Bump whats the answer
Bump whats the answer
mcguirkthejerk wrote:
Okay letsrun brainiacs, I have a tough question and I want to see who can solve it:
I want to know how many ping pong balls you can fit within a much larger sphere.
The larger sphere has a circumference of ~96in.
A ping pong ball has a circumference of ~4.9634in.
You do the math and as always, show your work.
Anyone who comes up with the correct answer is up for a prize.
Great thread, but the question is flawed.
The diameter of a ping pong ball is actually 4.94738011087 inches. I am amazed no one caught this obvious error.
Hey OP, is your first name Jim? And your son's name, Alex?
I made some mistakes here to get at 4229.Using brute force, I found an arrangement of 4638 balls that can fit inside the larger sphere.
rekrunner wrote:
So how's this math problem working out for you.
I went out and bought an ~96 in. (internal) sphere, and 6000 ping pong balls, measuring ~4.9634 in.
I found I could get as many as 5357 complete balls inside (and a little bit more), only if I cut up a bunch of them to fill in the gaps.
Otherwise, I only managed 4229 balls without using scissors, using a face-centered cube stacking, but I'm still checking.
This is relevant if you're trying at home or guessing how many for a prize or something (closest guess how many are in this gumball machine, e.g.)
If you do 60% efficiency, that comes out to 4510.
7517.72 is 100% efficiency
??? How do you get 7512.72? Depending on what you mean, shouldn't it be 7235.62?Why do you chose 60%? Seems rather arbitrary. We are not estimating gumballs. Or for that matter why "random close packing" over "thickest regular packing?I already found an instance with 4638 ping pong balls. 4510 doesn't seem very interesting.
transience wrote:
http://en.wikipedia.org/wiki/Random_close_pack#For_spheresThis is relevant if you're trying at home or guessing how many for a prize or something (closest guess how many are in this gumball machine, e.g.)
If you do 60% efficiency, that comes out to 4510.
7517.72 is 100% efficiency
Why the fuc would anyone want to know this shit. No wonder we haven't got a cure for cancer when scientists waste time doing this crap
Here is how the maximum density for regular hcp packing is derived, which I posted in another thread but which belongs here.
Consider a box of length 9sqrt(3)/2 and width 9r.
Arrange the first layer of spheres in alternating rows along the length such that there is a half sphere at the end of the 1st and 3rd rows, at the beginning of the 2nd and 4th rows, and the 5th row consists of 4 half spheres with a quarter-sphere at the end. Draw a picture if it's confusing. There is a little bit of space left over after the 1st and 5th rows, but that converges out as I'll explain later.
The 2nd layer can be constructed the same way, with the row of half spheres and quarter sphere at the top, and the other half spheres on opposite sides from the first row. And so on ad infinitum. Then the 3rd like the 1st, and so on ad infinitum.
Let m be the number of layers beyond 1. Then the volume of the spheres
U(m) = (4.5x4 + .5x2 + .25)x4pir^3/3 = 20.5x4pir^3/3 = 27pir^3
from my earlier post, the volume of the box containing the spheres is the area of the box times h.
V(m) = (2r + 2mrsqrt(2/3)) x (9rsqrt3)x9r/2
Then the density
D(m) = U(m)/V(m) = 27pir^3/(((2r + 2mrsqrt(2/3))x81r^2sqrt3)/2)
= pirm/((2r + 2mrsqrt(2/3)x(3sqrt3)/2)
as m --> inf this converges to
pim/(msqrt(2/3)x3sqrt3 = pi/(3sqrt2) ~ 0.74
Now, what about the little extra space in each layer above the top or bottom rows? Well, for a given length of the rows and number of layers, that extra space will remain constant if you increase the number of rows and thus the width of box, so the ratio of that space to the volume of the box for a given height converges to zero when the number of rows ---> inf. In general the density of this stacking method converges to pi/3sqrt2, though in this example it would be slightly lower.
The OP stating the circumference is approximately 96 inches, which means the diameter is approximately 31 inches. So the answer should be an approximation of no more than two significant figures.
This is how I would go about calculating it:
1) pick a maximum density regular stacking method
2) describe an arrangement closely inscribed in a sphere of some integer n ppb diameters wide, so that each part of the outer sphere is within 1.579in (one ppb diameter) of at least one ppb. At least one such arrangement must exist, pick one if there are more than one.
