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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

#11 bonanova

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Posted 23 August 2007 - 07:11 AM   Best Answer

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

#12 bonanova

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Posted 23 August 2007 - 07:23 AM

The moral of the story is: your girlfriend lied. Size DOES matter. ;)

Writersblock,

Still laughing ... I love it!

My only reply is ... ok I have two replies ...

[1] she never complained and
[2] the [apparently lacking] size spec is an important part of the logical solution.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#13 Writersblock

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Posted 23 August 2007 - 07:52 PM

Ok, so I still didn't get this so I spent like 4 1/2 hours doing some math. I think I get what you are saying now. The height of the "caps" depend entirely upon the radius of the sphere and the radius of the hole, and thus, so does the removed volume. I am still not convinced you are 100% right, but

I can see that because the length of the hole must always = 6, if you shrink the radius of the sphere = to the length of the hole, and then shrink the radius of the hole = 0, then the volume that is left will always be left, regardless of the size of the hole. That leaves us then with a sphere with a radius of 3, where your 36pi works. The only thing that changes is the "thickness" of the remaning donut, which at some point on the upper sizes would reach theoretical limits of thinness. Like if you have the earth drilled out so that only a 6 inch band remains. (Yes, I know it's not a perfect sphere but it suffices for an example of huge.

Again, my math is weak, so I can't prove this, but I think I smell what you are cooking logically.
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#14 Writersblock

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Posted 23 August 2007 - 11:19 PM

Wow. I haven't felt this dense since first year law school. I really need to bone up on my math skills. Here's what I did, and I capitulate that 36pi must be correct and bow to bonanova's quick mind.

My math - tell me if I screwed up anywhere.

Sphere volume = 4/3P(R*R*R) where P = pi and R = radius of the sphere
Cylinder Volume = P(r*r)L where P = pi and r = radius of the hole and L = length of the hole
Spherical dome Volume (Cap) = ((P(h*h)*(3R-h))/3) where P = pi and h = height of the cap and R = radius of the sphere [taken from http://www.monolithic.com/construction/formulas.pdf]

so the remaining volume = 4/3P(R*R*R) - (P(r*r)L + (2 ((P(h*h)*(3R-h))/3)))

Define h
2 caps height plus L = 2R so
2h+L=2R where L=6 so
2h+6 = 2R
2h= 2R-6
h= ?*(2R-6)

Define R
the original R must always be 1/2 L+h
If the hole length is 6 inches and goes through a sphere, then
R = 1/2 6 + h. so
R= 3+h

Define R in terms of h
h= ?*(2(3+h)-6)
h= ? * (6+2h-6)
h=h WTF? Infinite?

Try again

R=3+ (1/2*(2R-6)
R=3+R-3
R=R WTF? Infinite?

R-h=3
3+h-h=3
3=3 Infinite

So, regardless of
volume = 4/3P(R*R*R) - (P(r*r)L + (2 ((P((1/2(R-L))*(1/2(R-L)))*(3R-(1/2(R-L))))/3)))

I know that the relationship between R and h are dependant for an infinite set of numbers when L=6. Therefore if h=0 then R=L.
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#15 cpotting

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Posted 24 August 2007 - 12:59 AM

Here's the mathematical solution:
Radius of sphere = R, radius of hole = r, length of hole = 2L [so L=3], height of cap = h.

Since r*r = R*R - L*L and h = R-L, we can eliminate r and h and do everything in terms of R and L...



I intuitively knew what the answer would be, but I have been struggling to come up with the mathematical proof of it. Looking at your math, though, I am puzzled. How do reason that the area of hole's cross-section (r*r) equals the area of the sphere's cross-section (R*R) less the the square of half the hole's length?
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#16 bonanova

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Posted 24 August 2007 - 04:53 AM


Here's the mathematical solution:
Radius of sphere = R, radius of hole = r, length of hole = 2L [so L=3], height of cap = h.

Since r*r = R*R - L*L and h = R-L, we can eliminate r and h and do everything in terms of R and L...



I intuitively knew what the answer would be, but I have been struggling to come up with the mathematical proof of it. Looking at your math, though, I am puzzled. How do reason that the area of hole's cross-section (r*r) equals the area of the sphere's cross-section (R*R) less the the square of half the hole's length?


R*R = r*r + L*L because you can draw a right triangle where R is the hypotenuse [Pythagorus].

Hint: slice the thing along the hole's axis running vertically.
Go from the sphere's center horizontally to the surface of the hole - that's a distance r.
Go straight up to the top of the hole - that's a distance L [at right angles to r]
Go back to the center of the sphere - that's a distance R and is the hypotenuse.

