FabSwingers.com > Forums > The Lounge > Extremely tricky maths problem
Extremely tricky maths problem
Jump to: Newest in thread
I made a silver bracelet for my wife, by drilling out a cylindrical hole from the middle of a sphere of silver. The width of the bracelet, i.e. how high it is when laid flat, was 1 cm.
What is the volume of silver in the bracelet ?
It's a very tough problem  |
Reply privately, Reply in forum +quote
or View forums list | |
 |
By (user no longer on site)
over a year ago
|
"I made a silver bracelet for my wife, by drilling out a cylindrical hole from the middle of a sphere of silver. The width of the bracelet, i.e. how high it is when laid flat, was 1 cm.
What is the volume of silver in the bracelet ?
It's a very tough problem "
it doesn't make any sound at all unless you drop it |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
It doesn't sound as though we have quite enough data: the sphere of silver is 4/3 x pi x r^3. The bracket is 1cm thick so the cylinder drilled is (r-1) radius, which gives a volume reduction of {pi x (r-1)^2 x h, where h is the point where the cylinder meets the edges of the sphere} plus {the volume of the two extrusions such that the area of cross-section = pi x (r-1)^2}
Without the original radius (or radius of wrist) there isn't enough to go on, I don't think, or rather no way to do more than express the volume as a function of 'r'... |
Reply privately, Reply in forum +quote
or View forums list | |
"Depends on size of cylinder and drill"
Strangely , it doesn't , as the larger the diameter of the bracelet, the thinner the bracelet gets.
Therein lies the clue to solving this, when it appears there isn't enough data. |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"Surely you need to know the length of the bracelet or at least the number of beads?
I still don't know how it helps though  "
pythagoras but if it's a serious question which it isn't you need to know circumference and bore size |
Reply privately, Reply in forum +quote
or View forums list | |
|
By *tirluvMan
over a year ago
the right frame of mind -London |
in basic terms the 1cm height refers to th eintersection depth of the cylinder and sphere.
To get the radius of both the sphere and cylinder you would need to use high school trig to determine the angle and radius and residual arc length.
You would then calculate the volume of the residual mass (cross section of bracelet) and then multiply this by the circumfrence (once you have the radius) of the cylinder to give you the volume of bracelet.
To give you the volume of the sphere you would take the radius calculated and then apply the following formula: V= [(4/3)Pi]r{cubed} -if you really want me to do the maths I could probably do it but quite frankly, I'd rather eat cake
|
Reply privately, Reply in forum +quote
or View forums list | |
"It doesn't sound as though we have quite enough data: the sphere of silver is 4/3 x pi x r^3. The bracket is 1cm thick so the cylinder drilled is (r-1) radius, which gives a volume reduction of {pi x (r-1)^2 x h, where h is the point where the cylinder meets the edges of the sphere} plus {the volume of the two extrusions such that the area of cross-section = pi x (r-1)^2}
Without the original radius (or radius of wrist) there isn't enough to go on, I don't think, or rather no way to do more than express the volume as a function of 'r'..."
Very close. though in your formula h is in fact 1, and the radius of the cylinder is unknown. I think you have confused the width ( 1 cm ) of the bracelet with the thickness ( unknown ).
The key is that as the bracelet gets larger in diameter, the thickness reduces ( in fact remaining at the the same volume ). |
Reply privately, Reply in forum +quote
or View forums list | |
"in basic terms the 1cm height refers to th eintersection depth of the cylinder and sphere.
To get the radius of both the sphere and cylinder you would need to use high school trig to determine the angle and radius and residual arc length.
You would then calculate the volume of the residual mass (cross section of bracelet) and then multiply this by the circumfrence (once you have the radius) of the cylinder to give you the volume of bracelet.
To give you the volume of the sphere you would take the radius calculated and then apply the following formula: V= [(4/3)Pi]r{cubed} -if you really want me to do the maths I could probably do it but quite frankly, I'd rather eat cake
"
Eating cake is probably better lol
|
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"It doesn't sound as though we have quite enough data: the sphere of silver is 4/3 x pi x r^3. The bracket is 1cm thick so the cylinder drilled is (r-1) radius, which gives a volume reduction of {pi x (r-1)^2 x h, where h is the point where the cylinder meets the edges of the sphere} plus {the volume of the two extrusions such that the area of cross-section = pi x (r-1)^2}
Without the original radius (or radius of wrist) there isn't enough to go on, I don't think, or rather no way to do more than express the volume as a function of 'r'..."
