Monday Musing: Useless Calculations

I am a nerd. I used to be an engineer and so I like calculating stuff in my head. I hardly ever use electronic calculators, even when exact calculations of something are needed, preferring to do them by hand (try it, it can be a soothing thing) on chit’s of paper, backs-of-the-proverbial envelopes, etc. But a lot of the time, I am just calculating really stupid things for fun in my head, especially if I am sitting somewhere (doctor’s office, airport, porcelain throne, bed-before-sleep) with nothing to do. I also have other ways of amusing myself in such situations. For example, I might endlessly rewind and replay a conversation I had with someone over and over in my head, like a TV program, which I realize makes me weird but also remarkably patient with things like flight delays. But mostly, I calculate.

Full_moon_largeAnother thing I do is collect weird quantitative facts about stuff in my head (I have a pretty good memory for numbers; for other things… well, not so much–as unfortunately many people have found out upon meeting me for the second time! ;-). Quick, how much does a fully loaded 747 weigh? How much of that weight is fuel? How dense is gold compared to water? What is the radius of the moon? What is Avogadro’s number? I happen to know these and many other (mostly) useless things. I don’t know why, but I suck them up out of magazines and things like that, and some I remember from high school and college textbooks. (It helps that I am a big rereader of books.) I am also the type of person who reads his car manual from beginning to end, and idiotically remembers what the capacity of the windshield-washer-fluid tank is.

I use these useless things to calculate even more useless things (while waiting in the aforementioned doctors’ offices, airports, etc.). But I don’t calculate things exactly (most of the time), I just like to estimate stuff very roughly. Today, for example, I estimated (by looking while sitting on my balcony) that the amount of water flowing by in the river next to me (the Eisack) every minute is enough for everyone living in my city of Brixen to flush his/her toilet about 10 times each day (or enough for about 200,000 flushes). This was pretty simple to do:

  • Screenhunter_02_sep_12_1543Sometimes, the water management authorities dam up most of the water temporarily in the river, so I have seen the bottom of the river (or at least the larger rocks on the bottom–some water is always flowing), and so I can estimate the (higher today) average depth of the river just by looking at it. I’d say it’s about 2 feet.
  • The river looks about 50 feet across over here. (It’s wider in the photo at the right, which I took at a different spot.)
  • I timed a bit of driftwood floating down the river and in 10 seconds (one-thousand one, one-thousand two…) it went about 60 feet–it flows fast because of the steep downhill grade in this mountainous area–so about 6 feet per second.
  • I confirm my estimate of 60 feet in ten seconds in my head by noticing that the driftwood is floating just a tiny bit faster than a person walking fast in the same direction on the path next to the river. A fast walking person goes about 4 miles per hour, and 6 feet/second X 3600 seconds/hour = 21,600 feet/hour, and 21,600 feet/hour X 1 mile/5,280 feet = (approximately) 4 miles/hour. Checks out. Good.
  • The cross-sectional area of the river is 50 feet X 2 feet = 100 square feet.
  • The volume of water flowing by in a second is therefore 100 square feet X 6 feet = 600 cubic feet.
  • Newer commodes often have written on them the amount of water they use per flush. Most often I have seen the figure 6 liters/flush. Now, the problem is converting cubic feet to liters.
  • To do this, I think the following: I know that a cubic meter is 1000 liters. How many cubic feet are in a cubic meter? Well, I remember that there are about 3.3 feet in a meter, so 3.3 X 3.3 X 3.3 = (approximately) 36 cubic feet/cubic meter.
  • So, we have 1000 liters/cubic meter X 1 cubic meter/36 cubic feet = (very approximately) 30 liters/cubic foot.
  • Now 1 flush/6 liters X 30 liters/cubic foot = 5 flushes/cubic foot of water.
  • 5 flushes/cubic foot X 600 cubic feet/second (from above) = 3000 flushes/second.
  • 3000 flushes/second X 60 seconds/minute = 180000 flushes/minute of river flow.
  • 180,000 flushes/20,000 persons = 9 flushes/person, from a minutes worth of water flow, which I rounded up to 10 just ’cause it sounds better when I tell my wife this astoundingly impressive fact. 🙂 (Yeah, yeah, I know she’s sick of crap like this…)

Incidentally, it just occured to me as I write this that the amount of water flowing by every second (600 cubic feet) in the river weighs as much as about 18 Toyota Corollas (and this is not a very big river). I leave it as an exercise for the reader to convince him/herself of the approximate truth of this.

