A Diary of Numbers

Ballparking: Practical Math for Impractical Sports Questions

2012-01-18T16:01:00.000-08:00

I have a new book. It comes out in a few months. You should buy it. It will make you smile.

Fuzzy Numbers

2012-01-18T15:56:00.000-08:00

We're halfway through January, which means most people have probably given up on their New Years resolutions. For those who are still committed to going to the gym every day and losing that last three pounds, I salute you, but let me give you one bit of warning: Don't trust the nutritional labels on backs of packages. If you're counting calories, you may find yourself disappointed. One of the most difficult things to impress on students is that real world numbers aren't exact. There's always some degree of uncertainty. When you look on the back of that energy bar and it says "200 calories", it's probably not exactly 200 calories. Perhaps it's 201 calories. Or 197 calories. Or 234 calories. Just how far off is it? In a science lab, you would never report a real world number without providing some error range.¹ It may seem like I'm being a pedantic stickler here, but real world numbers are incomplete without an error range. Saying "This chocolate covered bacon cake is only 137 calories", is not as truthful as saying "This chocolate covered bacon cake is only 137 calories plus or minus 6,723 calories" when there's an enormous error range. Just how far off are nutritional labels?

To investigate how far off the reported numbers on a nutritional label are, I first bought a package of Mini Clif Bars.² Given the limited amount of equipment at my disposal, it would be difficult to measure the number of calories in each bar directly. However, it seems reasonable that the number of calories in a bar would be proportional to its weight, so you can estimate the caloric fluctuations by measuring the fluctuations in the mass. Each bar had a listed weight of 28 g and was said to contain 100 calories. The package contained 18 bars with weights ranging from 28.2 g to 33.3 g. The bars had a mean weight of 30.7 g and a standard deviation of 1.3 g. That's a 4% fluctuation away from the mean, which itself was almost 10% larger than the listed value. Scaling these fluctuations up to the listed value of 100 calories, we find that the actual number of calories for a Mini Clif Bar is likely closer to 109.5±4.5 calories.

A histogram for the different Mini Clif Bar masses (in grams)
shows significant fluctuations away from the mean.

Let me be clear. I'm not insinuating that the good people at Clif Bar & Company have devised a dastardly plot to plump up casual exercise enthusiasts. If anything, I'm kinda happy I got more food than I paid for. My only complaint is that nutritional labels don't include a range detailing how imprecise or "fuzzy" these numbers are. No matter how careful a manufacturer is, there are always the fluctuations that will creep in, and we would do well acknowledge this fact.

[1] The term "error range" is a bit misleading because the word "error" implies a making mistake. "Oops...I really didn't mean to drop that now bloody hatchet down your underwear. Please forgive my error." In reality, error ranges have nothing to do with making mistakes. All real world measurements have some degree of uncertainty, so perhaps it's better to think of them as "uncertainty" ranges rather than "error" ranges.

[2] Admittedly, a rigorous study should include more than one box and a variety of different food types, but doing one sample is enough to illustrate my point. Plus, I like Mini Clif Bars.

It's That Time of Year Again

2011-11-29T17:00:00.000-08:00

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...

2011-11-29T16:27:00.000-08:00

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

2011-10-02T15:08:00.000-07:00

"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

2011-09-12T05:40:00.000-07:00

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

2011-08-29T14:01:00.000-07:00

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

2011-08-19T14:19:00.000-07:00

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

2011-08-11T08:09:00.000-07:00

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?

2011-08-02T06:35:00.000-07:00

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

2011-07-29T11:30:00.000-07:00

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....

2011-07-29T11:26:00.000-07:00

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

2011-04-04T15:40:00.000-07:00

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

2011-03-21T07:31:00.000-07:00

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

2011-03-19T05:17:00.000-07:00

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

2011-03-13T07:57:00.000-07:00

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

2011-03-06T10:45:00.000-08:00

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)

2011-03-04T14:16:00.000-08:00

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

2011-03-02T12:00:00.000-08:00

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) ]

= 6×10⁵⁰photons

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

2011-02-24T13:31:00.000-08:00

“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

2011-02-23T11:03:00.000-08:00

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...

2011-02-16T10:52:00.000-08:00

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

2011-02-16T10:35:00.000-08:00

"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]

2011-02-09T10:25:00.000-08:00

"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)

2011-02-05T06:09:00.000-08:00

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"

2011-02-04T08:48:00.000-08:00

"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

2011-02-02T09:29:00.000-08:00

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!

2011-01-27T14:07:00.000-08:00

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

2011-01-25T13:24:00.000-08:00

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

2011-01-19T13:20:00.000-08:00

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.

Astronomical BS

2011-01-17T06:58:00.000-08:00

A few years back, a friend and I got into an argument about astrology. She started telling me how much of a Gemini I am and how I very clearly match all the personality traits. As an experiment, I decided to screw with her and see how she reacted:

Me: Actually, due to the precession of the equinoxes in the year I was born, I'm actually a Taurus. It's a very rare phenom. Most astrologists don't even know about it. My birth year was the only time it's happened in the last 153 years.¹

Her: Oh my God! That's so amazing! I've always thought you acted like a Taurus. Now I know why!

She then proceeded telling me how much of a Taurus I am and how I very clearly match all the personality traits. I've repeated this experiment several times with different people, always with similar results. No matter what sign I give, cognitive dissonance sets in, then there's rationalization, then I'm told how much of a [fill in the blank] I am and how I very clearly match all the personality traits.

Apparently, my little experiment was coincidentally prescient. In light of the latest astrological buzz, it turns out I actually am a Taurus.² Unbeknownst to people who follow astrology, the zodiac calendar has been wrong for years. In addition to the various dates being off, there's also a 13th astrological sign that was left out. How many astrology-believing Americans just had their signs switched?

Despite the absence of evidence or even a plausible mechanism³, many Americans still believe the stars control our personalities. According to Michael Shermer's Why People Believe Weird Things, a 1990 Gallup poll showed that 52% of Americans believe in astrology. There's very little overlap between the zodiac calendar and the actual positions of constellations in the sky. For this reason, about 90% of people's zodiac signs changed. This means that

(3.0×10⁸ Americans) · (0.52 believers per American) · (0.9 zodiac changes)
= 1.4×10⁸ zodiac sign changes.

That's 140 million people and a whole lot of cognitive dissonance. Thank you, Laurie for suggesting this one.

[1] At this point, I was basically making things up and throwing in ad-libed technical jargon. The actual "precession of equinoxes" has nothing to do with astrology.
[2] But, no, this does not justify my astrologist friend's assertion. As mentioned earlier, it doesn't matter what sign I give, astrologers will always say my personality matches.
[3] And, no, the fact that the moon controls the tides does not constitute a mechanism. Tidal forces are due to gravity, and as I show in How Many Licks?, the tidal force on you at birth due to the obstetrician is about 40 times larger than the largest tidal force of any heavenly body.

Seth MacFarlane: Family Guy Creator or Math Genius?

2011-01-13T05:43:00.000-08:00

On December 26, 2010, Family Guy creator and voice actor extraordinaire Seth MacFarlane tweeted, "If the T1000 had to kill everyone named Aiden Connor or Dylan Connor, he’d have tons of killing to do." How accurate is his estimate?

According to Book Of Odds, each name has the following probabilities:

The odds a male is named Dylan are 1 in 6,250 (US, 5/1990 - 9/1990).
The odds a male born in 2000 is named Aiden are 1 in 2,370 (US, 2000).
The odds a person's last name is Conner are 1 in 4,442 (US, 2000).

Assuming there are no weird correlations, this means the probability that a person is named "Dylan Conner" is 1 in 28 million and the probability that a person is named "Aiden Conner" is 1 in 11 million. Just considering the U.S. population, there are 150 million males roughly 5 and 12 of which would be named "Dylan Conner" and "Aiden Conner", respectively. If each Conner weighed 150 lbs, the total weight would be 1.3 tons.

Well done Mr. MacFarlane! [In poorly attempted Stewie voice] Victory is yours!

Lessons from Stick Figures Presents, "A Brief Interlude on Entropy"

2011-01-12T13:44:00.000-08:00

Plague Gauge

2011-01-11T14:50:00.000-08:00

"This is what the LORD says: By this you will know that I am the LORD: With the staff that is in my hand I will strike the water of the Nile, and it will be changed into blood. The fish in the Nile will die, and the river will stink; the Egyptians will not be able to drink its water."

— Exodus 7:17–18

Eeeew! How many people would it take to make a river of blood?

According to the Wikipedia, the Nile discharges 2830 m³ of water each second. You'd need an equivalent flow rate of blood to produce a Nile-sized river. According to the American Red Cross, you can give 1 pint of blood once every 56 days. To produce a river worth of blood, you would need a total of

# of people = (flow rate required) / (flow rate per person)
= ( 2830 m³/s ) / ( 1 pint / 56 days )
= 2.9 ×10¹³ people.

That's 4000 times more people than there are in the world today.

2011

2011-01-11T10:29:00.000-08:00

Unlike plants, cats, or small children, this numbers blog doesn't complain when I don't feed it for two months. Still, I've been feeling bad about not posting, so my New Years resolution is to start writing regularly again. More to come in a moment...

