Monday, May 31, 2010

Human Plants

My favorite Far Side cartoon depicts a small child hiding in the back of a beehive. The beehive has a rock-sized hole through its wall. In the foreground, a bee detective tries to figure out how a rock could have been thrown from the inside. The cartoon itself is not that funny, but there’s a great caption.
Artist: G. Larson
Medium: Ink on paper.
Title: It was late, and I was tired.

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

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

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

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

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

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

Solving for the maximum distance, we get,

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

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

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

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

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

Friday, May 28, 2010

Up

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

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

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

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

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

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

volume = (mass of the house) /  [(density of air) – (density of helium)]
= (300,000 kg) /  [(1.2 kg/m3) – (0.17 kg/m3)]
= 300,000 m3.

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

[1] In principle, I should also include the mass of the balloons here since heavy balloons will be more difficult to lift, but balloons are pretty light so I’m going to neglect their mass.
[2] To get this number, I assumed a cubic two-story house with dimensions 40 ft by 40 ft by 40 ft.  I assumed the walls and floors are 1.0 ft thick and that the house was made of some combination of wood and brick that had an average density of 1000 kg/m3.

Thursday, May 20, 2010

Hamster-Powered Mansions

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

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

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

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

Edit: After the comment I got below, I did some searching and found the hamster night light.

Wednesday, May 19, 2010

Special Guest: Sanjoy Mahajan

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

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

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

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

oil consumed = (# of fill ups) · (cost per fill up) / (cost per gallon)
= (2 fill ups per week per person) · (\$25 per fill up) / (\$3 per gallon)
= 17 gallons per week per person.

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

barrels per year = (barrels per week per person) · (# of people) · (# of weeks per year)
= (0.40 barrels per week per person) · (3.0×106 people) · (52 weeks per year)
= 6.2×109 barrels imported per year.

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

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

Tuesday, May 18, 2010

Licking Experiment

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

Wednesday, May 12, 2010

Skyscraper Farm

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

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

# of potatoes = (total area) / (area per potato)
= (257,211 m2) / (0.25 m2)
= 1.0×106 potatoes per year.

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

fraction = (Cal. per day produced) / [ (Cal. per day per person consumed) · (# people) ]
= (8.2×105 Cal. per day) / [ (2000 Cal. per day per person) · (8.0×106  people) ]
= 0.0051%

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

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

[1] That’s assuming large potatoes.

My second Book of Odds post

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

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

Tuesday, May 11, 2010

Special Guest Question from Desiree Schell

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

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

# of rolls = (fraction of players) · (# of people) · (rolls per player per game) · (# of games)
= (0.01 players/person) · (6.7×109 people) · (1000 rolls/player/games) · (50 games)
= 3.4×1012 rolls

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

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

[1] Some might quibble with my 1% assumption.  I would argue it’s a good order of magnitude estimate.  If I chose 10 people at random, there’s a good chance none of them are into role-playing, so the actual number is almost certainly less than 10%.  However, if I go through 1000 people at random, there’s almost certainly going to be one person into role-playing, so it’s probably greater than 0.1%.
[2] I cheated a little here.  This number seemed much too large, but I consulted my friend Josh who is more into gaming than I am, and he assured me that it’s a good estimate.  Thinking back to my days playing Risk (where you roll multiple dice for hours), I figured it’s good to within an order of magnitude.

Monday, May 10, 2010

If All Men Were Created Equal

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

According to Wikipedia,
“At 3.79 million square miles (9.83 million km2) and with over 309 million people, the United States is the third or fourth largest country by total area, and the third largest both by land area and population.”
Using this data, it’s easy to calculate the area per person,

( area ) / ( # of people )
= (3.79 million square miles) / ( 309 million people )
= 0.012 square miles.

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

Sunday, May 9, 2010

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

In honor of Mother’s Day, I’d like to do a problem suggested by my mom. She’s a huge Red Sox fan, and she’s curious about the Jordan’s Furniture promotion going on at Fenway Park. According to Boston.com,
“customers who buy anything at one of the regional chain's stores from April 1 through May 2 won't have to pay for their purchase if a Red Sox player hits the image of a Jordan's Furniture baseball on the company sign at Fenway Park between July 15 and the end of the season Oct. 3”
Mom would like to know, “What are the odds that a Red Sox player will hit the Jordan’s sign?"

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

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

probability of correct angle = (angular width of sign) / (total angular width)
= (0.0284 radians) / (1.57 radians)
= 0.018.

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

(0.996)47 = 0.83,

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

Happy Mother’s Day, Mom!

Thursday, May 6, 2010

Special Guest Dan Meyer

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

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

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

(10 sections) · (20 pages/section) · (6 columns/page)
· (90 lines/column) · (30 characters per line)
=  3.2×106 characters.

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

Tuesday, May 4, 2010

Elevator of Death

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

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

P = m g / A,

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

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

A · velevator = A’ · vair

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

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

P = ρ · vair2 / 2,

or, equivalently,

vair = ( 2 · P / ρ )1/2,

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

velevator = [ 2 · m · g / (A · ρ) ]1/2 (A’ / A)
= {2 · (2000 kg) · (9.8 m/s2) / [(4.0 m2) · ( 1.2 kg/m3)]}1/2 (0.5 m2) / (4.0 m2)
= 11 m/s.

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

We Have a Winner!

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

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

volume = (mass) / (density)
= (1500 million tonnes) / (1.0 g/cm3)
= 1.5×109 m3.

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

depth sunk = (total volume) / (area of the oceans)
= (1.5×109 m3) / (3.6×1014 m2)
= 4.2 microns.

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

Congratulations to our winner, Nathan Moore!