It's That Time of Year Again

It's that time of year again.  The holidays are here, and you can't flip through the channels without eventually hitting a Christmas special.  We watched Miracle on 34th Street yesterday, and I was struck by the scene at the end where all the mail addressed to Santa gets shipped to the courtroom where Kris Kringle is on trial. One of the clerks in the mail room says "there must be about 50,000" letters to Santa Claus that they have to get rid of.  How many letters get delivered to the court room?

In the movie, 21 duffel bags full of letters are brought in and dumped on the judge's desk.  Judging by how the men carry the bags, they certainly weigh between 2 and 200 lbs, so we can reasonable assume there are about 20 lbs of letters to Santa in each bag.  According to the United States Postal website, a standard letter weighs less than 3.5 ounces.  Assuming a 1 ounce letter, we can easily compute the number of letters delivered,

# of letter = (21 bags) × (20 lbs/bag) × (16 ounces/lb) / (1 ounce/letter)

= 7000 letters.

At about 7000 letters, the mail clerk was correct to within about an order of magnitude.  Not too shabby.

A Couple of Updates...

Fermi Lives at Harwoon Union

It's been a busy semester.  A couple of months ago, I received a very nice email from Lisa Therrien and several other teachers at Harwood Union Middle School in Duxbury, VT.  Apparently Fermi fever is catchy, because Ms. Therrien is teaching her students how to make order of magnitude estimates.  I got invited to listen to several of her students describe their estimations over Skype.  It was a very enjoyable experience, and I can truly say that Harwood Union Middle School has some very creative young mathematical minds.  I only wish more teachers incorporated this style of thinking into their math and science classes.


Estimation as a Skeptic's Tool

In addition to Harwood Union School, I also got invited to speak for Ann Arbor Science & Skeptics in October.  It was a great group, and we discussed how estimation can be used as a skeptic's tool. Here's a few of the cool estimates that were suggested by the group:

• How much hair throughout the entire world is grown in one day? (Enough to cover about 10 square miles.)
• What was Forrest Gump’s average speed as he ran across the country? (About 5 cm/s.) 
• If the Sun disappeared, how quickly would the temperature drop on Earth? (About 0.1°C per century if you’re talking about the average temperature of the whole Earth, or about 20°C per day if you’re talking about just the surface temperature.)

The only downer of the trip for me was that I had to admit my secret shame.  I, physicist Aaron Santos, have never actually read a Carl Sagan book.  It's a problem that I will hopefully be remedied in the near future, but for now, feel free to throw your rotten tomatoes.

Shark Blood

"Nom nom nom nom nom!"

I've been seeing this commercial a lot lately, so I got curious.  How sensitive would a shark's nose have to be in order to detect a drop of blood from one mile away?

Let's say you took a drop of blood and dumped it in the ocean.  Because of diffusion, that drop would eventually spread out in all directions, much like a drop of red food coloring in a glass eventually spreads out and makes the entire glass red.  If the blood spreads out equally in all directions, then it will eventually cover a hemisphere 1 mile in radius.  This means the blood will be distributed over a volume of

V = 2 π R³

=2 · (3.14159) · (1 mile)³

= 2.6×10¹⁰ m³.

A drop of blood is about 1 mL.  This means the density of molecules in a 1 mile radius drops by a factor of about

(1 mL) / (2.6×10¹⁰ m³) = 3.8×10^-17.

That's very tiny.  I'm not exactly sure what part of the blood the shark is supposed to smell, since there are many components that make up blood.  Let's assume he's smelling the red blood cells.   A healthy red blood cell count is in the range 4.6-6.1 million per μL = 5×10^-9 per mL.  If the blood spreads out uniformly over this distance, there will be

(3.8×10^-17) · (5×10^-9 per mL) = 1 red blood cell every 5 m³.

That's a very tiny amount.  I highly doubt a shark, or any other creature, could detect something with this small a concentration.  Besides, the MythBusters already busted this myth during shark week. 


Bonus Shark Blood Question #1: Even if a shark could detect concentrations this small, how long would it take a red blood cell to travel 1 mile?

Bonus Shark Blood Question #2: Where does a shark's blood come from if blood comes from bone marrow and sharks are cartilaginous fish?  (I know...this is not a numbers question, but I've always been curious.)

Mission Improbable

I watched Mission: Impossible yesterday and was reminded of two things: (1) there was a time when people thought Tom Cruise wasn't crazy and (2) there was a time when people thought the internet was.  Long before anyone dreamed of Facebook, Twitter, or rage comics, the internet was a pretty novel concept, and many people, myself included, weren't ready to dive in head first.  I haven't thought about this for awhile, but I'm pretty sure Mission: Impossible was the reason I decided to get a computer.  After all, if super spies used the internet, then it was probably pretty darn cool.  Who knows?  If I hadn't seen it, maybe I wouldn't be blogging this now.¹  Still, there's one thing that bugged me in my second viewing.  In the iconic scene where Cruise is dangling from a rope while trying to steal sensitive government data, an alarm will go off and ruin the mission if he touches the floor, makes a sound, or raises the temperature of the room by 1° F.  There were a few close calls on these fronts, but I was particularly skeptical of the temperature restriction.  How long would it take your body heat to raise the temperature of the room by 1° F?

Human bodies release heat at a rate of about P = 100 W.²  This heat will spread throughout the room, which has a volume of roughly 15 ft by 15 ft by 30 ft = 200 m³.   With a density of 1 kg/m³, the total mass of air in the room will be about m = 200 kg.  The specific heat of water is about c = 1 J/g·K.  From this we can fin the time it will take to raise the room by 1° F (~0.6 K),  

 t = m c ΔT / P

= (200 kg) · (1 J/g K) · (1° F) / (100 W)

= 20 minutes.