3) Calculate the volume of the arrangement for the greatest n less than 31/1.579, and the least n > 31/1.579. Or use 30 and 32, or some other interval based on your interpretation of "approximately 96." Express your answer as the 2 significant figure number between them (ex 4825 and 4967 would be 4900).
Brrrup wrote:
Why the fuc would anyone want to know this shit. No wonder we haven't got a cure for cancer when scientists waste time doing this crap
This has practical applications in the oil and gas field so that cancer scientist can get to and heat his lab among other things.
This is what happens when you can't get laid.
just a college runner wrote:
~96in.
~4.9634in.
96/4.9634= 20.7164437 ping pong balls.
so 20 full ping pong balls.
youre welcome
I know this is old...and I know we already know you're wrong. But your basic math isn't even right...The answer to your math problem is 19.34. What kind of horrible math are you doing??
mcguirkthejerk wrote:
Okay letsrun brainiacs, I have a tough question and I want to see who can solve it:
I want to know how many ping pong balls you can fit within a much larger sphere.
The larger sphere has a circumference of ~96in.
A ping pong ball has a circumference of ~4.9634in.
You do the math and as always, show your work.
Anyone who comes up with the correct answer is up for a prize.
Since I'm weak on the calculus....here's a crude estimate:
96 = 30.577 diameter for large sphere.
4.9634 = 1.5798 diameter small spheres.
The large sphere can contain a 21.6x 21.6 x 21.6 cube. That cube can contain 2197 full small spheres, which is 67.45% the volume of the L sphere. So there is 32.5% of the L sphere left over and could contain about 1058 S spheres....so I get about 3255 small spheres in the large sphere.
~3255
I meant 13x13x13 = 2197 in the cube...
Subfive wrote:
The large sphere can contain a 21.6 x 21.6 x 21.6 cube. That cube can contain 2197 full small spheres
It can contain 2197 full small cubes. That doesn't mean it can't contain more than 2197 spheres.[/quote]
~3255
Since the diameter is ~96, this would be ~3300.
Problem with the "9" button on the calculator.
Retardalert. wrote:
just a college runner wrote:~96in.
~4.9634in.
96/4.9634= 20.7164437 ping pong balls.
so 20 full ping pong balls.
youre welcome
I know this is old...and I know we already know you're wrong. But your basic math isn't even right...The answer to your math problem is 19.34. What kind of horrible math are you doing??
Any word on what the prize will be?
Seems kinda like a cop-out to use only 2 significant digits. Can you at least provide a response to the nearest 2 digits?Here's what happens when I follow your approach:1) I have a maximum density regular stacking method.2) I can describe how to arrange ping pong balls with this method.3) By "volume of the arrangement" do you mean number of ping pong balls? For a large sphere of exactly 19 balls wide, I can describe an arrangement of 4351 ping pong balls that fit inside. For a sphere of exactly 20 balls wide, I can describe an arrangement of 5121 ping pong balls that fit inside. What's the two significant digit figure between these, that apply to an ~96 in. sphere?Can you expand on your approach? Specifically, you avoid the most interesting part of the problem -- how to model the truncating of the ping pong balls by the larger sphere. What is it a function of?
Bad Wigins wrote:
Here is how the maximum density for regular hcp packing is derived, which I posted in another thread but which belongs here.
...
pim/(msqrt(2/3)x3sqrt3 = pi/(3sqrt2) ~ 0.74
The OP stating the circumference is approximately 96 inches, which means the diameter is approximately 31 inches. So the answer should be an approximation of no more than two significant figures.
This is how I would go about calculating it:
1) pick a maximum density regular stacking method
2) describe an arrangement closely inscribed in a sphere of some integer n ppb diameters wide, so that each part of the outer sphere is within 1.579in (one ppb diameter) of at least one ppb. At least one such arrangement must exist, pick one if there are more than one.
