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clop
2006-Apr-10, 01:17 PM
I'm fairly new to this community so I'm sure I'm breaking a whole heap of unspoken laws by posting this here, but there doesn't appear to be a forum devoted to mathematical and physical puzzles.

So.

You take a solid metal sphere and clamp it in a drill press, then core it straight down through the centre and out of the bottom using a drill bit. The resulting cylindrical hole through the sphere is 6 inches long.

What is the volume of the remainder of the sphere?

clop

antoniseb
2006-Apr-10, 01:32 PM
I'm moving this to off-topic babbling.

gwiz
2006-Apr-10, 01:36 PM
As you don't specify the hole diameter, the answer must not depend on it. With zero diameter the answer is 4*pi*3*3*3/3 = 36*pi cubic inches. Am I right?

antoniseb
2006-Apr-10, 01:51 PM
You take a solid metal sphere and clamp it in a drill press, then core it straight down through the centre and out of the bottom using a drill bit. The resulting cylindrical hole through the sphere is 6 inches long.

What is the volume of the remainder of the sphere?
You didn't give the diameter of the drill bit.

The initial volume of the sphere is slightly over 36*pi cubic inches.
The whole itself has a volume of 3/2*pi* d^2 cubic inches (where d is the diameter of the drill bit).

The actual radius of the sphere is the square root of ((d/2)^2 + 36).

So the actual answer (pardon the lack of math characters available to me) is:

(4/3 pi * sqrt((d/2)^2 + 36)^3) - (3/2 *pi* d^2)

Note I also included some extra parens for clarity.

Note also that as d approaches zero, the result approaches 36*pi, which is the volume of a six inch diameter sphere. As the diameter of the bit gets larger, the remaining volume grows because the radius required to create a six inch hole increases. Imagine an extreme case in which there is a sphere just slightly larger than 1000 inches, and you cut a whole in the center making a thin ring six inches wide. The volume of this ribbon would be much bigger than a six inch sphere.

Also note that it might take a special drill press to punch through six inches of some types of metal. Expect to keep pouring oil on the drill bit, and backing out pretty often.

SeanF
2006-Apr-10, 03:00 PM
You didn't give the diameter of the drill bit.
Doesn't matter.


The initial volume of the sphere is slightly over 36*pi cubic inches.
Slightly? The initial volume of the sphere could be much greater than 36*pi cubic inches, if the drill bit is sufficiently big, as you point out in your penultimate paragraph. :)


Note also that as d approaches zero, the result approaches 36*pi, which is the volume of a six inch diameter sphere. As the diameter of the bit gets larger, the remaining volume grows because the radius required to create a six inch hole increases. Imagine an extreme case in which there is a sphere just slightly larger than 1000 inches, and you cut a whole in the center making a thin ring six inches wide. The volume of this ribbon would be much bigger than a six inch sphere.
I bet you'll find that that thin ribbon would have a volume of 36*pi cubic inches. That's always going to be the answer, no matter how big the original sphere and drill bit are.

grant hutchison
2006-Apr-10, 03:04 PM
Since you don't mention the diameter of the sphere or the drill bit, and since I trust you to provide all the data I need, I have to deduce that the result is always the same, so long as the length of the sides of the hole is six inches: a narrow drill through a sphere just over six inches in diameter, or a broader drill through a larger sphere.

So, since I'm lazy, I make the calculation for a zero-diameter drill in a six inch diameter sphere :4/3*pi*3^3 = 113.1 cubic inches.

Grant Hutchison

Edit: Oops. SeanF posted while I was typing, but for some reason I didn't notice gwiz had got there before me, too.

Grey
2006-Apr-10, 03:23 PM
I bet you'll find that that thin ribbon would have a volume of 36*pi cubic inches. That's always going to be the answer, no matter how big the original sphere and drill bit are.Correct. It's a classic problem, assigned to most physics students somewhere in their careers. It's actually interesting to work out the formula without the assumption that the bit size does not matter, and then see how that value drops out when working through the algebra.

Gruesome
2006-Apr-10, 03:35 PM
I used a 6 inch diameter sphere and a 0.50 inch diameter hole. The volume of the remaining sphere is: 111.92128596529 cubic inches.

What do I win?

antoniseb
2006-Apr-10, 03:48 PM
I bet you'll find that that thin ribbon would have a volume of 36*pi cubic inches. That's always going to be the answer, no matter how big the original sphere and drill bit are.

In my first post I showed a calculation, but I left out a very importnat piece of the puzzle: the rounded end caps above and below the cylinder need to be counted in the removed volume:

Let's try it. Calculate the remaining volume for a sphere big enough for a four inch radius (8 inch drill bit) cylinder to be the mising volume. This gives a sphere radius of sqrt(3^2 + 4^2) = 5.

So the volume of the sphere is 4/3 * pi * 125 (~ 167 * pi). The volume of the Cylinder is 6 * pi * 16 (= 96 * pi).

NOW As it happens, the volume of the caps together is exactly the difference. So, I stand corrected.
edit: note that this required a little calculus.

farmerjumperdon
2006-Apr-10, 04:46 PM
I think all your answers might be flawed, or the wording of the question needs a touch of tweaking for the answers to be correct. The space occupied by the removed material is not a cylinder. A cylinder must be flat on the ends. The little curved bits that were removed at the very ends of the drilled hole can't be included as part of the cylinder, or you by definition do not have a cylinder.

I do not have the knowledge to know whether or not your math takes this into account - so I'll trust your answer. Just just tell me; in your solution, are the little curved bits that make up the beginning and end of the hole included in the area of the sphere left over? Or are they considered in with the cylinder - though they are not part of a cylinder?

