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# Hole in a sphere

Best Answer bonanova, 23 August 2007 - 07:11 AM

The volume of the spherical caps is given by:
[list]
where
[list]
[*] h = the height of the cap (difference between r and the distance from the centre of the[/*:m:1cc31][list] sphere to the centre of the circular end of the hole)

Kudos to cpotting for the cap formula.
Spoiler for Here's the mathematical solution
Spoiler for Here's the logical solution
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166 replies to this topic

### #51 roolstar

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Posted 08 February 2008 - 11:19 AM

That's the idea...

Is it?

Did Jkyle1980 post the right approach in drilling??

I think not!!

According to his picture, the diameter of the hole = diameter of sphere - 6 inch!!

Now that cannot ge right according to the answer and to the posts on this thread!

The 6 inchs in his picture cannot be seen from this angle, they represent the HEIGHT of the remaining part of earth once you put the earth bracelet flat on a plane!!
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### #52 bonanova

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Posted 08 February 2008 - 12:20 PM

Is it?

Did Jkyle1980 post the right approach in drilling??

I think not!!

According to his picture, the diameter of the hole = diameter of sphere - 6 inch!!

Now that cannot ge right according to the answer and to the posts on this thread!

The 6 inchs in his picture cannot be seen from this angle, they represent the HEIGHT of the remaining part of earth once you put the earth bracelet flat on a plane!!

Be nice.
He got the picture.
You got the dimensions.
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### #53 roolstar

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Posted 08 February 2008 - 01:52 PM

Be nice.
He got the picture.
You got the dimensions.

No disrespect intended at all!!

It's just that if Jkyle1980 uses this picture to make his calculations and reasonning, he will never get the 36pi answer...

And it's not at all because he's "slow" as he claims (and he's not), it's simply because according to his picture, the Radius is very relevant to the calculation of the remaining volume and the remaining bracelet of earth has a volume much bigger than 36pi!!

The 6 inchs are not reduced from the diameter of the drill, they are not the THICKNESS of the remaining crest, they are the HEIGHT of the remaining bracelet if it's laid on its side on a plane!!

If what I'm saying is wrong, I may be misunderstanding something very important...

Please take another look at the picture and tell me if it portraits the Original Puzzle.
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### #54 bonanova

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Posted 08 February 2008 - 04:36 PM

The 6 inches is along the axis of the hole - the height of the cylindrical part of what's removed.
All as you say.

And JK "got the idea" of how a 6" hole could be drilled thru the earth.
He labeled it wrong, but he got the idea.
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### #55 Jkyle1980

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Posted 08 February 2008 - 08:59 PM

None taken, roolstar. Actually, between the discussion over the last picture I finally get what 6" inch hole is being talked about. Thanks for the lengthy, yet helpful, explanation. Where can I mail the check for my tuition?
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### #56 roolstar

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Posted 08 February 2008 - 10:11 PM

None taken, roolstar. Actually, between the discussion over the last picture I finally get what 6" inch hole is being talked about. Thanks for the lengthy, yet helpful, explanation. Where can I mail the check for my tuition?

Sorry for the detour there guys...

I apparently didn't get that the picture portraits the misunderstood version that JK had at first...

To tell you the truth, I thought I was the one who didn't get it at first!!!

I even made some calculations about the "HEIGHT" of the cylinder in JK's picture and ended up with around 2Km!!!

ANyway, I promise I will post a new puzzle very soon to regain your respect
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### #57 Eugenia

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Posted 09 February 2008 - 03:00 AM

Eek, I don't understand the question. I know there is somethng wrong with my comprehension of the question, because then I would need the radius of the sphere. So right now, I will do some post digging and look for what i don't get. I'm so confused about what information I have exactly )=

Edited by Eugenia, 09 February 2008 - 03:02 AM.

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### #58 bonanova

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Posted 10 February 2008 - 07:19 AM

Eek, I don't understand the question. I know there is somethng wrong with my comprehension of the question, because then I would need the radius of the sphere. So right now, I will do some post digging and look for what i don't get. I'm so confused about what information I have exactly )=

It's a wonderful result, really. The answer is the same for any radius.
Except the radius must be at least 3" -- so that a 6" hole can be made.
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### #59 sunburntpixY

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Posted 22 February 2008 - 12:55 AM

So I have been reading this forum for almost a year and it took this problem to get me to register. i read all fo the responses and I STILL don't get it. Can someone draw a correct diagram showing hwo this might work for various R lengths or at least pointing to the different parts of the cylinder/hole and sphere being referred to in the problem? I haven't had any calculus at all, but I want to at least understand the problem well enough to comprehend the logic answer.

