Thursday, April 29, 2010

Bermuda Triangle

Everyone’s heard of the mysterious [cue echo-y scary voice] Bermuda Triangle.  If the Triangle were as dangerous as the stories would have you believe, there would be piles of shipwrecks jutting out of the water.  How many sunken ships would it take to fill the Bermuda Triangle?

According to various websites, the Bermuda Triangle covers 500,000 square miles with depths ranging anywhere from a few hundred feet to about 30,000 ft.  Assuming an average depth of 15,000 ft, we can compute the total volume we’ll need to fill with sunken ships

volume = (depth) · (area)
= (15,000 ft) · (500,000 square miles)
= 5.9×1015 m3.

Ships come in all different sizes.  Since it’s the sunken ship everyone knows, I’ll assume we’re dealing with Titanic-sized ships1.  According to Wikipedia, the Titanic had a length, beam, and height given by 269.1 m, 28.0 m, and 53.3 m respectively.  This gives a volume of approximately 4.0×105 m3.  We can then easily compute the number of Titanic-sized ships that would fit in the Bermuda Triangle,

# of ships = (volume of ocean) / (volume per ship)
= (5.9×1015 m3) / (4.0×105 m3 per ship)
= 1.5×1010 ships.

That’s 15 billion ships.  If you sunk one million ships per day, it would take 40 years to fill the Bermuda Triangle.

[1] Yes, I know there are no icebergs in Bermuda.  Work with me, people.  Work with me!

Book of Odds

I recently started contributing to Book of Odds.  If you like Diary of Numbers, check out my first BOO post here.

Quotes of the Day

These made me smile:

Person 1: I'm breathing Saganized air! My day just got better.

Person 2: The bad news: That Sagan molecule you breathed was sneezed by Hitler.

Person 3: That's OK. Hitler's sneeze was farted by Bach.

Person 4: Despite the title of his book, I can't seem to find the answer to the Lil' Kim song anywhere.

Wednesday, April 28, 2010

As Far as the Eye Can See

We’ve all heard the idiom “As far as the eye can see,” but just how far is this?  If you’re standing on the Earth, how far can you see?

There are many different ways to interpret my original question, so I should have been more specific.  As a friend wrote,

“I frequently see M31 (Andromeda galaxy) which is about 2 million light years. I did NOT see the GRB that was bright enough to be seen -5 billion light years- (visible only for a few seconds).”

I admit, I wasn’t thinking astronomically when I first posted the question.  Since the age of the universe is thought to be 14 billion years, it stands to reason that the absolute farthest the eye could see is 14 billion light years1.  I was thinking more terrestrially.  If we confine ourselves to the quasi-two-dimensional world that is the surface of the Earth, how far can the eye see?

Consider the Earth to be a sphere of radius RE = 6400 km.  The eyes of a standing person might be some distance H = 1.7 m off the ground.  By constructing a triangle (see diagram), the Pythagorean theorem predicts the equation

(RE + H)2 = RE2 + X2,

which can be rearranged to solve for the farthest terrestrial distance X one can see

X = [ (RE + H)2 - RE2 ]1/2
= [ (6400 km + 1.7 m)2 – (6400 km)2 ]1/2
= 4.7 km.

This is the distance at which ships start to sink on the horizon.  Note that the higher up you go (i.e. larger H), the farther you can see.  At the top of Mount Everest, you can see 336 km in each direction.

[1] A light year is the distance light travels in one year, so any object further away than this could not be seen since the light would not have had time to reach you.

Friday, April 23, 2010

Simon Says

Author, journalist, and TV producer Simon Singh1 has a great question for us today.  Dr. Singh has written numerous popular science books including Fermat's Enigma, The Code Book, and Big Bang: The Origin of the Universe.  His question is this:  

“How many razor strokes does the average clean-shaven grown up man make each morning?  People typically say 20 or 30, but when they count them they see that it is about ______.”  

