It is commonly believed that Archimedes devised a scheme using only sunlight and mirrors to burn Roman ships that were attacking Syracuse. The mirrors act as a parabolic reflector that focuses heat on the ship until the wood ignites. A group of MIT students and MythBusters tested whether this “Archimedes Heat Ray” was feasible and determined that it would have been far easier to burn the ships with more conventional weapons. As any 10 year-old boy with a magnifying glass and an ant hill can tell you, the Sun produces more than enough energy to set things ablaze under the right conditions. That said, perhaps the MIT-Mythbusters team could have gone after a more contemporary problem. How many mirrors would you need to melt an aircraft carrier in one second?

In order to melt an aircraft carrier, you first need to raise the temperature to its melting point. Aircraft carriers are probably made of steel or some other material with a similar melting point. The melting point of steel is different for different alloys, but according to Google it's generally in the vicinity of 1300°C. Since room temperature is about 25°C, this means we must first raise the temperature by 1275°C = 1275 K.

To raise the temperature, we must add heat to the aircraft carrier. The heat capacity (i.e. the amount of heat energy needed to raise one kilogram of a substance by 1°C) is 460 J/kg·K. An aircraft carrier weighs about 2.0x10^{8} kg. Using this info, we can compute how much heat we need to raise the steel in the carrier to its melting temperature,

OK, so we’re done, right? Not necessarily. So far we’ve only raised the temperature to the melting point of steel. At this point, it’s just really hot solid metal. In principle, we still have to add more heat to change the solid metal into molten metal. The amount of heat we need to induce the phase change is called the latent heat. To calculate the latent heat, we need the heat capacity (i.e. the amount of heat needed to transform one kilogram of a substance from one phase to another), which for steel is about 25.5 J/kg. We can then compute the amount of energy needed to make the phase change,

latent heat = (heat capacity) · (mass)

= (25.5 J/kg) · (2.0x10^{8} kg)

= 5.1x10^{9} J

This is a large amount of energy by most standards, but, as it turns out, it’s very small compared to how much we need to raise the temperature in the first place, so we can neglect it. In total, we’ll need about 1.1x10^{14} J of energy to melt the ship. All of this energy has to come from sunlight, which hits the Earth with 1360 W/m^{2} (this is called a solar flux.) From this we can use dimensional analysis to compute the total area of mirrors we would need to melt the air craft carrier in one second,

Recently, Chris Johnson of the Tennessee Titans challenged Usain "The Aptly Named" Bolt to a race. Bolt set a record in the 100m dash at the Olympics in Beijing with a time of 9.58 s, while Johnson ran a 4.24 s 40-time at the NFL combine. This got me wondering whether there was anything us physicists could predict about who would win in a 100m race. Who would win in a race, Chris Johnson or Usain Bolt?

Naively, one could compute Johnson’s time in the 100m by first calculating his average speed in the 40,

speed = (distance) / (time)

= (40 yd) / (4.24 s)

= 8.6 m/s.

Using this speed, we could then calculate his time in the 100m,

time = (distance) / (speed)

= (100 m) / (8.6 m/s)

= 11.6 s.

That’s barely a decent time for a high school student, never mind the fastest man in the NFL. Bolt would blow him away if this were his actual speed. Where did we go wrong? The problem is that this is only an upper bound on Johnson’s time. The reason is that we calculated his speed using his 40 time, and much of the first 40 yards of a sprint are spent accelerating from a low speed. If we assume Johnson accelerates uniformly for the whole run, we could calculate his acceleration,

acceleration = 2 · (distance) / (time)^{2}

= 2 · (40 yd) / (4.24 s)^{2}

= 4.07 m/s^{2}.

If we then assume he accelerates uniformly throughout 100m, we can calculate his new time,

t = 2 · (distance) / (acceleration) = 2 · (100 m) / (4.07 m/s^{2}) = 7.01 s.

That’s better than even the most unrealistic tricked out Madden player you can make. The reason is simple: since we assumed he was accelerating throughout the whole race, he has a much higher final speed then he would in reality. How much higher? About 64 mph!

Realistically, Johnson would probably reach his top speed somewhere between 10 and 40 yds and then maintain a roughly constant speed throughout the rest of the race. This being the case, he would almost certainly finish somewhere between these two bounds. Logic would seem to indicate that Bolt, typically regarded as the fastest man in the world, would still win handily. Unfortunately, our simple estimation is not precise enough to say who would win with any certainty. This is an important point. Sometimes we don't have enough background information to calculate definitive answers, and we can only calculate upper and lower bounds. Sadly, we'll have to deal with ESPN pundits endlessly debating who the faster man is.

