We have a winner for the End of Summer Contest! The question: how many people get stuck at the top of the Cedar Park roller coaster each year?

A stuck roller coaster is likely held in place by friction. From the looks of it I would guess a 2 pound weight added to the front of a stuck roller coaster wouldn't do much to budge it, but a 200 pound man hanging off the front seems like more than enough to get it rolling again. As such, I'll assume a 20 pound force is needed start a stopped roller coaster. This force is equivalent to the frictional force holding it in place.

There's a narrow range of heights near the peak at which the cars could stop and not accelerate because friction would be too strong. It looks like the radius of curvature for the track is about 10 meters give or take. The range of heights over which the carts would get stuck is only a small fraction of this. I'll guess 5 centimeters since it's likely bigger than 5 millimeter and smaller than half a meter.

The initial velocity of the cart will correlate with how far it makes it up the hill. Carts with speeds greater than some cutoff velocity will all make it over the hill. Carts with speeds less than some other cutoff velocity will all roll back down the hill. Carts with speeds between these two cutoffs will get stuck. Using conservation of energy (perhaps a dubious move since we're relying on friction to stop us) we can relate the initial velocity to the height of the cart will climb up the hill:

m v^{2} / 2 = m g h

Here, m is the mass of the carts, g is the acceleration of gravity, and h is the height climbed. The roller coaster might be 100 meters tall. Carts stopping at heights anywhere between 99.95 meters and 100 meters will get stuck. Using the energy equation, we can solve for the range of initial velocities for which this will happen. Leaving out the messy math, you'll find that this occurs for velocities between 70.009 mph and 70.025 mph.

As you can see, there's a narrow range of velocities over which the roller coaster will get stuck. What causes fluctuations in the initial velocity? If we assume each ride gets pushed with the same impulse, then the only fluctuating variable would be the mass of the riders. A roller coaster full of fatties is less likely to make to make it to the top. I'll assume the average person weighs 170 pounds with a standard deviation of about 40 pounds. If there are 50 riders, the standard deviation in weight would be about 3% of the mean total weight. I'll assume the deviation in the initial velocity is the same as that of the weight, i.e. 3% of the mean.^{1}

We still need to find the mean velocity. For the sake of keeping their customers happy, I'm going to assume that roll backs happen only 5% percent of the time. We can assume a normal distribution to find the mean for which this occurs. Skipping the messy details once more, I get about 73.6 mph.

Using this mean and standard deviation, we can then find the fraction of times a cart will get stuck at the top of the hill. Again assuming a normal distribution, I get about 0.1% of the time, which seems like a fairly reasonable estimate.

If the roller coaster operates 8 hours per day and each ride take 2 minutes, you'd have roughly 240 rides per day. Over a 100 day season, you'd have 24,000 rides. Roughly 24 of these would get stuck at the top.

[1] I'm assuming the carts are of negligible weight, which is probably stupid. I suspect it won't throw the number off too much as long as it's comparable to the weight of people.

10 million pounds of maple syrup! Am I the only one who thinks this number can't be right? If the thieves loaded up their truck with 3 tons on each trip, they would need to make 1700 trips!

On an unrelated note, contest winners will be chosen later this week.

I'd been at my new Gustavus office only a couple hours when someone stopped by and tacked my nameplate up. I could get used to this. Good schools really know how to treat their faculty with respect.

Today's question comes from physicist Brian Wecht. Brian is co-founder of Story Collider, a podcast that shares people's personal stories about how science has affected their lives. Brian is also a member of Ninja Sex Party.^{1}

Here's a picture of Brian in ninja form:

Brian's a very good ninja.

Brian asks,

What is the weight of all the facial hair grown by the world population of men on a given day? If you concentrated all that hair into one curly, villainous moustache, how long would that moustache be?

If I go a week without shaving, I'll have about half a centimeter of facial hair, which means my hair grows about 0.1 centimeters per day. Follicles are separated by about 1.0 millimeter, and they span an area of roughly 100 square inches. Assuming each hair has a thickness of 0.1 millimeters and the same density of water at 1.0 grams per cubic centimeter, I would grow about 60 milligrams of facial hair each day.

For simplicity, I'll consider myself to be a typical man. This is not necessarilly a safe assumption however, since, follically speaking, I'm much more Wolverine than Bieber. Still, it should be a safe assumption for an order of magnitude estimate. If that's the case and we assume 20 percent of the world population (~1.4 billion people) are facial-hair growing men, then the total weight of facial hair grown on a given day would be about 90 tons.

This brings us to the second part of Brian's question: How long of a villainous mustache can we make?

Ninja Brian will save us from this dastardly villain.

Mustaches take up much less area than beards. If you glue all the beard hair grown in one day to the ends of the mustache hair grown in one day, it would be as if the mustache grew about four times faster than normal, which amounts to roughly 0.4 millimeters of growth each day. Spread out over the 1.4 billion facial-hair-growing men, that would give a villainous mustache that's about 350 miles long. Snidely Whiplash would be proud.