Apparently, they would have time as long as they were fairly quick about it.³

[1] Admittedly, if I had never seen the movie, I couldn't be writing this post right now, but you get the idea.
[2] See the post titled "Igloos Rock!"
[3] I've neglected the fact that the room is temperature controlled.  As the temperature rises, the temperature control will turn on the AC to return it to normal.  This may have given them even more time.

And That's the Tooth

Yes, I watch absolutely terrible movies.

While watching Family Guy the other day, I saw this.  The Tooth Fairy has been played by everyone from Eddie Murphy to The Rock, but this one got me thinking: How many teeth does the Tooth Fairy collect each year?¹

Humans are born with 20 baby teeth.²  Since you live about 80 years, your rate of tooth loss is about

(20 teeth per person) / (80 years)
= 0.25 teeth / year · person.

Not all of the world believes in the Tooth Fairy, so it's a safe bet that she doesn't visit every newly gap-toothed child.  Assuming only 10% of the world (roughly 700 million people) have been visited by the Tooth Fairy, then the total number of teeth that have been collected this year would be

(10%) · (1 year) · (700 million people) · (0.25 teeth / year · person) = 1.8×10⁷

That's 18 million teeth, or roughly enough teeth to fill a refrigerator. Apparently, our cartoon Tooth Fairy is fairly accurate.


[1] In case you're wondering, most of my inspiration for estimations comes from watching cartoons.
[2] I'm not including wisdom teeth in this calculation since (a) the Tooth Fairy generally doesn't collect these, and (b) when removed, they're considered biological waste, so the dentist doesn't give them to you.  As if having surgery isn't bad enough, they don't even let you keep them a souvenir, which, incidentally, belonged to you in the first place.  I only discovered this after having shoulder surgery where the doctors removed some small bone fragments.  I was particularly upset that I couldn't keep the bone chunks, because I figured the Shoulder Fairy probably paid a lot more than the Tooth Fairy.  Alas, it wasn't to be.  My mom tried to convince me that it wouldn't have mattered anyway because there's no such thing as a Shoulder Fairy, but I think she was just trying to cheer me up.

Monorail Kitteh Physics

With the economy in the crapper, it's been suggested that the United States should invest in a large-scale public works project to reduce unemployment and kick start the economy.  Some have argued for the construction of a high speed rail similar to the ones found in Europe and Asia.  These trains utilize magnetic levitation or "maglev" technology for smoother and potentially faster travel.  China's high-speed line travels at a record 217 mph.  In contrast to a typical four hour Boston-New York bus ride, at this speed, you could cover the same distance in a little over an hour.  Sadly, the only American advances in rail technology over the last 10 years have been of the "kitteh" variety.  How large of a field would one need to levitate a monorail kitteh?¹ 

While current technology uses magnetic fields to levitate trains, the same effect could be accomplished using electric fields.  As anyone who's ever petted a cat and then shocked it by touching its nose can attest, cats easily become electrically charged.  Since like charges repel, one can levitate a cat by charging it and then placing a sufficient amount of like charges on the surface beneath the cat.  These charges will create an electric field E that will push up on the cat with a force

F = q · E,

where q is the net charge on the cat.  In order to levitate, this force must cancel the downward gravitational force on the cat,

F = mg.

Here, m ≈ 10 lbs is the mass of the cat and g = 9.8 m/s² is the acceleration of gravity.  Setting the two equal, one can solve for the electric field,

E = m · g / q.

You may notice that we've yet to specify to specify the charge q on our feline friend.  Estimating this value is not trivial, but one may arrive at a suitable estimate by using the following logic.  Let's say you rubbed a balloon on a cat. Some electrons will be transferred from the cat's fur to the balloon.  If you repeat this with a second balloon, you'll find that the two balloons repel each other.  In fact, they repel so much after you do this, that you can levitate one balloon about 10 cm above the other one.  The charged balloons will repel each other with a force similar to that between two charged point particles,

F = k · q² / r²,

where k = 9×10⁹ N · m²/C² is a constant and r ≈ 10 cm.  A balloon with a mass of 10 g feels a gravitational force of about 0.1 N.  Setting this gravitational force equal to the electrical force, we can solve for the charge on the balloon,

q =(F · r² / k)^1/2,

= [(0.1 N) · (10 cm)² / (9×10⁹ N m²/C²)]^1/2

= -3×10^-7 C.

That's about 2 trillion electrons worth of charges that have been transferred to the balloon.  Since that charge had to come from the cat, it has been left with a positive charge 3×10^-7 C.  Plugging this into the equation for the electric field, we get

E = m · g / q

= (10 lbs) · (9.8 m/s²) / (3×10^-7 C)

= 1.5×10⁸ V/m. 

That's 150 million volts per meter.  Air starts to spark at about 3 million volts per meter.  Unless you want lightning bolts shooting out of your cat Star-Wars-Emperor-style, it's probably not a good idea to try and levitate him with an electric field.


[1] My more sophisticated readers will no doubt recognize the similarities between monorail kitteh and the well-known jelly-toast rocket.  For those that are unfamiliar, the logic goes as follows.  It is a physical fact that jellied toast always lands jelly-side down.  Likewise, it is a physical fact that cats always land on four feet.  By tying or in some other humane way affixing the jellied toast to the back of the cat and letting it fall from a modest height, the pair must never touch the ground.  Presumeably, the cat-toast is left oscillating between jelly-side down and cat-side down states.  If one wishes to utilize this scheme as a means of transportation, simply wedge a bottle of diet coke between the cat and the toast and insert some Mentos into the bottle.  This should serve as a suitable propulsion mechanism.

Lighting the Way

Headlights: evil harbingers of environmental doom.