I've been thinking about the Hulk problem again. Specifically, where does all Bruce Banner's extra mass come from when he turns into the Hulk. It wouldn't be the first time the Marvel Universe violated the conservation of mass. As I discuss in How Many Licks?, there are serious—or perhaps not so serious—questions about how much Spider-Man has to eat to produce all that web. However, the Hulk's metamorphosis could, at least in principle, be due to some chemical reaction where he's absorbing mass out of the air. This begs the question: What volume of air would be absorbed when B. B. turns into the Hulk?

From before, we know that B. B. gains 350-580 kg. Since air has a density of 1.2 kg/m^{3}, we can compute the total volume of air absorbed:

volume = mass / density = (350 or 580 kg) / (1.2 kg/m^{3}) ~ 290-480 m^{3}

That's enough to suck out all the air in a 1,500 sq. ft. apartment. Rather than beating them up, the Hulk could suffocate them with his massive vacuum.

Here's part II of the problems I solved for Skeptically Speaking:

Problem 6:

Physicists say it's possible for all the air molecules in a room to spontaneously slide over into one corner, suffocating everyone in the room. What are the odds of that actually happening?

Solution:

Assuming an ideal gas^{*}, the probability of a molecule being on any side of the room is ½.The probability of n molecules all being on the same side is then (½)^{n}.The density of air is about 1.2 kg/m^{3} at sea level and 20^{°} C.A reasonably large room might be 10 m by 10 m by 3 m giving a total volume of 300 m^{3}.Multiplying the density of air by this volume gives 360 kg of air in total.Most of the air is made up of nitrogen with a molecular weight of 0.028 kg/mol.Dividing this into the total mass of air gives about 13,000 moles or n = 7.7x10^{27} molecules.

Therefore, the probability of finding all the gas molecules on one side of the room is ½^{ }raised to the 7.7x10^{27}th power. I’m not going to even attempt to write out this number.It is extraordinarily small.“How small?” you say.If you were to try to write out all the zeros, you would cover the surface of 10 million Earths.

^{*}A real gas is non-ideal. If you want to do this problem rigorously, you need to account for the fact that each molecule takes up a finite amount of space.

Problem 7:

If the Skeptically Speaking team got locked in the on-air booth, how long would it take for them to run out of air? We have 5 people in the booth every show, and the booth is about 10 ft by 6 ft.

Solution:

Technically, you’d never run out of air, but you would run out of breathable oxygen, which gets converted to CO_{2}.Assume you take a breath once every 3 seconds.According to Hypertextbook.com, human lungs have a volume of about 5000 cm^{3}.Oxygen makes up about 20% of air by weight.A quick web search shows that about 25% of the oxygen we breath in gets converted to CO_{2}.Using the density of air from the previous problem, we can compute the amount of oxygen lost per second:

Assuming the room has10 ft ceilings, the total volume is then ~17m^{3}.From this, the total starting mass of O_{2} can also be calculated:

(17 m^{3})

*(1.2 kg of air per m^{3})

*(0.20 kg of O_{2} per kg of air)

--------------------------------------

4.1 kg ofO_{2}

_{
}

Dividing the amount of O_{2} lost per second into the total amount of O_{2}, we get ~8200 seconds or roughly 2.5 hours.

Problem 8:

How likely is it that we inhale a molecule breathed by Carl Sagan?

Solution:

This is a pretty hard problem for a number of reasons. For starters, air molecules are constantly reacting to form new molecules. Second, it's not clear how long it would take Carl Sagan's air molecules to mix evenly across the globe. Third, Carl Sagan may have breathed in the same air molecules more than once.

To simplify, let's assume that air molecules get evenly distributed around the world very quickly, that Carl Sagan never breathed the same molecule twice, and that air molecules don't react. Strictly speaking, only noble gases won't react, but this assumption might be approximately true for nitrogen since it's fairly inert. Dr.Sagan lived 62 years and there's roughly 32 million seconds in a year. From this and the numbers used in the previous problems, we can estimate the number of molecules breathed by Dr. Sagan:

(5000 cm^{3} per breath)

*(1/3 breaths per second)

*(3.2x10^{7} seconds per year)

*(62 years per lifetime)

*(1.2 kg of air per m^{3})

*(10^{-6} m^{3} per cm^{3})