When I was an undergrad, I had this delusional desire to become an actor. As a male thespian with no experience and mediocre talent, I quickly discovered there's one sure way to get decent theatrical roles: audition for male parts at all girl schools. And so, I auditioned for student productions at Wellesley College where I met a variety of wonderful people. One of said people is my buddy Christine. Christine was (and is) super cool, and not only because she used to get me free food at the Wellesley cafeteria. Christine was also cool because she was taking physics.^{1} I distinctly remember one conversation I had with Christine about her physics class:

Christine: I'm taking physics at Wellesley.

Me: You should really see if you can take it at MIT instead.

Christine: [trying to be nice] Um...my dad thinks they do a better job teaching physics at Wellesley.

Me: [offended] What the hell does your dad know about physics?

Christine: Well, he did get his doctorate from MIT, and he just won the Nobel Prize in physics.

Judging from a map, the Eastern Seaboard of the United States appears to be about 2000 miles long. By comparing with other coastlines, we can estimate the total length of coastlines in the world to be about 50 times this.^{2} I'll assume one-third of the world's coastlines are sandy beaches.

Assuming the Cape Cod beaches I grew up near are fairly typical, they might extend about 200 feet up from the water. The depth of sand varies quite a bit from place to place. I've been on beaches where you'll hit rock before finishing the moat around your sandcastle, but many beaches have sand that extends much deeper. I'll assume the sand extends 10 feet deep on average since the actual number is likely to lie between 1 foot and 100 feet. From this, we can calculate the total volume of sand on the beach to be about 10^{10} cubic meters.

One type of sand.

Like beach depths, sand grains too vary over a size range that's greater than one order of magnitude. I'll assume 0.3 mm for the width of a sand grain since even the largest sand grains are each only a few millimeters in size. This gives a total volume of 0.03 cubic millimeters. From this and the total beach volume above, we can estimate that there are 10^{20} grains of sand on all the beaches in the world.

Just a quick note on NASA's Curiosity Rover that landed on Mars earlier this week. Someone on Reddit described it as being like sinking "an interplanetary hole-in-one." A typical golf hole might be about 500 yards with a cup that's about 4 inches wide, meaning that the angle of the golfer's shot can't be off by more than 0.006 degrees. The Curiosity Rover landed on a planet with a diameter of 6800 kilometers located 1.669 astronomical units away, meaning that the angle of NASA's "shot" can't be off by more than 0.0015 degrees. It's a great comparison, but even as unlikely an event as a hole-in-one underestimates the precision that NASA engineers obtained.

I normally like to come up with my own questions for contests. However, every once in awhile I get a question that's so good, I have to use it. My former student Ben sent me this question the other day, and it falls into this category.

You know the rules. I post a Fermi question below. To enter, estimate an answer and send it to “aaron at aaronsantos period com.” If your answer is closest to mine, I'll mail you a free signed copy of Ballparking: Practical Math for Impractical Sports Questions.^{1} Second prize receives a signed copy of my other book, How Many Licks? Or How to Estimate Damn Near Anything. Submit your entry on or before August 31, 2012. Don't worry…I won't spam you or share your email with any third parties.

Here's what Ben asked: I was at Cedar Point (the
amusement park in Ohio) with Che and David the other day and we had an
idea for another estimation problem that I think is pretty cool so I
thought I would send it to you. There's a ride at Cedar Point called Top Thrill Dragster - it looks like this :

Basically you get launched, climb the tower, and then tumble back down.
For a better idea of the ride, here is a passenger's point of view:

Occasionally, though, this happens:

It's called a "rollback". As you might imagine, this means that once in a blue moon, this actually happens:

Question is, how many times? I actually do know the real life answer to this, although it could change any day.
Alright, fellow estimators! How many times does the roller coaster get stuck on top each year?

[1] 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!

We have a winner for the "Look Out!" contest. Here's my estimate:

In order for the camera to break, the puck must be slapped against the boards at the exact spot where the camera is. Pucks are dumped against the boards more than twice and fewer than 200 times per game, so a reasonable estimate is about 20 hits per game. The puck itself needs to land in a space the size of a camera lens. The size of the lens depends on whether we're talking about a regular camera or a video camera.

This digital camera is a fair bit smaller...

...than this professional video camera.

For the sake of the problem, I'll assume we're talking about regular digital camera since that is what I was looking at during the game. From the picture above, the lens appears to have a diameter of roughly 3 inches giving an area of around 10 square inches. The circumference of a hockey rink is about 500 feet with a height of about 12 feet giving area of 6000 cubic feet. Assuming the pucks are distributed uniformly over the boards (not necessarily a safe assumption), you'll have about a you'll have about a 1 in 100,000 chance that the puck hits the camera. Since the cameraman was only there about 20 percent of the time, that decreases the probability that any given hit will break a camera to 1 in 500,000. Since there are 20 pucks hit hard against the boards per game and a 1 in 500,000 chance that any given puck will hit a camera, there must be a probability of 1 in 25,000 that any given game will feature a broken camera.

At present, each team plays 82 games per year. If we consider that many of the early years in NHL history had teams playing fewer than 50 games, we can estimate that on average about 60 games are played each year. Half of these might have a camera man. That gives 30 games with a camera per team per year. On average, there've been about 15 teams each year throughout the 90 years of NHL history. From this we can conclude that 40,000 games had a camera that could have been broken.