I’ve left Dr. Singh’s estimate blank.  Without counting, how many razor blade strokes does a man make each day?   

How many strokes you use clearly depends on how much facial hair you have, your shaving technique, and what kind of razor you use.  You almost certainly make more strokes with a dull single blade then you will with a brand new 5-blade razor2.  It’s difficult to measure the area of ones face since it’s not a flat surface, but a reasonable estimate3 is 7 in. by 8 in = 56 in2.  Some times you use strokes that cover several inches, other times your razor moves only a few millimeters.  Let’s assume the average stroke covers one square inch of area.  If you’re like me, you usually like to shave each area a few—let’s say 3—times to make sure it’s smooth.  From this we can estimate the total number of strokes,  

# strokes = (total area) · (strokes per unit area) 
= (56 in2) · (3 strokes per in2)
= 170 strokes.

This is within an order of magnitude of Dr. Singh’s estimate of 100 strokes.  For the record, I just counted and got 257 strokes with a fairly dull Gillette Fusion.  

Dr. Singh, thank you for both for the great question and for your heroic work fighting for free speech. 

[1] For those of you who read this earlier and are wondering why the attribution changed, it's because I screwed up.  I had written Dr. Singh and he was kind enough to reply with the above question.  Somehow I mixed up his email address with Simon Tofield's who had submitted a question earlier.  I'd like to give my sincerest apologies to both Simons for my stupid and very embarrassing mix up.     
[2] The Onion was fairly clairvoyant on this one.   
[3] I actually measured this to be sure.  It may seem large, but I’ve got one of those necks where you can’t quite tell where the facial hair stops and the chest hair begins.

Tuesday, April 20, 2010

Ice Bullet Theory

In principle, an ice bullet should be able to kill a person without leaving a trace since the ice would melt away destroying any evidence.  In practice, this is not the case as MythBusters demonstrated in a 2003 episode.  Will this always be the case?  What if the bullets started off really cold?  How cold would a bullet have to be to make it out of a gun barrel before evaporating?
Wow!  I’m impressed.  I like ntm’s analysis in the comments.  My only concern would be that not all of the energy goes into heating the bullet since some must be used to propel it forward.  I’ll consider the 45 g bullet as a test case.  Real bullets have speeds ranging anywhere from 180-1500 m/s.  We can use the formula for kinetic energy to determine how much of the 18,942 J goes into linear motion.  Plugging in the slowest bullet speed, we get
kinetic energy = (mass) · (velocity)2 / 2
= (45 g) · (180 m/s)2 / 2
= 730 J,
while the fastest bullet speed gives a kinetic energy of
= (45 g) · (1500 m/s)2 / 2
= 51,000 J.
In the case of the former, 730 J is barely a significant figure compared to the muzzle energy, which means that most of the energy will likely go into heating the bullet1.  In the latter case, there’s not even enough energy to propel the bullet forward never mind heating it.  If the muzzle energy is 18,942 J, then the absolute fastest a 45 g lead bullet can move is about 920 m/s, and that would mean all the energy goes into kinetic motion rather than heat.  For a 3.97 g ice bullet, the fastest speed achievable for a bullet absorbing all that energy is 3100 m/s.
Even if all the energy is transferred directly to kinetic energy, there is still air friction that will heat the bullet.  At high speeds, the frictional drag force is proportional to the velocity squared:
drag force = - (drag coefficient) · (air density) · (area of the bullet) · (velocity)2 / 2.
The air density is 1.2 kg/m3.  I’ll assume the cross-sectional area of the bullet is about 0.1 cm2 (i.e. about 0.3 cm by 0.3 cm.)  The drag coefficient for a bullet2 is about 0.295.  Plugging these numbers in, we find the total drag force is,
drag force = - (0.295) · (1.2 kg/m3) · (0.1 cm2) · (3100 m/s)2 / 2
= -17 N.
We can compute the total work done by air friction on the bullet by multiplying this force by the distance traveled.  Some guns have barrels that are as small as 10 cm, so the total work done on the bullet can be estimated by3
work = (force) · (distance)
= (-17 N) · (10 cm)
= -1.7 J.
If this is the only work that goes into heating the bullet, then its temperature change can computed as
temperature change = (energy) / [(mass) · (specific heat)]
= (1.7 J) / [(3.97 g) · (2.05 J/g·K)]
= 0.21 K.
This means the bullet could be 1 K (or equivalently 1°C) below the melting temperature and it would still fire! 
What gives?  Didn’t MythBusters bust this myth?  It turns out our assumption that all the energy went directly into the kinetic energy of the bullet is bupkis.  The truth lies somewhere between ntm’s result and my own.  If all the energy goes into translational motion, then the bullet will make it out of the barrel with no problems.  If all the energy goes into heating the bullet, it will melt almost instantaneously.  As ntm very accurately points out, you can only lower the temperature of the bullet to absolute zero, so there’s only so much we can freeze it.  Whether or not an ice bullet will ever work depends on how efficient your rifle is.  If we cool the bullet to zero temperature, it will be able to absorb 2200 J before it melts4.  Whether or not the bullet melts depends on how efficient the rifle is at transferring energy to the bullet.  So long as less than 12% of the muzzle energy is converted to heat, the bullet should make it out of the barrel fine.
At any rate, very good work, ntm.  I am suitably impressed.
[1] There is also some energy in the recoil of the rifle.  This should have much less kinetic energy than the bullet.  You can see this if you solve for the conservation of both energy and momentum.
[2] Since they’re made of different materials, it’s unclear whether or not ice bullets and lead bullets will had the same drag coefficient, but this is a reasonable order of magnitude estimate.
[3] Because of friction, the bullet will slow down so technically we can’t assume the velocity is constant over all 10 cm.  Because of this, our estimate here should be considered an upper bound for the work done on the bullet.
[4] We can also include the latent heat necessary to undergo a phase change, but this will be small.  