Lately, I've seen a lot of cars driving with their headlights on in the day time.   This seems like a waste of energy.  Aren't we supposed to turns the lights off when we're not using them?  Not only does it cost energy to produce the light, but the lights will actually exert a tiny force backwards on your car.  In principle, this will slow your car's acceleration by a tiny amount and cause you to use more gas.  How much gas do you waste by turning the on the headlights?

Most car lights look yellow.  Yellow light has a wavelength and frequency of λ = 570 nm and f = 5×10¹⁴ Hz, respectively.  Each photon of light that leaves the car carries with it some energy given by¹

E = h f .

If you have N photons emitted by the car, the car will have lost an amount of energy

ΔE = N h f .

Car lights consume energy at a rate of about P = 50 W.  This P is known as power, and it's the rate at which energy is transferred,

P = ΔE / Δt.

Assuming all the energy goes into creating yellow photons, we can find how many of particles of light leave the car each second by solving for N:

N = P Δt / h f, 

= (50 W) · (1 sec) / (6.63×10^-34 J · s) · (5×10¹⁴ Hz),

= 1.5×10²⁰ photons.

Each photon that leaves carries with it some momentum,

p = h / λ.  

As it leaves, the photon imparts some force on the car,

F = Δp / Δt.

The total force on the car is then

F = N h / λ Δt ,

= (1.5×10²⁰ photons) · (6.63×10^-34 J · s per photon) / (540 nm) · (1 sec),

= 1.8×10^-7 N.

That's a tiny force.  Just how tiny?  If you drive 3000 miles across the United States, it'll cost only 1 Joule of energy.²  One gallon of gasoline contains 1.3×10⁸ J of energy.  You would need to drive around the U.S. with your lights on 130 million times before you'd wasted a gallon of gas.

[1] Photons are particles of light.
[2] You can obtain this by multiplying the force times the distance.

Does Experience Matter?

 

The NFL is coming back, and many free agent veterans are on the move.  Sports talking heads often drone on about how much "experience matters" when it comes to sports.  There may be some logic to this—if you're having heart surgery, you certainly want to know your doctor's used a scalpel before—but it shouldn't make you feel good about your team overpaying some over the hill interception-prone QB just because he happened to win for a Super Bowl 14 years ago [See: Favre, Brett].  But if experience does matter, one should be able to see this statistically.  Is there a strong correlation between previous Super Bowl wins and future Super Bowl success?

According to ESPN The Magazine, the most experienced team has won the Superbowl 58% (25/43) of the time.  That's certainly more than half, but this could be due to random chance. After all, a 58% winning percentage over a 16 game season amounts to about 9 wins, which an average team could certainly achieve just by being lucky.  Assuming it is due to random chance, what's the probability that the more experienced team won 58% of the time?  To solve this, we need the binomial distribution  f (k; n, p) ,

where n = 43  is the number of trials, k = 25 is the number of success, and p = 0.5 is the probability of winning if the results were random.  The binomial coefficient (i.e. the weird "n over k" thing in the parentheses) is defined as

The function f (k; n, p) is the probability that a random event will produce exactly k success in n trials.  In this case, we want to know the probability that experienced teams have won at least 25 times in 43 trials.  Using Wolfram Alpha to sum up these probabilities, we find that there's an 18% chance of the more experienced team winning at least this many games.

In the last Superbowl, 29 former Super Bowl champs played for the Steelers and no former champs played for the Packers.  Clearly, it's not all about experience.

Angular Momentum Courtesy of Sandy

Randall Munroe, you truly are a geek god.

I got a great question from a reader named Sandy. After seeing the xkcd comic titled "Angular Momentum", Sandy wondered:

"...would [it] be possible to work out an estimate of how many people spinning anti-clockwise it would take to counteract the Earth's spin, assuming that the premise is sound."

Great question! First off, the premise is absolutely sound. Anyone who's watched a figure skater speed up when she pulls her legs in has observed angular momentum being conserved. Angular momentum conservation is fancy physics speak for "the universe likes to keep the total amount of spin-y things constant." In this example, as soon as people start spinning, the Earth's rotation will slow down, because the total amount of spin-y things has to be the same.

Admittedly, "spin-y things" is not the most descriptive term, so it helps to have a more formal definition.  It's more difficult to stop an object from spinning when the object is (1) massive and (2) rotating fast.  For this reason, you'd expect the angular momentum of a spinning object to depend on its mass and velocity.  You might also expect it to depend on (3) how the mass is distributed throughout the body, since it's easier to spin a broom stick along its axis (i.e. drill-bit style) than it is to spin it perpendicular to its axis (i.e. helicopter style).  These elements are combined to give an equation for the angular momentum,¹ 

angular momentum = (mass) · (velocity) · (distance from mass to rotation axis).

A 150 lb. person spinning around once per second would have an angular momentum of about 100 kg·m²/s.  The angular momentum of the Earth is roughly 7×10³³ kg·m²/s, so it would take about 7×10³¹ people to counteract the Earth's spin.  That's 10²² times the actual population, or enough people to cover the surface of the Earth 10 quadrillion times.  

I should point out that even if you stopped the Earth's rotation in this way, it would begin spinning again as soon as everyone stopped rotating, because, as stated earlier, the universe likes to keep the total amount of spin-y things constant.²

[1] I should note that this equation works for point masses rotating around an axis.  If you want the angular momentum for a something that's bigger than a single point, you need to add up a bunch of point masses.  For a spinning person, you could add up the angular momentum of each atom to get the total angular momentum.  In this problem, I've assumed the average atom in a spinning person is a 1 ft away from the axis of rotation and moving at 5 m/s.

[2] If my childhood experience with spinning carnival rides is any indication, I would not want to be around for the ensuing vomit parade that would inevitably follow this worldwide venture.

And...we're back....

I'm back!  Well, I'm for at least a little while until I inevitably get bogged down again.  I'll be posting a few nice estimates over the coming weeks, and we'll hopefully be having another contest or two. 

