Posted in Uncategorized, tagged estimation on September 28, 2010 |
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A small example, but I thought it was funny.

I am currently at Logan waiting for my flight to Heathrow. An hour or so ago, one of my friends asked me how long my flight is. I knew that my flight would depart at about 7:30pm and arrive at about 7:30am, but both times are local. So the actual length of the flight is about 12 hours minus the time difference – which I didn’t know!

But then I realized I could **compute** the time difference because I knew two other things – the average ground speed of a commercial airplane, and the circumference of the Earth. The average ground speed of a commercial airplane is about miles per hour, which I know from idly staring at that one channel that monitors the airspeed of a plane. The circumference of the Earth is kilometers (to an accuracy of better than one percent!), which I know from preparing for the Fermi Questions event at Science Olympiad. (This is a very handy number to know for certain types of estimates, such as this one.) Given this number, it follows that the velocity of the surface of the Earth is about kilometers per hour, or about miles per hour.

Now, suppose the flight takes hours. Then I have traveled a distance of miles, but at the same time I have crossed approximately time zones. So the difference in local times should be approximately . Setting this equal to and rounding to the closest integer, it follows that the time difference between Boston and London is hours (which it is) and that my flight will take hours (which it will).

An interesting idea this computation illustrates is that if you can estimate an integer (in this case, the number of time zones my flight will cross) with enough precision, you know it exactly. A more sophisticated variant of this idea is that a continuous function from a connected space to a discrete space must be constant.

(Full disclosure: I messed up the last step when I did this calculation the first time.)

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