Thursday, February 25, 2010

The One-Second Workout

Hitchhiker: You heard of this thing, the 8-Minute Abs?
Ted: Yeah, sure, 8-Minute Abs. Yeah, the exercise video.
Hitchhiker: Yeah, this is going to blow that right out of the water. Listen to this: 7... Minute... Abs. 

Scene from There’s Something About Mary, as quoted on IMDB.

We turn now to weight loss. Forget “Eight-Minute Abs” and “Ten-Minute Buns,” we have something that’ll even eclipse Ben Stiller’s sketchy hitchhiker friend. How fast would you have to run to burn 10 lbs of fat in one second?

If we learned anything from the cannibal estimation, it’s that biology is not my strong suit. Be that as it may, there’s still a fairly straightforward way to at least estimate an upper bound for this problem. In order to lose fat, we have to convert it into energy. There are 9.0 Calories per gram or about 3.8×107 J/kg in fat. That means there are about 1.7×108 J of energy in 10 lbs (~4.53 kg) of fat.

When we run, we convert chemical energy into kinetic energy. Kinetic energy is the energy of motion and can be computed using the formula,

energy = (person’s mass) · (speed)2 / 2.

If we start from the stop position and accelerate to some speed, the chemical energy we use will be at least as big as our kinetic energy. Not all of the chemical energy gets converted to kinetic energy. Since our bodies aren’t 100% efficient, some of the chemical energy gets converted into heat. Moreover, not all of the kinetic energy helps us move forward, since we’re also using some of it to swing our arms and legs. In short, we use more chemical energy than we get out, but we can use this fact to compute an upper bound on how fast we’d need to run to lose 10 lbs in a second. I’ll assume the person’s mass is 75 kg (~165 lbs).1 We can solve the equation above for velocity to estimate the speed a person would need to run to lose 10 lbs in a second,

speed = [ 2 · (energy) / (person’s mass) ]1/2
= [ 2 · (1.7×108 J) / (75 kg) ]1/2
= 2100 m/s

If you neglect the sonic boom and air friction that would surely kill most mere mortals, you could lose at least 10 lbs in one second by accelerating to 2100 m/s, or about Mach 6 (i.e. six times the speed of sound).2

[1] This is technically the average mass since the person is losing mass as he runs.
[2] Before you start burning fat, you use the chemical energy from ATP and glycogen. Rigorously speaking, this means you might have to run a little faster. However, given that you’d likely deplete your ATP and glycogen storage well before you’d lose 10 lbs in a single workout, I stand by this result as an upper bound.

Wednesday, February 24, 2010

Landing Face First in P

“Yay! It took all night, but I finally finished this term pappppppppppppppppp

Have you ever pulled an all-nighter only to sleep through the class for which your project was due? Ever worked so late you fell asleep facedown on your keyboard and then woke to an incoherent mess of random characters on the screen? If so, you probably had to delete quite a bit. How many incoherent characters would you accidentally type by using your keyboard as a pillow?

By holding down one key, you can gauge that a keyboard can type about 30 characters per second. If you get an average night’s worth of sleep (~8 hours), you’ll have held the key down for about 29,000 seconds. You can then compute the number of incoherent characters on the screen,

total # of characters = (# of characters per second) · (time)
= (30 characters per second) · (29,000 seconds)
= 870,000 characters.

You’d have close to one million characters, or about 200 single-spaced pages worth.

Well, it’s late. I should probably get some sleeppppppppp

Tuesday, February 23, 2010

Fermi Contest II

The rules are the same as the first contest. You can win a free copy of How Many Licks?. Here's how it works. I’m posting 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 How Many Licks?.* Submit your entry on or before March 31, 2010. Don't worry…I won't spam you or share your email with any third parties. Here’s the question:

Mario has appeared in over 200 video games as Nintendo’s main mascot. He has starred in TV shows, comic books, and feature films, but anyone who’s played one of his games knows it’s not all fun and games for the lovable plumber. He is constantly falling down bottomless pits, getting eaten by giant mutant mushrooms, and otherwise killing himself all while trying to save a princess who is apparently missing the part of the brain that allows one to avoid getting kidnapped. With that in mind, your estimation question is this: “How many times has Mario died in all of the videogames ever played?”

*NOTE: 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!

Comments and Corrections…

I’ve gotten a few comments and corrections from people, so I wanted to take the time to address some of those. First off, thanks to all the people who posted kind words. You made my day. Second, to all the people who like the blog but hate the color scheme, I’m going to try to fix it right after I post this. Also, thanks to all the people who expanded on what I wrote in the comments and/or emailed me related references. I always like hearing how this stuff relates to other things, and I really enjoyed the emails and posts telling me about the Biodome II project, Jules Verne’s Journey to the Center of the Earth, and all of the expanded estimations that people have done. Now on to the (gulp) corrections…

In my first post, I described estimations as being similar to science:

“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.)”