Thanks for reading and have fun estimating,

-Aaron

Hiatus

The semester is picking up and I'm working on a follow up book to How Many Licks?, so I'm taking a semi-hiatus from the blog world.  I'm going to try to post once every couple of weeks, but no promises.

Estimate with pride.

-ATS

Pi Day Pie

Check out the pie my wife made for Pi Day.

Pi i = Log(-1).  Get it?  If this appeals to your geeky and foodie side, please vote for it here!  Tell your friends!  Voting closes on March 23 (Wednesday at midnight), so vote ASAP.


UPDATE: We finished second!  Thanks to everyone who voted for us!

Death and Presidents

A few years back, an older friend told me a story about her grandmother.  It was November 1963, and her grandmother had just turned 105.  My friend went to visit her grandmother in the hospital.

Friend: "Grammy, the president's been shot!"
Grandmother: "Another one?!"

It was then that I realized her grandmother had lived (and was old enough to remember) every presidential shooting from Lincoln to Kennedy.  It blew my mind.

It's amazing how much can happen in a lifetime.  This point was driven home in a recent post on Reddit.  As of March 2011, our 10th president John Tyler (born 1790) had two living grand children!  This got me thinking: If you passed a baton from person to person, what's the least number of hands it could touch and reach the first even human?

According to Wikipedia, modern humans showed up about 200,000 years ago.  If you passed the baton from a centenarian to a new born that would live to be a centenarian, it would take only 2000 people.¹   This is roughly the number of people attending my high school.  If you wanted to go back to the time of Christ, it would only take 20 people. 

[1] Some people will no doubt quibble that the average life span was in the 30s for much of our history.  True, but this neglects two points.  First, we're doing an order of magnitude estimate and this fact won't change the answer by an order of magnitude.  More importantly, "life expectancy" is the average life span.  Life expectancy may have been low, but this was largely due to high infant mortality rates.  Back in the day, if you were lucky enough to make it to adulthood, you stood a good chance of making it to old age.  Since we're only looking for the oldest person at any one time, there's a good chance there was centenarian somewhere in the world.

Chalk Hell

Wright Laboratory of Physics Room 201 at Oberlin College.

"Begone, Ye Accursed Chalk!  I banish thee and thy vile and insufficient yield strength to the depths of Chalk Hell!!!"

Most colleges have at least one big lecture hall with chalk boards that slide up and down.  To make space for the boards to slide underneath, there's usually a small gap behind the wall.  This gap is Chalk Hell: a black hole out of which no chalk ever emerges.¹   How long would it take for Chalk Hell to over flow with chalk remnants?

A brief glimpse into the horror that is Chalk Hell.

Chalk Hell is about 1.0 m deep,  10 cm wide, and 3 m long giving a total volume of 0.3 m³.  Much of the chalk in Chalk Hell is about 0.8 cm thick and 6 cm long giving a total volume of about 3.8×10^-6 m³.  I drop at least 1 piece per lecture and over the course of a year there might be 300 lectures in a room.  From this we can estimate that there are

0.3 m³ / [ (3.8×10^-6 m³ per piece) · (1 piece per lecture) · (300 lecture per year) ]

= 263 years.

It would take over 2.5 centuries to before Chalk Hell flows over.

[1] Unless you duct tape the long erasers together to make a giant set of chop sticks that you can use to pick them out of Chalk Hell.

Death Statistics

There was an interesting post on Reddit a couple of days ago:

"After your death, what kind of statistics about your life would you like to know? ... Let's just say you get a sort of 'post-game' screen, like in a videogame. What kind of statistics (numbers, percentages, etc.) about your life would you like to know?"

I love this idea.  Here are my estimates for some of the statistics Redditors wanted to hear.  I’m assuming an 80-year life span. 

“Gallons of liquor consumed”: If you average 3 beer cans worth of liquor per week, you’ll consume 4.4 m³ in a lifetime. 

“Number of lies I told/was told”: Lies come in a variety of forms, but let’s not get too pedantic.  No matter how you define it, telling ten lies per day seems like a lot and telling one lie every ten days seems like too few.  I’ll assume people tell one lie per day on average.  This gives about 30,000 lies in a lifetime.  By symmetry, you might expect the number of lies told to equal the number of lies heard, but this is not necessarily the case because lies can be told in parallel.  For example, when a politician lies, he’s usually lying to huge crowds of people.  Assuming the average number of people who hear a lie is 10, you would hear about 300,000 lies in your lifetime.

“Number of biblical sins committed”:  Minor sins are pretty common.  If you average two sins per day, you’ll total about 60,000 sins.

“Percentage of time spent sleeping”: If you average 8 hours per night, you will have slept 33% of your life.

“Time spent on the Internet”: If you average 2 hours per day, you’ll spend about 7 years on the Internet.

“How many hours I've spent fapping”:  This varies a lot from person to person depending on gender and how fast your Internet connection is.  If you average 5 minutes per day, you’ll spend almost 100 days of your life fapping.

“Amount of semen expelled”:  Assuming each—er—deposit contains about 1 tablespoon, and you make a deposit once per day, you’ll have about 0.4 m³, or roughly about 1 bathtub full. 

“How much money I made in my lifetime (including all the money I earned when I was little)”:  The money you made when you were little is probably not a significant figure, but if you average a $50K salary from age 25 to 65 and manage to spend all of it, you’ll have spent $2 million.

“Time spent on videogames”.  If you average 2 hours per week, you’ll spend about 1 year of your life playing video games.  (Note: Hardcore gamers will average much more that 2 hours per week.)

“Time spent playing sports”: If you play a couple hours each week, you’ll have played a year worth of sports.

