## Six Minutes Off

Let me return, reindeer-like, to my problem, pretty well divorced from the movie at this point, of the stranded Arthur Christmas and Grand-Santa, stuck to wherever they happen to be on the surface of the Earth, going around the Earth’s axis of rotation every 86,164 seconds, while their reindeer and sleigh carry on orbiting the planet’s center once every hours. That’s just a touch more than every 5,091 seconds. This means, sadly, that the reindeer will never be right above Arthur again, or else the whole system of rational and irrational numbers is a shambles. Still, they might come close.

After all, one day after being stranded, Arthur and Grand-Santa will be right back to the position where they started, and the reindeer will be just finishing up their seventeenth loop around the Earth. To be more nearly exact, after 86,164 seconds the reindeer will have finished just about 16.924 laps around the planet. If Arthur and Grand-Santa just hold out for another six and a half minutes (very nearly), the reindeer will be back to their line of latitude, and they’ll just be … well, how far away from that spot depends on just where they are. Since this is my problem, I’m going to drop them just a touch north of 30 degrees north latitude, because that means they’ll be travelling a neat 400 meters per second due to the Earth’s rotation and I certainly need *some* nice numbers here. *Any* nice number. I’m putting up with a day of 86,164 seconds, for crying out loud.

Unfortunately, that extra wait — it comes to more nearly 6 minutes, 25.87 seconds — carries them something like 154,348 meters away from where the reindeer are, on their line of latitude. They may have a day to do something about the problem, but I don’t see them hiking 150 kilometers given that they start out without anything but the contents of their pockets.

Waiting for tomorrow is worse: after Arthur and Grand-Santa have sat up for two days, the reindeer will have made just over 33.84 orbits of the Earth, so they miss by 12 minutes and nearly 52 seconds and that’s clearly out. How long are Arthur and Grand-Santa going to have to wait for the reindeer to be in a reasonable range?

Well, after 13 days, the reindeer are going to have made 220.014 orbits, so the reindeer got to the latitude a mere 1 minute, 14.86 seconds before Arthur and Grand-Santa get to the right longitude. They miss each other by about 30 kilometers, but if Arthur at least can’t make 30 kilometers in 13 days he’s not applying himself. Things are even better if they can wait 66 days, or for the reindeer 1116.99772 orbits; then they’ll have missed the rendezvous by only 11.57 seconds, or just 4,630 meters. This brings them into March, admittedly, spoiling the deadline pressure of the movie, but I’m worrying about the mathematics and not the narrative requirements. Besides, maybe it’s Leap Year.

If they’re willing to utterly forget the movies and all other responsibilities, there’s an even closer miss coming up in a mere 3,061 days, when the reindeer will have finished 51,805.0006 orbits. They miss one another by just over three seconds, or 1.2 kilometers, and if you can’t move less than a mile in over eight and a third years you’re not doing much credit to the name “Santa”.

What’s at work here is that Arthur and Grand-Santa, and the reindeer and the sleigh, are going to come close to one another whenever there’s a particularly close match between the number of days Arthur’s waited times the length of that day and the number of orbits the reindeer have made times the time an orbit takes. If we wait long enough, the number of days divided by the number of orbits will hit on something tolerably close to the number of seconds in hours divided by the 86,164 seconds in Arthur’s day. Whenever the ratio between days and orbits gets to be very close to , or about 0.059 086 960 036 014, Arthur and Grand-Santa will be in hailing distance of the reindeer and sleigh.

This does mean it’s possible to specify just how close you want the two parties to get and figure out the minimum waiting time required. Or if you’re looking for a less challenging arithmetic puzzle, let Matlab or the equivalent do the hard work. (I suspect that figuring out the best combination of days and orbits is probably one of continued fractions, which I just look at from afar and smile at without approaching.) As I make it out, the two get to a really fantastic match, missing one another by a scant 23 centimeters, if Arthur and Grand-Santa can hold out for just … well, 1,837,779 days, and the reindeer hold out for 31,102,954.000 000 1 orbits. 5,031 years and six and a half months is a bit of a wait, though, and probably impractical for these purposes. But they’d be really, really close together.

And if they’re really patient, they can get closer yet, though they won’t top that in the first ten thousand years of waiting.

## Arthur Christmas and the End of Time | nebusresearch 3:35 am

onThursday, 7 February, 2013 Permalink |[…] Six Minutes Off | ne… on How Fast Is The Earth Spi… […]

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## nebusresearch | The Arthur Christmas Problem 6:18 pm

onWednesday, 24 December, 2014 Permalink |[…] “Six Minutes Off” shifts matters a little, by supposing that they’re not on the equator, which makes meeting up the reindeer a much nastier bit of timing. If they’re willing to wait long enough the reindeer will come as close as they want to their position, but the wait can be impractically long, for example, eight years, or over five thousand years, which would really slow down the movie. […]

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