Yeah, what he said. Took the words right out of my mouth. |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"I made a silver bracelet for my wife, by drilling out a cylindrical hole from the middle of a sphere of silver. The width of the bracelet, i.e. how high it is when laid flat, was 1 cm.
What is the volume of silver in the bracelet ?
It's a very tough problem "
Need to know the diameter of the hole you drilled out. |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"I made a silver bracelet for my wife, by drilling out a cylindrical hole from the middle of a sphere of silver. The width of the bracelet, i.e. how high it is when laid flat, was 1 cm.
What is the volume of silver in the bracelet ?
It's a very tough problem "
Stop being a miserable sod,and go out and buy her a bracelet |
Reply privately, Reply in forum +quote
or View forums list | |
"I made a silver bracelet for my wife, by drilling out a cylindrical hole from the middle of a sphere of silver. The width of the bracelet, i.e. how high it is when laid flat, was 1 cm.
What is the volume of silver in the bracelet ?
It's a very tough problem
Need to know the diameter of the hole you drilled out."
You don't, that's what makes it so tough. But someone has cracked it, see above. |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"I made a silver bracelet for my wife, by drilling out a cylindrical hole from the middle of a sphere of silver. The width of the bracelet, i.e. how high it is when laid flat, was 1 cm.
What is the volume of silver in the bracelet ?
It's a very tough problem
Need to know the diameter of the hole you drilled out.
You don't, that's what makes it so tough. But someone has cracked it, see above."
nobody cracked it we all knew the theory but without dimensions you can't work it out x |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
Well if it's true for all spheres then consider one that is 1cm high. The cylindrical hole would be infinitesimally small, so the volume of the bracelet is the volume of that sphere less zero. From the usual formula of a spheres volume with r = 0.5cm, we have volume of the bracelet = 4pi/3 x half cubed or pi/6.
|
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"Well if it's true for all spheres then consider one that is 1cm high. The cylindrical hole would be infinitesimally small, so the volume of the bracelet is the volume of that sphere less zero. From the usual formula of a spheres volume with r = 0.5cm, we have volume of the bracelet = 4pi/3 x half cubed or pi/6.
"
Smart arse |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
So if R is the radius of the sphere, n = R-1 is the radius of the drill.
In which case we are looking for:
V(sphere) - (V(cylinder)+2V(cap))
We can solve the depth of cylinder and height of cap as complementary proportions of R (d + 2c = 2R), so if we choose to work with x as d/2,
V(sphere) = 4/3 × pi × (n+1)^3
V(cyl) = pi × n^2 × 2x
V(cap) = 1/6 × pi × (n+1-x) × (3n^2 - (n+1-x)^2)
Which I was gradually simplifying, extracting pi from each for instance, and I suspect that as 1/3n^3 Is taken from 4/3n^3 we will result in a fairly round outcome based on a standard fraction of R. But if it's a choice between this and beer, mine's a pint... |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"Well if it's true for all spheres then consider one that is 1cm high. The cylindrical hole would be infinitesimally small, so the volume of the bracelet is the volume of that sphere less zero. From the usual formula of a spheres volume with r = 0.5cm, we have volume of the bracelet = 4pi/3 x half cubed or pi/6.
"
First of all he said bracelet!
But that aside, why is the hole considered to be infinitesimal small?  |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"Well if it's true for all spheres then consider one that is 1cm high. The cylindrical hole would be infinitesimally small, so the volume of the bracelet is the volume of that sphere less zero. From the usual formula of a spheres volume with r = 0.5cm, we have volume of the bracelet = 4pi/3 x half cubed or pi/6.
Smart arse"
the bracelet 1cm or 100mm in height could have been bored out to wrist size that wasn't explained and the lack of bore hole dimension means no calculation |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
I'll leave the question setter to give a more elegant answer, I've got to get off to the York Fab social, but a sphere with a hole in it even an infinitesimal one is still a bracelet. Although you would need a tiny wrist! |
Reply privately, Reply in forum +quote
or View forums list | |
|
By *ast_jjMan
over a year ago
Dublin and London |
"It doesn't sound as though we have quite enough data: the sphere of silver is 4/3 x pi x r^3. The bracket is 1cm thick so the cylinder drilled is (r-1) radius, which gives a volume reduction of {pi x (r-1)^2 x h, where h is the point where the cylinder meets the edges of the sphere} plus {the volume of the two extrusions such that the area of cross-section = pi x (r-1)^2}
Without the original radius (or radius of wrist) there isn't enough to go on, I don't think, or rather no way to do more than express the volume as a function of 'r'..."