“The percentage of how many people I came into contact with out of the people who lived on Earth at the same time as me”: There are about 7 billion people in the world right now.  From this, we might estimate that in your lifetime 14 billion people have been alive at one point or another.  Even if you average 100 new contacts per day, you’ll still only meet 0.02% of the people whose lives overlapped yours.

“Number of heartbeats and breaths taken”: Heartbeats and breaths happen about once per second and once every three seconds, respectively.  That means you’ll have about 2.5 billion heartbeats and 840 million breaths.

“Number of pens lost”: I lose one at least one pen per month.  That gives about 1000 pens lost.

“How many words I typed”: If you type at a computer for 0.5 hours per day and type 30 words per minute, you’ll have typed 26 million words.  (The 0.5 hours per day number might seem small, but I’m only counting the time spent actually typing.)

“How many letters I typed”: If each word averages about 5 letters, you’ll have typed 130 million letters.

“How many words I said”: If you talk for an hour per day at a rate of 3 words per second, you’ll utter 300 million words in a lifetime.

“How many hours I slept”: Most people sleep about 8 hours per night, which means 27 years of sleep.

“How many times I had sex”: If you average once per month, you’ll have sex about 960 times.

“How many times I ate burritos”: If you eat one burrito per week on average, you’ll eat 4200 burritos in a lifetime.

“How many times I said the word [expletive]”:  Some people never curse while others curse every other word.  If you lie somewhere in the middle, you likely curse about 5 times per day which would mean about 150,000 curses in your lifetime.

“Amount of time spent on Reddit”:  If you average 10 hours per week, you’ll spend almost 5 years on Reddit. 

“Number of witty remarks made”: It’s hard to define a metric for witty, but I like to think I make at least one quip per week giving a total of 4200 witty remarks.

“Amount of time spent in the air”: I’m assuming we’re talking about riding an airplane rather than time spent in the air jumping.  If you fly 4 hours out of the year, you’ll be in the air a total of 13 days.

“Fastest overall speed”: This question is not well defined, since speed has to be measured relative to some point.  If we’re free to measure in any reference frame, then your fastest speed would be close to the speed of light since we could always choose a reference frame that was moving close to the speed of light relative to you.

“Number of farts”:  This depends on your body chemistry.  Most people probably average at least 1 and less than 50 farts per day, so I’ll assume 5 per day on average.  This means you’ll have about 150,000 farts.

“Total volume of gas farted”: If each fart contains 50 mL of gas, the total volume of farts in one lifetime would be 7.5 m³.  That’s about the size of an office cubical.

“How many total inches of [penis] I have taken. Each sexual encounter counts (not only each partner)”: If you assume one sexual encounter per month and a 6-inch average, you’ll have “taken” 150 m of penis or about 1½ football fields.

“Total miles walked”: I calculated this in How Many Licks?  It’s about 23,000 miles.

“Number of laws I've broken”:  I jaywalk at least 5 times per week.  This means I will have broken the law at least 21,000 times.

“Number of times I said ‘I love you’”: Between family and my wife, I probably say it about 10 times per day which gives about 300,000 proclamations of love.

Death Star Physics Revisited (part II)

Saturn's moon Mimas looks eerily like the Death Star.

Given the very in depth analysis some of my readers sent me after the last Death Star post, I hesitate to follow up with this one since there's clearly some Star Wars physics that should to be debated before we continue.  However, I'm posting this any way because (1) I already typed it up and I'm pretty lazy about these kind of things and (2) if you allow my initial assumptions, there's still at least one more question that begs to be asked: how is all that laser energy stored?  Is there some giant car battery hidden somewhere?  After all, it would take 40 million years to collect enough solar power to blow up a planet.  Wouldn't one shot sap all the energy out of the Death Star?  According to relativity, one could very efficiently store energy as mass.  If that were the case, what percentage of the Death Star's mass would be lost with each shot?   

Luke mistakes the Death Star for a small moon, which suggests the mass of the Death Star is about that of a moon.  I'll assume the Death Star's mass is equivalent to that of Saturn's moon Mimas (M=3.7×10¹⁹ kg) since it kinda looks like the Death Star.  As we calculated before, the total energy required to blow up a planet is E = 2.1×10³² J.  According to relativity, the amount of energy stored in a piece of matter with mass m is given by the equation

E = mc²,

where c = 3×10⁸ m/s is the speed of light.  If we solve for the percentage of mass lost with each laser shot, we find

fraction of mass = m / M = E / (M c²)

= ( 2.1×10³² J ) / [(3.7×10¹⁹ kg) (3×10⁸ m/s)²]

= 0.0063%.

It's a very small percentage.  At this rate, the Death Star could blow up over 15,000 planets before it ran out of mass.  This of course assumes that the energy is directly transferred to the planet without loss.

Death Star Physics Revisited

 

I've been thinking more about the "Death Star Physics" problem from last year in which I estimated it would take about 2.1×10³² J of energy to blow up a planet.  It strikes me that I should have taken this problem even further.  How many photons were in the laser beam that blew up Alderaan?

As you can see from the video, the laser is green.¹ Green light has a frequency f = 5.6×10¹⁴ Hz.  Using Plank's constant h = 6.6×10^-34 J·s, one can compute the energy of a single photons by using the well known formula

E = h·f

From this we can compute the total number of photons,

number of photons = (total energy) / (energy per photon) =  ( E_tot ) / ( h · f ) 

 = ( 2.1×10³² J ) /[ ( 6.6×10^-34 J·s ) · ( 5.6×10¹⁴ Hz) ] 

= 6×10⁵⁰ photons

This is a huge number. It's a 6 with 50 zeros after it. I'd have to go back and check, but I think this is the biggest number I've posted on this blog.

[1] At least the visible part of the laser is green.  It's possible the are also UV or other high frequencies of light coming out of the laser.

The Bill O'Reilly Problem

 “Tide goes in, tide goes out. Never a miscommunication. You can’t explain that. You can’t explain why the tide goes in.”