Since there have been about 40,000 games with a camera and a 1 in 25,000 chance that a game with a camera will end a s a game with a broken camera, we can conclude that roughly 1.9 cameras have been broken. Since this is an order of magnitude estimate, any it's possible that anything up to 19 camera have been broken.

If you took all the living material (animal, plant, bacterial etc.) on the Earth, took it into space and made it into a ball:

1. What would be the gravitational pull on its surface?2. What existing solar system body would it be closest to gravitationally?3. What effect would it have on the gravitational pull of the Earth?

Brian has a boatload of cool books you should check out.

Let's take these one at a time.

1. Gravitational pull on its surface.
In order to solve this, it helps to know the total amount of biomass present on Earth. Biomass can mean different things, but here I mean the total amount of carbon that makes up a particular species. For example, the average human weighs about 50 kilograms (110 lbs) and there are seven billion of us worldwide, so our total mass would be about 3.5×10^{11} kilograms. Roughly 18% of the human body is carbon, so the total biomass would be roughly 6.3×10^{10} kilograms (70 million tons.) To answer Brian's question, we'll need to know the total biomass of all species on Earth.

I remember very little from high school biology, but I vaguely recall the food chain.

The food chain can be summarized fairly succinctly: "Big things eat little things."^{1} However, for this to be sustainable, the total biomass of the little things has to be much greater than the total biomass of the big things or else they would get eaten up too quickly. For example, land plants comprise 1000 times more biomass than land animals. Humans, being close to the top of the food chain,^{2} will compose a fairly insignificant percentage of the total biomass, so my estimation of the total human biomass will be of little use to us. Fortunately, Wikipedia provides a more convenient starting point:

"Apart from bacteria, the total live biomass on earth is about 560 billion tonnes C...."

Assuming other creatures have roughly the same percentage of carbon humans do, the total mass of all non-bacterial creatures would be roughly 3×10^{15} kilograms. How does this compare with the total mass of bacteria?

Wikipedia lists an estimate for the total number of individual bacteria as being roughly 5×10^{30}. Assuming a mass of 10^{-12} grams per bacteria, that would put the total bacterial mass at 5×10^{15 }kilograms, which brings the total mass of all living creatures to roughly 8×10^{15} kilograms (8 trillion tons.)

Most living species have a density close to water at around 1.0 gram per cubic centimeter. Molded into a sphere, our biomass ball would have a radius of about 13 kilometers. We can use Newton's law of gravity to find the gravitational pull at the surface of the ball,

g = G M / r^{2}.

Here, G =
6.67×10^{-11} N•m^{2}/kg^{2} is the gravitational constant, M =
8×10^{15} kg is the total biomass, and r = 13 km is the radius of the ball. Plugging in numbers, we find g = 3.2 mm/s^{2}. That's about 3000 times smaller than Earth's gravitational pull.

2. Closest (gravitational) body in the solar system.
Gravitationally, Pluto is the weakest planet (or "planet" according to some) in the solar system with g = 0.61 m/s^{2}, but even this is 200 times larger than the attraction that would be felt by our ball of biomass.

"I may not be a planet, but I'm still stronger than you, all of life."

Even moons are going to have a much stronger gravitational pull than our bio-ball. Gravitationally-speaking, the closest object to our little ball o' life would likely by an asteroid or a comet. For example, Halley's comet weighs about 10^{14 }kg where as asteroids can vary anywhere from 10^{11}-10^{21} kg.

What a ball of humanity might look like

3. Effect on Earth.
Unless it smashed into the Earth like the asteroid that took out the dinosaurs, our asteroid is going to have very little effect on the Earth. Even if it were touching the Earth, the gravitational pull would only increase by 0.03%. Apparently, life is only a very small part of the world!

Thanks for a great question, Brian! You can find out more about Gravity and Brian's other books on his website and follow him on Twitter.

[1] Having grown up in the former whaling capitol of the world, I am aware that there are notable exceptions to this rule.
[2] I suspect cannibalistic humans would technically rank slightly higher than non-cannibalistic humans.

At several points in my book characters travel to and from the south, either by barge on the Grand Canal or by horseback, and I've just very loosely estimated the time required by their journey. Could you do a more precise estimate of a journey from Beijing to Suzhou, taking into consideration possible routes and stops for the night?

...My other question concerns what it would have been like to live in Beijing at the time. Can you estimate what the approximate population would have been at that time, and what the population density would have been?

Google Maps lists the distance between Beijing and Suzhou as 1146 kilometers (712 miles). Driving nonstop from one to the other would take the better half of a day, about 15 hours, but her 18th century characters certainly weren't using cars. Let's consider the two modes of transportation Pauline lists.

