How many one inch spheres can fit in a 10x10x5 inch box?

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In summary, the conversation discusses the maximum number of one inch diameter spheres that can be packed into a box that is 10 inches square and five inches deep. There is a disagreement about whether the spheres should be placed directly on top of each other or if they should be placed in the depressions between four adjacent spheres. It is determined that the maximum number of spheres is 543, which can be achieved by placing the spheres between each other in six layers. Further discussion includes the concept of close packing and the possibility of extending it to a cuboid shape. It is suggested that the maximum number of spheres in a cuboid could be 706, but this is uncertain.
  • #1
extreme_machinations
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What Is The Maximum Number Of One Inch Diameter Spheres That Can Be Packed Into A Box 10 Inches Square And Five Inches Deep?
 
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  • #2
Well along the width of the base 10 spheres can fit same is the with the length of the base. Along the height of the box 5 sphere can fit. So the total no. of spheres is 10x10x5=500 spheres. That's what I think.
 
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  • #3
That looks to me to be much harder than a "E-4" problem! (I interpreted "E-4" as "fourth grade, elementary school!) Suppose you put the second layer of spheres not directly on top of the first layer, but "between"- that is each sphere in the depression between 4 adjacent spheres. "Sphere packing" in general is a very hard problem.
 
  • #4
I worked on it and found out that maximum no. of spheres in only possible if they are placed directly on to each other. For this particular problem if the spheres are placed as you have said than there are two possibilities. Either the no. of spheres that can be packed is (81+100+81+100+81)=443 or it is (100+81+100+81+100)=462. They both are less than 500. I can be wrong though.
 
  • #5
I would like to see some proof of this. Would analyzing the 2-D analog be any help? You know, circles in a square. I mean, just on top of one another, well, that's just doesn't seem to be a problem of any stature.

One concern I have DeathKnight is that you implicitly assume you can get only five levels. However, when you place the balls between each other, might you not be able to fit a sixth level?

Edit: Perhaps I shouldn't say implicitly. Have you in fact verified that you can't fit a sixth level?
 
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  • #6
Damn I was wrong. You can definitely add the sixth level which makes the no. of max spheres 543. I was making a simple error.
 
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  • #7
Well, I noticed something interesting when you consider the 2-D analog. Note the attached plot. I can't get 6 levels in 2-D but that's apparently because in 3-D, the height of the levels drop slightly due to the 3'rd dimension. This is enough it appears, to get the top one in. The next question then is what ratio of the dimension of the container to the dimension of the balls is needed to assure that the last level just fits?
 

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  • #8
DeathKnight said:
I worked on it and found out that maximum no. of spheres in only possible if they are placed directly on to each other. For this particular problem if the spheres are placed as you have said than there are two possibilities. Either the no. of spheres that can be packed is (81+100+81+100+81)=443 or it is (100+81+100+81+100)=462. They both are less than 500. I can be wrong though.

hexagonal close packing[placing spheres in the voids of the layer beneath ] and cubic close packing assures maximum usage of the available space in cubes which is = 74% . whereas in square close packing[placing spheres on top of each other and side by side as suggested by deathnight]
the maximum occupied volume is only 52.2%

im just not sure if this can b extended to a cuboid .
if the maximum occupied space in a cuboid is same as in a cube ,then the problem gets simplified .

available volume of cuboid = 74% of 500 ...eqn 1
volume of one sphere =4/3 {pie}{1/2}^3...eqn 2

now if we divide eqn 1 and 2 we get the number of spheres as[706].

which is greater than 543 , but I am not sure about this cos the packing efficiency fo cuboid could be different.
 
  • #9
Can Somebody Please Confirm This ?
 

1. What is mensuration?

Mensuration is the branch of mathematics that deals with the measurement of geometric figures such as length, area, volume, and angles.

2. What is "E-4" in mensuration question?

"E-4" typically refers to the fourth level or grade of mensuration questions, and is often used in educational settings to indicate the difficulty or complexity of the problem.

3. What topics are covered in E-4 mensuration questions?

E-4 mensuration questions may cover a range of topics, including but not limited to: perimeter and area of 2D shapes, volume of 3D shapes, surface area, and basic geometry concepts such as angles and triangles.

4. How can I improve my skills in solving E-4 mensuration questions?

The best way to improve your skills in solving E-4 mensuration questions is through practice. Make sure you have a strong understanding of basic mensuration concepts, and then work on solving a variety of problems at this level to build your skills and confidence.

5. Are there any tips for solving E-4 mensuration questions?

Some tips for solving E-4 mensuration questions include: carefully reading the problem and identifying what information is given, drawing diagrams or visual aids to help you visualize the problem, and breaking down the problem into smaller, more manageable steps.

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