Estimation Contest Round II

Check out the blog next week for a chance to win a signed copy of How Many Licks?

A Whale of a Time

Fox once aired a show called Man vs. Beast, featuring several of nature's greatest rivalries: world class sprinter Shawn Crawford vs. a zebra, hot dog eating champion Takeru Kobayashi vs. a Kodiak bear, and—my personal favorite—a team of little people competing with an elephant to pull a DC-10 airliner. This type of quality programming inspires mathematical minds. What other great match-ups could we have? For example, how many musclemen would it take to hold back a whale?

Balaenoptera musculus, the blue whale, averages 150 tons (1.5×10⁵ kg) and can reach a top speed of about 31 mph, which is roughly 14 m/s.  Assuming it takes 10 seconds to reach top speed, we can calculate the force the whale uses to accelerate itself using the formula,

force = (mass) · (change in velocity) / (change in time)

= (1.5×10⁵ kg) · (14 m/s) / (10 s)

= 2.1×10⁵ N.

Here, “N” is short for “Newton”, a unit of force. The musclemen would need to exert an equal and opposite force on the whale to hold him back. In the gym, it’s common to see some of the bigger men doing lateral rows with 200 lbs, which corresponds to 980 N.

From this we can calculate how many musclemen we’d need to hold back a whale,

number of musclemen = (force per whale) / (force per muscle man)

= (2.1×10⁵ N per whale) / (980 N per muscle man)

= 214 musclemen per whale.

It would take over 200 fairly strong people to hold back a whale. Check out Diary of Numbers tomorrow, as we’ll have a question from a special guest!

Asteroid Attack!!!

Every few decades, an asteroid comes dangerously close to Earth’s orbit. While an impact with the Earth would surely cause widespread destruction of civilization and an end to life as we know it, this outcome may be preferable to sitting through Armageddon, Deep Impact, Meteor, or whatever is the next in a long line of horrendously clichéd movies about giant space rocks hurtling towards the Earth threatening to, once again, destroy life as we know it. But if an asteroid were about to hit the Earth, could we save our planet by all jumping simultaneously? If everyone on the planet gathered on the streets of New York and jumped, how fast would the Earth be pushed out of the way?

There are 6.7×10⁹ people in the world and people weigh roughly 65 kg (~140 lbs) on average. This means the total mass of all humanity is given by,

total mass = (mass per person) · (# of people)

(65 kg per person) · (6.7×10⁹ people)

= 4.4×10¹¹ kg

If they all jump 0.3 m (~1.0 ft) off the ground, then we can compute the total potential energy from the jump using the formula below,

potential energy = (total mass) · (gravitational acceleration) · (height)

= (4.4×10¹¹ kg) · (9.8 m/s²) · (0.3 m)

= 1.3×10¹² kg.

Using the conservation of energy and the formula for kinetic energy, we can compute the speed at which they would leave the ground,

velocity = [2 · (kinetic energy) / (total mass)]^1/2

= [2 · (1.3×10¹² kg) / (4.4×10¹¹ kg)]^1/2

= 2.4 m/s

People leave the ground with a speed of about 2.4 m/s. The law of conservation of momentum states that when two things push off each other, the momentum of one thing is equal and opposite to that of the other thing. The momentum of an object is equal to its mass times its velocity, so we can compute the momentum of the people,

momentum = (mass) · (velocity)

= (1.3×10¹² kg) · (2.4 m/s)

= 3.1×10¹² kg·m/s.

The Earth’s mass is about 6.0×10²⁴ kg. From this, we can calculate the speed at which the Earth will recoil,

Earth’s velocity = (momentum) / (mass)

= (3.1×10¹² kg·m/s) / (6.0×10²⁴ kg)

= 5.2×10¹³ m/s.

The Earth would only move at about 5.2×10^-13 m/s. At that speed, the Earth would take an hour to move a distance of only a few atoms!

The Great Heater of Buffalo

Much of the northeast U.S. is currently digging its way out from all the snow. Snow is a big pain in many northern states. Upstate New York can get over 2.8 m (~100 in) of snow a year. Why don’t we just run a giant heater to melt it all? How much would it cost to melt all the snow in Buffalo?

The total area of Buffalo, NY is about 2.6×10⁹ m² (~1000 miles²). From the numbers above, we can compute the total volume of snow,

volume = height · area

= (2.8 m) · (2.6×10⁹ m²)

= 7.3×10⁹ m³.

Snow has a density of about 80 kg/m³. From this, we can compute the total mass of snow,

mass = density · volume

= (7.3×10⁹ m³) · (80 kg/m³)

= 5.9×10¹¹ kg.

The amount of energy it takes to melt a solid is called the heat of fusion. The heat of fusion of ice is 334 kJ/kg, meaning it takes 334 kJ of energy to melt 1 kg of ice. From the previous problem, the cost of energy is about 12.56 cents per kW·hrs. From this data, we can compute the total cost to melt one year worth of snow in Buffalo,

cost = (total mass of snow) · (heat of fusion of ice) · (cost of energy)

= (5.9×10¹¹ kg) · (334 kJ/kg) · (12.56 cents per kW·hrs)

= $6.9×10⁹.

At almost $7 billion a year, it would cost about 40% of NASA’s entire budget.

A Dry Calculation

Just finishing up the laundry and wondering, how much money would we save in our lifetime if we used clothes lines instead of electric dryers?¹

There are two ways to calculate this depending on whether or not you use your own electric dryer or one from a laundromat. For the laundromat, it generally costs around $1.25 to do a load of laundry. If you do a load of whites and one load of colors each week, then you spend $2.50 per week or $130 dollars per year. If you live for 80 years, you’ll waste $10,000 in your lifespan.

If you own your own dryer, it’s a little more complicated because you’re paying for electricity. A single dryer load takes anywhere from 30 minutes to one hour, so we’ll assume 45 minutes. If you do two loads each week, you’re running it for 1.5 hours. At least one reference lists a dryer’s power consumption as being about 4.0 kW. From the numbers above, we can calculate the total amount of energy used each week,

energy = (power) · (time)

= (4.0 kW) · (1.5 hours)

= 6 kW·hrs per week.

In Michigan, the cost of electricity is about 12.56 cents per kW·hr. From these numbers, we can compute the cost per week,

total cost per year = (number of kW·hrs per week) · (cost per kW·hr)

= (6 kW·hrs per week) · (12.56 cents per kW·hrs)

= 75.36 cents per week

= $39.18 per year.

Using the same lifespan as above, you would save about $3100 dollars in electric bills over a lifetime. In addition, you would save money by not purchasing a dryer in the first place. A quick Google product search shows that electric dryers cost around $400. Dryers have to be replaced about once every 8 years. This means you’d need to buy about 10 dryers in a lifetime, costing an extra $4000 dollars bringing the total savings to about $7100.

In short, dry your clothes on a line when you can. It’s good for the environment and it can save you thousands of dollars.

[1] While researching this problem I can across a very good site here.

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.