*(1 mole per 0.028 kg of air)

*(6.022x10^{23} molecules per mole)

-----------------------------------------------

~8.5x10^{31} molecules breathed in a lifetime

According to Hypertextbook.com, the mass of the whole atmosphere is 5.3x10^{18} kg. If we assume that air molecules have an average molecular weight equivalent to nitrogen's 0.028 kg/mol, then there should be about 3.2x10^{42} molecules in the atmosphere. This means that the probability P of a molecule being "Saganized" is then,

P = (8.5x10^{31}) / (3.2x10^{42}) ~2.7x10^{-12}

Here's where it gets a little tricky. The probability of a single molecule not being breathed by Sagan is (1-P). This is very close to 1, so you might be tempted to think it's very unlikely that to share any of your molecules with Sagan. However, the probability that none of the molecules you breath were ever breathed by Dr. Sagan is given by (1-P)^{n}, where n ~ 8.5x10^{31} is extremely large. Since the exponent is so large, the probability of never breathing a Saganized molecule will be very close to zero. In short, you breathe a lot of molecules breathed by Carl Sagan!

Problem 9:

I met my girlfriend online, and she's totally the one, but she lives on the other side of the continent. I've been wondering about the odds: not just that we would meet, but that we'd both be born at this time in history when the Internet exists and allows us to interact over such a large distance.

Solution:

I did a similar problem in How Many Licks? In it, I estimated the probability of finding Mr./Mrs. Right is about 1 in 200,000, but this is assumed that special someone was born at the same time in history. What if the love of your life was born in a different time period, like in the Christopher Reeve/Jane Seymour movie Somewhere in Time?

To start, I looked up the world population as a function of time. If you assume an entire population dies off after about 70 years (1 lifespan) and is replaced by an entirely new population, you can trace back about 1000 different to 70,000 B.C. With this assumption, you can calculate the total number of humans who have ever lived on the planet and then calculate the fraction of people living today. 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. As such, the chances of both lovers being born at the same time in history are much better than the chances that they are born in the same geographical location. This is very surprising, but it shows just how fast exponential growth is. In any event, the probability of finding that special someone is 1 in about 600,000.

That's all for now. Thanks again to Desiree, the Skeptically Speaking crew, and the audience for coming up with some great questions!

Thanks to Desiree and all the people at Skeptically Speaking who invited me on their show a few weeks back.In case anyone is wondering how I came up with the answers to the audience’s questions, I’m posting my estimations below.There are a quite a few, so I’m posting this in two parts.Enjoy!

Problem 1:

How much mass does Bruce Banner gain when turning into the Hulk?

Solution:

At first, I was thinking of the old Hulk TV show starring Bill Bixby as Bruce Banner.According to IMDB, Mr. Bixby actually turned into a green Lou Ferrigno (~275 lbs), so the mass gain is fairly small.Even if he turned into Shaq (~325 lbs), the most he would gain is about 90 kg (~200 lbs), but even that’s being generous.According to the Marvel Comics website, there are several different Hulks with weights ranging anywhere from 900-1400 lbs.Bruce Banner is listed as a slight 128 lbs.Doing the math, Bruce Banner will gain anywhere from 350-580 kg (~770-1270 lbs).

Problem 2:

How long would it take to find a needle in a haystack?

Solution:

This clearly depends on how big your haystack is.When I visit my wife’s family in Nebraska, we often see huge hay bales.They’re cylindrical and look about 1 m (~3.3 ft) in radius and about 3 m (~9.8 ft) in length.You can calculate the volume of a cylinder using the formula,

V = Pi R^{2} L,

where R is the radius, L is the length, and Pi=3.1415926...Using this formula, you’ll get about 10 m^{3} for the volume of a haystack.I figure you can sift through about one handful of hay every 30 seconds.If each handful is a cube 5 cm (~2 in) on a side, then the volume per handful is 1.25x10^{-4} m^{3}.You can then calculate the total time it would take:

(10 m^{3} per haystack)

*(1 handful per 1.25x10^{-4} m^{3})

*(30 s per handful)

---------------------------------------------

~2.4 million seconds or about 1 month

Problem 3:

How many gallons of ketchup get wasted every year in those little foil packets that get thrown in the garbage when you empty your tray?

Solution:

First off, there are much more creative things you can do with ketchup packets.However, since people are going to throw them out anyway, we should probably figure out how much environmental damage they cause.I assumed the average American gets fast food at least once per week. (Sadly, it may be a lot higher.)Whenever I go, it seems like they always give me two ketchup packets per tray.There are about 5 cm^{3} ~ 0.001 gallons per packet.There are about 300 million people in the U.S.