Death By Coconut

2010-11-16T11:14:00.000-08:00

In “I Know What You Did Last Summer of the Shark”, then Daily Show correspondent Stephen Colbert states that falling coconuts kill 150 people each year. You might assume that “death by coconut” was a purely random occurrence, but that might just be what the coconuts want you to believe. What's are the chances that the coconuts are out to get you?

According to Wikipedia, 54 million tonnes of coconuts were produced in 2009. From this, we know that if each coconut weighs 10 lbs, then roughly 1.0×10¹⁰ coconuts were produced. Now, we could quibble about the actual number. Some grow in the wild which might make the actual number larger. Some coconuts are picked instead of falling, so that might make the actual number smaller. You can argue either way, so let’s stick with this figure just to keep the problem relatively simple.

There are 6.7×10⁹ people in the world, each of which has about 1.5 ft² of area that the coconut could land on giving a total area of about 1.0×10¹⁰ ft². According to Wikipedia, the total land area in the world is about 1.5×10⁸ km². If people are randomly distributed across the land area of Earth, then the probability of being hit by a coconut is equal to the fraction of land area that people take up at any given time. Using Google’s calculator, we get

From this, we suspect that each year roughly 6 out of every one million people get bonked by a coconut.
It’s difficult to say how many people actually get killed by coconuts. From our calculation above and the world population, we can estimate the number of people that get hit by coconuts each year, but not everyone who gets hit from a falling coconut will be killed by it. Assuming all hits were fatal, we could calculate the total number of deaths by multiplying “hits per person” times the “total number of people”, to get

(6×10^-6 hits per person) × (6.7×10⁹ people) = 40,000 fatal hits.

If hits are only fatal 1% of the time, then 400 people will die from coconuts falling. However, this 1% statistic may be off substantially—possibly much more than an order of magnitude—so our estimate is very rough.

It’s difficult to tell how many coconut-induced fatalities will occur, but our estimate of “40,000 coconut hits” suggests that the number of deaths could be substantially higher than 150. If this were the case, then coconuts are certainly not out to get you since they kill fewer people than they would by chance.

We’ve assumed the following:

· Coconuts are 10 lbs on average.
· The mass of all the falling coconuts in the world each year is equivalent to the mass of coconuts produced each year.
· The probability of a falling coconut hitting a person is equal to the fraction of land area taken up by people.
· Only 1% of coconut hits are fatal.

These assumptions seem reasonable, but that does not mean they are necessarily correct. Coconuts are certainly between 1 and 100 lbs, so the first assumption seems decent. It’s possible the number of falling coconuts each year is off for the reasons stated above. Likewise, the percentage of fatalities could be off by several orders of magnitude. Equating the probability of a falling coconut hitting a person to the fraction of land area taken up by people is a reasonable first guess, but there are factors that might throw this assumption off. For example, perhaps more people live near coconut trees because people like to live in tropical climates.

While our estimate doesn’t have enough precision to answer this question conclusively, this example does illustrate an important point. As Weinstein and Adams describe in their book Guesstimation, estimates generally break up into three “Goldilocks” catagories: too big, too small, and just right. In this case, being “just right” means your estimated result is too close to call. When this happens, you need to put more effort into refining your estimate if you want to draw any conclusions. Refining the coconuts estimate to high precision is beyond the scope of what I can do in a silly blog post, but there is still a valuable lesson to be learned: In estimation as in life, there are times when even the best answer leaves a wide degree of uncertainty and it’s important to acknowledge when we don’t have enough information to draw a conclusion. That said, there are many examples where a test produces results that are so unlikely we can conclude they are not due to random chance.

Lucky Numbers

2010-10-24T16:47:00.000-07:00

Anna and I went out for Chinese food in Philadelphia today. As I looked at the lucky numbers in my fortune cookie, I couldn't help but wonder, "If everyone who ate Chinese food today played their lucky numbers in the lottery, what are the chances at least one of them would win?"

Both fortune cookies and lottery numbers usually show about 5 numbers that can range from roughly 1 to 50. The probability of picking the first number correctly is 5 out of 50. The probability of picking the second number correctly is 4 out of 49. The probability of picking the third number correctly is ... Multiplying these probabilities together, we can find the total probability of finding the right sequence of numbers¹,

P = [(5)! · (50-5)!] / 50! = 4.7×10^-7.

That's about one in two million. I generally go out for Chinese food about once per month, which seems like a reasonable amount for most people. Taking that as the average and using the fact that there are 3.0×10⁸ Americans, we can estimate the number of people that went out for Chinese today,

# of people going for Chinese = (prob. of going out for Chinese) · (total # of people)

= (1 day / 30 days) · (3.0×10⁸people)

= 1.0×10⁷people.

The probability that everyone will will pick the right numbers is P^10,000,000. Likewise, the probability of everyone picking the wrong number is (1-P)^10,000,000. The probability that at least one person will win is then just

1 - (1-P)^10,000,000

= 0.009.

There's about a 1% chance that if everyone played their lucky fortune cookie numbers at leat one would win.

[1] This is the well known binomial distribution.
[2] I'm assuming the fortune cookie's "lucky numbers" are random and uniformly distributed.

Special Guest Natalie Angier

2010-10-17T07:03:00.000-07:00

Today we're pleased to have a question from special guest Natalie Angier. Ms. Angier is a Pulitzer-prize winning science journalist for the New York Times. She has authored several books, most recently, The Canon: A Whirligig Tour of the Beautiful Basics of Science. She writes, "How many leaves are raked up nationwide on an average weekend afternoon in October?"

Before I begin, I have leaf-raking riddle for you. Without adding or rearranging the words, add punctuation to the following sentence to make it grammatically correct: "A boy raking leaves." The answer is below.

If you're going to rake leaves, you need leaves to rake. Some states like Arizona are desert-y and won't have many leaves to rake, but most places in the U.S. will have trees that shed. Even if you live in the right climate, you still need a yard with at least one tree in it. I'll assume that 1 out of 10 people owns a yard with a tree in it, since it's very likely that the actual number is greater than 1 out of 100 and less than 1 out of 1. Of these people, some will rake, but many will use a leaf blower or just let the leaves lie. Using a similar order-of-magnitude argument to the previous one, I'll assume 1 out of 10 people who have leaves rake them. The average leaf raker might rake his/her leaves once in October, and there are about 8 good leaf-raking October weekend afternoons each year. There are 3.1×10⁸ people in the United States. Combining these assumptions, we can estimate that,

# of people raking = (3.1×10⁸ people) · (0.1 tree owners per person)
· (0.1 raker per tree owner)· (0.125 chance of raking now)

= 390,000 people raking leaves each weekend afternoon in October.

But the question specifically asked for the number of leaves raked. This will clearly depend on the number of leaves a person has in his/her backyard. According to at least one source, a mature tree can have up to 200,000 leaves. To confirm this, I looked at leaves strewn across Tappan Square in Oberlin. The mean separation was about 6 inches between leaves. If you spread them out over a reasonably sized lawn (about 1/5 of an acre), you get about 200,000 leaves. If each raker rakes this many leaves one a weekend, there will be,

# of leaves = (200,000 leaves per raker) · (390,000 raker)

= 7.8×10¹⁰ leaves raked.

That's 78 billion leaves raked nationwide each afternoon in October. Thanks for the great question, Ms. Angier!

For those wondering about my earlier grammatical riddle, the correct answer is "A boy, raking, leaves."

A Relatively Good Calculation

2010-10-14T10:56:00.000-07:00

In 1905, Einstein published his special theory of relativity. The most well-known part of this theory is almost certainly the famous E=mc² equation that predicted a future with nuclear bombs and atomic energy, but this is not the only surprising prediction. The theory also predicted that objects shrink when they move really fast.¹ After hearing a professor describe this strange and fascinating phenomenon, I wondered two things. First, what was Einstein smoking? Second, if I ran really fast, would I be able to see atoms? How fast does a person have to run to be atom sized?

According to special relativity, the length of a moving object is equal to its original length times an extra factor

L' = L [1 – (v/c)²]^0.5.

Here, c is the speed of light, L' is the length of the object when it's moving, L is the length of the object when it's not moving, and v is the velocity at which it's moving. Our new length L' will be about 10^-10m or roughly the size of an atom. Our original length will be about 1.5 m. We can solve for v

v = c [1 – (L'/L)² ]^0.5

= (3.0×10⁸ m/s)[1 – (10^-10m / 1.5 m)² ]^0.5

= 0.9999999999999999999977778 c.

You would need to move very close to the speed of light to be atom sized.

[1] This phenomenon is called "length contraction".

Lieutenant Commander Banana Clip

2010-10-12T11:04:00.000-07:00

I saw this on Totally Looks Like the other day, and it certainly brought back a few repressed memories. When I was a kid, I definitely wore a banana clip pretending to be Geordi LaForge, and my family wasn't even big Star Trek fans. How many people in the U.S. wore banana clips over their eyes pretending to be Lieutenant Commander Geordi LaForge?

