Towards Sustainability in Space

As I was getting on the bus the other day, a fellow passenger stopped me and asked if I was the guy who wrote the “how to estimate anything” book. This was weird for two reasons: (1) I don’t exactly have a following so it’s odd that someone would recognize me and (2) how did this guy I’d never seen before know I took the bus at 7am? Putting the flattered/creepy feeling aside, I struck up a conversation with my would-be stalker. He asked if I knew any way to calculate how many plants you would need to survive in space¹.   You could easily calculate how many plants you’d need to produce enough food to survive,² but my bus mate wanted the answer to a more challenging problem: how many plants would you need to produce enough oxygen to survive?

To estimate an answer, we need to know the rate at which humans consume oxygen and the rate that plants give off oxygen. Previously, I estimated the mass of oxygen consumed each second to be about 5.0×10^-4 kg/s. Using the molecular weight of O₂ (~32 g/mol), we can compute the number of oxygen molecules lost each second,

rate of O₂ consumption = (5.0×10^-4 kg/s) / (32 g/mol)

= 0.016 mol/s.

These O₂ molecules are lost during respiration when they converted to CO₂ via the reaction,

C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O.

During photosynthesis, plants take this CO₂ from the air and convert it back oxygen,

6 CO₂ + 12 H₂O + photons → 2 C₆H₁₂O₆ + 6 O₂.

From the chemical equations above, we can see that for each O₂ absorbed, a CO₂ molecule is emitted. To maintain a constant ratio of O₂ to CO₂, we need to set the rates of these reactions equal to each other.

To compute the rate at which plants absorb CO₂, it’s helpful to have some experience growing plants. The basil in my apartment grows about a foot tall in a couple of months, and, when it’s this size, it weighs about 3 g. From this we can compute the growth rate of the plant:

growth rate = (mass added) / (time)

= (3.0 g) / (2.0 months)

= 5.7×10^-10 kg/s

While different species of plants grow at different rates, this seems like a reasonable order of magnitude estimate.

Carbon is the main building block of living things. Different plants have different percentages of carbon in them. If we assume plants have about the same percentage of carbon that glucose has, then they’ll be about 40% carbon by weight³.  Using this estimate, we can calculate the mass of carbon that gets absorbed into plant material each second to be 2.3×10^-10 kg/s. Since each of these carbon atoms comes from one CO₂ molecule, we can calculate the number of molecules absorbed each second by dividing this by the atomic weight of carbon (~12 g/mol),

rate of CO₂ consumption = (2.3×10^-10 kg/s) / (12 g/mol)

= 1.9×10^-8 mol/s per plant.

It should be noted that this result is the number of molecules absorbed per plant. Again, we need the rate of CO₂ consumption to match the rate of O₂ consumption. To calculate the number of plants we would need for this to occur, we can divide the CO₂ consumption rate into the O₂ consumption rate,

# of plants needed = (O₂ consumption rate) / (CO₂ consumption rate)

= (0.016 mol/s) / (1.9×10^-8 mol/s per plant)

= 8.4×10⁵ plants

To sustain breathable air for one person, we would about 840,000 plants. If each plant were to require one square foot of space, you would need about 20 football fields of space to grow them.

[1] Desiree, if more people start stopping me on the street and asking me to calculate stuff on the spot, I’m totally blaming you for it.

[2] This has presumably been solved by Gary Larson, Warner Bros, and a host of other cartoonists with the result being one coconut tree.

[3] At least one reference suggests this estimate is pretty close.

Can a Kicker Get Some Love?

It’s a big weekend in football with the Super Bowl today and the hall of fame selection announcement yesterday. By scrolling through the list of pro football hall of famers, you’ll notice that no full time punters and only one full time kicker (Jan Stenerud) have ever been inducted into the Pro Football Hall of Fame. These positions generally don’t require the strength that linemen need and definitely don’t get the glory that quarterbacks and receivers do, but they are still an important part of the game, so it’s odd that they’re so underrepresented in Canton. This is especially true of kickers since they comprise nine of the top ten scorers in NFL history. (The tenth, George Blanda, was a part time kicker.) What’s a kicker gotta do to get some love? What if they made every kick they ever attempted? Could they win an MVP then? If they could make a field goal from anywhere on the field, how many more wins could their team get?