1. Barge. I have to confess, I'm not all that familiar with barges, so I'm going to guess a bit. According to the Wikipedia entry for "water speed record," the fastest boat speed was about 42 kilometers per hours (26 mph) circa 1885. However, Pauline's story takes place about 100 years before this and her characters wouldn't have had access to propeller or steam technology. As such, I'm betting their barge moved appreciably slower. I'll guess 10 kilometers per hour. If the barge was much slower than this, walking would have been faster. Assuming they remained on the barge for the entire trip, it would take about a week to travel from Beijing to Suzhuo. This assumes the distance is a bit longer than the straight line distance since the Grand Canal is about 1700 kilometers long.

2. Horse. A horse's speed depends greatly on how much of a hurry it's in. According to the Wikipedia entry for "horse gait," a walking horse travels about 6.4 kilometers per hour (4.0 mph) where as a galloping horse can move fast as 48 kilometers per hour (30 mph). It's fair to assume the horse won't be sprinting at full tilt for a thousand kilometers in a row, so we should probably use its walking speed. Assuming 10 walking hours per day, a horse would travel about 77 kilometers each day and would complete the trek in about two weeks. This number is a bit misleading though because it assumes the horse is not carrying heavy cargo.

With a wagon, a horse can't travel nearly as fast.

If the horse pulls a wagon, it will travel appreciably less quickly. According to the Wikipedia entry for "Conestoga wagon," horses can travel about 25 kilometers (15 miles) each day, which would make the journey last about a month and a half.

Depending on the amount of time spent on the barge/horse and the amount of baggage brought along for the trip, the journey could take anywhere from one week to over a month. (Though this assumes they're not distracted by various love triangles and such.)

For the second question, I'll need to be careful. I tend to play fast and loose with numbers since I'm only going for order of magnitude accuracy, but I've gotten myself in trouble with populations before. As one reader pointed out in my Pinker post, population growth can be more accurately described by a geometric series:

Let's assume the fertile population in 1900 is "p". Let's also assume that a given generation basically dies out after 75 years. This is a realistic life expectancy. Then the generations of fertile (<25yr old) people looks as follows:

(1850) : p / m^2 (1875) : p / m (1900) : p (1925) : p * m (1950) : p * m * m (1975) : p * m^3 (2000) : p * m^4

If we follow his trend line to the year 1775, we get a population given by

population in 1775 = p / m^{7} + p / m^{6} + p / m^{5}.

Using my reader's values p = 0.782 and m = 1.565, we find that the world population in 1775 was roughly 170 million people.

As you can see from the chart above, the number is a bit off, but certainly within an order of magnitude. A more accurate number would be about 800 million people. At present, China is about 25% of the total world population. Assuming this percentage has remained roughly the same for the last few centuries, the population in China in the 18th century would be about 40 million people if you use my estimated number, though if you use the more accurate world population, you'll find it's closer to 200 million people.

If we want to find the population of Beijing specifically, we could assume the percentage of the Chinese population living there has remained fairly constant over the last few centuries. This is not necessarily a safe assumption because cities are much smaller than countries. The smaller size makes them more prone to fluctuations in population, but I'm hoping the end result won't be too far off. Beijing is currently home to 20 million people or roughly 1.5% of China's population. Using my 40 million people figure approximated above, we could estimate the Beijing population in the 1700s to be about 600,000 people. Using the more realistic population figure would raise the number to about 3 million people. According to the Wikipedia entry for "largest cities throughout history," Beijing grew from a population of 650,000 people in 1700 to 1.1 million in 1800, which would put my numbers very close to the actual value. It would seem that an 18th century Beijing would look like a 20th century Indianapolis, at least in terms of population.

Today's invited guest is Jon Haarr. Jon writes one of my favorite blogs, Toys and Bacon. Seriously, can there be a better combination than toys and bacon? I think not! But I digress.

Jon writes,

...I realized there's something about the
Transformers that always made me think. It's how they deal with weight,
density and size changes when they transform.
Soundwave is a transforming robot. In order to function as a spy for the
evil Decepticons, he transforms into a good old fashioned cassette player,
perfectly able to blend in among actual cassette players. The kicker is of
course that in his natural mode - as a robot - he's huge....

My questions are:
What kind of energy would go into these transformations?
Let's say that Soundwave does in fact weigh the same in robot mode as he
did when he was disguised as a cassette player. What would Soundwave's
structural integrity be in robot mode, and how would it affect his daily
work as an evil, giant, murdering machine?
And then, let us play with the idea that Soundwave is made of traditional
metal and that he maintains the weight from his natural, robot mode. How
would that affect his disguise mode, and what would happen if that cassette
player stood on someone's shelf?

Note the change in size between here...

...and here.

This is something that's bothered me about Transformers ever since I got Optimus Prime and Megatron toys for Christmas as a kid. In robot form, they're roughly the same size and would seem to be a fair match for each other in a fight. However, problems arise when they transform into disguised mode, after which Optimus becomes a "toy-sized" truck, whereas Megatron transforms into a life-sized handgun. Hasbro's never been a stickler for length scales,^{1} but this is a bit much. Even as a fairly stupid seven-year-old, I could tell by Toy Land standards Megatron was hideously deformed, and it was all I could do to convince the other toys not to mention it in front of him because he was probably pretty sensitive about it.