In that spirit, it would be in poor taste for me not to admit my slipups, screwups, messups, and other “ups” I can’t mention in a family-friendly blog. That said, here are my responses to some of your comments:

Mount Sinai…don’t you mean Ararat?

hwiersma, just looked it up and you’re absolutely right. Mount Ararat is where Noah is supposed to have landed; Mount Sinai is where Moses is supposed to have gotten the Ten Commandments. The original question from the Skeptically Speaking caller specifically asked for Mount Sinai, and I never bothered to check if it was right. Good eyes and thanks for the correction.

Could you even fit everyone in New York City?

I agree, Mabehr. There’d be very little chance of actually fitting the entire world population into an area the size of NYC unless they pile themselves into the buildings. But if you did stuff them all in buildings, how slow would the traffic be when everyone tried to get out…

Vampire vs. Zombie Hummingbirds

Jennifer, I was going to try to add something to the vampire vs. zombie hummingbird comment, but I don’t think I can. Some things are perfect just the way they are. J

Confessions of Mechanical Pencil User

I admit it. I cheated on the Simon’s Cat problem. I used a mechanical pencil. Phew! It feels so good to get that off my chest. In my defense, it was still an HB #2 pencil, and it made measuring exactly how much graphite I lost way easier than it would have been with a wooden pencil. It’s probably pretty easy to take sharpening into account, but I’m not sure it’s a big source of error. In principle, you’re only shaving off the sides of the pencil so the length should remain the same. (But, obviously, if you leave it in the pencil sharpener too long, the length will shrink.) I suspect the sharper point will lose material more quickly than a dull point, but I haven’t checked this. In regard to Thomas’s comment, I’m not sure why tracing the same line a hundred times should be different than drawing one long straight line. Is it because you change how much friction there is between the paper and the pencil? If so, I’m not sure that’s a significant source of error for an order of magnitude approximation, but it’s good to check just in case.

Giant Basil

I got a Facebook message from a friend saying, “A square foot is a lot of space for 3gm of basil.” I was overestimating a bit. I could make something up about needing extra room to actually go and pick the basil or needing to have extra basil in case some of it died, but in truth, I just wasn't thinking that hard about it. You could probably get by using a tenth of the space.

Passenger Pâté

This seemed to be the estimation that had the most problems. (Who knew cannibalism was such a touchy subject?) As Simkatu and Thomas pointed out, not all of a human body’s mass is made of useful calories. (By the way, Thomas, any day that I get to see an equation that uses “poop” as a variable is a happy one.) I got an email from Dr. Jeremy S. that described this even more precisely:

“First off, you neglect to factor the fraction of body mass that is water. The calories per gram of protein and fat is per dry weight. Approximately 50-70% of body mass is water depending on age and gender. Further, you can estimate fat calories by multiplying average body mass by average fat percentage. As far as total body protein is concerned, my guess is that you can probably sneak by taking the remaining body weight and multipling by calculating by the percentage of weight that is neither fat nor water. (This is probably not a perfect assumption as some organs such as liver and bone marrow* are fatty and other organs are tough to take without significant preparation (I.e. Intestines.) Also, not eating bones is a big mistake. Fat rich marrow is a great source of calories as our frugal recent ancestors knew (ask your grandparents, they probably ate marrow and loved it. My grandma would spread it on toast.)”

Anyway, those are my corrections. I may revisit some of these problems later and go into more detailed corrections, but it’s getting late and I have a background to fix and an estimation contest to post. Until my next set of corrections, remember, these are only order-of-magnitude estimates, and goofy order-of-magnitude estimates at that.

P.S. Thomas, I’ll try to get on the Mercury flywheel problem soon.

Monday, February 22, 2010

A quick update...

I'm having some computer trouble, so unfortunately I won't be posting an estimation today. Tomorrow, I'll be revealing the question for our second estimation contest, so check in then for your chance to win a free signed copy of How Many Licks? I'll also provide some quick corrections and try to address some of the comments people have left. Thanks for reading!

Saturday, February 20, 2010

Simon’s Cat Calculation

Today’s question comes from the very talented maker of the Simon's Cat videos. Simon Tofield’s popular animations currently have over ten million views on Youtube. With the prevalence and ease of today’s computer animation tools, it’s rare to find high-quality, hand-drawn animation, particularly if you’re looking for something as funny and true-to-life as the Simon’s Cat videos. Hand-drawn animation takes a lot of work and a lot of pencils. How many times could Simon Tofield draw Simon's Cat using a typical HB pencil before the pencil ran out?