(2 packets per week per person)
*(52 weeks per year)
*(3.0x10^{8} people)
*(0.001 gallons per packet
---------------------------------------------
~15 million gallons of ketchup

That’s enough ketchup to fill a football field 40 ft high.

Problem 4: How much wood could a woodchuck chuck if a woodchuck could chuck wood?

Solution: I hate this question. If I answer it, will you guys please stop asking it? OK, here goes…
A quick search of the Interwebs shows that woodchucks (a.k.a. groundhogs) consume up to 0.5 kg per day, and they can live up to 6 years in the wild or 10 years in captivity. I’m assuming we’re talking about wild woodchucks because if I had a captive woodchuck, I’d like to think I’d feed him something tastier than wood. From these numbers, we can calculate the amount of wood chucked:

(0.5 kg per day)
*(365 days per year)
*(6 years)
----------------------------
~1100 kg (~2400 lbs)

That’s over a ton of wood chucked.*

* Note: I’ve assumed “chucking” means “eating.” Google also lists “to throw away” and “to vomit” as alternate definitions.

Problem 5:

During the "Great Flood," how much rain would have to fall each day, over 40 days, to reach the top of Mount Sinai?

Solution:

According to Wikipedia, Mount Sinai is 2285 m (~7500 ft) high.Dividing the height by 40 days, you find that you'd need an average of 60 m of rain per day to reach the top of the mountain.As point of reference, Seattle, which is known as “the rainy city," gets about 3.6 m per year.Even if none of it seeped into the ground, it would take 15-20 years to reach the top of Mount Sinai.

Check out tomorrow's blog for the rest of the Skeptically Speaking calculations.

Thank you to everyone who entered our Fermi Estimation Contest this past January.Congratulations to winner Alex D, who received a free copy of How Many Licks?Several contestants asked for my solution, so I’m posting my results below:

I assumed the snowman was made of three spheres of radii R_{1}=30 cm, R_{2}=45 cm, and R_{3}=65 cm.The volume of the snowman is then given by:

V = (4Pi / 3)*(R_{1}^{3} + R_{2}^{3}+R_{3}^{3}) ~ 1.6m^{3}

From Hypertextbook.com, there are roughly 100 crystals per snowflake and 10^{18} molecules per crystal.One water molecule weighs 2.992x10^{-26} kg.From these facts, we can calculate the mass of a flake:

(100 crystals per flake)

*(10^{18} molecules per crystal)

*(2.992x10^{-26} kg per molecule)

-----------------------------------

3.0x10^{6} kg per snowflake

According to Wikipedia, the density of snow varies considerably from 8% the weight of water for newly fallen snow, to 30% after it settles, to 50% in late spring.Since the density of water is about 1000kg/m^{3} and the snowman is pretty densely packed, I chose 500 kg/m^{3}.

Using these numbers, we can calculate the number of snowflakes:

(1.6 m^{3})*(500 kg per m^{3}) / (3.0x10^{6} kg per snowflake) ~ 2.7x10^{8} snowflakes

My name’s Aaron Santos. I’m a geeky physicist interested in improving science education. I graduated from MIT with a bachelor’s degree in physics and got my PhD in physics from Boston University. Between undergraduate and graduate school, I worked as a Boston Public School teacher at Boston Latin Academy.

Since leaving teaching, I’ve developed a strong desire to communicate science to the general public. Toward this end, I recently wrote How Many Licks?: Or, How to Estimate Damn Near Anything, a book of (hopefully) humorous and thought-provoking Fermi approximations.

So what’s A Diary of Numbers?

It’s a blog featuring various calculations I’ve done. They’ll be similar to the types of calculations you’ll find in How Many Licks? Things like:

How many calories are in the Stay Puft Marshmallow Man?

How many times can you wash your favorite t-shirt before it turns entirely into dryer lint?

How much would it cost to run the entire U.S. with solar energy?

Why write a blog?

Several reasons:

It’s quicker to write a blog than it is to publish a book (much, much quicker).

The capitalist in me says it’ll help sell more copies of my already published book.

Hopefully, people will read it, get really excited about doing math and science, and the world become a more enlightened place.

Is this really going to enlighten the world?