Where was I going with this? Ah, yes! Soundwave. Let's consider three cases:

1. Soundwave Maintains His Robot Weight. At first glance, this might seem like a fair assumption. After all, he's a robot first and foremost, so it makes sense that he'd keep his robot weight when disguised. However, as Jon pointed out to me, "It's obvious that he's not that heavy in his 'disguise', because we see
people pick him up." Just how heavy would he be? In the cartoons, he looks like he's at least as tall as a two-story building. Conservatively, this would put his height around 20 feet. Assuming he's made of iron with a density of 7.9 g/cm^{3} and a thickness and width of 4.0 feet and 5.0 feet, respectively, he would weight roughly 100 tons. Even if you had the entire Ukrainian weightlifting team at your disposal, you still couldn't pick him up, so you can kiss those IKEA shelves goodbye. To be fair, this isn't nearly the densest material known to man,^{2} but he's still going to have a hard time convincing anyone that he's just a plain ol' cassette player. This brings us to option #2....

Soundwave...one bad robot you don't wanna mess with.

2. Soundwave Maintains His Cassette Player Weight. Perhaps Shockwave is small-boned—er—alloyed. Just how dense would he be if this were the case? If I assume a cassette player has a mass of 2.0 pounds and use the dimensions above, then in robot form Shockwave would have a density of 80 grams per cubic meter. That's incredibly light. You might think that a light breeze would be enough to knock him over, but it's even worse that that. If he maintained his cassette player weight, he'd be less than half the density of helium, so you'd have to tie a string around him to make sure he doesn't go floating off into the stratosphere. The only way Soundwave is on remotely good scientific ground is if the Transformers discovered some novel unknown physics on Cybertron, which brings us to option #3...

But who said he's just an evil killing machine?

3. Soundwave Maintains a Constant Density While Transforming. Although this is what the cartoons seem to suggest, the change in size would represent a clear violation of mass conservation.

Or would it? I've dealt with the problem of spontaneous weight gain previously in The Hulk Revisited. In that problem, I hypothesized that Marvel's Hulk gains weight by adsorbing air molecules with the result that he would suck out the equivalent of all the air in a 1500 square foot apartment upon transfiguring, instantaneously killing everyone (friend and foe alike) in said apartment with the resulting vacuum. However, Jon's question is unique in that he specifically mentions energy. One could imagine that Soundwave, during his transformation from 1980's cassette player to soulless killing machine, converts some excess energy to mass. From above, we know roughly 100 tons of mass is unaccounted for during the transformation. Perhaps Soundwave converts this mass to some new form of energy via Einstein's mass-energy equivalence equation E = mc^{2}. Here, c = 3.0×10^{8} m/s is the speed of light. Using this equation, we find that the excess mass is equivalent to about 8.2×10^{21} J of energy. That's about 80 times the energy that the entire United States uses in a year! I suppose it's possible an evil villain like Shockwave has this much extra energy lying around, but if he does, I imagine he'd find a more terrifying use for it than turning himself into an object whose most destructive power consists of playing Chris de Burgh's "Lady in Red" at moderately high volumes.

Sorry for the lack of updates. I just finished moving to Minnesota and am still sans internet and personal belongings. I'll have some fun guest questions posted soon. In the meantime, don't forget to enter the Look Out contest for a chance to win a free copy of Ballparking. Also, check me out on the Skeptically Speaking podcast this Friday.

Monel metal is a very hard alloy of nickel, copper, and iron. It is extremely corrosion resistant and is excellent for many wet applications. However, if not isolated from other metals in salt water environments, it can cause corrosion. In 1915, a ship was built with a hull entirely of Monel, with the expectation of exceptional durability. However, the 215-foot-long, 34 foot wide Sea Call had to be scrapped after just six weeks of use. While its monel hull was fully intact, the steel frame of the ship deteriorated beyond use from electrolytic interaction with the monel in the salt water environment.

Question: How much of an electrical current did the corrosion produce? Could it have lit a light bulb? Many light bulbs?

If you've ever built a lemon battery, you know that two different metals can pass electric current between each other when connected by a salt bridge. This works because of a chemical reaction that occurs when the metals are exposed to the ions in the salt bridge. In the case of the lemon battery, the metals are usually a copper penny and a zinc-coated nail, while the salt bridge is the lemon.^{1} In high school chemistry, you may have learned that this type of interaction is known as an oxidation-reduction or redox reaction.

In principle, I could look up the oxidation-reduction potentials for Monel and other common metals that appear in the sea, and from this I could calculate the number light bulbs one could light up. However, I'm in the middle of packing and moving to a new state, so, in the spirit of order of magnitude estimates, I'm gonna wing this a bit.

At 215 feet long and 34 feet wide, we have a surface area of at least 680 m^{2}. A typical atom has a radius on the order of angstroms (~10^{-10} m), so there are roughly 10^{23} atoms directly in contact with the water. The corrosion must have been visibly noticeable, so I'm betting it ate through about a millimeter of the hull.^{2} This thickness is equivalent to roughly 10^{7} atoms giving a total of about 10^{30} oxidized/reduced atoms during a six week time span. If each atom gave up one electron, you would have a total current of

That's 44 kiloamperes of current. A normal 100 W bulb plugged into a standard 120V North American wall socket receives roughly 0.8 amperes of current. Distributed evenly, this could have lit about 55,000 light bulbs.