To solve this problem, we need to know two things: (1) how much graphite is in a typical HB pencil and (2) how much graphite is used on a typical drawing of Simon’s Cat1.  An HB pencil is the same as the #2 pencil used for standardized tests, and these are about 18 cm (~7 in) long. A Simons Cat drawing2 will use different amounts of graphite depending on how big its drawn. Ill assume the original sketch is as large as the picture of the cat as viewed on my laptop when I watch the video in full screen. If thats the case, then the lines that make up Simons Cat should be about 10 cm (~4.0 in) if you stretched them out.

Now comes the difficult part. How much graphite gets used in a 13 cm long line? To get a rough idea, I took the #2 pencil on my desk and went back and forth with it 100 times on a piece of scrap paper 12 cm long. This is equivalent to drawing a 1200 cm long line. When I started there was about 3.0 mm of graphite showing and when I finished, there was 2.0 mm showing. From this, we can compute the amount of graphite used per cm drawn,

graphite used per cm drawn = (1.0 mm graphite) / (1200 cm drawn)
= 0.00083 mm of graphite used per cm drawn.
Using this, we can estimate how many Simon’s Cat drawings can be made out of a typical HB pencil,
# of cats = (graphite per pencil) / [ (cm per Cat) · (graphite used per cm) ]
= (18 cm) / [ (10 cm per Cat) · (0.00083 mm of graphite used per cm) ]
= 2200 Simon’s Cats.
Simon can draw roughly 2000 Simon’s Cats with a single pencil. This is a little larger than the number of frames in a one-minute movie.
Thank you, Simon. I like your cat.

[1] In How Many Licks?, I did a similar problem calculating how long of a line you could draw with a pen.
[2] I’m assuming we're talking about just the cat and not all of the background drawings.

Thursday, February 18, 2010

Adam Savage and Digging Holes to China

Today’s question comes courtesy of MythBuster and geek superhero Adam Savage. Adam was nice enough to write not one, but two questions:

How many licks to get to the center of a Tootsie Pop?
Where do you have to dig a hole through the center of the Earth to reach China?

Since I’ve already addressed the first one in How Many Licks?, I’ll focus on the latter. First off, there are a couple of ways to interpret this question. Adam specifically asked “where” you have to dig a hole. The snarky answer is, of course, “Through the center of the Earth, dummy,” but it’s never a good idea to offend geek superheroes, especially ones that are nice enough to write you back and have a propensity for blowing things up. But if we want to end up in Beijing and pass through the center of the Earth, then we’ll have to start at the opposite end of the Earth and travel a distance equal the Earth’s diameter (~1.3×104 km). China’s longitude and latitude are 39.55º N and 116.25º E. The exact opposite position on the globe would be given by 39.55º S and 63.75º W. By plugging these coordinates into Google Maps, you can see that this would put us on the east side of Argentina.

I’d like to expand on Adam’s original question because there’s a lot of interesting physics in this problem. First off, let’s say we build a giant elevator shaft through the Earth and you jump inside. Would you make it all the way to China? The answer is almost certainly no. The net gravitational force gets smaller and smaller the closer you get to the center since you’re getting closer to having an equal amount of mass pulling on you from both sides. Once you pass the center, gravity start pulling in the opposite direct so that eventually you’ll be digging up instead of down. (This might make the second half of the trip easier so long as you can avoid the falling debris.) But what happens exactly in the middle? Will you float there with no net gravitational force on you?

In addition to these interesting physics questions, there’s also a practical question: how long it would take? There are obviously many technical difficulties. First, there’s the issue of pressure. Much like deep-sea diving, as you travel further into the Earth, the pressure increases to levels where life cannot survive. Second, there is extreme heat with temperatures reaching as high as temperatures on the surface of the Sun. Then, of course, there are the mole people. Putting all these issues aside, we can make a very cartoon-y estimate in the following way.

If we only want one person going through the hole, the hole will only need to be about 1.0 m2 wide. Using the diameter of the Earth, we can compute the total volume of dirt to be dug,

volume = (area) · (height)
= (1.0 m2) · (1.3×104 km)
= 1.3×107 m3.

If we can dig one shovel full (~1.0 L) every five seconds, or, equivalently, 0.2 L/s, we can compute how long it would take to dig a hole to China,

time = (volume) / (volume per second)
= (1.3×107 m3) / (0.2 L/s)
= 2100 years.

If you started back in the time of Christ and never took a break, you’d still be digging today.

Adam, thanks so much for taking the time to contribute a great question!