Holy shit...my head hurts...  |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"Well if it's true for all spheres then consider one that is 1cm high. The cylindrical hole would be infinitesimally small, so the volume of the bracelet is the volume of that sphere less zero. From the usual formula of a spheres volume with r = 0.5cm, we have volume of the bracelet = 4pi/3 x half cubed or pi/6.
"
Ah. But the bracelet needs to be 1cm "deep" at any point, so the initial sphere we are talking about has radius 1 (for no hole to be drilled, but the bracelet to meet its criteria).
Thus the volume remaining after (not) drilling is 4pi/3.
Whether that holds true as the sphere expands is something I'm not entirely convinced is true, but I would be more convinced that 4pi/3 cm^3 is accurate |
Reply privately, Reply in forum +quote
or View forums list | |
|
By *ast_jjMan
over a year ago
Dublin and London |
"So if R is the radius of the sphere, n = R-1 is the radius of the drill.
In which case we are looking for:
V(sphere) - (V(cylinder)+2V(cap))
We can solve the depth of cylinder and height of cap as complementary proportions of R (d + 2c = 2R), so if we choose to work with x as d/2,
V(sphere) = 4/3 × pi × (n+1)^3
V(cyl) = pi × n^2 × 2x
V(cap) = 1/6 × pi × (n+1-x) × (3n^2 - (n+1-x)^2)
Which I was gradually simplifying, extracting pi from each for instance, and I suspect that as 1/3n^3 Is taken from 4/3n^3 we will result in a fairly round outcome based on a standard fraction of R. But if it's a choice between this and beer, mine's a pint..."
Now I'm with you on the pint... |
Reply privately, Reply in forum +quote
or View forums list | |
"Well if it's true for all spheres then consider one that is 1cm high. The cylindrical hole would be infinitesimally small, so the volume of the bracelet is the volume of that sphere less zero. From the usual formula of a spheres volume with r = 0.5cm, we have volume of the bracelet = 4pi/3 x half cubed or pi/6.
"
That is exactly the elegant solution to this. The full maths as per Twisted flame above leads you to this by the long route.
It's a lateral step to conclude that as you're not given the diameter, you don't need it, so can alter the diameter to whatever value makes it easy to work out. |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"I'll leave the question setter to give a more elegant answer, I've got to get off to the York Fab social, but a sphere with a hole in it even an infinitesimal one is still a bracelet. Although you would need a tiny wrist!"
Wouldn't that be a ring? |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"Well if it's true for all spheres then consider one that is 1cm high. The cylindrical hole would be infinitesimally small, so the volume of the bracelet is the volume of that sphere less zero. From the usual formula of a spheres volume with r = 0.5cm, we have volume of the bracelet = 4pi/3 x half cubed or pi/6.
First of all he said bracelet!
But that aside, why is the hole considered to be infinitesimal small? "
What’s the second word from the end mean?  |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"It doesn't sound as though we have quite enough data: the sphere of silver is 4/3 x pi x r^3. The bracket is 1cm thick so the cylinder drilled is (r-1) radius, which gives a volume reduction of {pi x (r-1)^2 x h, where h is the point where the cylinder meets the edges of the sphere} plus {the volume of the two extrusions such that the area of cross-section = pi x (r-1)^2}
Without the original radius (or radius of wrist) there isn't enough to go on, I don't think, or rather no way to do more than express the volume as a function of 'r'..."
Superb! |
Reply privately, Reply in forum +quote
or View forums list | |
|
By (user no longer on site)
over a year ago
|
"Well if it's true for all spheres then consider one that is 1cm high. The cylindrical hole would be infinitesimally small, so the volume of the bracelet is the volume of that sphere less zero. From the usual formula of a spheres volume with r = 0.5cm, we have volume of the bracelet = 4pi/3 x half cubed or pi/6.
First of all he said bracelet!
But that aside, why is the hole considered to be infinitesimal small? What’s the second word from the end mean? "
the penultimate word means, da da da |
Reply privately, Reply in forum +quote
or View forums list | |
» Add a new message to this topic
A bit of lateral thinking there I suspect !
Care to enlighten the others ?
