SOLUTION: I have a 6 inch diameter ball. How many balls will fill a room 10ftx12ft with a 8ft ceiling (10x12x8=960)? This room has no furniture or obtrusion's to consider. I can visualize

Question 1064392: I have a 6 inch diameter ball. How many balls will fill a room 10ftx12ft with a 8ft ceiling (10x12x8=960)? This room has no furniture or obtrusion's to consider.
I can visualize 8 balls occupying a 1ftx1ftx1ft space if the balls are set in straight rows from one another. 8 balls/cubic foot of space.
8 balls x 960 = 7,680 balls
BUT if I move the second level of balls over so they nestle into the depressed areas found between the first row of balls I am overwhelmed trying to figure that out.
I have looked online and see there is rounded questimate of .74, but I am not sure if this is what I need?
Do I simply take 8 balls/cubic ft x 960 cubic ft divide by .74 = ?
= 10,378 balls
Thanks

Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
When packing spheres into a container,
there is always "wasted" empty space between the spheres.
There is also empty space between the spheres and the boundaries (floor, walls, ceiling) of the container.
The volume of a sphere with 6 inch diameter (3 inch radius) is

.
A cube with 1 foot edges has a volume of

.
With the packing you visualized first,
the fraction of the space filled with balls is
.
You would fit only balls.

The alternate packing you propose is a better option,
especially for filling a large room with relatively small balls.
The height used up for one layer of balls is rather than .
For a 1 cubic foot container, even with that strategy,
you cannot achieve a very efficient packing of balls
with a diameter of .

For the most efficient packing, the maximum theoretical fill ratio is,
as you found, .
You can approach that ratio, if the size of your container
is large compared to the volume of 1 sphere.
The size of your container is

.
If you could fill 74% of that space with balls, you would fill

.
The number of balls that fit in that "fillable" space is
(rounded).
So, may be a good estimate.

For a more accurate calculation,
accounting for additional space wasted along floor, ceiling and walls,
we can calculates how many layers of balls fit into ,
at per layer.
That would be layers.
That is 18 layers, with some wasted space above the top layer.
Layers 1, 3, 5, etc would have .
Layers 2,,4,6,etc would have .
with 18 layers, you would have
.
That is better than in layers of per layer,
but it is far less than the estimate
RELATED QUESTIONS
the scale on a blueprint is 1/2 inch for every 6 feet. the height of the living room... (answered by edjones)

Madison Duke is wallpapering a laundry room 8ft by 8ft by 8ft high. How many square feet... (answered by kac123kac)

Hi Can someone please help with question If a room has a width of 5 m, a length of 6... (answered by addingup,jim_thompson5910)

Could you please help me with this question?:(Mady has an infinite number of balls and... (answered by richard1234)

i have a swimming pool that is 12 ft by 4 ft want to fill it with gum ball how many gum... (answered by richwmiller)

Can someone please explain the problem If a room has a width of 5 m, a length of 6 m,... (answered by greenestamps)

I have a word problem I need help with. A dog is tied to a 8ft. rope tied to a stake in... (answered by elima)

How many tiles, which measures 1x1 would I need to fill a 10x14... (answered by ehku,DWL)