Can we Build a Bridge to China only using Bricks

Sometimes asking silly questions leads us to discover not so silly, and very unexpected, answers. Here the question is this, suppose we are given a pile of bricks, just bricks mind you, no cement nor mortar, and we are asked to take the bricks and stack them one on top of the other, so that they don't fall over, but reach from here to China. Take a look at fig. 1 to see what I mean. I'm sure your immediate reaction is, "that's crazy," I know mine was. Probably your second reaction is, "Well, if Henry is writing about this, I'd better be careful." Sometimes it is wise not to rely on your first impulse. |

All we need to attack this problem this problem is to understand the concept of "center of mass." Intuitively, if you have some irregularly shaped object, the center of mass is the point in the object where you could balance it on the head of a pin. For example, the center of mass of a human body is somewhere around the solar plexus. If you remember as a kid balancing yourself on top a fencepost, the spot that you stuck your belly on is pretty close to your center of mass. Now take a look at fig 2. What is the center of mass of this system? Well, that's simple, it is just halfway between the two balls in the picture. If the balls were connected by a rod, we could balance the two of them on the tip of a pin at this point. | |

Now take a look at fig 3 and suppose that the two masses are not the same. The expression for the center of mass is a little more complicated now, but if you look at it, you can see that it is just what we call the "weighted average." This is what you would expect, as the center of mass should shift towards the heavier object. What does all this have to do with building a bridge to China? Well, in order to keep the bricks from toppling over, we need to make sure that the center of mass of the bricks that are on top of the bottom brick don't move past the edge of that bottom brick. |

Let's see how that works out for the first few bricks. It may
seem a little strange, but we are building this bridge from the
top down, at least during the planning phase. Once we have a
plan, we can build it from the bottom up, as you would expect.
Anyway, the topmost brick can protrude halfway over the brick
beneath it, since the center of mass of the top brick is in the
exact middle of the brick. Let's assume, to keep things simple
that each brick is 1 foot long. Also lets call the point where
the topmost brick starts 0. Thus the location of the
* COM*(Center Of Mass) of the top brick is 1/2. Now the
edge of next brick that we put underneath the first two should be
at the center of mass of the two bricks above it. Where is this?
Well, the
* COM* of the first brick is at 1/2, and the center of mass
of the second brick is at 1/2 ( left edge of the brick + right
edge of the brick) = 1/2(1/2 + (1/2 + 1)) = 1 so the
* COM* of the two brick system is 1/2(
* COM* of brick1 +
* COM* of brick 2) = 1/2(1/2 + 1) = 3/4. Let's see if we can
figure out the general case. Suppose we have already stacked up n
bricks, where n is some number. We want to put brick number n+1
at the
* COM* of the n brick system, and figure out what the new
* COM* is for the new n+1 brick system. Lets call the
position of the
* COM* of the n brick system C(n). To find C(n+1) we need to
take the weighted average of the n brick system and the new 1
brick system whose left edge is at C(n) (the old
* COM*). Thus we have:

C(n+1) = 1/(n+1) * (n*C(n) + 1/2(C(n) + (C(n)+1))). ^^^^^^ right edge of new brick ^^^ left edge of new brick ^^^^^^^^^^^^^^^^^^^ COM of new brick ^^^^^ weight of COM of n bricks * position ^^^^^^ Divide by total weight of the (n+1) brick system ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ weighted average

We can simplify the expression on the right hand side a lot, first by noticing that 1/2(C(n) + (C(n)+1)) = C(n) + 1/2. Then we have that n*C(n) + C(n) + 1/2 = (n+1)*C(n) + 1/2. Thus finally we get:

C(n+1) = 1/(n+1) * ((n+1)*C(n) + 1/2) = C(n) + 1/(2*(n+1))

What this means is that the
* COM* of the n+1 brick system is located 1/(2*(n+1)) over
from the
* COM* of the n brick system. Another way of saying this is
that after putting down n bricks, we have moved the center of
mass from 0 to 1/2 + 1/4 + 1/6 + 1/8 ... + 1/(2n). Another way of
writing this is to take out the common factor of 1/2 and then the
rest becomes 1/2 * (1 + 1/2 + 1/3 + 1/4 ... + 1/n).

Now the question of whether we can reach China boils down to
adding up a lot (a lot) of numbers, namely the (infinite) series
1 + 1/2 + 1/3 + ... (It's always those ...'s that kill you) Now
the idea of adding up an infinite number of numbers has been
around since the ancient Greeks. You might have heard of Zeno's
paradox, where he "proves" that if you shoot an arrow at a tree
the arrow never reaches the tree. How does he "prove" this? Well,
Zeno said: consider the time it takes for the arrow to get
halfway to the tree. Now cut that time in half and the arrow
again is halfway between where it was and the tree. Cut that time
in half and the arrow is again only halfway between where it was
and the tree. Continuing in this way we can see that the arrow
never reaches the tree. What Zeno was trying to do is add up the
series 1/2 + 1/4 + 1/8 + 1/16 + ... 1/(2^n). He knew that no
matter how many terms you add to this series, the sum never
reaches the number 1. It gets as close as you like, but it never
gets there. Mathematicians say that the series
** converges** to 1, which is a fancy way of saying
that it gets as close as you like, but it never gets there. What
about our series, namely 1 + 1/2 + 1/3 + ..., does it
** converge** to something, and if so what? I fired up
my trusty computer and added the first 100 terms of the series.
It came up with 5.17... After 1000 terms the answer was 7.48...
So maybe it
** converges** to 10, eh?

Before we get to the answer, I hope you will all agree that
the series 1+1+1...+1 doesn't
** converge** to any number, it just continues to get
bigger and bigger. Similarly the series 1/2 + 1/2 + 1/2 ... keeps
growing, it just takes twice as long. Now lets look at the first
two numbers in our series, namely 1 and 1/2. Each of them is
bigger than 1/2, so our series is bigger than 1/2 + 1/2. Now lets
look at the next two terms, which are 1/3 and 1/4, both of which
are greater than or equal to 1/4, and there are 2 of them so
their sum is greater than 2 * 1/4 = 1/2. Now lets look at the
next 4 terms, which are 1/5 + 1/6 + 1/7 + 1/8. Each of them is
greater than or equal to 1/8, and there are 4 of them so their
sum is bigger than 4 * 1/8 = 1/2. We can continue in this way
always taking twice as many terms and noticing that their sum
will alway be bigger than 1/2. Thus the sum of our series is
bigger than the sum of the 1/2 + 1/2 + 1/2 ... series, which
we've already admitted keeps on growing. What this means is that
if we place our bricks according to this plan, we can span any
distance we like, even going all the way to China. As Gomer Pyle
used to say, surprise, surprise surprise.

One final note. We might ask how many bricks it would take to
get to China. Well, if each brick is 1 foot long, I can tell you
that using this scheme it would take about half a million bricks
to go 10 feet. Maybe it's not the most practical scheme, but hey,
I was a math major not an engineering major, and it is possible,
** IN THEORY!**

Nadine and Henry wish you a theoretically perfect but practically wonderful New Year. Adios.

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