How many balls does it take to fill a 16x16 ft room

How many balls does it take to fill a 16x16 ft room?

2.5k views

How many balls does it take to fill a 16x16 ft room?

| 2.5k views

+1 vote

Equation:

No of basketballs that can be accommodated in a 16X16 room is = ( Available Volume of the room / Volume of each basketball )

Assumptions and Facts:

1. We are assuming that, its the basketball we are talking about in this case study. Size of the ball is assumed to be of radius 15 cms.
2. The room is assumed to be of a cube size. So height, width and length all are assumed to be 16 ft.
3. The best possible fit ratio is assumed to be 75%. So only 75% of the room is available for fitting basketballs in it.
4. 1 ft = 30 cms (fact)

Process:

Volume of a Room can be calculated by formula = Length * Breadth * Height

Total space in room in cms cube (A) = 16 * 16 * 16 * 30 * 30 * 30

Calculation of available space in room (B) = 0.75 * A

Volume of each ball = Volume of each sphere = (4/3) * (22/7) * cube of radius

Volume of each ball (C) = (4/3) * (22/7) * 15 * 15 * 15

Dividing B by C, you should get around 6300, which means we can fit in approximately 6300 basketballs in a room of dimensions 16 ft X 16 ft X 16 ft

by (19 points)
+1
As there's always a space between the balls, I think it would be much easier to treat the balls as cubes in the first place, thus not calculate sphere volumes.
Clarify the question:

Is the room empty or it has furniture.

Are the balls arranged properly or just thrown in randomly

What kind of ball it is - golf, tennis, basketball?

Assumptions and Facts:

Room is empty

Ball - Tennis ball

Balls will be stacked perfectly

Equation:

No. of balls = Volume of room/ Volume of ball

Computation:

Volume of room = 16*16*12 = 3072 cubic feet (assuming empty) = 3072*1728 = 5300000 cubic inch

Volume of ball - assuming square volume of side 2.5 inch - 15.62 cubic inch

Edge cases:

Windows

Furniture

Other equipment

Ceiling top layer

Final computation:

5300000/15.62  = 3,39000 approx (Tennis balls with empty room stacked properly)

if furniture = 307000 approx
by (13 points)

Assumptions/Clarifications:

- Room is rectangular in nature with dimensions: a X b X c units. Balls are perfectly spherical with radii r units.

- Room has no content (furniture, windows), anything that will not let its volume be 'abc cubic units'.

- Interviewer wishes to know the maximum number of balls I could fit.

- Radius of ball(r) is insignificant compared to room's dimensions(a,b,c). This is important so that max number of balls I could fit will be very significantly greater than a little bit of space left at the ends because of effective radius not being a multiple of a particular dimension. Imagine fitting balls with diameter just greater than half of the length of the largest edge. No matter the packing, you can only fit 1.

- you already don't know from the lattice packing science that says spheres can be tightly packed at best 75% density. A fact already used by someone to answer this question.

Packing and Calculation:

Now, we need to strategise a packing of balls so that we fit maximum.

Imagine laying out  balls along a bottom floor edge, say 'c'. The best fit would be to place them in a line side by side. Number of balls I can fit is x = (c/2r).

Now place another line adjacent to it. You want to place it such that a ball in the second line falls between the gap of the 2 balls in the first line for the densest fit. By simple 2D geometry(height of equilateral triangle), the line connecting radii of second line is say h = (root(3)*r).Therefore the number of such lines I can lay out to fill floor is say y = b/h = b/(root(3)*r). Also ignoring the r distance between edge most lines and edge(insignificant compared to b as assumed). (You can ignore the math for the interview I guess).

Now you have a 2D plane of balls on the floor. Next obvious step is to plan the next 2D plane of balls on it to fill the height. This of course needs to be done in such a way that a ball in the second plane is in between the gaps of 3 adjacent balls of the first plane for the tightest packing. Skipping the geometry and math involved, lets say the the distance between the centers of balls of these 2 planes turns out to be some h. The number of such planes I can do is a/h = z(say).

Therefore the number of balls is x*y*z.

by (17 points)

Clarifications:

1. The height of the room?

2. The room is occupied or furniture

3. Height of the room?

4. Only one layer of the floor is filled or the balls can be stacked upon each other?

Assumptions:

1. Capacity is utilized to the fullest

Only one layer and balls are not stacked upon each other:

- This case is fairly simple all you need to do is

1 feet = 30.48 cms

The diameter of the basketball ~25 cms

16 feets = 30.48 * 16 = 488 cms

The number of balls in one column = 488/16 = 19

The number of columns = 488/16 = 19

Some of the space remains unoccupied, as the entire space to fit a ball is not available

The answer = 30* 30 = 900;

Additional height required to stack another layer:

(refer the calculations in the attached image)

~21.65

Height of the room  average ~ 10 feet = 304.8 cms

Number of layers in the stack possible =

First layer requires complete space 304.8 - 25 = 279.8

Number of subsequent layers = 254.8/21.65 = ~11.76

With 11 as the ball must occupy the space completely:

= 11 + 1 (bottom layer) = 13 layers and 361  balls in each layer

=  11* 361

= 3971

With 13 as the ball as the earlier unoccupied space would allow to bottom balls to spread and allows some space in height:

12 + 1 intital layer = 13 * 361

= 4693 balls

My recommendation - it will easily release enough height attributed to the Spread due to the incomplete space in 2D and 14 layers would be possible hence answer is 4693

by (23 points)

Assumptions

• Height of the room = 12 ft
• Voids present between balls
• The room is empty
• Maximum efficiency of packing without pressing the ball

#Volume of the room calculation

Volume of the room = 16 ft x 16 ft x 12 ft = ~3000 cu. ft.

#Effective volume of a ball caculation

When balls are stacked on top of each other, voids are left between.

• Let us take the smallest unit of a pack of balls which can be replicated on all sides to fill the room.
• If one ball is replicated, there will be less efficient packing
• If two balls are replicated, it is same as if one ball is replicated
• However, if three balls are stacked like the image below, we will have maximum efficiency

Volume of this unit = Volume of the cuboidal shaped container = Length x Width x Height

Width of one ball = ~ 6 cm

Length = Width of left bottom ball + Width of right bottom ball = 12 cm

Width = Width of one ball (sideways) = 6 cm

Height = Width of the bottom ball + Effective width of the top ball = 6 cm + 0.8 x 6 cm = 10.8 cm = ~ 11 cm

Volume of the unit = 12 cm x 6 cm x 11 cm = ~ 800 cu cm

Total Number of Units = Volume of the room / Volume of one unit

Total Number of Units = 3000 cu. ft. / 800 cu. cm

1 ft = 30 cm approx.

Total Number of Units = 3000 x 30 x 30 x 30 cu cm / 800 cu cm = 81,000,000 / 800 = ~100,000

Since one unit contains three balls, total number of balls will be thrice the number of units

by (33 points)