“Be kind, for everyone you meet is fighting a hard battle” - Often attributed to Plato but likely from Ian McLaren (pseudonym of Reverend John Watson)

Saturday, May 31, 2014

Stopping time

Image credit: http://www.mastermarf.com
I recently completed a business trip to Thailand entailing some 33 total hours in the air. My client won't reimburse business class, let alone first class, so "economy plus" was the best that I could do. On Delta, that gives me 4" of extra legroom and a few extra degrees of recline in the seat. That doesn't sound like a lot but it makes a huge difference!

In any case, while on such flights, I bring along work and recreational reading and, typically, attempt to get some sleep, I also tend to give some thought to something to do with the flight. For example, I've made several posts about using the accelerometer in my smart phone to determine acceleration and takeoff speeds and distances. On another trip, I used the pressure sensor in my Galaxy Note 3 to estimate the cabin altitude of the aircraft in cruise. While there are apps that will automatically give you altitude based on the pressure sensor, I hadn't (and haven't) yet installed one so it was an interesting little exercise in calculus, fluid mechanics, and the ideal gas law to estimate that the cabin altitude was about 6,700 feet.

On the most recent trip, I was wondering about my relationship to the sun as we chased it (westbound) or rushed toward it (eastbound). It occurred to me that, for any speed up to about 1040 m.p.h., there's some latitude where that westbound speed would keep the sun at the same position in the sky. At the north or south pole, if the sun were visible at all, you wouldn't even have to move, you'd only have to rotate.

In any event, it was a pretty trivial exercise in spherical coordinates to find the appropriate latitude at a speed of 500 m.p.h. (a reasonable speed for the B747-400 at altitude flying into a typical not too strong headwind). For any speed, the equation is ~ \theta =\arccos\left(\frac{24v}{2 \pi R}\right)~, where v is the speed in miles per hour and R is the Earth's radius in miles (of course, any units of speed, i.e., length/time, and length will work so long as they're consistent and the "24" is changed to the number of whatever time unit that's used that comprises a day). Substituting 500 for v and 3,957 for the radius, I find that a direct westbound path at about 61.14 degrees north (or south) latitude will make time stand still. Or, at least, it will make the sun stand still in the sky. Were I to walk at 3 m.p.h., I'd need to be at about 89.8 degrees north (or south) latitude. This would be around 11 miles from the north (or south) pole.

Of course, I'd be jolted forward a whole 24 hours each time I crossed the International Date Line so, sadly, I can't use this method to stay Forever Young.


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