Apparently, I was not alone. It strikes me that role playing as Geordi takes three things: a certain level of maturity, Star Trek knowledge, and banana clips. I chose the phrase "a certain level of maturity" carefully. I'm pretty sure most 2-year olds and 90-year olds didn't do this. You need just the right amount of maturity. With too much, you just think it's silly. With too little, you can't see how genius the banana clip Geordi visor really is. I suspect about 1/3 of the population has the appropriate maturity level.

After maturity, you need Star Trek knowledge. As I said earlier, I wasn't even that big of a Star Trek fan, but I still rocked the Geordi visor. That said, I suspect at least 1 out of 10 people had my level of Star Trek knowledge.

The final ingredient is, of course, banana clips. These were fairly ubiquitous in the late 80s and early 90s when Star Trek the Next Generation was popular. As a reasonable guess, I would say at least at least 1 out of 10 people had access to banana clips at this time.

Using the assumptions above, we can estimate that about 1 out of every 300 Americans, or about one-million people pulled off the Geordi look. LeVar Burton, you deserve a commission from the banana clip companies.

[1] Doh! "Jordi LaForge" should be spelled "Geordi La Forge." Thank you, to my student Kara Kundert for the correction!

Safety in Numbers

2010-09-30T10:38:00.000-07:00

It’s commonly said that airplanes are the safest mode of transportation. It’s true that more people die in car crashes than plane crashes each year, but most people also drive more often than fly. On a "per trip" basis, which mode of transportation is safer?

The question did not specify whether we were considering just crashes in the U.S. or in the entire world. Assuming crashes are equally likely in all parts of the world, the “fatalities to trips” ratio should be about the same in both cases. Just make sure you are consistent (i.e. don’t divide the number of U.S. plane crash victims by the total number of flights in the world.)

According to “Ask a Scientist” ¹, in the U.S. there are

“…about 40,000 deaths per year in automobile accidents vs. about 200 in air transport.”

You can check these numbers against other references to make sure they are accurate. To answer the question, we need to estimate how often a person flies and how often a person drives. On average, we might guess that an American flies about once per year. This is reasonable since you’d certainly expect people to fly more than once every 10 years and less than once per month. In contrast, most of us drive (or ride in) a car about twice per day, even if it’s just to get to and from work or school. Since there are 3.1×10⁸ Americans, this means there are

# flights per year = (1 flight per American per year) × (3.1×10⁸ Americans)

= 3.1×10⁸ flights per year,

and

# car trips per year= (2×365 car trips per Amer. per year) × (3.1×10⁸ Americans)

= 2.2×10¹¹ car trips per year.

The fraction of deaths is then just,

Fraction = (# deaths per year) / (# trips per year)

= (200 deaths per year) / (3.1×10⁸ plane trips per year)

6.5×10^-7 deaths per trip

= (40,000 deaths per year) / (2.2×10¹¹ car trips per year)

1.8×10^-6 deaths per trip.

You are 2-3 times more likely to die in a car trip than a plane trip, so according to our numbers plane travel is still safer. Given the precision of the estimation, it’s possible that other reasonable assumptions would come up with the opposite result, since both figures are within an order of magnitude of each other.

[1] You can find this stat and a nice discussion of the topic at here. Their numbers are different partly because they’re talking about fatalities per mile not fatalities per trip.

I Can Haz Estimation?

2010-09-26T15:54:00.000-07:00

If you’re like me, there’s nothing you like better after a long work day than kicking back and looking at some grammatically incorrect cats. And who better to provide all your LOLcat needs than The Cheezburger Network? In honor of my favorite intertubes diversion, how many pictures of cats are on the entire Internet?

This is a difficult problem. That being the case, we expect our estimate will likely be very different from the actual number. In cases like these, it is helpful to determine what the wrong answers are. To do this, you need to calculate upper and lower bounds. To start, let’s estimate how many people are cat owners. You might guess that 10% of people own cats. Is this reasonable? Well the actual number is certainly less than 100% of people, and very likely greater than 1% of people. If you asked a 100 of your friends and family, at least one of them probably has a cat. The next question you could reasonably ask is how many cats do cat owners typically have. Some people have 10 cats, while others only have one. A reasonable guess is 2 cats per owner.

Now comes the tricky part. What fraction of people put pictures of their cat on the web? Many of you are tech savvy and have accounts on Facebook, Flickr, etc. If you have a cat, you very likely have at least one picture of it on the web somewhere. But what about your parents, grandparents, and friends that may not be as tech savvy as you? What about people in developing nations that may not have Internet access? If you average over everyone including people in other countries, what percentage of cat owners will post pictures on the web? To be safe, lets say 10% again. Is this reasonable? As before, the actual number will certainly be less than 100%. Will it be great than 1%? Possibly, but there are a lot of people that don’t like putting their information up on the Internet and the ones that do might not put their cat on the Internet. To be safe, let’s put the lower bound at 0.1%. We can summarize what we know in a chart like the one below:

	Upper	Guess	Lower
% of people that own cats	100%	10%	1%
Cats per owner	10	2	1
% of owners that put cat on web	100%	10%	0.1%
Pictures per cat	100	10	1
World population	6.7x10⁹	6.7x10⁹	6.7x10⁹
# of cat pictures	6.7x10¹³	1.3x10⁹	6.0x10⁴

Our upper and lower bounds are very far apart (i.e. 9 orders of magnitude). The answer certainly lies between them, but is there any way we can tighten these bounds?

Let’s try calculating the answer in a different way. According to Netcraft, there are 2.5x10¹⁰ web pages as of a few years ago. As an upper bound, we might say that at most, every page has 10 cat pictures on it. Even in this extreme case, there would only be 2.5x10¹¹ cat pictures, so we’ve effectively reduced our upper bound by a factor of 100.

Now lets see if we can adjust our lower bound. A quick Google image search for the word “cat” pulls up about 1.5x10⁸ results. Looking at the pictures on the first page, we can see that the vast majority of the images are of cats, but occasionally you get one that’s not a cat. This is very rare. In fact, it seems much more likely that Google is missing cat pictures since it doesn’t have access to private photos on sites like Facebook and Flickr. For this reason, we could probably take the Google result as a lower bound. Just to be extra careful, let’s say the lower bound is 1/10^th the number of “cat” images found by Google.

Our final results:

Upper Bound -- 2.5x10¹¹ cat pictures

Actual Estimate – 1.3x10⁹

Lower Bound -- 1.5x10⁷

My Eyes Are Buggin’

2010-09-21T17:31:00.001-07:00

I had an interesting conversation with one of my physics students recently. She had a dance class right before she came to lab.

Her: I hate dancing with glasses.

Me: Because they fall off when you’re doing the turn-y things?

Her: Yes, it’s so much easier with contacts.

Me: You know, you can calculate how fast you’d have to turn to get the contacts to pop out.

How fast would a dancer have to spin to get her contacts to pop out?

Contact lenses weigh about 0.05 g and are about 1.0 cm² in area. They are held on mostly by suction. Suction occurs because there is a partial vacuum created under the contact as it's pulled away from the eye. To pop out, a contact needs to overcome air pressure, which for our atmosphere is about 100 kPa. As the dancer spins, her contacts move in a circle with roughly a 10 cm radius. The force to keep an object in uniform circular motion is given by

force = (mass) · (velocity)²/ (radius).

The maximum this force can be before the contact pops out is determine by atmospheric pressure

force = (pressure) · (area).

Solving for the velocity, we get

velocity = [(pressure) · (area) · (radius) / (mass)]^1/2

= [(100 kPa) · (1 cm²) · (10 cm) / (0.05 g)]^1/2

= 140 m/s.

A dancer would have to rotate about 1400 per second to get her contacts to pop out.

Smile! You’re the Monocle Smile Contest Winner!

2010-09-21T17:11:00.001-07:00

Hello, Ladies. How are you? Fantastic. We have a winner!

The Question: If he worked nonstop for the rest of his life, how many personalized Youtube videos could Mr. Mustafa make?

Born February 11, 1974, Isaiah Mustafa is currently 36 years sexy. He’s young and he’s in shape, the latter of which makes longevity more likely. As best I can tell, he only has one major thing going against him: he played NFL football. However, his career was fairly short and he was a wide receiver, so he’s not as likely to have the same health issues as a 12-year 350 lb lineman. One simple way to solve this is to subtract his current age from his life expectancy. According to Wikipedia, Americans are 38^th in the world for life expectancy with and average lifespan of about 78.2 years. Using Wolfram Alpha, we know that Mr. Mustafa was 36.6 years old on the contest deadline. From this we can compute how many years he has left,

78.2 years – 36.6 years = 41.7 years.

Each personalized video appears to be about 30 seconds in length. If he worked nonstop¹ making 30-second videos for 41.7 years, he would have a total of about 44 million videos.

Congratulations to our winner, Joey Reichert!

[1] Some might argue that we haven’t taken into account sleeping. True. If he did sleep, our final number would only be one third as big. But the original question did say “nonstop.” “But wouldn’t he die younger if he never slept?” you ask, suspiciously. Also true. But we could always film him while he slept, and which of you ladies out there wouldn’t appreciate that?

A Cure for Global Warming?

