The Bucket Escape Trick

   I haven't been posting much lately since cable problems deprived me
of TNT and with it, W^3. This Saturday however, I paid a visit to
Kris to see the most recent episodes. Since she suggested someone else
should trash the escape-from-the-well scene, I guess I'll do it as the
resident geek in the group.  In fact, I'm thinking of how to present
this scene to some of my students as a physics problem. If any of you
survived introductory high school or college physics courses I'd
appreciate your feedback as to how you would react to something like
this in a class. Even if you were smart enough to avoid physics
courses, I hope I can make this an amusing tangent.  As a kid, I
always loved Isaac Asimov's essays where he'd apply some bit of
physics to an everyday situation that most of us never thought about
and come up with some amusing conclusions. I guess this will be my
attempt to emulate the great master, but I fear my effort won't be
nearly as entertaining as his.  So bear with me, Gentle Reader.
Warning: this turned into a very long post. You might want to
dump this to a file to peruse later or just skip to the end!
   Recall the scene: Jim and Artie are trapped in a 30 foot deep well.
Jim places the powder from a couple of cartridges under a bucket, stands
on the bucket and ignites the powder. The resulting explosion propels him
to the top of the well. I think it's fun to look at just what this implies.
   First, I wondered how fast Jim was blown upward. This is easy to
figure. The blast gives Jim a kinetic energy (energy of motion) of 
1/2 m v^2 where m is his mass and v is his initial velocity. This is just
the definition of kinetic energy - take it on faith. Would I lie?
   As he rises, gravity slows him down. If he starts out with too
little energy he'll stop and begin falling back before he reaches the
top. At a minimum, he must have just enough energy to reach the top of
the well as gravity slows him to a stop. How much energy is that? To
reach a height of h requires an energy equal to mgh where m is mass
again and g is the "acceleration due to gravity" which is determined
by the gravitational attraction to the Earth. g is a constant (almost)
for all objects on the Earth and is equal to 32 ft/sec^2. The product
of mass and g is just the quantity we call weight and is expressed in
pounds in the horrible English system of units.  Again, a physics
course would explain where this relation comes from, but we'll just
grab it and use it.  If we assume Jim weighs about 200 lbs, that means
to reach the top he needs an energy of

E = m g h
where mg = 200 lbs and h = 30 ft. Plugging in these numbers gives

E = (200 lbs) (30 ft)
E = 6000 ft lbs 

to reach the top of the well.  By comparing this energy to his kinetic
energy we can find his initial velocity.
   To repeat, the energy Jim has available is 1/2 m v^2 while the energy
he needs is mgh. To successfully reach the top of the well these two
must be equal, so we say

1/2 m v^2 = m g h.

   We know everything in this equation but v. Solving for v and
plugging in the numbers gives us

v = 43.8 feet/sec 
or, in more familiar units:

v = 29.8 miles/hour.

Let's just round it off and call it 30 miles/hour.

   So far, so good. Travelling 30 miles/hour is impressive but not
unreasonable. Jim West should be able to handle that easily. But we
haven't said anything about how he acquired that velocity. Just how
reasonable is it that the gunpowder would propel him upward at 30
miles/hour? We need to know a little more about the performance of
typical 19th century gunpowder.
   I'm not sure exactly what type of weapons Jim and Artie usually
carried, but I have some data for the 1860 Colt Revolver which was a
very popular weapon, so I'll take it as being typical.  According to
my sources, an 1860 Colt Army Revolver in .44 caliber fired a lead
ball weighing 140 grains at a muzzle velocity of 736 feet/sec when
loaded with 28 grains of black powder.  What the heck is a grain, you
ask? This is yet another unit of weight in the *&^%$#^! English
system. There are 7000 grains in a pound, but don't worry about it -
I'll generously provide the unit conversions.

Digression on units of measurement:
(This could be worse. If all this calculation drives us to drink,
Asimov once pointed out that the English system offers us volume
measured not only in pints, quarts, fifths, and gallons, but also
bushels, pecks, and the ever-popular firkin. Then, we must remember
that a "standard" firkin of any liquid is a quite different size than
a "firkin of ale". Furthermore, we have a still different volume for
the special unit known as a "firkin of ale in London". Presumably the
Londoners don't want to be caught dispensing ale in the same size
containers as those peasants in the countryside! Now if I hear one
more student tell me that they prefer the English system because the
metric system is "too complicated"...I think I'll throw a firkin of
ale at them, in or out of London!)