We’re going to use the 0-16 2008 Lions as our test case. Since we’re assuming their kicker, Jason Hanson, is so good that he makes every attempted field goal, we have to add 3 points for every missed field goal he had in 2007. In addition, we’re assuming he can make field goals from anywhere, eliminating the need for a punter. (Sorry, Nick Harris, you’re fired!) This means that for every punt in a game, the Lions would have gotten 3 points. Below is a breakdown of how the outcome of each game would change:

Original Score	# of punts + missed FGs	New Score	Result
34-21	6	34-39	W
48-25	6	48-49	W
31-13	5	31-28	L
34-7	8	34-31	L
12-10	9	12-37	W
28-21	4	28-33	W
25-17	6	25-35	W
27-23	6	27-41	W
38-14	5	38-29	L
31-22	3	31-31	T
32-20	7	38-41	W
47-10	9	47-37	L
20-16	2	20-22	W
31-21	4	31-33	W
42-7	4	42-19	L
41-21	7	31-42	W

The new Lions would have been 10-5-1. (The tie came because there’s no way to tell who would have won in overtime.) We’ve learned that the perfect kicker could net his team as many as 10.5 extra wins. Sadly, we’ve also learned that the 2008 Lions were so terrible that even if we spotted them 3 points every time they touched the ball, they still wouldn’t win the division and would barely make the playoffs.

Peanut Chairs Cost Twice as Much as Silverdome

In 1975, the 80,300-seat Pontiac Silverdome opened after taxpayers spent $55.7 million to bring the project to fruition. This past November, city officials sold it for $583,000, barely 1% of its original price. It goes without saying the people of Michigan deserved more for their investment. (I’m looking at you, Lions.) Some have said this was “sold for peanuts,” but I beg to differ. What costs more: the Silverdome or the number of peanuts needed to build a life-size replica of the Silverdome?

Let’s just consider the 80,300 chairs. You can construct a reasonably sized chair out of one square meter of peanuts. Peanuts are about 1.0 cm thick and 2.5 cm long, meaning we could construct a chair out of about 4000 peanuts. A single peanut weighs about one gram, and 4000 peanuts weigh about 4.0 kg. A 0.5 kg jar of peanuts costs as little as $2.00. From this info, we can calculate how much it would cost to build just the Silverdome chairs out of peanuts,

(number of chairs) · (mass per chair) · (cost per mass)

= (80,300 chairs) · (4.0 kg per chair) · ($2.00 per 0.5 kg)

=$1.3×10⁶

Peanut chairs cost twice as much as the city got for the Silverdome. Sorry, Pontiac, you didn’t even sell it for peanuts.

Reinforced Helium Balloons

Getting concrete to float on water is a common contest for first year engineering students. By molding the concrete into a boat shape, one can make it so that the mass of water displaced is greater than the mass of the concrete. When this happens, the buoyant force will allow the concrete to float on the surface of the water. Buoyancy is also what causes helium balloons to float in air. That said, I’d like to propose an even more challenging contest for advanced students: getting a concrete balloon to float in air.   How large of a concrete balloon would you need in order for it to float when you filled it with helium?

Consider a spherical balloon of radius R. The surface area A of the balloon is given by the formula,

A = 4 π R².

Since concrete is brittle and likely to crack, the balloon will need to be fairly thick so it won’t leak. If we assume its thickness T is 1.0 cm, then we can approximate the total volume V of concrete as,

V ~ 4 π R² T.

According to Hypertextbook.com, the density ρ of concrete is 2400 kg/m³. From this we can compute the total mass M of the concrete,

M = 4 π R² T ρ.

To get the balloon to float, we need the net mass displaced to be at least equal to the mass of the balloon. Put another way, the mass of air displaced by the balloon has to be equal to the mass of the concrete balloon plus the helium inside. The density of air is ρ_air =1.2 kg/m³ and the density of helium is ρ_He= 0.1786 kg/m³. The total volume of the balloon is given by,

V = 4 π R³ / 3.

By using the buoyancy condition described above, we can set up an equation to find the radius of the balloon:

(Mass of Concrete) + (Mass of He) = (Mass of Displaced Air),

(4 π R² T ρ) + (4 π R³ ρ_He / 3) = (4 π R³ ρ_air / 3).

This equation can be solved for R:

R = 3 T ρ / (ρ_air - ρ_He)

= 3 · (0.01 m) · (2400 kg/m³) / (1.2 kg/m³ – 0.1786 kg/m³)

= 70 m.

That’s a balloon 140 m in diameter or roughly one and a half football fields.

Hot Water and Popcorn Not Included

One of the great joys of life in Ann Arbor is Briarwood Dollar Movies. The price of a movie ticket ranges anywhere from 50¢ on a Tuesday to $1.75 every other day. Many of us would kill to have a movie theater-sized TV screen in our living room—so why not live in the movie theater? What costs more: a month’s rent or a month’s worth of non-stop, back-to-back movie tickets?

