Jump to content


Welcome to BrainDen.com - Brain Teasers Forum

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account.
As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends.

Of course, you can also enjoy our collection of amazing optical illusions and cool math games.

If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top.
If you have a website, we would appreciate a little link to BrainDen.

Thanks and enjoy the Den :-)
Guest Message by DevFuse
 

Photo
- - - - -

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


  • Please log in to reply
166 replies to this topic

#161 b1soul

b1soul

    Newbie

  • Members
  • Pip
  • 2 posts

Posted 16 March 2011 - 07:21 AM

the answer is the volume of a sphere with a diameter of 6 (I can't remember the formula for spherical volume)

my logic: if the remaining volume of the sphere is the same no matter the diameter of the cylinder, reduce the cylinder to a line with a length of 6 inches, the remaining volume is the sphere with a 6 inch diameter

Edited by b1soul, 16 March 2011 - 07:22 AM.

  • 0

#162 Amanda:D

Amanda:D

    Newbie

  • Members
  • Pip
  • 1 posts

Posted 03 May 2011 - 02:17 AM

Don't read unless you're sure you want to.: D

Spoiler for Answer

  • 0

#163 googon

googon

    Junior Member

  • Members
  • PipPip
  • 78 posts

Posted 03 May 2011 - 02:31 AM

its the normal length of a sphere with 6 as a diamator
logic: with a flat side the diamator of the cylandir could be the diamaotr of the largest circle that can fit on the side but a sphere has no flat sides so the cylandir can not have a diamator
  • 0

#164 auran

auran

    Newbie

  • Members
  • Pip
  • 1 posts

Posted 12 May 2011 - 11:31 PM

The problem is people are thinking of a drill bit and the distance the 'point' travels. For a sphere the size of the earth they are envisaging a drill (of any size but lets say 1mile diameter) and seeing the point touching the north pole and drilling down and exiting the south pole - which is an 8000mile distance.

Think instead of a hole saw (google image hole saw if you dont know what one is - imaging one is essential to the problem) - it isnt a 'solid' drill bit - it is only a perimeter. Now think of a hole saw almost the size of the earth such that when you lower it down over the earth it only cuts through 6 inches of material. You can see that the smaller the hole saw the more material it would cut through - if your hole saw was 1mile wide it would bore through all 8000 miles... if your hole saw is 4000 miles in diamater it cuts through less material. if your hole saw is 8000 miles wide less 6 inches it only bores through a minute amount of material.

This is the core of the problem - people think the length of a hole (that passes exactly through the center of the sphere) is measured from the two most distant points - ie north pole to south pole when it isnt - it is the shortest distance - if you could stand on that huge ring left after hole sawing the earth the 'thickness' as you measure down is 6 inches.
  • 0

#165 seikenguy

seikenguy

    Newbie

  • Members
  • Pip
  • 1 posts

Posted 19 October 2011 - 12:04 AM

Hello, thank you for the awesome puzzle! Now, having read a lot of the discussion, I have come to think that there are actually 3 answers possible:

1,2) If the length of the hole is equal to the diameter of the sphere (6 inch hole, 6 inch diameter sphere) then the answer is either 36pi or 0. That is because in order to drill such a hole in such a sphere the diameter of the drill must be either 0 inches or 6 inches (again must think about the length of the INSIDE of the sphere after the hole's been drilled out. Like standing inside the hole - if there's anything visible left, then the hole is less than 6 inches).
So, if the diameter of the drill is 0 inches, the remaining volume of the sphere equals to the original volume of the sphere (36pi), as nothing was drilled away. If the diameter of the drill is 6 inches, the sphere is, of course, completely drilled away with nothing left, so you can only imagine the hole, and the volume is thus 0.

3) If the length of the hole is smaller than the diameter of the sphere (6 inch hole, 7 inch diameter sphere) then the answer is impossible to figure out logically, without computing it. And if the problem would have no answer that is possible to figure out logically, thus it would have no answer at all, given it is a logic puzzle.

So actually, there's three answers to this problem, given the aforementioned reasoning - 36pi, 0 and No Answer.

Please, correct me if my reasoning somewhere is wrong.

p.s. Actually, while writing this post I have lost my line of thinking several times, and rewrote it accordingly =p This is the final version that I can finally read without getting confused myself... Made my night *sleepy*.
  • 0

#166 DarwinII

DarwinII

    Newbie

  • Members
  • Pip
  • 1 posts

Posted 14 December 2011 - 03:17 AM

Since it (almost) does not matter what size the sphere is, just use the easiest one. A six inch shere would be nice. Then the hole would be zero diameter and all the sphere would be the remainder. Use the formula 4/3 pi r^3 with a radius of 3 inches.
  • 0

#167 cakecookies2

cakecookies2

    Newbie

  • Members
  • Pip
  • 5 posts

Posted 15 December 2012 - 01:48 AM

IT SAID DRILLED THROUGH SO THE AMOUNT REMAINING IS 0


  • 0




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users