Long before we had cars we had horse drawn carriages. Doesn’t this mean that we should also have had bird drawn planes? How many pigeons would it take to carry a 747?

According to Wikipedia, the Boeing 747 has a maximum take-off weight of 442,000 kg (~975,000 lbs). Since birds build nests, we know that a pigeon is strong enough to carry a twig. Wood densities vary, but they’re roughly about 500 kg/m^{3}. This means that a 5 mm thick 10 cm long twig would weigh about 1.3 g. From this we can compute the number of pigeons needed to carry a 747.

# of birds = (mass to be carried) / (mass carried per bird)

= (442,000 kg) / (1.3 g per bird)

= 3.4×10^{8} birds

It would take about 340 million pigeons. This is much more than we could realistically harness to a plane, which is probably good given what pigeons typically do to your car.

Check in tomorrow to find out who won the Mario Estimation Contest.

Whether it was the prom, a first date, or an important job interview, everyone’s had some special event ruined by a zit. For times like these, it’d be great if there were a Hoover vacuum^{1} attachment that could suck the pore open again. Is this physically possible? How much of a pressure difference would you need to vacuum off a zit?

Many zits can be popped with just your fingers, but we’re really interested in the diffcult-to-pop zits^{2}. For that reason, we’ll consider zits that are at least as strong than as the maximum force our fingers can produce. It’s difficult to know one’s finger strength a priori, but this site suggests that a healthy thumb should produce a maximum force of about 200 N while a healthy index finger may produce 100 N of force. This force squeezes on the zit causing the internal pressure to increase and the skin to deform. The deformation may absorb some of the force so that the pressure inside the zit is less than it would be otherwise. For simplicity, I’m going to neglect this effect and assume the force gets applied directly to the zit over an area of about 0.25 cm^{2}. From this, we can estimate the pressure needed to pop a zit.

pressure = (total force) / (area)

= (300 N) / (0.25 cm^{2})

= 1.2 ×10^{7} Pa.

That’s about 2000 psi^{3}. It’s also about 120 times larger than atmospheric pressure, meaning that even with a perfect vacuum there’s no way to build up enough of a pressure difference for this to work^{4}. This makes intuitive sense as anyone who’s tried to vacuum a zit off can tell you.

[1] I imagine it will look something like that tube at the dentist’s office that sucks up your face.

[2] Why use a Hoover attachment when you can just use your fingers?

[3] Good call, Boomer!

[4] Mr. Wizard did an experiment similar to this where they tried to see how far you could suck water up a straw. I’ve been unable to find video of this online, but here’s one just for nostalgia’s sake.

A friend suggested posting problems the day before the answers so that readers have a chance to solve them on their own. I like that idea so I’m going to try it out today. Feel free to post your solutions in the comment section.

The WWE (formerly the WWF) has a huge following. This modern day morality play took in $475.16 million in revenues last year, but there’s one way to improve on this already booming business: incorporate anthropomorphic Japanese monsters battling it out in a cityscape arena. You could even have flying monsters like the great Mothra. In real life, how fast would Mothra have to beat her wings to stay afloat?

According to Wikipedia, Mothra weighs about 22,000 tons in adult form with a wingspan of 250^{ }m. From this wingspan, we can estimate the total area of Mothra’s wings to be about 63,000 m^{2}. Assuming her wings move 100 m down with each thrust, she will push a total volume of 6.3×10^{6} m^{3} of air downward. Using 1.2 kg/m^{3} as the density of air, we can compute the total mass of air propelled down with each thrust,

air mass = (density) · (volume)

= (1.2 kg/m^{3}) · (6.3×10^{6} m^{3})

= 7.6×10^{6} kg.

Pushing the air down will result in an equal an opposite force up on Mothra. This force must at least balance the force of gravity if she is to fly. The force of gravity can be computed as follows,

gravitational force = (mass) · (gravitational acceleration)

= (22,000 tons) · (9.8 m/s^{2})

= 2.0×10^{8} N.

Using dimensional analysis, we can construct a formula for the upward reaction force needed to keep Mothra afloat and set this equal to the gravitational force,

gravitational force = (air mass) · (range) · (frequency)^{2}.

We can then solve for the frequency,

frequency = {(gravitational force) / [(air mass) · (range)]}^{1/2}

= {(2.0×10^{8} N) / [(7.6×10^{6} kg) · (100 m)]}^{1/2}

= 0.5 Hz

This number is a lot smaller than I expected. For example, a hummingbird’s wings oscillate at 50-200 Hz. Since Mothra is so much larger than birds and insects, I expected she’d need to beat her wings really fast just to hover. Her mass—hence the gravitational force pushing down on her—will grow as her length cubed. The reaction force I calculated also grows as her length cubed since it’s proportional to the volume of air she pushes down. That may be true. After all, insects’ wings do appear to beat more frequently than birds’ wings, but I’m still a little bit skeptical. If it were correct, then large flying creatures would only need to beat their wings at the same frequency as small flying creatures to stay afloat, and that would mean that large birds could fly just like small birds, albeit with greater energy requirements since it would take more energy to move their massive wings. At any rate, I would like to hear readers’ opinions to see if my answer is crazy.