—Bill O'Reilly

 
Political pundit Bill O'Reilly elevated himself to meme status this past month because of his assertion that you can't explain the rise and fall of the tides.  Most people probably don't know the exact mechanism, but even in our Are You Smarter Than a 5th Grader? society, people should at least recall that there was some explanation learned in middle school science.  After all, it was over 300 years ago that Newton figured out the tides were caused by the Moon's gravitational field.  How does the Moon produce the tides and can we use physics to estimate just how big they should be?

We live in a gravitational field.  At the surface of the Earth, the field causes all masses to accelerate to the ground at a rate of 9.8 m/s².  But the Earth is not the only object that has a gravitational field: all masses are attracted to each other by gravity.  Right now, the mass of your body is exerting a very slight pull on the pyramids of Egypt due to gravity.  While your gravitational field is very weak, massive things like the Moon can create very large gravitational fields.  We can show the gravitational fields due to the Moon and the Earth schematically like this:

Gravitational field lines around the Earth (blue) and Moon (orange).

At each point in space, an arrow tells you what direction an object placed there would be pulled.  As you can see, the more massive (i.e attractive) Earth has more arrows pointing towards it.  Only arrows very close to the Moon point towards it.  The reason is that gravitational fields get weaker as you get further away, so if you're very close to the Moon you'll be pulled more towards it than the Earth.  This decrease in strength as you get further away is the reason there are two tides per day.  The Moon pulls very hard on the part of the Earth that is facing it.  The water there will shift a little closer to the Moon creating high tide.  Why is there a second tide 12 hours later?  In addition to pulling on the water, the Moon also pulls on the center of the Earth.  It pulls a little harder on the center of the Earth than it does on the water at the opposite end of the Earth because the center is closer.  This effect elongates the surface of the water giving it an oval shape as seen below:

The gravitational pull of the Moon effectively stretches the Earth.  (Not to scale.)

This is a nice qualitative description, but is there a way we can compute the height of the tides?  Yes!  It turns out that fields are not the only useful way to describe gravity.  There's also something called "gravitational potential".  Masses tend to move from areas of higher potential to areas of lower potential.  The potential V at some point due to a mass M is given by

V(r) = -G M / r,

where r is the distance between the point and the center of the mass and G = 6.67×10^-11 N·m²/kg² is the fundamental gravitational constant of the universe. If water in the oceans can flow to a lower potential, it will.  For this reason, we know that the gravitational potential should be the same everywhere on the surface of the ocean, because if it wasn't water would flow until it was.  We can write the potential due to the Earth and the Moon as

V = -G ( M_Earth/ r_Earth + M_Moon/ r_Moon),

where M_Earth/Moon and r_Earth/Moon are the mass and distance from the Earth/Moon, respectively.  We can plot lines of equal potential much like you can plot lines of equal height on a topographic map: 

Contour plot showing lines of equal gravitational potential.  The white circle on the left represents the Earth while the dark speck in the center represents the Moon. 

Notice the the potential lines are slightly warped by the Moon giving the characteristic oval shape one observes in the tides.  We can look up the mass of the Earth (M_Earth=6.0×10²⁴ kg) and the mass of the Moon (M_Moon=7.4×10²² kg).  When the Moon is directly overhead, r_Moon is roughly 3.7×10⁸ m and r_Earth=6.4×10⁶ m.  From this, we can calculate the gravitational potential at the surface of ocean to be  V=-6.25445×10⁷J/kg.  Halfway around the Earth, the height of the water will be lower but the potential must be the same.  If you calculate this height using the values listed and a little geometry, you find the difference between high tide and low tide is about 22 m.  In the Bay of Fundy, which has the largest tides in the world, the water rises and drops about 16 m, so our estimate is very good considering we didn't consider land masses or other complicating factors.

The Fruit Hunter

My in-laws got me a copy of Adam Leith Gollner's The Fruit Hunters for Christmas.  I'm just a few chapters in, but there's already a few good knowledge nuggets,

"The sweetness issue actually went to the United States supreme court in 1893. They ruled that tomatoes are vegetables because they aren't sweet," 

and an estimation,

"There are an estimated 240,000 to 500,000 different plant species that bear fruit.  Perhaps 70,000 to 80,000 of these species are edible; most of our food comes from only 20 crops."

 How long would it take to try every fruit?

Most people eat three meals per day or 1095 per year.  If you ate one new species of fruit at every meal, it would take about 70 years to try every edible fruit.  Even if I start now and live to be 100, there is literally no way I will every try every fruit.

Beginning a New Semester...

It's been a busy beginning to the new semester.  Fortunately, I've had a couple very nice people send me interesting estimations links.  Enjoy!

Courtesy of Joey:

"If we were to take all that information and store it in books, we could cover the entire area of the US or China in 13 layers of books..."

 
Courtesy of Sean "The Cool Physics Guy":

The amount of snow in Boston is equivalent to about 0.49 Shaqs or 0.60 Nate Robinsons.

Louis CK Is Dead Wrong

"Out of all the people that ever were, almost all of them are dead."

—Louis CK

Louis CK (very funny guy) does a nice stand-up routine, but I couldn't help noticing that one line in his opening.  As I calculated in Skeptically Speaking...(part II), there's an unexpectedly large percentage of people that are still among the living:

"Surprisingly, I computed about 30% of people who have ever lived are still living. I was a little skeptical of this, so I decided to look it up and found that other people have done this problem as well and they estimated 5.8%. Since these numbers are easily within an order of magnitude of each other, they're likely close to the actual result."

No matter which result you take, it's clear that not "almost all" of us are dead.  This counter-intuitive result stems from the fact that population growth can happen exponentially.  Put another way, there simply weren't that many people back in the day, but they had a lot of sex so there are a lot more of us now.  Admittedly, it's probably a bad sign that I'm more irked by this mathematical inconsistency than I am by the whole having sex with a dead kid thing.

