Recently, we were asked "If room
204 in the Harris building was removed of all desks, chairs, tables and any
other removable objects, how many tennis balls would we be able to compact
into that room?" Several guesses were made from 40,000 to 5 million
tennis balls.
Once divided into two groups, each group was given a shoebox (not the same size though) with two tennis balls. Armed with those items, a calculator, paper, pencil and nothing else, we set out to find how many tennis balls it would take to fill up Harris 204.
At first we thought about figuring out how many tennis balls the shoebox would hold. But the shoebox didn't hold exactly 3 by 6 balls with two levels of balls. Just by looking and guessing, it seemed to hold 2.5 by 5.25 balls with 1.75 levels, and those aren't the easiest numbers to work with. So we decided "Why measure in shoe boxes when we need to find out how many tennis balls at the end anyways?" We scrapped the shoebox and continued on to find out how many balls line up against the walls.
We took into consideration that
area of the room was the length of the room times the width times the height.
So we could easily measure how many tennis balls ran across the length, width,
and height, multiply and have our answer. Easy enough. However, the
room wasn't a perfect rectangle since a corner of our room was cut out for
the door and then another little portion of the room was taken up by a support
column on one side of the wall. We had to subtract these areas from
our total area. After we've subtracted those areas, we were left with
the following perimeter, each side of which we named A, B, C, D, E, F.
(See diagram 1).
We measured the number of balls each length was by marking a sheet of paper with the distance of four balls. We would count how many sets of four we had then account for the leftover parts by measuring with one tennis ball. After "measuring" how many balls ran across each wall, we found that wall
A = 149.5 balls
B = 129.25 balls
C = 28.5 balls
D = 20 balls
E = 3 balls
F = 6 balls
We also found that the height of the room, H, was equal to 44.5 balls. Then it was just an easy equation to solve.
(AB - CD - EF)H = total number of balls in the room.
Since AB gives you the area of the room but you want the area without areas CD and EF, you just subtract those two areas (or the balls in areas CD and EF) multiplied by the height of the balls and that gives you the total number of balls in the room.
However, then we realized this
would only work if the balls stacked nicely on top of each other in neat
rows and columns. Since they don't stack nicely but one ball will fall
in between the space between the two balls underneath it, more balls can
actually be put in. We also realized that if the balls fall into the
crevice between the two balls underneath it, every other row would have one
less ball than the one above or below it.
Not only did
we have to consider the difference in the row totals, but our hieght would
be different since the balls feel lower down. Originally we measured as though
two balls would stack vertically. However when considering the compactedness,
the two balls will be a 60 degree angle. This is demostrated in disgram two. Notice x is our original
height, but in reality the height will be the altitude of the equilateral
triangle formed by three balls. So we had to discover the proportion of the
height we had. The y is found by the sin of the 60 degree angle. Therefore
our original height, x, only accounted for approx. 86.6% of the real height,
y. So our new hieght for the room became H=52.539 balls.
Now we go back to our original equation (AB - CD - EF)H and find area of each part individually.
AB = ABH - (H/2)A - (H/2)B
* Since every other row has one less ball than the row above or below it, we are skipping every other row with the total number of balls in each row. The height tells the number of rows we have stacked on one another. So if every other row is missing a ball, then H/2 accounts for the balls that needed to be subtracted from our original area.
CD = CDH - (H/2)C - (H/2)D
EF = EFH - (H/2)E - (H/2)F
And finally...
AB - CD - EF = total number of balls in the room.