Wednesday, February 17, 2010

Can you pass the Phil please?

Q: A plane crashes at the U.S.-Canada border. Where do you bury the survivors?
A: You don’t bury survivors.

But if you were the only survivor, how long could you survive by eating the other passengers?

We’re going to assume it’s winter and the bodies stay frozen so they don’t decay. A Boeing 747 can carry 416 passengers, meaning there are 415 potential menu choices. Assuming our cannibal doesn’t eat the bone, he’ll be getting all of his calories from the rest of the tissue. A normal-sized person is about 68 kg (~150 lbs). Skeletons weigh about 14% of the total body weight, leaving about 59 kg of edible other stuff. Since protein is 4.0 Calories per gram and fat is 9.0 Calories per gram, we’ll assume that the average caloric density is 6.0 Calories per gram. From this we can compute the total Calories in each person,

Cal. per person = (edible mass of person) · (Cal. per g)
= (59 kg) · (6.0 Cal per g)
= 3.5×105 Cal. per person.

Multiplying by 415 people, we can compute the total number of Calories,

total Cal. = (Cal. per person) (# of people)
= (3.5×105 Cal. per person) · (415 people)
= 1.5×108 Cal.

Assuming a 2000 Calorie per day diet, we can compute just how long it would take to eat and successfully digest all the other passengers,

time = (total # of Cal.) / (# Cal. per day)
= (1.5×108 Cal.) / (2000 Cal. per day)
= 200 years.

It would take 200 years to eat all the passengers. So, if you’re the only survivor in a plane crash, look on the bright side. At least you wouldn’t starve to death.

Tuesday, February 16, 2010

Chris Moore and Vampire Hummingbirds

This question comes courtesy of the very funny and talented Christopher Moore, author of The Stupidest Angel, Fluke, Lamb, and many other hilarious books. Chris wants to know…

“If your body could be reduced to its pure sugar content (sure, you can use equivalent carb calories), how long would it take 1000 hummingbirds to drink you as nectar?”

There are many interesting hummingbird facts on the Web. According to one source:

“Hummingbirds feed on a variety of flower nectars with caloric values that may vary from 10 to 82 calories per meal (1/100 fluid ounces). We found that when using a relative rich sugar solution, a three-gram male Ruby-throated Hummingbird ate five meals an hour.”

Taking the mean, we’ll say that hummingbirds consume 46 Calories per meal or roughly 230 Calories per hour. There are 4.0 Calories per gram in a carbohydrate. From this we can compute the mass of carbohydrates consumed by a hummingbird in an hour,

mass eaten per hr = (Cal. eaten per hr) / (Cal. per g of carbs)
= (230 Cal. consumed per hour) / (4.0 Cal. per gram)
= 58 grams of carbs per hour per hummingbird

This means that 1000 hummingbirds would consume 58,000 grams of carbohydrates per hour.

Figuring out how many carbohydrates are in the human body is a trickier question. Many diet websites suggest getting anywhere from 40-65% of your daily calories from carbohydrates, but the human body doesn’t store anything close to this amount. According to at least one source, the human body is about 61.8% water, 16.6% protein and 14.9% fat, leaving at most 6.7% for carbohydrates. In general, humans store energy in glycogen, also called animal starch, instead of table sugar. I’ll assume this is what the hummingbirds are after. According to Wikipedia, glycogen makes up about 8% (~100-120 g) of our liver, but only about 1-2% or our muscles mass. Muscles make up about 42% of male human body, which can weigh about 82 kg (~180 lbs). From this we can estimate the total mass of glycogen in the human body,

total mass = (mass in liver) + (mass in muscles)
= (110 g) + (0.01)
-->· (0.42) · (82 kg)
= 450 g

Now we can compute the total time it takes for a hummingbird to devour all the carbohydrates in a human,

time = (g of carbs per human) / (g of carbs eaten per hour)
= (450 g per human) / (58,000 g per hour)
= 28 s

It would take only about 28 seconds for hummingbirds to suck all the nectar out of a human body.

This, of course, assumes we have normal hummingbirds rather than vampire chupacabra-eque hummingbirds, which would devour our entire body like a cartoon piranha. How long would it take for 1000 vampire chupacabra-eque hummingbirds to devour a human body? Assuming they eat just as fast as normal humming birds, we can use the same math above,

time = (g of total human) / (g eaten per hour)
= (81 kg) / (58,000 g per hour)
= 1.4 hours

It would take about one hour and twenty-three minutes to be eaten by vampire hummingbirds. Thanks so much to Chris Moore for taking the time to submit a Fermi problem. Check out his very funny books at or follow him on Twitter @TheAuthorGuy.