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Here is one more example: recently (on a train) I wondered how much the air in the Empire State Building weighs. Here is how I went about estimating the answer:

  • Empire_state_buildingI read somewhere that a north-south NY city block is about a 20 of a mile. I confirm this rough figure in my mind by thinking that Manhattan is about 12 miles long and the northernmost streets are numbered around 215 or so. Since there is a bit of Manhattan below 1st street, I figure 200 blocks divided by roughly 10 miles gives a nice round number of 20 blocks per mile. Good.
  • A mile has 5280 feet, so a 20th of that is half of 528 feet, about 250ish feet. (It’s a rough calculation!)
  • It seems to me that the area of the footprint of the building (from having seen it many times) is probably close to the square of a city block (it actually is more rectangular, with the north-south dimension a bit less than a block and the east-west one a bit more), so let’s just say 250 X 250 feet, which is 62500 square feet, or roughly (remember, I have to keep this stuff in my head! And I’ll round up this time, since I rounded down last time) 70,000 square feet.
  • It’s a little broader at the bottom floors and tapers sharply starting at the 86th through the 102nd floors, I think, so I’ll just say it is 90ish stories.
  • Let’s say 10 feet (surprise, a nice round number!) of height for each floor, so multiplying by the area of the footprint, we get 10 X 90 X 70000 = 900 X 70000 = 63,000,000 cubic feet of internal space. You with me?
  • I’ll say about a sixth, or roughly 13 million cubic feet of this is probably taken up by solid stuff including people, internal supports, furniture, etc., so we’re left with a nice round number: 50 million cubic feet of air.
  • Now I just happen to know that the  density of air is about 0.08 pounds per cubic foot (at sea level and normallish temperatures), but even if I didn’t, I just remembered reading somewhere that air is about 800 times lighter than water, and knowing the density of water I could have figured it out easily enough.
  • So, the weight of all the air in the Empire State Building is… 0.08 X 50,000,000 or 8 X 500,000 which equals… (drumroll, please) 4,000,000 pounds!

Which, as it happens, is 2,000 Toyota Corollas, or ten times the weight of a fully loaded Boeing 767 (by now you know not to ask why I know this!), like the one which crashed into the World Trade Center. Each tower of the WTC was bigger than the Empire State, so it is interesting to note that the weight of each of the planes that struck it (the other plane was slightly smaller), was less than a tenth of just the weight of the air inside the building.

What’s surprising about such estimates is how often they are very close to the reality. This is especially true in a multi-step approximation, where over- and underestimates at various steps tend to cancel each other out, usually resulting in something not too far off from the truth. To convince you of this, I emailed my friend, the mathematician John Allen Paulos, and asked him to estimate the weight of the air inside the Empire State Building. I told him he could look up the density of air, but nothing else, and to tell me his reasoning. This is what he wrote back:

Here’s my quick back of the envelope rough calculation of the weight of the air in the Empire State Building:

The building is about 1200 feet high and at ground level it a large square which then tapers as the building rises. I guess that on average it is about 200 feet by 200 feet. This gives us 48,000,000 cubic feet for its approximate volume. Since the density of air at sea level is about 1.2 kg/cubic meter or, translating into English units, roughly 2.5 pounds/35 cubic feet, the approximate weight of the air in the building is 48,000,000 x 2.5/35 or about 3.4 million pounds, somewhere around 3 or 4 million pounds.

The thing to notice here is that while John’s individual assumptions are significantly different from mine (for example, my estimate of the area of the footprint of the building, 70,000 square feet, was 75% greater than his estimate of 40,000 square feet), in the end things kinda’ even out and my answer of 4 million pounds is less than 20% greater than his answer of 3.4 million pounds.

But how can we know the actual figure? We cannot. We can only get closer and closer approximations by measuring things more and more accurately (the volume, not just of the building, but of everything in it, which must be subtracted). It’s not like there’s an easy way to pour the air out of the building and weigh it!

The fun in doing these estimates is in NOT looking anything up, and instead trying to answer questions by using, along the way, what we do know to estimate everything we need to know to answer our question.