But Was It...MURDER???? [1]

"Meanwhile, Dr. Johnson’s cursory examination revealed the body was not quite cold; he concluded that death had occurred three to four hours earlier."

—E. J. Wagner's "A Murder in Salem" from the Nov. 2010 issue of Smithsonian

While stuck in Philadelphia International Airport after missing a connecting flight, I came across E. J. Wagner's "A Murder in Salem" in the Smithsonian magazine.  I was struck by the simple physical analysis implicit in Dr. Johnson's examination of the murdered man's body.  It seemed Sherlock Holmes-ian.  How much would a dead body cool in four hours?

In principle, a dead body loses most of its heat through the skin.²  There is a well known physics equation describing the amount of heat flowing through a material from a hot temperature to a cold temperature.  In symbols, this equation is written as

dQ / dt = (k · A / d) · ( T_hot -T_cold )

Here, dQ is the heat energy that passes through the surface in some small amount of time dt.  The variables k, A, d, T_hot, and T_cold represent the thermal conductivity, the surface area of the material, the thickness of the material, and the temperatures of the hot side and cold side, respectively.  Immediately before death, a healthy human body has a temperature of 37 °C (~98.6 °F).  The average April temperature in Salem, MA is 8.7 °C (~47.6 °F), but room temperature is 25 °C (~72 °F).³   I'll assume the temperature difference between a healthy body and air in the room is about 12 °C.  A large percentage of the energy our bodies use gets converted to heat.  We consume about 2000 Calories per day or about 100 W, so we can estimate the heat energy as

dQ / dt = 100 W.

Dividing dQ/dt by the temperature difference, we can estimate the ratio of k·A/d

k · A / d = ( dQ / dt ) / ( T_hot -T_cold )
= (100 W) / (12 °C) 
= 8.3 W/°C

The thermal conductivity equation is helpful, but we still need to know how much heat energy is stored in the dead man's body.  An object's capacity for storing heat is called the "heat capacity".  This heat capacity can be described by the equation

dQ = m · C_p · dT

where dQ is the heat energy lost when temperature decreases, m is the mass of the body that stores the heat, C_p is the heat capacity, and dT is the change in temperature.  The heat capacity of a human body is about 3470 J/kg·°C and I'll assume the deceased man weighed 60 kg.  Now you might think we could just solve these equations to obtain the time it takes for all the heat to flow out of the body, but there's one problem.  Since the body's cooling, the temperature difference between the body and the air is always changing.  To deal with rates of change, we generally use calculus.  To do this, we combine the two equations and integrate to get

T_final =  T_air + (T_{healthy body} - T_air) exp(- C t),

where,

C =k · A / [ m · C_p · d ]

= ( 8.3 W/°C)  / [ (60 kg) · (3470 J/kg·°C) ]

= 0.00004 Hz.

Plugging in, we can plot the body temperature as a function of time:

We get a final body temperature of 32°C.  This is still above room temperature but not yet cool, consistent with Dr. Johnson's observations.

[1] Given that Captain Smith had his head bashed in by a blunt custom-made club, I suspect the answer is yes.
[2] According to at least one reference, skin has a thermal conductivity of about 0.3 W/m·°C.  I tried using this number directly and estimating human skin's area and thickness, but I kept getting screwy answers.  Presumably, other tissues (fat, muscle, etc.) and clothing also effect the thermal conductivity which might explain this.  As you'll see, there's a nice estimation trick you can do using the fact that we consume about 2000 Calories per day to get around this problem.  
[3] They didn't have the best heating systems back then, so the air may have been colder.  Fortunately, the qualitative results don't depend on the exact number.

Punxsutawney Phil Stats (Courtesy of My Sister Laurie)

Here's a good Facebook discussion by my sister and her friend on the statistical accuracy of Puxatony Phil:

Sister: Just spent the last few minutes calculating Puxatony Phil's statistical accuracy. Of 114 years of forecasts, he has only been 39% accurate. Unfortunately, this means his shadow is not at chance, but instead is actually statistically *inaccurate* at p < 0.02 two-tailed. Based on his prognosis today, this makes me sad.

Sister's Friend: I'm not sure your statistical approach is valid, Laurie. I don't think long and short winters are equally likely, and Phil seems to have a bias towards seeing his shadow. We need to do a signal detection analysis.

Sister: Here are the actual stats in case anyone would like to run their own analyses: Sees Shadow- Phil was right 37 of 99 times. No Shadow- Phil was right 7 of 15 times

Sister's Friend: So early springs are more likely than long winters (69/114), and Phil tends to see his shadow more often than not (99/114). Since a shadow is supposed to predict a long winter, this accounts for the negative relationship. But that doesn't make Phil a useful anti-guide: d' = -0.4, which is pretty near chance.

Sister: Great work, [Friend]. Also nice to see that early springs are slightly though not statistically more likely than late winters.

For reference: http://www.lifeslittlemysteries.com/punxsutawney-phil-weather-prediction-accuracy-1292/

The Curse of "Aaron"

"I feel your pain, Aaron Rodgers."

There are many perks associated with being named "Aaron" (e.g. you're almost always first in line in kindergarten), but there's two major drawbacks that come with the name:

• people constantly ask you how your name's spelled even though it's a very common name because people are stupid and can't quite figure out that a word can begin with two A's.
• pocket dialing.

Being the first name in everyone's phone means being the first person pocket dialed.  But pocket dialing is not nearly as bad as its evil bastard cousin: pocket texting.  It wouldn't be so bad if it was only the occasional blank text, but having your cell phone company charge you 10¢ for every button pushed by an errant buttock starts to add up.  How much money do Aarons spend each year for butt dialed texts?

