The sphere

Seán · Guest

Here's one for ya.
In any sphere over 6cm diameter it's possible to drill a hole that goes right through the centre of the sphere and is exactly 6 cm deep. The diameter of the hole will vary depending on the diameter of the sphere. So for large spheres you will be left with a large wedding ring shape that is exactly 6 cm deep.

What is the volume of the "wedding ring"?

Guest · Posted: Sat Sep 24, 2005 12:06 am Post subject:

It is volume=4/3¶r3

tricky · Guest

a large hole Question

Duh · Guest

No, you're wrong. If a sphere is 7 cm in diameter then in order to drill a hole to the center you would need to drill 7 cm deep. and the size of the hole is not going to change if you use the same size tool to do the job.

Sean · Guest

Remember the depth that is being discussed is the depth of the "wedding ring" shape that is left after the hole has been drilled. So if you have a sphere 7cm diameter and you drill any diameter hole through the center, the depth of that hole is going to be less that 7cm. The diameter of the hole will vary in order to leave you with a hole exactly 6cm deep.

Riddleman · Site Admin Joined: 18 May 2005 Posts: 598

I have no clue

hduncan · New Member Joined: 20 Jun 2006 Posts: 5

if you drill a hole in anything you get a cylender for a hole. the volume would be 6cm times the area of the circle that is covered by the drill bit. for example if the drill bit is a one inch drill bit then the area of the circle covered by it is pi X r^2. or pi X .5^2. which equals .785 in ^2. then simply multiply by the depth of the hole which is 2.36 in (6cm) and you would get 1.852 in^3.

the variable is not the constant 6 cm depth. the variable is the area of the circle created by the drill bit.

GameMaster · Site Regular Joined: 20 Apr 2006 Posts: 55

Your explaination of the problem is flawed...

Remember the depth that is being discussed is the depth of the "wedding ring" shape that is left after the hole has been drilled.
"So if you have a sphere 7cm diameter and you drill any diameter hole through the center, the depth of that hole is going to be less that 7cm. The diameter of the hole will vary in order to leave you with a hole exactly 6cm deep."

"6 cm deep" means (or at least I and the few people who responded have read it to mean) that the bit will be 6 cm long.... That's the depth of the hole (or the height of the cylinder removed from the spere). It's independant of the diameter of the hole....

The volume of the remaining, can be found by intergration (or with the formula for volume of a sphere) will be:

The area of any sphere s, with a cylinder c removed will be:
2*Intergral from 0 to radius of s : As(x)dx - Ac
where As(x) = Π * (x/2)^2 and Vc = 6 Π r^2 and r < radius of s
So, we start to integrate....

= Π * ( 1/2 *integral of x^2) - Ac
= Π * (1/2 * (x^3)/3) - 6r^2 | from x=0 to x = radius of s, call it R.
= Π (R^3/6 - 6r^2)

The "depth" of the whole is still 6cm. The volume of the hole is still 6Πr^2. The only thing that changes the VOLUME of the remaining peice... The DEPTH is constant.

Riddleman · Site Admin Joined: 18 May 2005 Posts: 598

not good at math

OliverWestby · New Member Joined: 14 Jan 2007 Posts: 1

the volume of the wedding ring shaped piece left over is 36pi or 113.097

deryk · Guy Who Comes A Lot Joined: 19 May 2006 Posts: 159

i don't think there is enough information to even answer this. Your can drill a 6cm deep whole into any sphere with a diameter greater than 6cm. As the riddle states: the larger the sphere the wider the drill whole will have to be to still create a depth of 6 cm, and thus the closer to a wedding ring shape there will be. maybe i just don't know enough math yet but wont the volume of the wedding ring change depending on the size of the original sphere?

i know everything · Posted: Tue Feb 13, 2007 10:41 pm Post subject:

deryk is right, the math involved in this is seriously flawed, the size of the cylinder would not be determined, there is not information, the drill might be the same, or the size of the drill bit you use might fluctuate drastically, so therefore, i can conclude that this is a dumb, (sorry) riddle.

theintellectual · Site Regular Joined: 26 Oct 2006 Posts: 55

I don't think this is considered a riddle............................

deryk · Guy Who Comes A Lot Joined: 19 May 2006 Posts: 159

well i thought about it again. since the thickness of the ring shape would get thinner and thinner as the original sphere gets bigger and bigger maybe there is a direct relationship and perhaps one could solve it like gamemaster did. I asked some friends and they say u could use calculus and thats what gamemaster seems to have used. I wont take calc till nxt year so i dunno still Razz

i know everything · Posted: Sat Mar 03, 2007 4:15 am Post subject:

but the rate at which the "ring" shape is enlarged is not set. I know everything, (hence the name,) so TRUST ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
please?

deryk · Guy Who Comes A Lot Joined: 19 May 2006 Posts: 159

having read a number of gamemaster's answers i have reason to believe that he or she knows a lot more than everything lol

i know everything · Posted: Sun Mar 04, 2007 4:10 am Post subject:

even if it sounds good, the riddle is still flawed. The rate at which the cylinder is not fixed or specified, so
= Π * ( 1/2 *integral of x^2) - Ac
= Π * (1/2 * (x^3)/3) - 6r^2 | from x=0 to x
= radius of s, call it R.
= Π (R^3/6 - 6r^2)
will not work.
Even though i have no idea what that means.

deryk · Guy Who Comes A Lot Joined: 19 May 2006 Posts: 159

lol hahahahaha ok

BobaFunk · New Member Joined: 06 Aug 2009 Posts: 28

you cannot assess a definitive volume of the wedding ring without giving the specifics of the size of sphere. i have, however dissected the equation with that unknown. first find the volume of the sphere.

V=4/3Pi^2

with the information of a hole 6cm deep, we need to find chord length, which is at a distance of 3cm perpendicular to the circle center. Using Pythagorean Theorem and the radius as the hypotenuse, we can calculate chord length

r^2=L^2+H^2 (L= chord radius length, H=height)

with that, we can find the entire length of the chord. this length would be the diameter of drill.

2L=a (chord length)

now we need to find the volume of the both the cylinder using (a) as the new radius

V=Pi*a^2*H

and the spherical cap

r-H=x (distance from perpendicular chord length to arc of sphere)

V=Pi*x(3a^2+x^2)/6

knowing now that there are one cylinder and two spherical caps, it's a matter of subtracting volumes from the original sphere size.