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## Mathematics in real life

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 $600$ 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 $4 \times 10^4$ 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 $1600$ kilometers per hour, or about $1000$ miles per hour.

Now, suppose the flight takes $x$ hours. Then I have traveled a distance of $600x$ miles, but at the same time I have crossed approximately $\frac{3}{5} x$ time zones. So the difference in local times should be approximately $\frac{8}{5} x$. Setting this equal to $12$ and rounding $\frac{3}{5} x$ to the closest integer, it follows that the time difference between Boston and London is $5$ hours (which it is) and that my flight will take $7$ 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.)

### 7 Responses

1. on October 1, 2010 at 8:08 am | Reply Patrick Hulin

As far as I know, though, planes tend to travel in great circles (since those are geodesics). So equatorial distance is the measurement you want, but you don’t want the plane flying east-west.

• on October 1, 2010 at 8:19 am | Reply Qiaochu Yuan

Again, this was a pretty rough calculation. This is also part of what cancels with the subtleties mentioned in hernan’s answer.

2. on October 1, 2010 at 12:37 am | Reply Peter Cameron

Another back-of-the-envelope calculation: how far can you see from the plane if the weather is clear? I was reminded of this when a maths teacher of my acquaintance asked me about practical applications of Euclidean circle theorems. Multiply your height above sea level (in metres) by 13 and take the square root; the result is the distance to the horizon in kilometres. (For feet and miles, replace 13 by 3/2.)

3. on September 29, 2010 at 9:20 am | Reply hernan

It appears that your calculation assumes that the plane is flying strictly east/west , no?
More to the point, your “circumference of the earth” is equatorial, but is that relevant here? Shouldn’t it be measured along the paralell along which your are travelling?

• on September 29, 2010 at 11:26 am | Reply Qiaochu Yuan

You’re right. I made several simplifying assumptions, both because this was a spur-of-the-moment calculation and because I was lazy. But the two subtleties you describe affect the answer in opposite ways and so approximately cancel. (This is why I get the right answer for the number of time zones but not for the distance from Boston to London.)

4. on September 28, 2010 at 7:48 pm | Reply Adrian Petrescu

Perhaps I missed it, but shouldn’t the rotation of the Earth factor in? You approximated the angular velocity of the earth at 1600 km/h, but then I don’t see how it factored in to the distance you traveled. Or are you just assuming it’s negligible? (Which it can’t possibly be, in 7 hours even a plane that just hovers still in the air will be translated by over 1/4 of the circumference of the Earth, right?)

• on September 29, 2010 at 6:02 am | Reply Qiaochu Yuan

Sorry, I misspoke; when I said “airspeed” I meant “ground speed.”