2010-09-14T17:44:00.000-07:00

I came across the above photo on Reddit the other day (original link here). As one Redditor so aptly put it, “The stupid, it burns!!!” Interestingly, there are multiple layers of stupid here. Like a giant stupid onion. Leaving aside the fact that water dumped into a sink would eventually wash back into the ocean, how many buckets of water would each Earthling need to dump to cancel out the rising sea levels?

According to the Wikipedia entry for "Current sea level rise", predicted values for the ocean rise due to global warming range anywhere from 90 to 880 mm. I'll take 90 mm and just calculate the lower bound. The total area of the oceans is roughly 3.4 ×10¹⁴ m², meaning that the total volume is about 3.0×10¹³ m³. One bucket might hold 50 L and there are 6.7 ×10⁹ people in the world. Form this, we can compute the total number of buckets needed,

# of buckets = (total volume) / [ (volume per bucket per person) × (number of people)]

= (3.0×10¹³ m³) / [ (50 L per bucket per person) × (6.7×10⁹ people)]

= 90,000 buckets.

Even neglecting the fact that the water will just wash back into the ocean, this method would still require every person on the planet to dump 90,000 buckets of sea water down their sink. As Shakespeare might once have said, "The stupid, it doth burn."

Bounding Beck

2010-09-10T12:45:00.000-07:00

Sarah Palin said there were “hundreds and hundreds of thousands of people.” Glenn Beck put the estimate at 500,000 people. Congresswoman Michelle Bachmann stated that she wouldn’t let anyone tell her there were less than 1 million people. AirPhotosLive.com estimated only 87,000 people based on aerial pictures it took. How many people were at the Glenn Beck rally?

As you can see from the photos on AirPhotosLive.com, most of the people assembled along the Lincoln Memorial Reflecting Pool. According to Wikipedia, the pool is approximately 618 m long by 51 m wide. Judging from the photos, attendees seem to cover an area between 3 and 4 reflecting pools wide, or roughly between 150 m and 200 m. Multiplying by the reflecting pool length, this means the total area taken up by the attendees is between 92,700 m² and 125,000 m². To be sure we’re not forgetting anyone, we’ll count the attendees that spilled over near the Washington monument and those that got a front row seat. This looks like no more than 20% of the total number of people, so we’ll say they cover a total area of at least 90,000 m²and at most 150,000 m².

Looking at the aerial photo, there’s a whole lot of green grass you can see between people. This makes sense since most people don’t like to be squished and you need some extra room so that people can move around. Each person needs at least 1.0 m² of space to feel comfortable. Judging from the sparse parts of the crowd, there’s appears to be no more than 10 m² of space between people on average. The average density certainly lies between these two extremes. From these densities, we can use the equation

number of people = (density of people) × (area)

to compute an upper bound,

number of people = (1.0 person per m²) × (150,000 m²)

= 150,000 people

and lower bound,

number of people = (0.1 people per m²) × (90,000 m²)

9,000 people.

As you can see, the “87,000 people” estimate from AirPhotosLive lies right in the middle of the two extremes. It is certainly a reasonable estimate, and it brings up an important point. One way to test the accuracy of your estimate is using upper and lower bounds. By using the largest and smallest realistic numbers in your estimation, you can put a cap on what the actual number must be. We know there’s certainly more than our lower bound of 9,000 people and certainly less that our upper bound of 150,000 people.

By using upper and lower bounds, we also see that our politicians and pundits are either lying or possess a certain degree of innumeracy. Of these two options, I sincerely hope it’s lying. I’ve grown accustomed to politicians bending the truth, but I’m absolutely terrified by the thought that people who have gained this much influence can’t do middle school math. This does beg one to ask, “Why even bother lying about numbers like this?” For one, false claims like this are easy to debunk even with mediocre math skills. In addition, 87,000 people at a rally is nothing to sneeze at, so why exaggerate?

Finally, I should note that I had some hesitation when writing this post. This is not intended to be a political blog, and I’m not here to spout my personal beliefs. The fact is that no matter what side of the political spectrum they’re on, rally organizers tend to exaggerate numbers to make it seem like they’re more popular than they really are, and it’s our duty as rationally thinking citizens to call them on it.

On Birthdays and Monkey Poo

2010-08-19T15:31:00.000-07:00

It’s my sister Laurie’s birthday today, and much as it pains me to say it, it’s never easy being the four-years-younger brother of a super genius. When you’re graduating middle school, she’s graduating high school. When you’re graduating high school, she’s graduating college. When you’re getting a BS from MIT, she’s getting a PhD from Harvard. When you publish your first book, she’s giving a TED talk and getting tenure at Yale. Even today as she grows ever older and more decrepit, turning an age I can remember our mom and dad being, I can still feel that familiar tingle of our (usually) healthy sibling rivalry. If there's one thing I take solace in, it's that I've never had to worry about getting monkey poo in my hair. What’s the probability that at some point in her career as a primate researcher, Dr. Santos has gotten monkey poo in her hair?

Cayo Santiago is one of the places my sister does research on primates. Located southeast of Punta Santiago, Puerto Rico, the 600 m by 400 m (240,000 m²) island is home to about 950 Rhesus monkeys. As can be seen from the photo on the right, Cayo is a fairly lush island featuring lots of trees monkeys can climb.¹ I’ll assume a typical monkey spends about 50% of its time climbing, meaning that at any moment there are about 475 monkeys in the trees.

Each year, Laurie generally spends about 2-4 weeks on Cayo, and she has done so for about 10 years. If she works 8 hours per day, she’ll have spent up to 93 days on the island. Assuming a Rhesus monkey’s digestive schedule is similar to that of a human, each monkey probably “uses the bathroom” at least once per day. From this and the information above, we can estimate the total number of high altitude monkey poos taken while Laurie has been on the island

(# of poos per day per monkey) · (# of monkeys) · (# of days)

= (1.0 poo per day per monkey) · (475 monkeys) · (93 days)

= 44,175 poos.

From a bird’s—er—monkey’s-eye view, her hair will appear to have about 1.0 ft² of area. Our monkey bombardier will have to hit this spot if Laurie is to spend the rest of her night shampooing. From this we can calculate the probability of a random poo landing in her hair,

P = (area of hair) / (total area)

= (1.0 ft²) / (240,000 m²)

= 3.9×10^-7.

The probability of a poo not landing in her hair is, of course, 1 – P. The probability that none of the 44,175 monkey poos have landed in her hair is (1 – P) ^44,175 ~ 0.98. This means there’s a 2% chance that a monkey has poo-ed on my sister’s hair at least once.

Happy Birthday, Laurie! You’re as good a science role model and an even better sister than I could reasonably have hoped for.

[1] For simplicity, I’ll assume that 100% of the island is covered in trees.

The Monocle Smile Contest

2010-08-07T11:13:00.001-07:00

Look at your math. Now back at mine. Now back at your math. Now back to mine. Sadly, yours isn’t mine. But if you enter the Diary of Numbers Estimation Contest then your math could seem like it’s mine. Look down. Back up. You’ve won a book featuring the math your math could look like. Anything is possible when you enter the Diary of Numbers Estimation Contest. I’m on a blog.

The Question: Isaiah Mustafa (aka the sexy Old Spice man) reached meme status because of his work in a very funny Old Spice commercial. Recently, he created a series of humorous personalized Youtube videos featuring his character from the original commercial. If he worked nonstop for the rest of his life, how many personalized Youtube videos could Mr. Mustafa make?

Rules: You can win a free copy of How Many Licks? To enter, estimate an answer to the question below and send it to “aaron at aaronsantos period com.” If your answer is closest to my estimate¹, I'll mail you a free, signed copy of How Many Licks? To be eligible, you must submit your entry on or before September 15, 2010. Don't worry; I won't spam you or share your email with any third parties.

[1] I know, I know. How do I know my answer is correct? I don’t. I make no pretenses that my answer is correct or even close. Your answer may very well be a better estimate than mine. In fact, your estimate may even be exactly right and you still may not win the contest if somebody else's answer is closer to mine. Sorry about that. This is the best way I could come up with to pick a winner and I'm not changing it now. Like any good game, there's an element of luck required even if you do have great skill. With that disclaimer out of the way, good luck and happy calculatings!

TED

2010-07-30T15:19:00.000-07:00

Check out my super smart sister giving her TED talk.

20,000 Kissing Minutes

2010-07-27T17:47:00.001-07:00

A new Dentyne commercial claims we spend 20,000 minutes kissing. What kind of estimator would I be if I didn't check their math? How many minutes does the average person spend kissing?

Couples often kiss each other several times per day¹. If we assume the average day consists of 30 seconds of kissing and the average persons spends 60 years in a relationship, then we can estimate the number of minutes spent kissing in a lifetime:

time kissing = (time kissing per day) · (days per year) · (years in relationships)

= (30 s per day) · (365 days per year) · (60 years in relationships)

= 11,000 minutes.

That’s an average of 11,000 minutes spent kissing in a lifetime. The commercial’s claim is well within what one might reasonably expect. Good job, Dentyne and “stub”.

[1] It’s can be a lot more if they make whoopee.