Thanks!
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### #60 bonanova

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Posted 22 February 2008 - 08:05 AM

sunburntpixY,

Thanks for the inquiry to my puzzle.

I don't have a diagram that i can post. But this may be helpful...

It will be helpful to visualize the following ideas if you draw a circle on a piece of paper and think of it as a cross section of a sphere. The center of the circle is the center of the sphere. Now draw parallel lines -- equal distances from, and either side of, the center of the circle. Make sure the lines go all the way through the circle. The lines represent a cross section of a hole drilled through the sphere. The distance between the lines is the diameter of the drill. The area between the lines is a cross section of the hole. The area outside the lines but inside the circle is a cross section of the portion of the sphere that remains after the hole is drilled. And, most importantly to understand, the length of the lines between the points that they intersect the circle, from end to end, is the length of the hole. Note that the length of the hole is less than the diameter of the circle [sphere]. Repeat. The length of the hole is not the same thing as the diameter of the sphere. Finally, note that if you had drawn the lines farther apart -- go ahead and draw another pair of lines that are farther apart -- the length of the hole decreases.

OK - now think through these statements.

[1] For there to be anything left of the sphere, the diameter of the drill better be less than the diameter of the sphere.

[2] For the hole to be 6 inches long [length is measured in the direction the drill traveled while it was making the hole] the sphere had to have been at least 6 inches in diameter. You can't drill a 6-inch hole through a 4-inch-diameter sphere. But you can drill a 6-inch hole through a 7-inch-diameter sphere.

[3] In all cases, we are talking about drilling a hole completely through the sphere. You can look through from beginning of the hole to its end, and see all the way through the sphere. You could string the sphere onto a rope, so long as the rope is thinner than the drill, of course. And, in all cases, what we mean by the length of the hole is the distance between the beginning and the end of the hole. You have to see this length as being different from the diameter of the original sphere.

[3] If the diameter of the drill is negligibly small, a negligible amount of the sphere is removed, and the remaining volume is essentially the original volume: [4pi/3]r^3 and the length of the hole is essentially the diameter of the sphere. Only in this case is the length of the hole the same as the diameter of the sphere. Think of a pearl, with the tiniest of holes, string on the slenderest of threads. In the case of a vanishingly small [diameter] drill, the length of the hole approaches the diameter of the sphere, and the remaining volume approaches the initial volume.

[4] If the diameter of the drill is negligibly less than the diameter of the sphere, the remaining portion of the sphere is a tiny band of material at the "equator" of the sphere - think of the drill entering the sphere at its "North Pole" and exiting the sphere at its "South Pole", and the length of the hole - the height of the remaining volume - is now much less than the diameter of the original sphere. In the case of a vanishingly small difference between the sphere's diameter and the drill's diameter, the remaining volume approaches zero - the entire sphere has been drilled away - and the length of the hole [obviously] approaches zero as well. This would be the case if you drew your two lines tangent to your circle instead of intersecting the circle. If you really understand this point, you can see how a 6-inch hole could be drilled [somewhat disastrously to be sure] through the Earth! Remember, the length of the hole is defined as the height of the remaining portion of the sphere, not as the length of the drill. Or think of it this way: if you crawled inside the hole and painted the inside surface of the hole, the painted surface would be a circular cylinder. The length of that cylinder is the length of the hole.

Now if you can visualize all of that, you understand what "length of the hole" and "remaining volume" mean.

The puzzle is cryptically worded. Intentionally. It's meant to make the solver think through all of these things so that the question is even understood. That's part of the challenge of the puzzle.

If you can visualize the "length of the hole" then the question at least is understandable:

If you drill a 6-inch hole through a sphere, what is the volume of the remaining portion of the sphere?

And, if you've read this thread, you've heard it asserted that in every case - for every sphere whose diameter is at least 6 inches - the remaining volume is the same. A person with calculus skills can compute that volume and see that the answer can be expressed in terms of the length of the hole - only. An amazing result!

But another person - one who does not like tedious calculations - simply takes note of the fact that an answer has been requested even though the original diameter of the sphere is not given.

The only way that can be a reasonable request is if the answer does not depend on the original diameter. The person who reaches that conclusion smiles, rubs his/her hands together, and computes the answer for the simplest case: a 6-inch-diameter sphere. Remember, we said that's the case where the hole has zero diameter and thus removes zero volume, so that the remaining volume is the original volume:

For a 6-inch diameter sphere, V = [4pi/3]r^3 = 36pi cubic inches. And that's the correct answer.

Hope that makes this thread easier to follow. You're not alone in asking about what it means.

- bn
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