Monday, April 19, 2010

Those Screwy Celtics

In a 1991 episode of the sitcom Cheers, Boston Celtics forward Kevin McHale becomes obsessed with knowing how many bolts are in the Boston Garden basketball court, and the obsession ruins his game.  If only Kevin had a copy of How Many Licks? How many bolts are in the basketball court at the old Garden?
An NBA basketball court is 94 ft by 50 ft.  As you can see, there are about 10 floorboards running across the width of the court and 18 running down its length giving a total of 180 floorboards.  If each floorboard is held down by 4 bolts (one at each corner), there will be 720 total bolts. 
A quick Google search shows that there are actually 988 bolts.  Presumably, my estimate is lower because I only counted the playing surface and not the several meters floorboards that make up the sidelines.

Tuesday, April 13, 2010

And Boy Are My Arms Tired

I’m traveling until Thursday night.   I should be able to post consistently again once I get back.  In the mean time, try calculating this: How fast would you half to flap your arms to fly?

The physics of this problem is essentially the same as the “Beat It, Mothra” problem, so I’ll use the same formulas.  A person’s wingspan and arm width are about 2.0 m (~6.6 ft) and 10 cm, respectively.  This means the “wings” have a total area of about 0.2 m2.  I’ll assume the arms/wings move 0.30 m (~1.0 ft) down with each thrust, so that each push propels 6.0×10-2 m3 of air downward.  As before, we can compute the total mass of air moved with each thrust,

air mass =  (density) · (volume)
=  (1.2 kg/m3) · (6.0×10-2 m3)
= 72 g.

For a 75 kg person, the force of gravity is

gravitational force = (mass) · (gravitational acceleration)
= (75 kg) · (9.8 m/s2)
= 740 N.