I like to consider myself a fairly average Aaron.  I get about 1 errant text per month, or about 12 errant texts per year.  Some Aarons have unlimited texting and others don't have cell phones, so I'll assume only 10% of Aarons pay for texts.  Judging by my Facebook friend list, about 1 in 200 or roughly 1.5 million Americans are named "Aaron".  This means the total money spent by Aarons for pocket texting is

(10%) · ( 1.5×10⁶ Aarons ) · (12 texts per year per Aaron) · (10¢ per text)

= $180,000.

That's $180,000—four times what the average high school teacher makes—given to cell phone companies by Aarons because of pocket dialing.

Skee Ball Game Theory

As a kid, I always wanted to win the stereo at the Dream Machine arcade in the North Dartmouth Mall.  The stereo was worth 10,000 tickets.¹  How long would it take to win that many tickets and how much would it cost?

Anyone who's spent time in an arcade knows that the best way to get tickets is through skee ball.  In addition to giving out lots of tickets, you could steal−er−acquire even more by pulling the tickets out of the the slot very quickly.  Usually one skee ball game would net you about 5 tickets.  Back in the day, it cost 25¢ to play and you might get $2 allowance each week from mom. At this rate, it would take 

(10,000 tickets) / [ (5 tickets per 25¢) · ($2 per week)]
= 5 years,

and a total cost of

(10,000 tickets) / (5 tickets per 25¢)
= $500.

Given that it was at most a $100 stereo, I was probably better off spending the money on baseball cards.

[1] I was about 10 at the time, so my memory could be flawed on the exact number.

Igloos Rock!

A good portion of the nation is digging out from under the snow right now.  It seems like the perfect time for an igloo problem.  How warm is it in an igloo? 

Human bodies undergo various chemical reactions and in the process give off heat.  By "heat", I should be clear that I mean thermal energy not temperature.  If you only have your body to keep warm (i.e. no hand warmers or space heaters), then the amount of heat produced in a day should be equal to about 2000 Calories since this is the amount of chemical energy in food that one eats in a day.  This 2000 Calories per day is equivalent to about 97 W of power.  A good portion of this energy will eventually turn into heat, but where does this heat go? 

If heat doesn't somehow escape the igloo, then the temperature inside would keep increasing and the Inuit would burn themselves to death.  Fortunately, heat escapes igloos in many different ways.  For example, as the air inside heats up, the air molecules begin to bounce around very fast.  When they collide with the walls of the igloo, some of their energy gets absorbed and the igloo molecules start to wiggle around a little bit.  As the igloo molecules wiggle around they hit other igloo molecules and this process continues until eventually some of the igloo molecules on the outside surface start jiggling too and bumping the cold air outside.  This whole process ultimately results in heat being transferred out through the walls of the igloo.  Now if the room is going to stay the same temperature (i.e. have the same thermal energy), then the extra energy being produced by thermal processes has to be equal to the energy lost through the walls.¹   This means a one person igloo will have 96 W of power flowing through its walls.  There's a nice equation that describes how energy flows through a material, 

P = k · A  · (T_hot − T_cold) / x, 

where P is the power passing through, k is called the thermal conductivity, A is the area of the material, x is the thickness, T_hot and T_cold are the temperature of the hot and cold ends, respectively. The thermal conductivity of snow is about 0.16 W/m·°C.  On a cold day, the outside temperature can be as low as −45 °C (−49 °F).  The walls might be 0.3 m (~1 ft) thick.  If we make our igloo 2 m (~6.5 ft) in diameter, then the total area of the walls would be 3.1 m².²   Solving for T_hot we find that the temperature inside the igloo is

T_hot = P  · x / [k · A] + T_cold

= (96 W) · (0.3 m) / [ (0.16 W/m·°C) · (3.1 m²)] − 45 °C

= 13 °C.

That's a toasty 13 °C or 55 °F.  According to Wikipedia, igloos can be anywhere from −7 °C (19 °F) to 16 °C (61 °F) when warmed by body heat alone.  So if you're stuck in the cold, warm up by building yourself a nice little igloo hut.

[1] I'm assuming the energy lost through the floor and through a small door is negligible.

[2] This assumes the area of the walls is roughly that of a hemisphere of diameter 3 m.

Time Lapse Aging

There's something cool about time lapse photography.  It seems several folks on YouTube have put together time lapse photographs of themselves aging over time.  By taking 1 photo per day every day for years, you can put together a little animation that shows you aging.  From what I've found, the longest stretch of time over which someone took photos like these is 17 years.  If you took one photo of yourself per day for your whole life, how long would the video last?

 

Films generally run around 30 frames per second.  A person can live over 100 years or 36,000  days. This means you would show a total of 36,000 pictures.  If each picture being shown for 1/30th of a second, the video would last 20 minutes. 

Sadly, unless someone started doing this at least 30 years ago, I won't be able to see a full life progression sped up in my lifetime.

Split Ended

I hate split ends.  They make me sad.  Fortunately, my wife Anna doesn't mind cutting them off for me.  How long will it take to cut off all my split ends?

Sadly, there are no hair care products that can remove splits ends.  I could just shave my head and eliminate them rather quickly, but I'd like to keep my rather luxuriant flowing hair so I'm just going to consider how long it will take to go through each hair individually.  In How Many Licks?, I estimated that there were about 250,000 hairs on a (non-bald) human head.  Roughly 5% or 12,000 of my hairs have splits ends.  If my wife can sift through 10 hairs per minute and cut all the split ends off, then it would take about 17 straight days of sifting to remove all the split ends.  

We could go further with this.  There's some question about what to do with the hair she's already looked at so that she doesn't end up checking it twice.  Moreover, my computer science friends might suggest that a more efficient search algorithm could decrease the time requirement immensely, but I'll leave that to the nice people at Google and Bing.