Special Guest Saul Griffith

2010-07-20T19:53:00.001-07:00

This week's question comes from special guest, Dr. Saul Griffith. In addition to being a 2007 MacArthur Fellow, Saul has received numerous awards for his inventions including the National Inventors Hall of Fame, Collegiate Inventor's award, and the Lemelson-MIT Student prize. When not inventing, he co-authors children's comic books called HowToons about building your own science and engineering gadgets. He is also a columnist and contributor to Make and Craft magazines. Saul writes,

"How many shot glasses of oil per day per person for the average American's personal contribution to the gulf oil spill?"

According to Wikipedia, the Deepwater Horizon oil spill released up to 16,000 m³ of oil per day. There are roughly 300 million Americans. One shot glass contains about 60 mL of volume. From this we can easily compute the contribution per American,

# of shot glasses = (vol. of oil per day) / [ (# of Americans) · (vol. per shot glass)]

= (16,000 m³ per day) / [ (300 million people) · (60 mL)]

= 1.0 glass per day per person.

The oil spill was basically the equivalent of having every American dump a shot glass of oil into the gulf every day. Thanks for the question, Saul!

Hamster Contest Winner

2010-07-19T17:28:00.000-07:00

We have a winner for our “Hamster-Powered Mansions” estimation contest. The question: How many buff hamsters would it take to completely power a mansion?

Mansions come in a wide range of sizes, and each will have different energy requirements. According to Factcheck.org, Al Gore’s mansion used 191,000 kilowatt-hours in 2006. Dividing by one year, we can compute his average power consumption to be 22,000 W.

From this MAKE magazine video¹, we can see that a hamster is at least powerful enough to light up an LED, but this is only a lower bound since he might be able to power even more LEDs if we connected them in the circuit. According to Otherpower.com, their hamster Skippy had no trouble lighting up 6 LEDs, and they estimate he should be able to power 200. Being fairly conservative, I’ll assume our hamsters can power 50 LEDs. According to Wikipedia, the voltage drop across one LED is anywhere from 1.5-4.5 V and the current should be between 1 and 20 mA. Assuming the LED acts like an Ohmic resistor,² we can estimate the electrical power created by a wheel-spinning hamster,

power = (# of resistors) · (current) · (voltage drop)

= (50) · (10 mA) · (3.0 V)

= 1.5 W per hamster.

From this and the power consumption of Al Gore’s mansion given above, we can estimate the number of hamster’s you’d need to power a mansion,

# of hamsters = (power per mansion) / (power per hamster)

= (22,000 W) / (1.5 W per hamster)

= 15,000 hamsters.

You’d need about 15,000 hamsters to power a mansion. Congratulations to our winner Bryan Merrill. Bryan will be receiving a free copy of How Many Licks? Keep reading for our next contest.

[1] Speaking of MAKE magazine, check out Maker Faire in Detroit July 31 and August 1.

[2] I’ve assumed LEDs require 3.0 V and 10 mA.

Comma Chameleon

2010-07-06T15:13:00.001-07:00

Anyone who reads Diary of Numbers regularly knows I’m far from a grammar nazi. Despite my best efforts, my grammar is typically atrocious. There is, however, one thing I’m very particular on. Even though both of the following are equally correct, I much prefer the latter:

(1) “Dorothy was afraid of lions, tigers and bears.”

(2) “Dorothy was afraid of lions, tigers, and bears.”

Recently, something struck me: since both usages are perfectly correct, I’ve been wasting printer ink all my life. How many cartridges of ink do you waste in your lifetime by always including the extra comma?

This is a tricky problem because the amount of printer ink a person uses varies considerably depending on his/her career, hobbies, etc. If you’re a novelist, you’ll probably be printing a lot more commas than a rodeo clown. Moreover, the problem’s a little wishy-washy because of how I originally phrased it¹. A better way to phrase the question would have been, “How many cartridges of ink would you waste in your lifetime if everyone included the extra comma?”

Some days I print 100 pages, other days I print none. On average, I probably print about one full page per day². Lists appear about once every 10 pages³. This means I’d be printing one superfluous comma every 10 days. At this rate, I’d produce 2900 needless commas over an 80-year lifespan.

According to HP’s website, the average HP LaserJet Q2612A Black Print Cartridge yields 2000 pages worth of ink. A single comma requires about one-fourth the ink of a letter and according to one of my MS-Word documents, there are about 2000 letters per page. This means one page worth of ink is equivalent to about 8000 commas. From this and the info above, we can estimate the number of printer cartridges one will waste in a lifetime,

# ink cartridges = (# ink cartridges per page) · (# pages per comma) · (# of commas)

= (1 ink cartridge per 2000 pages) · (1 page per 8000 commas) · (2900 commas)

= 0.00018 ink cartridges.

Over the course of your lifetime, you will waste 1/5,000^th of an ink cartridge by including the extra comma. Even if everyone in America included the extra comma, we’d waste only about 680 printer cartridges each year.

After posting this problem, a good friend, who had previously worked for Yale University Press as a copy editor, told me that Yale never included the extra comma to save ink!

[1] Sorry about that.

[2] This is more than printing one page every ten days and less than printing 10 pages per day, which seems like reasonably good bounds for and order of magnitude estimate.

[3] Again, this estimate is reasonably good if you consider upper and lower bounds. A list certainly does not appear on every page but you’ll probably see more than one on a hundred pages of text.

A Mime Is a Terrible Thing to Waste

2010-07-02T14:42:00.000-07:00

I like mime. There. I said it. Just to be clear, I’m not talking about the guys who stand on street corners painted gold pretending to be statues just to get money. Those guys are creepy¹. I’m talking about real mime. The kind that requires actual talent. I’m talking about the Marcel Marceau kind of mime.

About ten years ago, an acting teacher suggested I see Marceau perform in town that night. Knowing the type of reputation mimes had, I hesitated, not wanting to associate with their type. Ultimately, I decided to go since Marceau was 77 at the point², and I was unlikely to get another chance to see him perform if I changed my mind³. I wish I could say I had an instant conversion to becoming a mime-lover during the show, but my reaction was more of a “meh.” It was a cute show, definitely entertaining and at certain points even laugh-out-loud funny. It was even good enough for me to admit that not all mimes were bad. But this was supposed to be the best mime in the world, and with that kind of reputation, I expected something better than a bunch of funny and occasionally thought-provoking skits.

I went back to work on Monday. I’d told a postdoc about the show beforehand and she’d chided me about it, but she asked me how it went:

Her: How was the show?

Me: It was all right. Pretty funny actually. There was one bit with a mask maker where he gets a smiley mask stuck on his face. There was also one with a kid in a park where he kicks a ball through a window. In the background, there are these two old ladies watching and gossiping. I think my favorite one was this dating service that keeps sending him women who are either too tall or too fa—

Her: Wait. I thought you said it was just him.

Me: [Confused] It was.

That’s when I realized the magic of Marcel Marceau. None of those other people were there. There were no set, no props, no text nor sound of any kind. Not even a lighting change. Just a silent 77-year-old man alone on a stage.

In honor of the great Marcel Marceau, see if you can calculate how many fewer words he said during his lifetime because of his career choice.

People usually say about 3 words per second when talking. The only times mimes don’t speak is during performance. Marceau might have averaged 5 shows per week during his 60 years of performing. Each show is about 1.0 hour long. If we assume only 5% of his time was spent talking, we can estimate the number of extra words he could have spoken in his lifetime,

words = (yrs performed) · (wks per yr) · (shows pr wk) ·(hrs per show)

· (fraction of time talking) · (words per s)

= (60 yrs) · (52 wks per yr) · (5 shows pr wk) ·(1.0 hr per show) · (0.05) · (3 words per s)

= 8.4×10⁶ words.

Marceau would have said about 8.4 million extra words had he not been a mime.

[1] I’m convinced they’re the reason real mimes are listed below jugglers and magicians on the street performer totem pole.

[2] As another member of our class put it, “He’s still alive?!”

[3] Plus, he was in a Mel Brooks movie. How bad could he be?

Lost Ring

2010-06-21T14:43:00.000-07:00

I got a question from my sister this past week:

“I recently lost my engagement ring. (It's not the first time I lost it, but I think this time it's really gone. Sigh.). But, I’m not alone—as I just learned that Jupiter lost a band recently… How many times bigger is Jupiter's band than my engagement band?”

OBERLIN

Jupiter’s radius is about 71,000 km.¹ My sister’s ring’s radius was likely about 1.0 cm. This would make Jupiter’s ring (or band) about 7 billion times bigger if we’re talking about its linear dimension. However, the word “bigger” has several meanings. If we’re talking about how many times larger the volume of the ring is, then we need to cube the linear dimension. In this case, Jupiter’s ring is about 3.6×10²⁹times bigger. As antoniseb has pointed out, there’s also the issue of density. If we want to talk about how much more massive the ring is we need to know the ratio of Jupiter’s ring’s density to the density of gold. The density of gold is 19.3 g/cm³. Jupiter is a gas giant with a low average density of about 1.33 g/cm³. The upper atmosphere where the ring is located is probably much less dense. Assuming it has the same density as air at the surface Earth (~0.0012 g/cm³), Jupiter’s band will be about 2.2×10²⁵ times bigger.

Sorry about the ring. If you want to make sure you don’t lose your next ring, you can always do what I did.²

[1] Since the ring is not at the equator, it will actually be a little smaller than this.

[2] See picture above.