We can solve for the frequency using the equation from the Mothra problem,

frequency = {(gravitational force) / [(air mass) · (range)]}1/2
= {(740 N) / [(72 g) · (0.3 m)]}1/2

= 185 Hz

This mean you’d have to flap your arms up and down at least 185 times each second in order to just get off the ground.  I’ve simplified both this and the Mothra problem considerably.  As Rich pointed out in the Mothra comments,

“…wouldn't there be a force downward when Mothra lifts her wings up? I haven't really studied how moths wings work but I would assume that the upswing at least creates some downward force, requiring a faster wing beat in order to compensate for those downward forces.”

It’s a very good point that we neglected in both problems.  As I understand it, the down stroke (i.e. the one that pushes the air down and the animal up) is done with the wings spread out as much as possible so that the maximum amount of air is pushed down.  The up stroke occurs closer to the body so that less air is propelled.  There are reasonably good videos of these dynamics here and here.  If the two strokes were the exactly same, they would cancel each other and there would be no net up force.  In these problems, I’ve assumed the upstroke is negligible.  This assumption may be good for an order of magnitude estimate, but for something more precise you definitely need to include the effect the up stroke. 

Sunday, April 11, 2010

Happy Birthday, Anna

It’s my wife’s birthday tomorrow.  I won’t say her age here, but if she lives to 100, how many birthday candles will she have blown out?

It seems two readers have beat me to the punch.  Well played.  They’ve both assumed she has one candle on her first birthday, two on her second, three on her third, etc.  By making two columns of numbers as A has done, you can see that each pair of numbers on a line sums to 101:

1 +100 = 101
2 + 99 = 101
49 + 52 = 101
50 + 51 = 101

We can then multiply by 50 rows to get the answer,

 50 x 101 = 5,050 candles

Well done, Zachary and A!

Friday, April 9, 2010

Death Star Physics

There’s a lot of Star Wars physics one could quibble with.  Take the Death Star.  How, exactly, is a giant laser supposed to blow up a planet?  How much energy would it take to blow up Alderaan?

To calculate how much energy you’d need to blow up Alderaan, you first need to know what’s holding it together.  Unless the Star Wars universe is vastly different from our own, Alderaan is likely being held together by gravitational forces, i.e. all the different chunks of matter that make up the planet are attracted to each other by Newton’s law of gravity.  In order to blow it up, you need to impart at least enough energy to overcome this attraction.  Fortunately, physics has a well-known solution for this problem.  Most introductory courses on electricity and magnetism teach you to calculate the energy required to assemble a uniformly charged sphere with the result1

E = (3 / 5) k Q2 / r,

where E is the energy, k is a constant, Q is the total charge on the sphere, and r is the radius of the sphere.  This can be derived from Coulomb’s law for the force acting on charged particles.  Since Newton’s law has the same inverse square dependence on distance as Coulomb’s law, the formula can be modified to calculate the gravitational energy holding together a sphere of uniform mass2,

E = - (3 / 5) G M2 / r,

where G = 6.67×10-11 N·m2/kg2 is a fundamental constant and M is the mass of the sphere.  According to Wookieepedia, the radius of Alderaan is 6,250 km and its gravity is “standard,” making it slightly smaller than Earth with a mass of about 5.7×1024 kg.  Using these numbers, we can estimate the amount of energy needed to blow up Alderaan,

E = (3 / 5) G M2 / r
 = (3 / 5) (6.67×10-11 N·m2/kg2) · (5.7×1024 kg)2 / (6,250 km)
= 2.1×1032 J

As Marvin the Martian once said, “Where’s the kaboom?”  If you harvested all the solar power that falls on the Earth from the Sun, it would take about 40 million years to collect enough energy to blow up Alderaan.

[1] See, for example, The Feynman Lectures on Physics, Vol. II, Chapter 8.
[2] Strictly speaking, if Alderaan is like Earth, it will not have a uniform mass distribution, but this assumption makes the math a lot easier.

Tuesday, April 6, 2010

Busy Week

I’ll be traveling a bit for the next two weeks, so my posts may be a little sporadic.  When I get back, I’ll try to address all the questions people have left in the comments and I’ll post a few of the Fermi questions readers have sent me.  In the mean time, I’m going to try to answer one of Thomas’s questions from a while back:

If we took a cubic meter of the sun and dropped it into the ocean, how much water would boil away?