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Suppose you have a baterium cell of a kind which divides into two every minute. (Normal bacteria like E. Coli divide about twenty times slower than that, but it’s just an example.) Now you put this cell into a large jar (with lots of bacterium food) at 11 AM. In one hour, at 12 noon, the jar has just completely filled with bacteria. Can you work out the time between 11 AM and 12 noon when the jar was half full? Can you estimate it? Go ahead and keep the figure in your head. I’ll give you the answer later.

Meanwhile, let me say a few words about doubling times. Let’s say you have an investment which is earning 10% interest per year. How long will it take for you to double your money?

There is a very simple little rule which works quite well in approximating doubling times for rates of growth between -25% (that’s “minus” 25%) and 35% or so (and very accurate for single digit percentage rates of growth), which goes like this: just divide 70 by the percentage rate of growth, and you have the time needed to double the quantity. (The reason this works is a little complicated and would require me to explain stuff I don’t want to get into at the moment.)

So, what is the answer to the question above: how long will it take to double your money if it is growing at 10% annually? The answer is simply 70 divided by 10, or 7 years. Say a country’s population is growing at the rate of 2% annually. How long before it doubles? 70 divided by 2, or 35 years! This rule is very useful in doing the rough mental estimates that I like to do.

I’ll give one last example: I read somewhere recently that the total energy consumption of the world is currently approximately 5 X 1020 Joules per year, and worldwide energy consumption is increasing at a little over 2% annually. (This rate is expected to go up, not down, in the next couple of decades. China’s energy consumption has been growing at double-digit rates!) The following question occured to me: at this rate, how long will it take before we outrun the total amount of energy which is coming in from the sun? (Fossil fuels are just a stored form of this solar energy, and renewable forms of energy like wind power, are also just a small subset of the total radiant energy we receive from the sun daily.) Here’s how I went about estimating how long it would take:

  • Screenhunter_04_sep_12_1554I know (I did some research on solar panels a few years ago) that the total radiant power coming in from the sun per square meter is about 1400 Watts (1 Watt of power is a Joule of energy per second).
  • Half the world’s surface (the side facing the sun) receives energy at this rate. What is the area of this region? Well, it is just a circular cross section of the Earth, and the radius of the Earth is about 6,000 kilometers.
  • The area of a circle is Pi X radius X radius, which is 3 X 6000 X 6000, or approimately 100 million square kilometers, in our case.
  • One square kilometer is 1000 meters X 1000 meters = 1 million square meters, so we have a total area receiving solar energy of 100 million square kilometers  X 1 million square meters/square kilometer, or 100 trillion square meters.
  • 100 trillion square meters X 1400 Watts/square meter = 1.4 X 1017 Watts of power, or 1.4 X 1017 Joules per second.
  • So in a year we have 60 X 60 X 24 X 365 seconds or approximately 60 X 60 X 20 X 400 = 28,800,000, or about 30 million seconds = 3 X 107 seconds.
  • 1.4 X 1017 Joules/second X 3 X 107 seconds/year = roughly 4 X 1024 Joules of total radiant energy from the sun every year.
  • Let’s just round it up to 5 X 1024 Joules. Remember, our current world wide consumption is 5 X 1020 Joules annually, or only 1/10,000th(!) of the total radiant energy of the sun that falls on the Earth every year. This seems a tiny fraction, but consider:
  • At 2% annual growth in worldwide energy consumption, we double consumption every 35 years (by the approximate doubling time rule given above).
  • How many times do we need to double consumption to reach 10,000 times our current level? This is just log2 (10,000). I know that 214 is 16,384 (I was a programmer!) and this is more than the factor of 10,000 that we need. So let’s just say we need 14 doublings.
  • At 35 years/doubling X 14 doublings, we get 490 years.

In other words, given our current worldwided energy consumption, and the fact that it is growing at more than 2% per year, if it were to continue to grow at that rate, we will have outstripped ALL the energy coming in from the sun in less than 500 years! Pretty shocking, no? And if we took into account the solar energy that is absorbed by the atmosphere before reaching the surface of Earth, and things like that, we have MUCH less time during which we can sustain 2% growth in energy consumption. I know very little about economics, but I wonder if economic growth rates are related to energy consumption rates in any straightforward way. (Robin?) If so, this points to a cap on economic growth as well. So that’s my nerdy column for today.

Oh, and yes, the answer to the bacteria question: the jar will be half full at 11:59 AM. Just think about it for one minute!

All my previous Monday Musings can be seen here.

Have a good week!

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