Thanks for a great question, Matt! You can find out more about Matthew Frederick on his website and blog. You can also check out the 101 Things I Learned series here.

[1] Fun Fact: Alessandro Volta, after whom the electrical unit the "volt" is named, is credited with inventing the first battery after allegedly using his tongue as the first salt bridge. I'm not sure how one makes a discovery like this. I can just imagine Volta saying in a Homer Simpson-esque voice, "Mmm...metal alloys. Ooooh."

[2] Note: This is a big assumption, which I'm not entirely sure is correct. The final answer can change dramatically depending on this number. For example, if the corrosion was only a micron deep, the final result will be about 55 bulbs. Feel free to play around with different thicknesses to see how many bulbs you could light.

EDIT: Matt just informed me that, "The steel frame corroded; the Monel was intact!" Doh!

It's been over a week since I did my Reddit AMA, and I want to say thank you to every one who asked questions. As evidenced by the giant spike in page views, it had a bit of an effect on my blog viewership.

So thank you to everyone who asked questions (even I didn't get a chance to answer yours.)

There were, however, a few questions that I said I'd get back to people on, and I wanted to address those.

(1) One of the questions I got and couldn't answer at the time was the following:

A physicist! I've been waiting for one. I've been wondering this for a while, but can't come up with a solid answer.

If I was in space and I attached an LED light to one corner of a cube, is it possible for me to push/toss/throw/rotate the cube in such a way along a linear path that the LED light's pattern would never repeat itself (aka, there would never be a period)?

Sadly, I still don't have a good answer to this one. The best I can do for now is say that another Redditor provided me with this link to his work on the chaotic motion of rigid bodies. It seems pretty cool, but I'm too bogged down in summer research (and moving to a new state!), so I can't devote a whole lot of time to this one.

(2) Yum_Krill (an appropriate username if I ever heard one) asked,

How large would a creature have to be to make a sound like The Bloop? The largest creature on earth, the blue whale, would not be able to.

I had to look up the Bloop.^{1} Apparently, it's some weird animal-like deep ocean sound that is both loud and low-frequency:

According to at least one source, the Bloop reaches about 246 decibels. For context, normal human conversation occurs around 60-70 decibels, the loudest rock music reaches about 150 decibels, and blue whales can reach about 188 decibels.

Decibels are a logarithmic scale, which means that a human conversation at 60 decibels would have a power ratio of 10^{6} and a blue whale singing at 190 decibels would have a power ratio of 10^{19}. The Bloop would have a power ratio of about 10^{25}, or roughly a million times bigger than a blue whale.

Whatever the Bloop is, it presumable has some object that mechanically oscillates to produce sound. The amplitude of this oscillation grows as the square root of the power. Since the Bloop's power is 10^{6 }times bigger than the blue whale's, its amplitude of oscillation would be about 1000 times larger than a blue whale's. If the rest of the Bloop remains proportional in size, it should be about 30 kilometers long and weigh close to 180 billion metric tons.

(3) In response to a message about a flying Pegasus, I replied

We need to consider two things here: wing area and wing flapping rate. I did a similar problem for Mothra's wingspan. Horses weigh about 500 kg, which gives a downward gravitational force of about 5000 N. If you assume her wings flap 2 meters down and do so once every second, she'd need winds that were about 1000 m2 in area. A 2 meter wide wing would need to be about 5 football fields long.

After posting the problem, I got a message that read as follows:

You say Horses couldn't possibly fly unless they had unrealistic, impossibly long wings (up to 5 football fields long), then how come Pterosaurs weighted up to 400 pounds could fly at blazing fast speeds with reasonably sized wings?

I understand the difference in weight, but what was the main difference here to allow this other heavy animal to fly easily, is it wing-flapping speed?

I was unaware of overweight pterosaurs when I was doing the problem, but I still had reason to be concerned with my result because of the earlier problem I did on Mothra. However, in an effort to keep up with the questions, I was working pretty fast and (stupidly) had my normal skeptical filter on silent. That said, I think I know what wrong here.

A horse weighs about 500 kilograms, which means it feels a downward gravitational force of about F = 5000 Newtons. You need at least this much force to have liftoff.

To generate this force, Pegasus pushes air down. The mass of the air pushed depends of the density of the air (ρ =1.2 kg/m^{3}) and the volume V of air swept out with each downward thrust:

m = ρ V = 2 ρ L w h

where L is the length of the wing, w is the width of the wing, and h is the vertical height the wing is thrusted downward. The factor of 2 comes in because there are two wings.

The factors affecting thrusting force are the frequency of flapping f (in Hertz), the distance h the wings are thrust down (in meters), and the mass m of the air pushed (in kilograms). Using dimensional analysis, we can estimate the magnitude of the upward force as

F = 2 ρ L w h^{2}f ^{2}.

This can be solved for the length,

L = F / (2 ρ wh^{2}f ^{2}).