Happy Father’s Day

2010-06-20T06:57:00.000-07:00

Happy Father’s Day to all the dads out there. My Dad suggested a problem the other day: How long would it take to wear down a tire by 0.75 inches if you’re driving a 55 mph?

I’ll assume a tire tread starts about an inch thick. By the time 0.75 inches have worn off, it’s time to get them replaced. Tires presumably wear down because of friction between the rubber and the road. When tires rotate, this frictional force from the ground pushes the car forward while an equal and opposite force pushes back on the road. The faster you drive, the greater the frictional force is and this presumably causes more wear and tear. However, if you’re driving at a constant speed, you probably don’t burn as much rubber as you would by slamming on the breaks every five seconds. It’s not immediately clear which of these effects causes greater erosion, but you’re generally advised to change tires roughly every 50,000 miles¹. If that’s the case for driving at 55 mph, then it’s easy to compute how long it will take before your tires have worn down by 0.75 inches:

time = (distance) / (speed)
= (50,000 miles) / (55 mph)
= 38 days.

It should take a little more than a month to wear off 0.75 inches of your tires by driving at 55 mph.

Thanks, Dad. Happy Father’s Day!

[1] You can also test it by sticking a penny head first in the tread. If you can see Lincoln’s head, it’s time to change tires.

A New Magnetic Train

2010-06-14T06:13:00.000-07:00

As I was driving the other day, I noticed the little dashboard compass needle spinning whenever the car turned. Perhaps there’s a more energy efficient way to travel with a compass. How fast could you go from Brooklyn to Manhattan by sitting on the point of a giant compass needle?

***********WARNING: Math Ahead***********

Imagine a giant compass needle whose pivot is centered halfway between Manhattan and Brooklyn. Initially the needle points south so that its tip lies in Brooklyn. The needle would have to be about 8.0 km (~5.0 mi.) long. It would need to carry people, so it might look somewhat like a subway car, making it at least 3 m (~9.8 ft) in diameter.¹

The torque, or rotational force, on a magnet is given by the equation,

torque = (magnetic moment) · (magnetic field) · sin(angle).

The angle in question is the angle between the magnetic moment and the magnetic field. This angle would change as the needle turns. On average, the sine of the angle would be about 0.63 as the needle passes from Brooklyn to Manhattan. The magnetic field would be the Earth’s magnetic field, which has a magnitude of roughly 5.0×10^-5 T.² The needle could be a large permanent magnet (like the ones on refrigerators), but these are not as strong as electromagnets and aren’t able to flip the north and south poles like an electromagnet can³. An electromagnet can also be turned on and off, so you could control when our magnetic transporter was about to leave the station. We can make an electromagnet by wrapping tons of wire around the needle in a helical fashion and then applying a current through it. The magnetic moment for a cylindrical coil is given by,

magnetic moment = (number of coil turns) · (current) · (area across the cylinder).

From the diameter above, the area would be about 10 m². If the wire is 1.0 mm thick and there are 10 rows wrapped around each other, then the 8.0 km needle would have about 8 million turns in it. I’ll assume the current in the coils is 1.0 A.⁴

The time it takes the needle to rotate to Manhattan is given by the equation,

time = [2 π · (moment of inertia) / torque]^1/2.

The moment of inertia of the needle can be computed using a formula found here. Plugging in the values for our needle, we can calculate the moment of inertia to be about 3.4×10¹² kg·m².⁵

By combining equations and plugging all the numbers in from about, we find that it would take about 1.1 days to get from Brooklyn to Manhattan traveling by giant compass needle.

**************************************************

If you made a giant compass needle to harness the Earth’s magnetic field and tried to use this to travel, it would take you a day to get from Brooklyn to Manhattan. Since the subway costs about $2.00 and takes less than an hour, building a giant compass is probably not an efficient use of time or money.

There’s another interesting dilemma in this problem. Nothing in life is free, so you’d suspect that even our poorly functioning magnetic transporter must also sap energy from somewhere. Much like oil, it’s possible the source of Earth’s magnetic field would also eventually run out if we made lots of giant magnetic transporters. How long do you think it would take to sap away the Earth’s magnetic field?

[1] It would also be highly magnetic, so you couldn’t bring your credit cards.
[2] The “T” stands for “Tesla,” a unit of magnetic field.
[3] If you couldn’t flip the poles you’d never be able to get back to Brooklyn in the same way you left.
[4] I don’t have a good reason for assuming this value of current other than the fact that one of my intro physics books uses this value in a problem involving a copper wire. It would be interesting to see how this problem would work with superconducting wires, but you’d have to cool them down to very low temperatures, so the passengers might become a wee bit cranky and hypothermic.
[5] You can estimate this by assuming the needle is a 1.0 ft thick cylindrical shell made of iron.

Special Guest: Maryellen Hooper

2010-06-01T19:22:00.000-07:00

Today’s question comes from the very funny comedian Maryellen Hooper. Ms. Hooper has appeared on The Tonight Show with Jay Leno and in her own Comedy Central special, Lounge Lizards. In 1998, she was awarded “Female Comedian of the Year” at the American Comedy Awards. She writes,

“I've decided to ask a question on behalf of my son, Nate and his best friend, Noah. They're 5 & 6 years old…How many Legos would it take to build a ladder to the Moon?”

***********WARNING: Math Ahead***********

I’ll assume we’re using the standard 8-peg Lego brick. A single Lego is 9.6 mm tall, 32 mm long, and 16 mm wide¹. By multiplying these together, we can calculate the Lego brick’s volume to be about 4.9 cm³. To find the total number of Legos needed, we have to estimate the dimensions of the ladder.

The Moon is about 380,000 km away from the Earth. I’ll assume the sides and rungs of the ladder are both 6.0 cm wide and 6.0 cm thick. I’ll also assume the rungs are 60 cm (~2.0 ft) long and that adjacent rungs are separated by 30 cm (~ 1.0 ft). With this rung separation, there will be about 1.3×10⁹ rungs. From this we can compute the total volume of the ladder.

volume = (# of sides) · (volume of the sides) + (# of rungs) · (volume of the rungs)

= 2 · (6.0 cm · 6.0 cm · 380,000 km) + 1.3×10⁹ · (6.0 cm · 6.0 cm · 60 cm)

= 5.5×10⁹ m³.

To find the number of Legos we’d need to build the ladder, we just need to divide the volume of a single Lego into volume of the ladder,

# of Legos = (ladder volume) / (Lego volume)

= (5.5×10⁹ m³) / (4.9 cm³ per Lego)

= 1.1×10¹⁵ Legos

**************************************************

A Lego ladder to the moon would require 1.1 quadrillion Legos!² Using prices from Lego’s website, this would cost about $51 trillion dollars. At his richest, Bill Gates was still 500 times too poor to afford this. Thanks for the great question, Maryellen!

You can find Maryellen’s tour dates and clips of her doing standup on her website. You can also buy her comedy CDs Fixer Upper and Dignity Under Duress.

[1] Apparently, the good people at Lego are too good for the metric system, so they actually made up a “Lego Unit”. One Lego Unit is equal to 1.6 mm.

[2] Yes, “quadrillion” is a real number.

Human Plants

2010-05-31T10:52:00.000-07:00

My favorite Far Side cartoon depicts a small child hiding in the back of a beehive. The beehive has a rock-sized hole through its wall. In the foreground, a bee detective tries to figure out how a rock could have been thrown from the inside. The cartoon itself is not that funny, but there’s a great caption.

Artist: G. Larson
Medium: Ink on paper.
Title: It was late, and I was tired.

After rereading the estimation below, I believe the same title applies…

How close to the Sun would you need to be to become a plant? I should probably elaborate. There have been many great advances in biotechnology. If scientists can grow ear-shaped cartilage on the back of a mouse, why not grow a food-making plant inside a human? If this worked, it would stop world hunger. We’d be mobile symbiotes walking around producing our own (very) local food using only the green energy of the Sun. But is there enough energy from the Sun to do this. From a conservation of energy standpoint, how close to the Sun would we need to be to absorb all the energy we need?

***********WARNING: Math Ahead***********

If the FDA is to be trusted, the average person should intake about 2000 Calories per day, which is a total power consumption of about 100 W. The solar flux reaching the Earth’s surface is 1370 W/m². The maximum amount of body area that can absorb sunlight at any given time is about 1.0 m², meaning that on Earth we’d be able to absorb 1370 W of power. If we could convert 100% of that power into useable food energy, we would have about 13.7 times more energy than we would need. But how far can we get from the Sun and still meet our energy needs?

To compute just how far away from the Sun we could go and still absorb enough energy to survive, we need to know how the density of solar energy decreases as you go further out. The energy density decreases as the inverse radius squared (~1/r²). This is easy to see if you remember that the total amount of solar energy available from the Sun is the same at all distances but the area over which that energy spreads out grows as the radius squared. From this we can construct an equation to compute the farthest distance we plant-a-noids could go away from the Sun,

power needed = (solar flux at Earth) · (body area) · (Earth-Sun distance)² / (max distance)²

Solving for the maximum distance, we get,

max distance = [(solar flux at Earth) · (body area) · (Earth-Sun distance)² / (power needed)]^1/2
= [(1370 W/m²) · (1.0 m²) · (1.0 A.U.)² / (100 W)]^1/2
= 5.5×10⁸ km.