   Anyway, where was I? Oh yes, we have the mass and velocity of the
bullet fired from a typical 19th century gun. We can go back to the
old 1/2 m v^2 formula for kinetic energy and determine the energy of the
bullet. It turns out to have a kinetic energy of about 170 foot lbs.
This is less than 3% of the 6000 foot lbs we said Jim needed to get to
the top of that well!
   Where can we get more energy? I said that 170 foot lbs was the
energy of a single bullet propelled from Jim's revolver. That bullet
was propelled by a charge of 28 grains of powder.  If we use twice as
much powder we should get twice as much energy, right? Since we need
about 33 times as much energy it figures that we require 33 times as
much powder.  That comes to 924 grains, which is about two ounces of
powder. Certainly not an unreasonable amount to carry, but I don't
recall seeing Jim pull out 33 cartridges for powder!
   So, I said I was going to answer Kris' challenge to show how bogus
the escape-from-the-well trick was and I seem to have failed to do so.
If Jim needed only two ounces of powder, what's the beef? The fact is
that we've overlooked a crucial point. We computed how much powder
would provide an explosion with an energy equal to what Jim needed.
However, Jim could only utilize a *small* fraction of that available
energy. Why? In a firearm, the bullet is confined by the chamber and
barrel so that the hot gases from the burning powder can't escape.
They push the bullet along the barrel and, as long as the bullet
doesn't exit the barrel, they keep pushing the bullet and accelerating
it to higher speed. Once the bullet leaves the barrel, the hot gases
escape, there's no longer any force to push the bullet, and it just
keeps going with the same speed with which it left the muzzle. That's
why rifles and cannons have long barrels; to keep the gas pushing the
projectile for as long as possible so it can acquire as much energy
from the gas as possible.  If the barrel is too short, the projectile
leaves the barrel before it has had time to get all the energy of the
expanding gas. The muzzle velocity is then much less than we would expect.
   How would that bucket behave when the powder was ignited in it? We
can make a rough estimate in the following manner. We need to know how
much force the powder exerts on the bucket. To estimate that, let's go
back to the revolver momentarily.  In the revolver, the force is
related to the mass and acceleration of the bullet by Newton's famous
equation F = m a.  To solve this for the force we must know the
acceleration.  The acceleration, a, is just the amount by which the
velocity changes in one second. We know the bullet starts out with
zero velocity and leaves the muzzle a short time later with a velocity
of 736 feet/sec.
   Knowing the velocity changes by 736 feet/sec in a distance of 0.5
feet allows us to determine that the acceleration is 542,000
feet/sec^2. (Actually, the force and acceleration are constantly
changing as the bullet travels down the barrel, but I'm just
calculating a constant average value of acceleration to simplify life.)
The force exerted on the bullet is just F = m a which gives F = 340 lbs.
Quite a lot of force on such a small object!
   This force of 340 lbs comes from the pressure of the gas. Recall
that pressure is just force/area. The bullet has a radius of 0.22 inch
which means that it has an area of 0.61 in^2.  Dividing the 340 lb
force by this small area tells us that the pressure inside the gun is
a whopping 81,000 lbs/ft^2, or in more familiar units, 556 lbs/in^2
(often written 556 psi).
   How does this relate to Jim's bucket? First, Jim has 33 times as
much powder. If it all burns quickly enough, it will produce 33 times
as much gas. That would produce 33 times as much pressure IF it were
confined in the same volume as the revolver. Of course, it isn't. The
bucket has a much larger volume for the gas to fill, and that will
cause a corresponding reduction in the pressure. Let's say the
bucket is 10 inches wide and 10 inches deep. That means it has a
volume of 1.8 ft^3. This is 3400 times the volume of the gun barrel!
We'll decrease the previous value of pressure by this amount to
account for the larger volume of the bucket but we'll also multiply
the value by 33 since Jim used 33 times as much powder. Thus, the
initial pressure from the burning powder is

P = (81,000 lbs/ft^2) 33/3400
P = 786 lbs/ft^2.

   That must have been one well built bucket! I would think 786 lbs/ft^2
would tear a tin bucket to pieces, not to mention what it would do to
Jim. I wouldn't want to get hit with a 1 foot wide rock that weighed
786 lbs, but then I'm not Jim West of course.
   Since the cross sectional area of the bucket is pi r^2 and r is 5
inches (half the 10 inch diameter), we find the surface Jim stood on
had an area of 0.54 ft^2.  The upward force is just this area times
the pressure, which gives us a value of 424 lbs. Subtracting 200 lbs
needed to counter Jim's weight leaves 224 lbs of net force to propell
Mr. West upward. This force acting on his mass will give him an upward
acceleration of 36 feet/sec^2. Unfortunately, this acceleration only
lasts for a very short time.
   Just how long does this force act? Not long. As we observed
earlier, as soon as the bucket starts to move the gasses escape from
underneath it and the pressure drops drastically, imparting less
velocity to Jim than it would otherwise.  This means that in reality
Jim would need even more powder which would produce an even bigger
blast, taking even more of a toll on Jim and that poor bucket!
   In conclusion, (at last!!!!!) I'm sure the *real* Jim West would
never try a stunt like this, would you Mr. West? After all, Jim
was a military man and all the stuff I've calculated here would be
learned by any freshman at West Point in his introductory courses.
Well, maybe that guy Grant wouldn't have mastered it, but he was a
poor student who flunked several courses and barely managed to
graduate near the bottom of his class. What ever happened to him anyway?

-- Paul (the Shotgun-Physicist of Louisiana)
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Paul L. Fisher, KB5KBX               "It is difficult to say what is
Dept. of Physics & Astronomy          impossible, for the dream of yesterday
Louisiana State University            is the hope of today and the reality
Baton Rouge, LA 70808                 of tomorrow."
phone: (504) 388-828                --Robert H. Goddard
e-mail: fisher@rouge.phys.lsu.edu
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