If we use, as I did previously, the values F = 5000 N, ρ =1.2 kg/m^{3}, w = 2 m, h = 2 m, and f = 1 Hz, you find that the wings must must be 260 meters, or roughly two and half football fields long. I had forgotten the two, but that's not the only problem.

There's two other potential problems I foresee with this. First, dimensional analysis is great for figuring out how certain physical quantities scale with other physical quantities, but if you're looking for actual numbers you can be off by quite a bit. Putting that aside, there's still a second problem. Whenever you do an estimation, you should go back and check to see if your original assumptions make sense. Now if your wings are five footballs long, it doesn't make sense that they're thrusting only two meters downward. In the video below, they appear to be thrusting about 4 meters, which reduce the result to roughly a 65 meter wingspan.

Moreover, if I up Pegasus's wing flapping rate to 3 flaps per second, we further reduce the wingspan to a much more believable 7 meters. I apologize if I mislead any Pegasus fans out there.

I'm not sure what this says about pterosaurs. I know very little about them except that they varied in size from 10 inches (~25 centimeters) to over 40 feet (~12 meters). Just from physical considerations, I would have guessed the largest of these were more like ostriches (i.e. flightless), but maybe there's some evidence to the contrary.

There's an important take home point here. Whenever you reach a conclusion, whether it be for an estimation or a general conclusion about life, you always need to go back and ask what assumptions you made to get there. You've got to make sure the assumptions you made still make sense by the time you get to the end!

Thanks once again, Reddit. If there are more questions you want answered, message me. I can't guarantee a response (I'm still reading some from last week), but I can guarantee that if I see your question and it intrigues me, I'll answer it on the blog and give you credit.

How much do blue-collar criminals steal per year compared to white-collar criminals? How much do blue-collar criminals make per year on average? How much money do we spend on the justice system in totality? How much economic gain do we get from it? In other words, how much are we spending on the system and how much crime in economic terms are we preventing?

...was defined by sociologist Edwin Sutherland in 1939 as "a crime committed by a person of respectability and high social status in the course of his occupation."

In contrast, blue-collar crime is committed by individuals of lower social status. Using these definitions, there's not a sharp cutoff between blue- and white-collar crimes. Even if these terms were well-defined, getting precise statistics on how many of these crimes are committed is difficult since many of them go unreported. For this reason, I'll need to make a lot of assumptions in this problem.

According to at least one source, there are roughly 2.3 million people presently held in American prisons. If we use data from the Bureau of Prisons, we find the relative percentage of prisoners incarcerated for various crimes:

As mentioned above, there's some arbitrariness surrounding the definitions of blue- and white-collar crimes. Given the context of Dr. Ariely's question, I'm only going to consider crimes that have an inherent financial component.^{1} I'll further assume that drug, robbery, burglary, larceny, and property offenses all fall under blue-collar crime, while banking, insurance, counterfeit, and embezzlement offenses fall under white-collar crime.^{2} Using this categorization, over 50% of crimes would be considered blue-collar, which is in stark contrast to the 0.4% of crimes that are considered white-collar. According to Sutherland, "less than two percent of the persons committed to prisons in a year belong to the upper class," which is consistent with my assumptions. It's worth noting that these figures only count crimes where the crook was convicted and sentenced to serve time. There are undoubtedly a large number of crimes where the perpetrator has been either not apprehended, acquitted, or given a sentence that did not involve incarceration. Below is a chart illustrating the clearance rate (i.e., the fraction of time charges are filed for reported crimes) for various crimes:

Even these figures are fraught with peril. For example, robberies are more likely to be reported than money laundering, if for no other reason than the fact that a victim is immediately aware that a robbery is taking place when there's a gun pointed at his face.

For simplicity, I'm going to pull some numbers from Wikipedia and make some very basic assumptions. According to the Wikipedia entry for "Crime in the United States", in 2009 roughly 3466 crimes were committed for every 100,000 people.^{3} Spread over a population of 300 million people, this means about ten million (~1×10^{7}) crimes are committed each year. Roughly 50% of these (~5×10^{6}) can be considered financially-motivated blue-collar crimes, while roughly 0.5% of these (~5×10^{4}) can be considered financially-motivated white-collar crimes.

I suspect the amount a blue-collar criminal steals each year varies quite a bit from person to person. A professional crook might be hitting a new house every day, but shoplifters and other petty thieves are less likely to be raking in the dough. You could steal anything from one dollar to over a million, but the average take, at least for professional crooks, is much more likely to be in the thousands. For concreteness, let's say $1000 per crime. If only to avoid being caught, I suspect the average criminal is not likely to rob more than 10 times a year (certainly fewer than 100). From these numbers we can conclude that the average blue-collar criminal steals at most $10,000 a year, with one-time offenders stealing much less and Ocean's 11-esque pros looting much more. Using this average as an upper bound, we can compute the maximum total amount our five million blue-collar criminals steal each year to be about $50 billion. Since this is an upper bound, we can reasonably argue that a realistic number is more likely in the one billion to $20 billion range.