Probably not. I’m not even sure who’s going to read it. But even if only a few people get a kick out of reading it and learn to enjoy thinking scientifically, I’ll feel it’s been worth my time. At the very least, the blog should provide a good forum for organizing my thoughts.

So you want people to become scientists?

I think everyone should be a scientist. I don’t mean everyone should have jobs where they have to walk around wearing white lab coats. I mean everyone should use the scientific method: make hypotheses about the world, see if there’s any evidence contradicting your hypotheses, and make new hypotheses that fit that evidence. The scientific method is useful everywhere, from high tech labs to programming your VCR clock. (Yes, I still have a VCR.)

Wait a minute! You said you’re concerned about science education. Why not write a science blog instead of a number blog?

I view estimation as being more science than math. First of all, it requires a certain degree of intuition about the real world that’s not necessary in pure mathematics. More importantly, estimation has all the rudimentary parts of science: You hypothesize about how big (or small, or expensive, etc.) something is, then you gather evidence by calculating the answer in a bunch of different ways. In this respect, your hypothesis is falsifiable. In estimation, much like science, you can never say what the answer is but you can certainly show what it isn’t. You get progressively closer and closer to a correct theory by making better and better estimates. There’s even peer review. (To anyone who doubts this last point, I suggest you do an estimate and then show your friends. There’s a good chance they’ll be harsher than most referees.)

But aren’t we learning science and math in school?

No. There’s already a lot written on the problems of science and math education. The main problem is we aren’t really doing science so much as learning about the discoveries other people have made by doing science. There are a lot of good books that teach people about scientific discoveries, but very few teach people how to think scientifically. Even the books that teach “experiments” often miss the point. Usually you’re just replicating some neat experiment, and while that’s generally pretty good for demonstrating science principles, it’s very different than coming up with your own experiments (or thought experiments) and trying to figure out answers using the scientific method. These books are entertaining and certainly educational, but I don’t think they do justice to the struggles, the ups and downs, the creativity—in short, the “process” that real scientific thinking entails. For these reasons, I think estimation is a better way to teach people to think scientifically.

But scientific discoveries are important! You’re just solving goofy math problems!

True, I am just solving goofy math problems. Some might argue that I should just write about science experiments and describe the process. That’s a pretty passive way of learning, which is usually not as good as actively learning. My hope is that the goofy problems here will inspire readers to solve their own estimation problems, rather than just read the blog. In this way, people will actively be doing science rather than just reading about science that other people have done.

All right, but wouldn’t it be better to write about actual experiments? At least then people will be learning real science.

There’s a practical reason for this: not everyone can afford atomic force microscopes and big hulking lab equipment. If you want to train someone to think scientifically, estimation is a fairly cost-effective way of doing so because anyone can do it anywhere at any time. More importantly, this is a way of doing “real science.” A theoretical physicist does these sorts of calculations and thought experiments all the time (albeit with more advanced mathematical techniques). As far as using the scientific method, theoretical science is just as much a science as is chemistry, biology, or any other experimental science.

But a lot of these calculations are really just using common sense. How is this helping people become better thinkers?

I marvel at this statement because so many people think of scientists in this way. Doing science is using common sense. We’re not robot super geniuses. We’re just trained to think critically about things. It’s not like Einstein’s brain was made fundamentally different than the rest of ours, he was just doing things that made sense given what people knew about the world.

Getting back to the main point, these simple estimation problems can be a good way to develop scientific thinking skills. This is true no matter how advanced your background is. While the calculations I do here are certainly a lot less rigorous than what you’ll probably find in a published paper, this is purely a pedagogical choice: having fun goofy problems allows anyone to get involved without being intimidated by big scary equations.

All right, so tell me how to estimate.

I can’t. The main complaint I’ve gotten about How Many Licks? has been from people who wanted to learn some secret magical trick that will show them how to calculate anything. Unfortunately, there’s no magic bullet, formula, or procedure that will work every time. Estimation, like science, is something of an art. You can estimate just about anything, but doing so requires hard work and a lot of thinking. As I said earlier, estimation is a lot like science.

Well, at least give me some general guidelines.

That’s what the problems are for. The main compliment I’ve gotten about How Many Licks? was from people who enjoyed the problems, and that’s what this blog is about. By seeing a bunch of examples, hopefully you’ll be able to pick up some general tricks and methods of solving problems. Feel free to discuss, disagree with, and argue about any of the problems I present here. Ultimately, the best way to learn estimation—and scientific thinking—is to be actively engaged in it, and if you’re learning enough to disagree with my results I’ll feel like my work here is done.