As a disclaimer, I know very little about astrophysics, plasma physics, and nuclear physics.  That said, I think I can still come up with a decent estimate even with my limited knowledge.  I’m assuming Thomas means a cubic meter of the Sun’s core, which is both denser and hotter than its surface.  According to Wikipedia, the Sun’s core has a temperature of 1.57×107 K and a density of 1.5×105 kg/m3.  The core is too hot for atoms to be stable, so it’s basically a bunch of free protons, neutrons, and electrons swishing about.  Physicists call this phase of matter “plasma.”  When we dump a cubic meter of this plasma into the ocean, it will cool as it gives off heat to the surrounding water.  As it cools, the protons will bind with electrons to form atoms—I’ll assume they all form hydrogen atoms—and these hydrogen atoms will then form hydrogen molecules.  Each of these phase changes releases some heat1 into the water.  To compute how much energy is given to the water, we need to know both the energy gained by cooling and the energy gained by phase changes.

From the density, we know that the total mass of one cubic meter of the Sun’s core is about 1.5×105 kg.  Since the atomic weight of a hydrogen atom is 1.008 g/mol, we can compute the total number of hydrogen atoms we’ll get out of the chunk o’ Sun,

number of atoms = (mass) / (atomic weight)
= (1.5×105 kg) / (1.008 g/mol)
= 1.5×105 mol.

This is equivalent to 9.0×1031 hydrogen atoms.  The ground state of a hydrogen atom is –1300 kJ/mol (~-13.6 eV).  From this we can compute the amount of energy released in the formation of hydrogen atoms,

energy released = (energy per atom) · (# of atoms)
= (1300 kJ/mol) · (1.5×105 mol)
= 2.0×108 kJ.

It takes two hydrogen atoms to form a hydrogen molecule.  This means the 9.0×1031 hydrogen atoms in our glob will form 4.5×1031 molecules or 7.5×104 moles of molecules.  Each H2 molecule has a binding energy of 436 kJ/mol, so the total heat released by the formation of molecules is

energy released = (energy per molecule) · (# of molecules)
 = (436 kJ/mol) · (7.5×104 mol hydrogen molecules)
= 3.3×107 kJ.

There’s also energy released in the cooling process.  I’ll assume both the plasma and molecular hydrogen are well approximated by an ideal gas, which has a specific heat of 20.79 J/K·mol at constant pressure2.  From this, the temperature change, and the number of moles of hydrogen atoms, we can estimate the heat released by cooling,

energy released = (specific heat) · (# of atoms) · (temperature change)
= (20.79 J/K·mol) · (1.5×105 mol) · (1.57×107 K)
= 4.9×1010 kJ.

As you can see, the heat released by cooling is much larger than either of the phase changes indicating that this is the only significant factor.  This means about 4.9×1010 kJ of energy is used to boil ocean water. 

To determine how much water boils away, we need to know water’s specific heat (~4.186 kJ/kg·K) and its latent heat of vaporization (~2260 kJ/kg).  If the ocean starts at room temperature (~298 K = 25 °C), it will have to increase by about 75 K in order to boil.  Using this data, we can estimate the total mass of water that would be boiled away3.

mass = (energy added) / [(latent heat) + (specific heat) · (temperature  change)]
= (4.9×1010 kJ) / [(2260 kJ/kg) + (4.186 kJ/kg·K) · (75 K)]
= 1.9×107 kg.

It would boil away about 20 million kilograms of water.  That’s about a cube of water 27 m on a side.

[1] Physicists call this “latent heat.”
[2] The specific heat of an ideal gas at constant pressure is equivalent to 5R/2, where R is Boltsmann’s constant.
[3] For simplicity, I’m assuming all the heat gets transferred directly to the water that gets boiled off (i.e. the rest of the ocean stays at room temperature.)