**************************************************

The furthest a human-plant hybrid could go from the Sun would be about 500 million kilometers1. That’s somewhere between Mars and Jupiter.

“Wait a minute,” you say. “You assumed the plants were 100% efficient. Shouldn’t you assume a more realistic efficiency?” Really? You were fine with genetically engineering some sort of human-shaped lichen, but thermodynamic efficiency ruffles your feathers? Carnot would be so proud. It’s absolutely true the efficiency matters here. In fact, you might be able to tell from my phrasing of the question that I originally thought Plant Man would need to be a lot closer to the Sun. If you were only 20% efficient at converting energy to food, you would need to be closer to the Sun than the Earth is to produce enough energy. In real plants, only a small percentage of the energy absorbed goes into creating food. My number is only an upper bound. If you want to impress me, see if you can calculate how close to the Sun you’d actually after to be.

[1] In hindsight, Plant Man reminds me a lot of Nuclear Man from Super Man IV: The Quest for Peace. Spoiler alert: This is the worst Superman movie ever made.

Up

2010-05-28T16:45:00.000-07:00

You can learn a lot from Pixar’s film Up. For example, there are apparently some circumstances when it’s OK for a creepy old man to make off to South America with an adolescent boy. But what physics did we learn from Up? How many balloons would it take to lift the old man’s house?

In Reinforced Helium Balloons, I calculated how much helium you’d need to lift a concrete balloon. The physics is the same here. The helium in the balloons is lighter than air, so there’s a “buoyant” force that pushes the balloons up. Buoyancy arises because the gravitational force from the Earth pulls more strongly on heavier objects, so the heavier air will be pulled closer to the Earth forcing the lighter helium to be squeezed upwards. This buoyant force can be approximated by the equation¹,

force = [(density of air) – (density of helium)] · (volume) · (gravitational acceleration).

In order for the house to float, this force must be at least as large as the gravitational force on the house, which is given by

force = (mass of the house) · (gravitational acceleration).

By setting the buoyant force equal to the gravitational force, we can calculate the total volume of helium needed to float the house. Assuming the house weighs about 300,000 kg,² we can compute the total volume of air needed,

volume = (mass of the house) / [(density of air) – (density of helium)]

= (300,000 kg) / [(1.2 kg/m³) – (0.17 kg/m³)]

= 300,000 m³.

If each balloon took up a cubic foot of space, it would require 11 million balloons to lift the old man’s house. That means you’d need a cluster of balloons with a diameter of about 250 balloon widths. Judging from this photo, it seems Pixar had way too few balloons, but in principle this idea could work. For example, just this past week Jonathan Trappe set a world record by becoming the first cluster-balloonist to cross the English Channel. For those that are skeptical, check out this article in Wired.

[1] In principle, I should also include the mass of the balloons here since heavy balloons will be more difficult to lift, but balloons are pretty light so I’m going to neglect their mass.

[2] To get this number, I assumed a cubic two-story house with dimensions 40 ft by 40 ft by 40 ft. I assumed the walls and floors are 1.0 ft thick and that the house was made of some combination of wood and brick that had an average density of 1000 kg/m³.

Hamster-Powered Mansions

2010-05-20T17:08:00.000-07:00

It’s about time for another estimation contest. The rules are the same as before…

Rules: You can win a free copy of How Many Licks? To enter, estimate an answer to the question below and send it to “aaron at aaronsantos period com.” If your answer is closest to my estimate¹, I'll mail you a free, signed copy of How Many Licks? To be eligible, you must submit your entry on or before July 15, 2010. Don't worry; I won't spam you or share your email with any third parties.

Question: Rotating a turbine is common way to generate electrical energy. You can find turbines in wind farms, the Hoover Dam, and bicycles. While watching reruns of Family Guy, I came up with this question: How many buff hamsters would it take to completely power a mansion?

Special Guest: Sanjoy Mahajan

2010-05-19T15:51:00.001-07:00

We here at Diary of Numbers love all things estimation. For this reason, we consider a special honor that today’s question comes from Sanjoy Mahajan. Dr. Mahajan is the Associate Director at the Teaching and Learning Laboratory at MIT and is also the author of Street-Fighting Mathematics, The Art of Educated Guessing and Opportunistic Problem Solving. Dr. Mahajan would like to know,

“How much oil does the US import (in barrels/year)?”

Note: This should be an easy answer to look up, but as Dr. Mahajan notes, it is “an interesting question to find methods that don't require looking up anything.” See if you can solve this without looking up anything.

Answer: Per Dr. Mahajan’s request, I did not look up any figures for this calculation. (I did, however, look up a conversion factor from “barrels” to “gallons.”) To simplify the calculation, I assumed that the volume of oil imported is roughly equivalent to volume of gasoline used by cars. Oil is used in applications aside of cars, but not all gasoline comes from imported oil, so I'll assume these two effects cancel each other. I’ll further assume the average American fills up on gas about 2 times per week and he pays about $25 per fill up. Since gas is about $3 per gallon, we can compute the total amount oil each person consumes each week,

oil consumed = (# of fill ups) · (cost per fill up) / (cost per gallon)

= (2 fill ups per week per person) · ($25 per fill up) / ($3 per gallon)

= 17 gallons per week per person.

That’s about 0.40 barrels per week for each American. There are about 300 million Americans. From this, we can estimate the total number of barrels imported each year,

barrels per year = (barrels per week per person) · (# of people) · (# of weeks per year)

= (0.40 barrels per week per person) · (3.0×10⁶ people) · (52 weeks per year)

= 6.2×10⁹ barrels imported per year.

That’s an estimate of 6.2 billion barrels per year. According to the U.S. Energy Information Administration, the U.S. imports 9,783,000 barrels/day, which is equivalent to 3.6 billion gallons per year. Our estimate is off by less than a factor of 2. This is partly due to luck, but it’s also due to the power of the Fermi estimation. Thanks for a great question, Dr. Mahajan.

You can order a copy of Dr. Mahajan’s Street-Fighting Mathematics on Amazon.

Licking Experiment

2010-05-18T07:55:00.000-07:00

How Many Licks? According to this, I was within an order of magnitude.

Skyscraper Farm

2010-05-12T17:41:00.000-07:00

When I was in Chicago a few months back, the Museum of Science and Industry had an exhibit describing one man’s idea to convert skyscrapers into farms that could supply big cities with fresh locally grown produce. If you converted the Empire State Building into a giant greenhouse, what percentage of New York’s population could you feed on a daily basis?

This clearly depends on what types of plants are being grown. Some foods are denser in calories than others. If all you’re growing is celery, you won’t be able to feed as many people as you could by growing potatoes. A potato plant might produce 10 potatoes each year while taking up 0.25 m² of area. The Empire State Building has 102 stories and a total floor area of 257,211 m². From this we can compute the total number of potatoes that would be produced each year by our skyscraper farm,

# of potatoes = (total area) / (area per potato)

= (257,211 m²) / (0.25 m²)

= 1.0×10⁶ potatoes per year.

Given that the population of New York City is about 8 million people, our result already doesn’t look very promising unless New Yorkers have figured out a way to survive one octant of a potato a year. Assuming New Yorkers have the bodily functions of typical humans, they’ll each consume about 2000 Calories per day. If each potato has 300 Calories¹, the farm will produce 3.0×10⁸ Calories per year or equivalently 8.2×10⁵ Calories per day. From this, we can compute the fraction of New York’s population we can feed,

fraction = (Cal. per day produced) / [ (Cal. per day per person consumed) · (# people) ]

= (8.2×10⁵ Cal. per day) / [ (2000 Cal. per day per person) · (8.0×10⁶ people) ]

= 0.0051%

Note, that’s not a fraction: it’s a percent. It’s 0.0051 percent. It would take 20,000 Empire State Building-sized skyscraper farms to make enough produce to feed New Yorkers for a year. That’s a far cry from the estimated 150 30-story building that would be needed according to the article referenced. Even if by some miracle of genetic engineering, we could get each plant to produce 10 potatoes a day rather than a year, we’d still need 50 Empire State Building-sized skyscraper farms to feed the population of New York.

This is one of those calculations I desperately wanted to work out. I love the thought of green energy-efficient buildings producing fresh food for city folk. I’ve checked my math twice, but can’t find a mistake. The fact that Zachary and ntm get similar results convinces me that the estimate is at least close. Even being very generous, I just don’t see how the numbers can work out it.

[1] That’s assuming large potatoes.

My second Book of Odds post

2010-05-12T17:27:00.000-07:00

Check out my second Book of Odds blog post discussing the oil spill:

"Let’s face it. Nothing ruins a perfect beach day more than a disgusting, oily, highly flammable seagull swooping in and trying to steal your French fries..."

Special Guest Question from Desiree Schell

2010-05-11T16:54:00.001-07:00

Today’s question comes from Desiree Schell, the host of the very entertaining and thought-provoking Skeptically Speaking. Desiree wants to know, “How many dice rolls in all the role-playing games in an average year?”