Movies last any where from one to three hours. If we take two hours as the average, then we’d need to see about 12 movies per day. A typical month has about 30 days in it, and four of these days will be Tuesdays. Using this info, we can calculate the cost of living in the theater,

(12 movies per day) · (4 days) · (50¢ per movie)

+ (12 movies per day) · (26 days) · ($1.75 per movie)

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

= $570

This is less than many people pay for rent in Ann Arbor. Now if only there were a place to shower...

Trust the Midas Touch

According to myth, Dionysus granted King Midas the ability to turn anything he touched to gold. All this seemed like a good idea until he accidentally turned his food, drink, and daughter into gold. (No word on how Mrs. Midas felt about any of this.) This fable seems to be missing an important ingredient: shouldn’t he be turning air into gold as well? How much air would Midas turn to gold each second?

Knowing some simple physics will help us immensely with this problem. First off, air molecules are constantly in motion, which means they’re constantly bouncing off old Goldy McGoldypants. Each time they bounce off him, the air molecules turn into gold and presumably precipitate into gold dust that falls to the ground. Each air molecule is moving with some kinetic energy—the energy of motion—where the kinetic energy KE is given by,

KE = m v² / 2,

where m is the mass of the molecule and v is the velocity or speed of the particle. From thermodynamics, it is known that the mean kinetic energy of a molecule is given by,

KE = 3 k_B T / 2,

where, k_B = 1.38×10^-23 J K^-1 is Boltzmann’s constant and T=298 K is room temperature. Most of the air is made of nitrogen gas, which has a mass of 4.6×10^-26 kg. From this info, we can solve for the average speed of the particles,

v = (3 k_B T / m)^1/2 

= [ 3 · (1.38×10^-23 J K^-1) · (298 K) / (4.6×10^-26 kg) ] ^1/2
= 520 m/s.

At that speed, we’d expect a lot of air to collide with Midas. If Midas is 1.8 m tall and 0.6 m wide, then he has a cross sectional area of 1.1 m². The number of collisions will be roughly equal to the number of molecules passing through this area. Using the velocity above and the density of air, we can compute the mass of molecules bouncing off Midas each second,

mass = (density of air) · (cross section area) · (speed of molecule)
= (1.2 kg/m³) · (1.1 m²) · (520 m/s)
= 690 kg/s.

If the air-to-gold reaction obeys mass conservation, Midas would be generating 690 kg each second. Admittedly, his lungs would almost instantly fill up gold. 

Counterpoint: Your Fermi sense may be tingling. This number seems too large. After all, this would mean that all the molecules in a fairly large room (i.e. a room with a volume of about 500 m³) would collide with Midas in one second! That seems very unlikely. We may have simplified too much. The reason is that while each molecule is moving at about 520 m/s, they are constantly bouncing off of each other so that their net speed is much less than this. (To learn more, click here.) Complicating things even more is the fact that Midas is essentially creating a vacuum by removing molecules from the air. In addition to sucking more molecules towards him, this will decrease the pressure, which may in turn lower the temperature of the room. In any event, it’s a much more complicated problem than we’re treating here. If you think you can do a better job, feel free to post your own solution. I’m curious to see what others come up with.

Helium Luggage

Airline luggage fees have gone out of control lately, and this has lead to a very predictable (and annoying) result: it takes forever to get on a plane because everyone is trying to store all their luggage as carry-ons. Many people would gladly pay the extra money for a ticket so that they wouldn’t have to deal with this frustration. I’d much rather go back to the good old days when we only had to pay a fee if our bags were over 50 lbs. Plus, I could finally use my plan of filling my luggage with helium to beat the system. How much extra weight can you add by filling your suitcase with helium?

My suitcase has dimensions of roughly 1.5 m by 0.75 m by 0.5 m (~4.9 ft by 2.5 ft by 1.6 ft), giving a total volume of 0.56 m³. About half of this is filled with clothes, toiletries, and an alphabetized stack of faculty rejection letters, but that leaves about 0.28 m³ of interstitial space filled with needlessly heavy air. Air has a density of 1.2 kg/m³, but helium only has a density of 0.17 kg/m³. Using these numbers, we can compute the amount of mass we would lose by switching the air with helium:

mass lost = [ (air density) – (helium density) ] · (volume)

= [ (1.2 kg/m³) - (0.17 kg/m³) ] · (0.28 m³)

= 0.29 kg

Ha! I just fit an extra 0.63 lbs in my suitcase. Take that, System!