3) Calculate the volume of the arrangement for the greatest n less than 31/1.579, and the least n > 31/1.579. Or use 30 and 32, or some other interval based on your interpretation of "approximately 96." Express your answer as the 2 significant figure number between them (ex 4825 and 4967 would be 4900).
rekrunner wrote:
For a large sphere of exactly 19 balls wide, I can describe an arrangement of 4351 ping pong balls that fit inside. For a sphere of exactly 20 balls wide, I can describe an arrangement of 5121 ping pong balls that fit inside. What's the two significant digit figure between these, that apply to an ~96 in. sphere?
without an explanation of just what "~96" means from the OP, I would just say 4700 +/- 400 and add that it's a guesstimate, not a confidence interval.
you avoid the most interesting part of the problem
-- how to model the truncating of the ping pong balls by the larger sphere. What is it a function of?
The OP said you can't cut the PPB's. Do you mean calculating where each PPB touches the sphere? The volume between the PPB's and the sphere?
If you're packing the spheres regularly, it's the maximum diameter of the structure that matters, because
1) the smallest sphere of that diameter will not fit around a structure with a greater maximum diameter
2) a smaller sphere won't fit around the structure even though it may be smaller than its maximum diameter on other axes, and
3) there are (probably), for a given regular stacking method, no two "spherically complete" arrangements with the same maximum diameter but different volumes. By "spherically complete" I mean the diameter on all other axes is within 1PPB of maximum, i.e. no PPB's are "missing" from the structure.
How many patterns did you get for 19 and 20 balls wide? There should only be a few possible maximum diameters on [19in, 20in]. If, for example, you knew the sphere was exactly 19.313 PPB's in diameter, you could find the largest possible maximum diameter less than 19.313 PPB's of a "spherically complete" hcp stack, count the PPB's in that stack, and voila, that's how many will fit. I would calculate it, and prove 3), but it would probably take a long time and give me a headache.
Or maybe you are talking about random stacking and I was this long-winded for nothing.
What about this: use volume of sphere equation...with radius of 9.67156. The radius is half of the number of ping balls that will fit in the diameter of a sphere of 96 in.
I get ~ 3789 ping balls fitting in the 96 in sphere.
If you insist on interpreting "~96" as it was written, why not attempt to solve the problem for a 96.00" sphere (merely 4 significant digits), then express your answer as an approximation ("~") with two significant digits?
I wasn't asking you to interpret the OP, but rather for you to further clarify your method further, assuming at least 2 significant digits. If my two figures, from your large sphere as a multiple integer PPB approach, are 4351 and 5121, what can your term "the two significant digit figure between these" possibly mean? What is the right 2 digit number between 43 and 51? The numbers I gave are the result of infinite precision, the large sphere being expressed in exact terms of the ping pong ball diameter. 4700 +/- 400 is one significant digit, and you are now reluctant to commit to that? If you don't want to play, just be honest. Don't hide behind the OP and meaningless significant digits.
When I say "truncate", I think of removing a fractional part, so that we are left with positive integers. I'm not cutting the balls, in order to count them, but cutting them from the count.
Let's try to make this more clear.
A basic naive approach to the solution is to divide the volume of the large sphere by the volume of a ping pong ball, and we get a very generous theoretical upper bound of 7235 balls.
If we apply the ~0.74 factor for the known maximum density of sphere packing, which you have independently derived, we can tighten this upper bound to 5357 balls.
Yet, when I do simulations, manipulating a few variables, the most I could find was 4638. My search was lengthy, but not exhaustive, (and not necessarily error-free).
This yields a significant difference: 5357-4638 = 719 balls.
When I talk of modeling the "truncated" balls, I mean attempting to count all of the fractional parts of the missing 719 balls. In this "sphere packing" within a sphere problem, what is the expected number of balls to be "truncated" from the count, and what fraction of these balls would have been inside the larger sphere?
All of your attempts to calculate maximum densities, and letting the "inside" layers of balls go to infinity, marginalize the "missing" balls to zero.
In a large sphere of exactly 19 ping pong balls in diameter, I am able to find 3,145,278 possible arrangements of maximum density sphere packing. Only 3 of those are HCP. FCC is 2 more.
In a large sphere of exactly 20 ping pong balls in diameter, I am able to find 12,582,912 possible arrangements of maximum density sphere packing. Only 3 of those are HCP. FCC is 2 more.
Furthermore, all of these possibilities I can find are centered at the origin of the large sphere. I can also imagine innumerable offsets in every combination of the x,y,z axes. You can imagine now why I say my search was not exhaustive.
If you are interested, for every one of these arrangements, I can calculate how close the closest ping pong ball was to the large sphere, and how far the farthest one was.