it's size relative to the diameter of the hole, and the hole being closed at the bottom, there is an air compression factor to be considered which should be noticeable when the diameter and ball are close in size. The air below the ball not having an easy exit flow around the ball will be compressed and will slow the ball considerably after enough air has been compressed below. A very small ball will still have to contend with this problem because of its envelope which slightly increases its mass and diameter (the dai of the ball and the diameter of the air hugging it). Because of this piston effect I think if the ball dia. was just slightly smaller than the hole dia., then the ball would hit bottom dropping like a feather. Everything is "Cause and Effect"

You are quite right. If the ball and the hole diameter become comparable, there will be all sorts of nasty side-effects! For the original problem, we specified a small lead ball, suggesting (but not specifying) that it is much smaller than the hole diameter.

In any case, the intention of this challenge was to bring all the effects to the surface for consideration. So thanks for this one!

I'd reckon there's two ways to tackle this. One is to figure the biggest factor which I'd guess to be terminal velocity and use this only whilst frantically hoping that all the other factors have a negligible effect and that no-one actually puts it to the test. The second option is to sit down and calculate every effect but I'd say you'd run a risk of your lead ball reversing direction half way down and shooting you through the calculator. That way lies madness! Is Jorrie a professional sadist or does he pose these questions for amusement??

Quoting nutwood: "The second option is to sit down and calculate every effect but I'd say you'd run a risk of your lead ball reversing direction half way down and shooting you through the calculator."

Engineers always hope for such an effect to jump out of the sums - ruined calculator and perhaps also a sore hand/head or so, but anti-gravity would be a great reward!

Lets send it on to the MYTHBUSTERS to find out.

I know this might be too little, too late, the last post being on Sept. 19, but a lot of answers ignore the Coriolis effect, even though it is specifically mentioned in the challenge.

A few did say that Coriolis would make the ball contact the side of the hole. Some also said that this contact would cause friction which would slow the ball. However, if the ball is round and smooth (we assume that by the definition of "ball"), then any contact starts the ball rolling, and rolling friction of a round smooth ball is almost zero, right?

I believe that Galileo did an experiment with two balls. One that he dropped a vertical distance and one that rolled down an inclined plane, which had a drop of the same distance. While the two balls achieved the same final velocity (Galileo's experiment probably was not long enough and his instruments not accurate enough to take into account air friction differences), the rolling ball did take longer, because its path was much longer for the same average velocity.

Now in the challenge question, if the ball contacts the wall due to Coriolis effect and it has a Coriolis acceleration other than vertical (up or down the hole), won't it begin rolling in a spiral path around the perimeter of the hole? We assume the bore hole is smooth also allowing the ball to spiral almost frictionless. This spiral path is analogous to Galileo's inclined plane. Given the extreme length of the hole, air friction does play a part, and the ball should have the same terminal velocity in spiral motion as it does in free fall, if there were no Coriolis effect. The spiral path is longer, so for the same final velocity, and therefore same average velocity, the time to travel the spiral path is also longer, just as in Galileo's experiement. Unfortunately, the math involved is too complex for me.

One thing does bother me though. I am surprised no one else picked up on this. That is the wording of the challenge question in one sentence near the end:

"The hole was drilled roughly 15 degrees (1000 miles) east of the geographical South Pole, so one has to take into account Coriolis and many other effects."

Shouldn't the wording be "north of the geographical South Pole"? If I am at (or near) the geographical South Pole, and have a GPS that always keeps me headed in the correct (geographic, not magnetic) easterly direction, until I have traversed 15 degrees (of longitude, since it is "east"), or 1000 miles, don't I keep walking in circles around the pole for a very long time? How can you go any direction except north from the South Pole?

Remember the old joke/riddle, "If you walk due South one mile, then turn and walk due East one mile, then turn again and walk North one mile, and end up at the spot you started from and a bear walks by, what color is the bear?"