Never mistake a “hockey league for people who have never played hockey before” for a “hockey league for people who have never skated before.” Concussions ensue. During my ill-advised attempt at athleticism, I was struck by one question in particular: If we padded up a morbidly obese fat guy and shoved him in the goal, could he block every shot? How fat would a hockey goalie need to be to completely block the goal?

Consider a cylindrical goalie.^{1} Since an official NHL hockey goal is 4 ft tall and 6 ft wide, a 6.0 ft tall goalie would be more than tall enough to cover the top of the goal. To span the width, he must have a minimum radius of 3.0 ft. Using the formula for the volume of a cylinder, we can estimate his total volume,

volume = Ï€· (radius)^{2}· (height)

= 3.14159 · (3.0 ft)^{2}· (6.0 ft)

= 4.8 m^{3} (~170 ft^{3}).

Some people can float in water while others sink^{2}, so our bodies must have a density close to that of water, roughly 1.0 g/cm^{3}. From this, we can compute his total mass,

mass = (density) · (volume)

= (1.0 g/cm^{3}) · (4.8 m^{3})

= 4,800 kg.

At over 5 tons, our heavyweight would need to weigh about eight times more than the world’s largest man.

[1] Something about this statement makes me extraordinarily happy.

I did pull-downs at the gym the other day and felt unusually strong. I was lifting 300 lbs like it was nothing, and then, despite my ego’s warning, I decided to check out the equipment because it seemed too easy. It turns out the weights had not one but two pulleys on them! This means I was actually only lifting one quarter of the total weight. How many pulleys would a person need to lift the Empire State Building?

Pulleys can provide additional force to help move an object. By attaching a pulley to a weight as shown here and pulling the rope with some force, you get double that force on the weight because there are two ropes pulling up on it. If you add N pulleys to the weight, you will get 2N times as much force. The trade-off for adding a pulley is that you now have to pull the rope twice the distance that you want to lift the weight. For example, if you’re using a single pulley to lift an object 1.0 m off the ground, you need to pull the rope 2.0 m. If you want to do the same thing but with two pulleys attached to the object, then you have to pull the rope 4.0 m.

The Empire State Building weighs about 7.3×10^{8} lbs (~365,000 tons). A reasonably strong person can lift 100 lbs. To lift the Empire State Building, we’ll need the force applied by the pulleys to balance the weight of the building,

(weight of the Empire State Building) = 2 · (# of pulleys) · (force applied).

From this, we can solve for the number of pulleys,

# of pulleys = (weight of the Empire State Building) / [2 · (force applied)]

= (7.3×10^{8} lbs) / [2 · (100 lbs)]

= 3.7×10^{6} pulleys.

You’d need almost 4 million pulleys to lift the Empire State Building. Even if you could fit them all, you’d still have a lot of rope to pull. Just to lift the building 1 m off the ground, you’d need to pull the rope 4,500 miles, roughly the distance between New York and Athens, Greece.

Today’s question comes from special guest Paul Bloom. Dr. Bloom is a professor in the Psychology Department at Yale University. He has authored several books including his most recent, How Pleasure Works, which will be published this June. Dr. Bloom wants to know,

“How much time does the average person spend doing up his seatbelt? How does it compare to the time lost by summing up the increased chance of death and injury by not doing up one's seatbelt?”

Most of the time, it only takes about 2 seconds to buckle or unbuckle a seatbelt^{1}. If you drive to and from work or school each day, then you will buckle/unbuckle at least 4 times per day. By using these numbers and assuming an 80-year lifespan, we can compute the total amount of time spent buckling/unbuckling a seatbelt in one’s lifetime,

total time = (buckles per day) · (time per buckle) · (days per year) · (years)

= (4 buckles per day) · (2 s per buckle) · (365 days per year) · (80 years)

= 2.7 days.

According to Book of Odds, the odds of dying in a transportation accident this year are about 1 in 6,279. This being the case, the odds of not dying in a car accident this year are then 6,278 out of 6,279. By again assuming an 80-year lifespan, we can compute the odds of never dying in a car accident,

prob. of never dying in accident = (prob. of not dying in accident this year)^{# of years}

(6,278 / 6,279)^{80} = 98.7%.

This means our chances of dying in a car accident are 1.0-0.987 = 1.3%. About half of these deaths happened to people that weren’t wearing seatbelts. Because half the deaths occurred to people wearing seatbelts, you might be tempted to think that seatbelts don’t protect you, but that’s not the case because people wear seatbelts 83% of the time. As an example, let’s say there are 10,000 people. Of these people, 17% or 1700 people don’t wear a seatbelt and 83% or 8300 people do wear a seatbelt. Of the 10,000 people, 1.3% or 130 people die in an auto accident. Half of this 130 or 65 people were not wearing a seatbelt and the other 65 people were. This means the probability of dying without a seatbelt is given by,

prob. of dying w/o seatbelt = (# of deaths w/o seatbelt) / (# of people w/o seatbelt)

= (65 deaths without seatbelt) / (1700 people without seatbelt)

= 3.8%.