EDIT - Never mind - just saw the previous post - didn't notice it.

The Supreme Canuck
2006-Apr-10, 04:53 PM
Yep, antoniseb took care of it:


In my first post I showed a calculation, but I left out a very importnat piece of the puzzle: the rounded end caps above and below the cylinder need to be counted in the removed volume:

Let's try it.

Edit: Criminey. You edited while I posted!

Gruesome
2006-Apr-10, 05:28 PM
Calculations? Who calculates things these days? I just built a 6" sphere, ran a 0.5 inch diameter hole through it and let the computer tell me the volume. Seems a whole lot easier than dealing with pi and all that schtuff.

The_Radiation_Specialist
2006-Apr-10, 05:31 PM
Calculations? Who calculates things these days? I just built a 6" sphere, ran a 0.5 inch diameter hole through it and let the computer tell me the volume. Seems a whole lot easier than dealing with pi and all that schtuff.

cool! what/where software do you use?

antoniseb
2006-Apr-10, 05:36 PM
I just built a 6" sphere, ran a 0.5 inch diameter hole through it and let the computer tell me the volume.

That would give you the wrong answer, since the hole in that sphere would be only 5.96 inches long.

Gruesome
2006-Apr-10, 05:50 PM
That would give you the wrong answer, since the hole in that sphere would be only 5.96 inches long.

The computer says it's 5.98 [5.979130, but who's counting]. So I offset the sphere's surface 0.01 to make the hole 6.0 inches long and I get a revised volume of 113.05209961108 cubic inches.

Gruesome
2006-Apr-10, 06:07 PM
cool! what/where software do you use?

Unigraphics. Currently in S. Carolina working on gas turbines.


P.S. It should be noted that any calculations involving pi are expressly undeterminable, since the exact value of pi has an infinite decimal expansion.

tofu
2006-Apr-10, 06:25 PM
(the diameter of the drill bit) Doesn't matter.

So just to be clear, if I had a 6 inch diameter sphere, and I used an 8 inch diameter drill bit, thus milling away the entire sphere, leaving nothing behind, the volume of that nothing would be the same as the volume of the original sphere?

SeanF
2006-Apr-10, 06:30 PM
So just to be clear, if I had a 6 inch diameter sphere, and I used an 8 inch diameter drill bit, thus milling away the entire sphere, leaving nothing behind, the volume of that nothing would be the same as the volume of the original sphere?
Well, I didn't say that the diameter of the drill bit doesn't matter, I said that the fact that clop didn't give the diameter doesn't matter. For any given diameter sphere, there's no more than one drill bit diameter that will leave a six-inch long hole.

And an eight-inch drill bit on a six-inch sphere doesn't do it.

:)

antoniseb
2006-Apr-10, 06:33 PM
an eight-inch drill bit on a six-inch sphere doesn't do it.
An eight inch drill bit on a ten inch sphere will do it exactly.

tofu
2006-Apr-10, 06:38 PM
For any given diameter sphere, there's no more than one drill bit diameter that will leave a six-inch long hole.

oh I get it now! thanks

WaxRubiks
2006-Apr-10, 08:15 PM
If the diameter of the drill doesn't matter then I choose zero(or 1/infinity)for the diameter and get 36*pi, is this right?

clop
2006-Apr-10, 10:47 PM
Well done everyone. It is 36pi cubic inches. You're all too clever for me.

clop

mike alexander
2006-Apr-10, 11:47 PM
Ah, now I get it. (I'm no good at it until I can visualize it). As the diameter of the cylinder increases relative to the diameter of the sphere, the relative volume in the ring portion decreases, but the relative volume of the spherical caps increases in exact proportion. Right?

Irishman
2006-Apr-13, 04:01 PM
farmerjumperdon said:
I think all your answers might be flawed, or the wording of the question needs a touch of tweaking for the answers to be correct. The space occupied by the removed material is not a cylinder. A cylinder must be flat on the ends. The little curved bits that were removed at the very ends of the drilled hole can't be included as part of the cylinder, or you by definition do not have a cylinder.

Except the problem doesn't exactly state that the removed bit is a cylinder.


You take a solid metal sphere and clamp it in a drill press, then core it straight down through the centre and out of the bottom using a drill bit. The resulting cylindrical hole through the sphere is 6 inches long.

It states that you core a sphere. To core a sphere with a drill bit, you cut from top through bottom, removing the curved bits of metal in those zones. The resulting hole is described as "cylindrical". How else would you describe the hole? It is a circular hole that runs straight through, with smooth walls. It is cylindrical, without being a cylinder. Maybe that's some word quibbling. I think the problem statement is clear - core a sphere. The remaining hole is 6 inches long.


Frog march said:
If the diameter of the drill doesn't matter then I choose zero(or 1/infinity)for the diameter and get 36*pi, is this right?

Sure. A 0 diameter drill bit means that the sphere is exactly 6 inches diameter. Math works fine. If the drill bit diameter is 1/infinity, then you need to build that drill bit. ;) Undefined means you don't know how big your drill bit is, so you don't know how big to make your sphere.

The size of the drill bit is linked to the size of the sphere. And vice versa.

WaxRubiks
2006-Apr-13, 05:36 PM
Sure. A 0 diameter drill bit means that the sphere is exactly 6 inches diameter. Math works fine. If the drill bit diameter is 1/infinity, then you need to build that drill bit.


sure, just take any drill bit and grind it down to nothing.