When I answered her question, I hadn't proved 36pi is true for all spheres.
I just guessed. [It's really a quite surprising result!] But, for all she knew
I did all this math in my head in 15 seconds! We were colleagues, and I would
place her IQ somewhere north of 160 - high enough to think it could be
done, and ... high enough that she usually left me in the dust in our work.

She was impressed. It was an moment to savor, and I thought I'd share the story.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#17 Writersblock

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Posted 24 August 2007 - 06:29 AM

Well thanks for sharing. I enjoyed crunching my brain on it.

IF anyone is interested, I think the easiest way to think about it is this:

L is the length of the cylinder and is a constant at 6" and R is the radius of the sphere, r is the radius of the cylinder's head, h is the height of the dome on top and bottom of the cylinder that is drilled out. Ok, we are defined.

L must stay constant, so as R expands toward infinity, r and h, and thus the volume of the cylinder and the 2 domes, must expand proportionatly so as to keep the original shape a sphere, and to keep L constant. Thus, for every sphere, where L is constant at 6, the remaining volume is the volume of a sphere where L = 2R. B))
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#18 mwdoperator

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Posted 24 August 2007 - 10:11 AM

I have been looking at this one for two days. If you drill a 6 inch hole through a 6 inch sphere, well you just cant do it. Right?
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#19 cpotting

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Posted 24 August 2007 - 11:24 AM

Okay - I've been at this for a while. Thought I would try to find a way to show that the volume of the remainder of the sphere is constant. This would prove the "intuitive" answer that the volume is equal to a sphere 6" in diameter. However, I seem to have proven the opposite . This is only the third time in 27 years that I have used my high school calculus, and I was hoping someone out there may see my error and correct me.

[list]p = pi
D = the diameter of the sphere
d = the diameter of the hole
R = the radius of the sphere (.5D)
r = the radius of the hole (.5d)
H = the height of the hole (end to end)
h = the height of the hole (centre to end = .5H)[list]

After drilling, the remaining ring shape is equivalent to a barrel of unknown width (D), 6" high (H) and with the central cylinder portion removed.
The formula for a barrel with sides bent to the arc of a circle
[list]= pH(2DD + dd) / 12
= (1/12)pH(8RR + 4rr) [list]

The formula for the cylindrical portion that was removed
[list]

Therefore the formula for the volume of the ring shape is
[list]

and since RR = rr + hh (thanks bonanova - I see that now), we have
[list]
which expands to
[list]

Now, if we take the derivative of this function, we should see the rate of change in volume with respect to the radius of the hole. This should be 0, meaning that the volume remains constant. Unfortunately, that is not what I am getting.
[list]rr
(constants have been bolded including H and h [which = 6 and 3])
dx®/dy = 2r + 0 + 2r - 2r
= 2r [list]
This means the volume varies with the radius of hole. That would mean that the volume is not always equal.
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#20 bonanova

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Posted 24 August 2007 - 06:02 PM

[list]p = pi
D = the diameter of the sphere
d = the diameter of the hole
R = the radius of the sphere (.5D)
r = the radius of the hole (.5d)
H = the height of the hole (end to end)
h = the height of the hole (centre to end = .5H)[list]

After drilling, the remaining ring shape is equivalent to a barrel of unknown width (D), 6" high (H) and with the central cylinder portion removed.
The formula for a barrel with sides bent to the arc of a circle
[list]= pH(2DD + dd) / 12
= (1/12)pH(8RR + 4rr) [list]


cpotting,

kudos for finding all these neat [cap and barrel] formulas.
That's does all of the calculus work.

I found it useful to put everything in terms of R and h.
You can do this by noting that RR = rr + hh [Pythagorus]

V[barrel] = ph [8RR + 4rr] /12 = ph [8RR + 4RR - 4hh]/6 = 2ph [RR - hh/3]

V[cylinder] = height x area = Hprr = 2ph [RR - hh]

V[barrel] - V[cylinder] = 2ph [RR - hh/3 - RR + hh] = 2ph [2hh/3] = 4phhh/3.

Recalling that h=3,

V[barrel] - V[cylinder] = 36p [a constant].

Looking at your derivation, everything is correct.
If you add like terms in your expression for V,
you'll see that all the rr terms add up to zero.

Your derivative expression is correct, also,
except that you should include the constants (1/12)pH8, etc...
[if y(x) = Cxx, then dy/dx = 2Cx, not just 2x]
You'll see again that the r-dependent terms add up to zero,
and the derivative is zero.

OK?
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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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