Likewise, the probability of dying with a seatbelt is given by,

prob. of dying with seatbelt = (# of deaths with seatbelt) / (# of people with seatbelt)

= (65 deaths with seatbelt) / (8300 people with seatbelt)

= 0.8%

As you can see, the probability of dying without a seatbelt is about 4 to 5 times greater.

If we assume that people of all ages are equally likely to die in a car accident^{2}, then we can estimate that on average a person loses about 40 years of his/her life after dying in a car accident. From this, we can calculate the average difference in life expectancy between people that do and don’t use seatbelts,

average time lost =

(prob. of death w/o seatbelt – prob. of death with seatbelt) · (average time lost per death)

= (0.038-0.008) · (40 years lost per death)

= 1.2 years

As you can see, for a total cost of 2.7 days spent buckling and unbuckling your seatbelt, you’re expected to gain over a whole year on your life expectancy, so buckle up!

Dr. Bloom, thanks for a fun and challenging question.

[1] I say “Most of the time” because we’ve all been in a cab, rent-a-car, or some other unfamiliar vehicle where it can take upwards of a minute to figure out how to buckle a seatbelt. I’m assuming these situations are rare enough that we can neglect them in this calculation.

[2] As I found out later, this turns out not to be the case.

I had a great time at the APS March meeting in Portland this past week, but I was struck by a bit of irony. Several of the talks had to do with energy technology and sustainability, and yet the whole idea of having thousands of physicists travel in planes to a remote conference seems at odds with this goal. Given the advent of high speed internet, Skype, etc., the technology certainly exists to have speakers present their talks as a live streaming video in the same way that Ignite and TED do. In addition to being more environmentally friendly, it would be more cost effective since APS wouldn’t have to rent a huge conference center and members wouldn’t have to pay for hotels and airfare. For just the flights, how much of a carbon footprint did the APS meeting create?^{1}^{}

In the meeting program, there are about 10 talks listed per page for about 600 pages. Most speakers only gave one talk, meaning there are about 6000 people at the conference, and that doesn’t even include all the organizers, staff, sales people, and other physicists who did not present a talk. Some attendees live in Portland and don’t have to have to fly, while others flew half way across the globe. A quick scan of the participants reveals that many live in the U.S., so as a rough estimate of the average travel distance, I’ll assume the average flight was from Omaha, NE to Portland, OR. Since physicists don’t generally like to waste money, I’ll assume most got a cheaper flight with one layover. According to one carbon footprint calculator, this contributes about 1,962 lbs of CO_{2} per attendee. From this, we can estimate the total carbon footprint to be roughly 12 million lbs of CO_{2}. It would be much more efficient—and probably cheaper—to supply every physicist with a video camera and have the talks streaming over the web.

[1] Other people have considered this problem before. If you’re interested, you can find a good discussion here.

If you’ve ever shaved off a full beard before, you may have been tempted to try out a few different facial hair configurations in the process. Maybe you start by shaving it into some pork chop sideburns, then reduce it to a goatee, then shave a little more until it’s a fu manchu before finally trying out the Charlie Chaplin look. In principle, how many different beard combinations are there?

To solve this, we need to first know how many hairs are in a beard. Obviously, this will vary from person to person. Robin Williams’ bird nest will have a lot more hairs than Sidney Crosby’s pathetic excuse for a playoff beard. I’ll assume a typical beard covers a total area of 30 cm (~1.0 ft) by 10 cm (~ 4.0 in) or, roughly, 0.030 m^{2}. On my face, individual hairs are separated by about 0.5 mm on average, meaning there are 0.25 mm^{2} per hair. From this data, we can estimate that there are roughly 120,000 hairs on a reasonably large beard.

To simplify things, I’ll say that a specific beard is defined by which hairs are shaved and which are not.^{1} That said, a hair can only be in one of two states: shaved or not shaved. With this overly simplified definition, we can easily calculate the maximum number of possible beards:

# of possible beards = 2^{120,000 }= 4.0×10^{36,123}

That is a huge number^{2}. Just writing it out would take about 15 pages.

[1] In principle, you could differentiate beards not only by which hairs are shaved and not shaved but also by how long each hair is. If any of you readers want to try enumerating this, I’d be interested to see what you get.

[2] You might be tempted to say that this number is way too big. You’re right. We have a very strict definition for what constitutes different beards. After all, if you pluck one hair out of a goatee, it’s still a goatee. There’s a pretty deep question here: how many hairs do you have to remove before it’s no longer a goatee. This question applies not only to hairs forming beards, but also to molecules forming complex structures. In physics, we try to define something called an “order parameter” that can tell what kind of structure a bunch of atoms are forming, but these definitions are always somewhat arbitrary. I’d be curious to see if anyone could define a reasonably good beard order parameter.