White collar criminals are a different story. While fewer in number, they can use their fortunate disposition to gain control of much more money. A list of the top ten white-collar crimes shows thefts ranging from Martin Frankel's $200 million (~$2×10^{8}) swindling of insurance companies to Bernie Madoff's $65 billion (~$6.5×10^{10}) ponzi scheme looting, but these are extreme cases.^{4} A more typical example might be Martha Stewart, who "avoided a loss of $45,673 by selling all 3,928 shares of her ImClone Systems stock...." If we assume a $100,000 (~$1×10^{5}) take, then our 50,000 white-collar criminals would rake in about $5 billion (~$5×10^{9}) each year.

There's a problem with this estimate. As one can easily see, Madoff alone took in more than ten times my estimate.^{5} What gives? While crimes like Madoff's are rare, they're significant because of the sheer volume of money involved. Significant rare events like these can be kryptonite to Fermi estimators. Moreover, as I pointed out earlier, blue-collar crimes are less likely to be reported. As such, my estimate is probably only good as a lower bound, so I'm going to need to up it a bit. Since we're talking order-of-magnitude estimates, it makes sense to bump my number up by a power of ten. This would give about $50 billion (~$5×10^{10}) stolen each year by white-collar criminals. This is within an order of magnitude of the same numbers listed by the Wikipedia entry for "White-collar crime":

While the true extent and cost of white-collar crime are unknown, the FBI and the Association of Certified Fraud Examiners estimate the annual cost to the United States to fall between $300 and $660 billion.

It is estimated that white-collar crime cost the United States from $200 to over $300 billion every year. This is staggering compared to the estimated $15 billion to $20 billion in damages that blue-collar crime inflicts.

As you can see, my numbers are within a power of ten of these other estimates, but there is, admittedly, a huge error range associated with these figures.

Let's try to put these numbers in context. The 2013 budget for the Department of Justice is set at $27.1 billion. Roughly $8.5 billion (~$8.5×10^{9}) of this is spent on prisons/detention facilities. Federal prisoners make up about 6% of the total incarcerated population, or about 138,000 people. As such, each federal prisoner would cost

In 2007, around $74 billion was spent on corrections. The total number of inmates in 2007 in federal, state, and local lockups was 2,419,241. That comes to around $30,600 per inmate.

In 2005, it cost an average of $23,876 dollars per state prisoner. State prison spending varied widely, from $45,000 a year in Rhode Island to $13,000 in Louisiana.

Even if we're building fairly efficient prisons at a net cost of $30,000 per prisoner per year, imprisoning blue-collar criminals costs the American taxpayer anywhere from 3 to 30 times more money than they would lose by letting these criminals run free. In contrast, letting 50,000 white-collar criminals run free would cost taxpayers 30 times more than it does to them lock up. This number skyrockets to 440 times as much money if we assume our white-collar criminals are stealing $660 billion.

A more efficient justice system would prosecute fewer of these guys...

...and more of these guys.

Economically, it seems drastically more efficient to lock up the high-class crooks and leave the petty thieves to themselves. However, this doesn't address Dr. Ariely's last question. Taken by itself, incarcerating blue-collar criminals is not cost-effective, unless you assume the mere threat of incarceration has prevented anywhere from 3 to 30 times more crimes. More on this in a moment. For now, let's naively assume the rate of crimes committed doesn't depend on the type of punishment doled out. If that's the case, taxpayers lose on average between $20,000 and $29,000 for every incarcerated blue-collar criminal. In contrast, taxpayers save between $70,000 and $13 million for every incarcerated white-collar criminal. Given five million (~5×10^{6}) blue-collar criminals and 50,000 white-collar criminals, the taxpayers would lose at most

($29,000) × (5×10^{6}) − ($70,000) × (5×10^{4})

= $140 billion

and potentially save

($29,000) × (5×10^{6}) − ($70,000) × (5×10^{4})

= $500 billion.

Given the huge error ranges, it's a little bit of a wash to say whether or not the justice system is cost effective for financial crimes. One thing that might shift the balance toward being a good system is the notion that the threat of punishment has prevented crimes from happening. Sadly, I don't really have a good way of estimating the number of crimes that would have occurred without a justice system. All I can really say is that you'd need to have prevented about five million crimes to break even.

So what have we learned from this? First off, whoever said crime doesn't pay was grossly misinformed. Second, the Department of Justice would do well to focus less on petty thieves and more on the white-collar criminals who got us into the financial mess we're currently in.

Dr. Ariely, thank you for a very challenging question. I feel like I learned a lot doing this one.

[1] For example, rape, murder, and jaywalking are all crimes, but there's no money exchanged, so I'm excluding these.
[2] Drug related offenses can be either possession (which does not have a financial component) or selling (which does have a financial component).
[3] It's unclear if this number represents the number of crimes committed or the number of crimes reported to police. I'm assuming the former for simplicity, but if it is the latter, one would need accurate clearance rates to compute the number of crimes committed.
[4] Seriously, who trusts a guy named Bernie Madoff? As in, "Bernie made off with my life savings!"
[5] Even if we take the $50 billion upper bound figure for the amount that all blue-collar criminals made in a year, it's still less than the amount that was stolen by one Bernie Madoff.