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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
Go to the full post

166 replies to this topic

### #31 Jkyle1980

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Posted 04 February 2008 - 12:36 AM

So we have spheres of various size. We have drills of various circumference, but all only 6 inches long. I'm still not getting it. How can a six inch long drill with a diameter of any size (1/2 in. or a mile) dirll completely through the earth. I'm picturing a small 1/2 hole and a larger 1 mile depression but still only 6 inches deep.
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### #32 bonanova

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Posted 04 February 2008 - 08:40 AM

So we have spheres of various size. We have drills of various circumference, but all only 6 inches long. I'm still not getting it. How can a six inch long drill with a diameter of any size (1/2 in. or a mile) dirll completely through the earth. I'm picturing a small 1/2 hole and a larger 1 mile depression but still only 6 inches deep.

Visualize the material that's removed from the sphere by the drill.
It comprises two spherical caps sandwiching a right circular cylinder.
Define the length of the hole to be the height of the cylinder.
That's the same as if you climbed inside the hole and measured its length from end to end with a measuring tape.
Note that because the end caps have height of their own, the length of the hole is less than the diameter of the sphere.

Also note that as the diameter of the drill approaches the diameter of the sphere, the length of the cylinder [length of the hole] approaches zero. That is, all the sphere is removed: comprising two hemispherical end caps and no cylinder. This means that any sphere of diameter greater than 6 inches can have a 6 inch hole drilled completely through it.

To understand the puzzle is to understand that it could have been worded worded more helpfully. See next.
But that's part of the puzzle - to understand the condition of "6-inch hole drilled through a sphere", even tho the puzzle is worded [accurately, but] not helpfully:

Drill a hole through a sphere. Make the diameter of the drill large enough that the inside cylindrical surface of the hole measures exactly 6 inches in length. Now calculate the volume of the portion of the sphere that remains = volume of: (sphere - cylinder - 2x [end caps]).
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
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### #33 roolstar

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Posted 05 February 2008 - 10:33 AM

So we have spheres of various size. We have drills of various circumference, but all only 6 inches long. I'm still not getting it. How can a six inch long drill with a diameter of any size (1/2 in. or a mile) dirll completely through the earth. I'm picturing a small 1/2 hole and a larger 1 mile depression but still only 6 inches deep.

If you look at a doughnut cut through a sphere ( or a regular doughnut), it's height (when you put it on the plate) is about 5 cm.
Now the diameter of the original sphere (before the hole) is actually the diameter of the doughnut itself (the circle) maybe 15 cm!

The same with a hole through earth: The height of the doughnut cut is 6 inch, but the sphere has a much bigger diameter.

And the drill is not 6 inch long at all, the drill can actually be as long as you like but it's width (or radius) has a MAXIMUM* to be calculated in a way to leave a 6 inch doughnut in the sphere in question.

In the case of the earth, its radius has to be a small fraction less than that of the earth.
And in a 3 inch radius sphere, its width has to be 0 and the doughnut's height will then be 6 inch with no hole in the middle!
Of course with a sphere with less than 6 inch diameter, the puzzle doesn't hold.

* Why MAXIMUM? Well, the drill itself can be a regular drill with regular dimensions, but it's going to take a long time to drill through the earth and leave a 6 inch doughnut from the planet, not to mention there will be no place to stand!

Edited by roolstar, 05 February 2008 - 10:36 AM.

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### #34 Jkyle1980

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Posted 05 February 2008 - 09:12 PM

Maybe this has already been posted. A friend asked me this a while back, and I answered her in less than a minute. She said I was a genius. But I said there were two ways to arrive at the answer, and I simply chose the easier way.

A 6-inch hole is drilled through a sphere.
What is the volume of the remaining portion of the sphere?

The hard way involves calculus. The easy way uses logic.

I feel like the kid getting off the short bus. However, I am determined to understand this. Where in the original post does it state that the hole will leave only 6 inches left around it? It says "A 6-inch hole". I read that as either a hole 6 inches deep or a hole 6 inches in diameter. Either way, it's not very big and will take away essentially nothing from an object the size of the earth. Now you guys are making it sound as if (from a 2D standpoint) there is a circle with all but six inches cut out. I am seeing something that looks more like a bracelet than a doughnut.

(I do appreciate the patience with the slow kid, too.) Am I just completely misunderstanding the wording of the question?
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### #35 roolstar

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Posted 05 February 2008 - 09:47 PM

I feel like the kid getting off the short bus. However, I am determined to understand this. Where in the original post does it state that the hole will leave only 6 inches left around it? It says "A 6-inch hole". I read that as either a hole 6 inches deep or a hole 6 inches in diameter. Either way, it's not very big and will take away essentially nothing from an object the size of the earth. Now you guys are making it sound as if (from a 2D standpoint) there is a circle with all but six inches cut out. I am seeing something that looks more like a bracelet than a doughnut.

(I do appreciate the patience with the slow kid, too.) Am I just completely misunderstanding the wording of the question?

Not at all Jkyle1980 !

I followed your posts on this forum very closely and know exactly how "SLOW" you are!

You are completely right (at least I think so), I was completely mislead by the same fact as well: It was not so clear in the OP.

But once I read (in an later post) that the cylinder left AFTER the drilling is 6 inch high, I saw the effect of drilling the 6 inch hole through earth more clearly...
And it does look more like a bracelet too!
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### #36 bonanova

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Posted 05 February 2008 - 11:12 PM

I'm certain the question was meant to be a bit demanding.
It was recalled by a friend from an MIT entrance exam.

It requires visualizing the drilled sphere.
Once you see it, drilled, the height of end caps is a non-issue.
The hole has an unambiguous length.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #37 Jkyle1980

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Posted 06 February 2008 - 02:34 AM

I dropped out of Calc III years ago because I sucked at visualizing three dimensions. I am still just drawing a blank. When you guys are talking about the cylindar, are you meaning what is left of the original sphere or the piece that is removed from the sphere? Because if it is the former, I am just way out of it. (This may also be why I went to Tulsa instead of MIT. Go GOLDEN HURRICANE!!!)
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### #38 bonanova

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Posted 06 February 2008 - 05:30 AM

I dropped out of Calc III years ago because I sucked at visualizing three dimensions. I am still just drawing a blank. When you guys are talking about the cylindar, are you meaning what is left of the original sphere or the piece that is removed from the sphere? Because if it is the former, I am just way out of it. (This may also be why I went to Tulsa instead of MIT. Go GOLDEN HURRICANE!!!)

The cylinder is what is removed from the sphere.
Think of beads on a string.
The string is the cylinder.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #39 Jkyle1980

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Posted 06 February 2008 - 06:10 AM

Okay, beads on a string. I like that. It's simple. I can see it. Now is the six inches the length of the string or the diameter of the string? Or is it the width of the bead from the hole to the outside?
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### #40 bonanova

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Posted 06 February 2008 - 02:59 PM

Length.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

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