![]() Down to zero, if the smallest size of a simple container is less than the diameter of the sphere for example, even if you had cubic miles of space in a tube the diameter of five inches, you can't fit a single six-inch sphere in it.Īlthough each void is relatively small compared to the size of a sphere, there are lots of them – remember, each sphere touches 12 other spheres! –, totaling about 26% of the container volume. ![]() When the container is not immense compared to the spheres, more volume (relatively speaking) is wasted between the spheres and the container walls, so the real $N$ you can achieve is less than what the above predicts. The larger the container is compared to the size of the spheres, the closer you can get to the limit given above. The rest of the volume is "wasted" in the small voids between the spheres, and in the voids between the spheres and the container walls. Finding the formula itself is easy enough to do, the real problem is finding out where the numbers come from ( $N \, \le \, \frac$$spheres in the container. Would anyone mind clarifying why the maximum density is 74%?Įdit: I had some great responses, thank you very much for them! A few problems though: My own knowledge is only equivalent to an advanced class high schooler, so that's probably why this goes over my head. However, I have yet to really understand how they derive this number. the maximum density the sphere's can occupy in the close packing method is around 74%. From what I've gathered from previous Stack Exchange questions etc. However, having my own doubts about the correctness of the method I did some research on my own and came across close packing. And so this is sometimes the event in question, right over here, is picking the yellow marble. The probability of picking a yellow marble. So they say the probability- I'll just say p for probability. ![]() ![]() Initially my friend was working with a crate method, namely turning the spheres into cubes whose length was equal the sphere's diameter, and finding out how many cubes fit (with some estimates, this is easy enough). Find the probability of pulling a yellow marble from a bag with 3 yellow, 2 red, 2 green, and 1 blue- I'm assuming- marbles. Obviously, since a soccer ball is a sphere, the resulting question is the title. The assignment is: Find how many soccer balls fit inside a cylindrical building. A jar contains 5 blue cubes, 4 blue spheres, 2 green cubes, and 7 green spheres. You can often write recurrence relations for counting problems like this, but here it’s far more work then needed, and you seem to have found a simple solution.So this is on the behalf of another friend for a school assignment. ![]() If you reach the right-most face in the red area, you’ve run out of red candies first, so to answer your question, you want to know how many paths (going only right, up, or away) there are from the origin to the upper-right box that reach the red area.Īlternatively, you would want to know what fraction of the $\frac$ sequences do not correspond to paths that lie within the following figure. In the picture below, a candy-sequence would correspond to a path of adjacent cubes from bottom left to top right that moves right, up, or away from the viewer. Note, this only holds because the balls are being randomly selected. So the probability of getting a blue ball is 7/21 or 1/3. If you think of picking red, blue, and green as corresponding to taking steps in the $x$, $y$, and $z$ directions on a three-dimensional lattice, each way of drawing candies corresponds to a path from $(0,0,0)$ to $(10,20,30)$ of steps that go one unit in a positive direction each. There are 7 blue balls and 21 balls altogether. The following is almost certainly overkill, but what the heck. 4 Blue Cubes Second container: 4 Red Cubes 3 blue Cubes There are 4 outcomes and they are as followed: (R,R), (R,B), (B,R) And (B,B) The question is asking about the probability of one cube is red out of the 2 containers. ![]()
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