Judging from my fairly eclectic group of friends, I would estimate that about 1% of people play role-playing games fairly consistently¹. By “consistently”, I mean they play about once a week or about 50 times per year. I’m assuming the majority of players play consistently. Games will differ in how often a player rolls, but it’s not uncommon for a player to roll about 1000 times². From this, we can easily compute the number of rolls in an average year:

# of rolls = (fraction of players) · (# of people) · (rolls per player per game) · (# of games)

= (0.01 players/person) · (6.7×10⁹ people) · (1000 rolls/player/games) · (50 games)

= 3.4×10¹²rolls

That’s 3.4 trillion rolls. If you add up the gravitational potential energy from all these rolls, it’s about a third of the energy released by a nuclear bomb. Thanks for the question, Desiree!

You can hear Desiree on Skeptically Speaking Fridays at 6pm MST, or you can follow her and the show on Twitter at @teh_skeptic and @SkepticalRadio, respectively.

[1] Some might quibble with my 1% assumption. I would argue it’s a good order of magnitude estimate. If I chose 10 people at random, there’s a good chance none of them are into role-playing, so the actual number is almost certainly less than 10%. However, if I go through 1000 people at random, there’s almost certainly going to be one person into role-playing, so it’s probably greater than 0.1%.

[2] I cheated a little here. This number seemed much too large, but I consulted my friend Josh who is more into gaming than I am, and he assured me that it’s a good estimate. Thinking back to my days playing Risk (where you roll multiple dice for hours), I figured it’s good to within an order of magnitude.

If All Men Were Created Equal

2010-05-10T16:08:00.000-07:00

The United States Declaration of Independence states, "We hold these truths to be self-evident, that all men are created equal...." This statement is not exactly self-evident. One might even say it’s demonstrably false. Even leaving aside an individual’s genetic variability, it doesn’t take a trust fund baby to see that we’re not all born on the same economic and social footing. If we were all economically equal, how much land would we own?

According to Wikipedia,

“At 3.79 million square miles (9.83 million km²) and with over 309 million people, the United States is the third or fourth largest country by total area, and the third largest both by land area and population.”

Using this data, it’s easy to calculate the area per person,

( area ) / ( # of people )

= (3.79 million square miles) / ( 309 million people )

= 0.012 square miles.

That’s about 7.8 acres for each person. Well played, Zack. For extra points, see if you can calculate the rate at which it’s shrinking due to population growth.

Mother's Day, Red Sox, and Jordan's Furniture

2010-05-09T09:21:00.000-07:00

In honor of Mother’s Day, I’d like to do a problem suggested by my mom. She’s a huge Red Sox fan, and she’s curious about the Jordan’s Furniture promotion going on at Fenway Park. According to Boston.com,

“customers who buy anything at one of the regional chain's stores from April 1 through May 2 won't have to pay for their purchase if a Red Sox player hits the image of a Jordan's Furniture baseball on the company sign at Fenway Park between July 15 and the end of the season Oct. 3”

Mom would like to know, “What are the odds that a Red Sox player will hit the Jordan’s sign?"

There are 35 Red Sox home games between July 15 and the end of the season. As of today, the Sox have hit 42 homeruns through 31 games, which suggests they’d hit about 47 through 35 games.

If we assume homeruns have an equal probability of being hit to left, center, and right field, then we can calculate the fraction of homeruns hit towards the Jordan’s sign. Since the foul lines are at right angles to each other, homeruns can only be hit in one quadrant of a circle. Any other ball that left the park would be foul. A quadrant of a circle has an angle of Pi/2~1.57 radians or 90°. The Jordan’s sign is 12 ft long and at 421 ft from home plate, so it spans and angle of about 0.0284 radians. From this we can compute the probability that a homerun is hit at the correct angle,

probability of correct angle = (angular width of sign) / (total angular width)

= (0.0284 radians) / (1.57 radians)

= 0.018.

This means a homerun has about a 1.8% chance of being hit in the direction of the Jordan’s sign. However, this doesn’t mean it will be hit at the correct height. It’s difficult to say exactly what fraction will be hit at the right height, but as a rough estimate, I’ll assume one of every five homeruns will be at the right height to hit the sign. This means 0.4% of homeruns will hit the sign, or, equivalently, 99.6% of homeruns will not hit the sign. If there are 47 homeruns, the probability that none of them will hit the sign is

(0.996)⁴⁷ = 0.83,

or, about 83%. This means there’s about a 17% chance that one of the homeruns will hit the sign.

Happy Mother’s Day, Mom!

Special Guest Dan Meyer

2010-05-06T17:38:00.000-07:00

Dan Meyer is a blogger, tweeter, and teacher. He recently gave a very nice TED talk on math education. Dan writes,

“I'm very curious how many individual characters I'd find in the Sunday issue of the New York Times.”

I’m assuming Dan means “characters” as in letters and punctuation rather than “characters” as in Snoopy, Marmaduke, and Sarah Palin. If so, then the calculation is straightforward. The Sunday Times has about 10 different sections (Arts, Book Review, Sports, etc.) Each section has about 20 pages. Using today’s front page as a guide, we can see that there are about 6 columns per page. Each line has about 30 characters and each column has about 90 lines. Multiplying these together we get,

(10 sections) · (20 pages/section) · (6 columns/page)

· (90 lines/column) · (30 characters per line)

= 3.2×10⁶characters.

There are about 3 million characters in the Sunday issue of the New York Times. Thanks for the question, Dan!

Elevator of Death

2010-05-04T17:08:00.001-07:00

Many people have a fear of elevators because they’re worried the cable might break. Most elevators have safety mechanisms that make riding extremely safe, but what if it did go into free fall. What is the terminal velocity of an elevator in an elevator shaft?

Let’s consider what’s going on physically. What happens if the elevator were flush against the walls of the elevator shaft so that no air from underneath could escape? When you first sever the cables, the elevator begins to fall. As it falls, it compresses the air underneath and this air starts building up pressure. The pressure eventually grows large enough to counteract the gravitational force pulling the elevator down. At this point, the elevator stops its descent and rests on a cushion of air. This will happen when

P = m g / A,

where P is the difference between the air pressure above and below the elevator, m = 2000 kg is the mass of the (maximum occupancy) elevator, g = 9.8 m/s² is the acceleration of gravity, and A = 4.0 m² is the cross-sectional area of the elevator shaft.

Now imagine there’s a small area A’ = 0.5 m² on the sides of the elevator shaft that the air underneath can leak through. This area is pretty small so only a small amount of air can pass through it at any given time. The elevator still falls until the pressure counteracts the gravitational force, but this time the elevator is falling with a constant velocity. The rate at which some volume of air passes through the leak must be equal to the rate at which the elevator’s volume sinks down. Written mathematically,

A · v_elevator = A’ · v_air

where v_elevator is the terminal velocity of the elevator and v_air is the speed at which air passes through the leak.

Air will flow from the bottom to the top because of the pressure difference P. Using Bernoulli’s equation, we can estimate the air speed in the leak,

P = ρ · v_air² / 2,

or, equivalently,

v_air = ( 2 · P / ρ )^1/2,

where ρ = 1.2 kg/m³ is the density of air. Using the equations above, we can solve algebraically for the terminal velocity of the elevator,

v_elevator = [ 2 · m · g / (A · ρ) ]^1/2 (A’ / A)

= {2 · (2000 kg) · (9.8 m/s²) / [(4.0 m²) · ( 1.2 kg/m³)]}^1/2 (0.5 m²) / (4.0 m²)

= 11 m/s.

That’s about 25 mph. In a car crash, that’s enough to do serious damage, but in many cases it isn’t fatal. This makes sense. According to the Wikipedia entry for elevator, “in 1945 when a B-25 bomber struck the Empire State Building in fog, severing the cables of an elevator cab, which fell from the 75th floor all the way to the bottom of the building, seriously injuring (though not killing) the sole occupant - the female elevator operator.”

We Have a Winner!

2010-05-04T15:31:00.001-07:00

We have a winner for our “fish out of water” estimation contest. The question: How far would the oceans sink if we took all the fish out?

According to Wikipedia, the biomass of marine fish is 800-2000 million tonnes, so I’ll assume 1500 million tonnes. Biomass is the total amount of mass a set of organisms has. Fish (or at least live fish) don’t float, nor do they sink. This suggests their density is about equal to that of water 1.0 g/cm³. From these numbers we can compute the total volume of fish,

volume = (mass) / (density)

= (1500 million tonnes) / (1.0 g/cm³)

= 1.5×10⁹ m³.

The ocean comprises about 71% or 3.6×10¹⁴ m² of the Earth’s surface. From this, we can compute how much the oceans would sink by taking all the fish out,

depth sunk = (total volume) / (area of the oceans)

= (1.5×10⁹ m³) / (3.6×10¹⁴ m²)

= 4.2 microns.

That’s about 240 times smaller than a millimeter. This may seem small, but if you’ve ever been fishing, you know that fish are much more rare than you’d expect. Most of the time you look over the side of the boat you’ll see nothing.

Congratulations to our winner, Nathan Moore!

Bermuda Triangle

2010-04-29T18:07:00.000-07:00

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 m³per 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

2010-04-29T17:57:00.001-07:00

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

2010-04-29T17:38:00.001-07:00

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.