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It is a little known fact that Santa Claus is in fact a vampire. Why else would he spend so much time near the North Pole and travel only at night? Yes, he fears daylight!

His problem is that during the summer, there is plenty of daylight on the North Pole. This is why, every equinox, he migrates to the other pole. In March he flies to the South Pole, and in September, he flies back to the North Pole.

To avoid daylight, he has to complete the trip during the night. He could leave the North Pole at 18:00 GMT (6pm), fly down along the Greenwith meridian and reach the South Pole at 6:00 GMT (6am). He would have to fly 20000 km in 12 hours at ~1666 km/h. Santa could easily fly that fast. It was proven he flies faster than the speed of light to deliver all the presents in one night. But he wants to relax and enjoy the view.

So the question is: What is the minimum speed Santa needs to fly to travel from the North Pole to the South Pole while avoiding all daylight.

It is assumed the Earth is a perfect sphere with perimeter 40000 km, the Sun is over the equator and shines over exactly 1/2 of Earth surface. The altitude of flight is negligible.

Florian F
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2 Answers2

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Santa can travel at approximately 761.15 kph all the way. At 18:00 GMT he leaves hugging the line of sunset (momentarily he will be travelling directly south, but as he drops to lower latitude he has to fly at an increasing angle towards the west so viewed from above the north pole his progress would vaguely resemble a spiral). Eventually, at a latitude of approximately 62.8 N, Santa will find he is travelling almost due west, and can only just keep up with the line of sunset. At that point, he starts to swing gently south (at any particular latitude L, he wants to follow a direction d = 270 - acos(0.46*cos(L)) - this points in a more southerly direction as he gets closer to the equator). Initially he will only slowly fall back from the line of sunset, but the further south he goes the more local time he loses until, as he crosses the equator (at a heading ~ 207 degrees), the local time will be midnight. He then "mirrors" his course so far in the southern hemisphere. If his heading is correct he will finally get to the south pole just on local sunrise. Any minor mistakes and Santa is toast.

Santa can only calculate his exact minimal speed by accurately solving an integral equation (if he has similar maths skills as me he will have to do this numerically), so he could reduce his speed a little below my estimate with a more accurate calculation. But the closer to the limiting speed he travels at the longer the journey will take (at 762 kph it already takes him about 20 days to make the trip), so unless he has lots of time to spare he may want to travel a little faster than the limit. Also, if he does decide to travel as close as possible to the limiting speed he will need to take into account the fact that the orientation of the sun to the north-south pole line will be changing during his trip, so he will need to solve an even more complicated integral (and leave the pole some time before the spring solstice).

The path near the pole is simple - he can just follow the sunset line (this part is reminiscent of the lake-monster problem). But the next section is a bit more complicated. Consider his passage between some latitude L and L-dL, the optimal angle to cross that latitude will be that which, for any specific speed he is travelling, loses him the least of his head-start to the murderous sunrise. Locally, this can be modeled by solving the equations below. I think that shows that the optimal path after the break-away is not a straight line but a vaguely s-shaped curve. enter image description here

enter image description here

I am adding a plot showing Santa's path (I have reversed the route so he is going from the equator to the north pole, 'cos that's the way i did the maths). Note that at 762 kph, the sunrise line catches up by pi/2 = 6 hours over the route. So he can mirror the route to do the complete passage from north to south. Takes approximately 20 hours, but he can take as long as he wants circling in the vicinity of latitude ~1.09 radians=63 North enter image description here

Penguino
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    At 27N Santa will have to fly faster than 761.15 km/h to keep up with the speed of the sunset line. – Florian F Sep 15 '14 at 08:27
  • @FlorianF He may not have to if he has enough time left. You're the OP, so you'd know, but it doesn't strike me necessarily as wrong. – Neil Sep 15 '14 at 09:17
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    But you describe a path where Santa is hugging the sunset line down to 27°N. To do so he needs to fly at 1485 km/h. That doesn't work. You should make it clear at what point Santa leaves the sunset line. – Florian F Sep 15 '14 at 10:11
  • @Florian F I made an error in the latitude - the 27 degrees was from the north pole rather than from the equator, so he breaks away at 63 N (actually ~62.8). See my added justification above regarding why (although there is not a 'kink') he can go slower if he doesn't traverse the equatorial region in a straight line. – Penguino Sep 15 '14 at 21:40
  • @Florian F I don't think that is correct. V is a function of phi, not theta. In a infinitesimal patch at latitude phi Vlocal is effectively constant Vlocal=V0cos(phi). The differential eqn is used to determine what is the optimal angle theta to cross any particular latitude. i.e. I am calculating the angle (theta) that produces the minimum 'loss of head-start (D)' in crossing any particular latitude (phi). Theta will be a function of phi, but that is irrelevant to the differentiation. D is a function of U, theta and phi, but I can do a partial differential in theta to find its minimum. – Penguino Sep 16 '14 at 21:25
  • Right... comment deleted. – Florian F Sep 16 '14 at 21:26
  • @Florian F I have added a plot (to my answer) to show his path at almost minimum velocity (~762 kph). Ther eis a small mistake - you need to subtract pi/2 from the red line (so he starts with the sun pi/2 behind and ends with it just catching up as he reaches the pole). – Penguino Sep 16 '14 at 21:46
  • I have to confess that I agree with eBusiness that the optimal path should be a straight line (great circle). That is why yours looks suspicious to me. No offense. I think your integral shouldn't add to pi/2 because you are adding horizontal distances that are not on the equator. You should project D on the equator by dividing it by cos(phi) before converting to angles and integrating. – Florian F Sep 16 '14 at 22:08
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    From the new plot it seems that Santa takes around 13 hours to get from equator to 63°N. 13h*762km/h ~= 10000km, that is exactly enough to win another 6 hours of night, but as he takes 13 hours to do so the net loss would be 7 hours, not 6 hours like your graph tells. – aaaaaaaaaaaa Sep 16 '14 at 22:25
  • @eBusiness your argument that 10,000 km would win 6 hours would be relevant if he was traveled on the equator. But he is also moving North. The north distance doesn't win him time as it is perpendicular to approach of sunrise, but as he moves north the east distance he needs to travel to win time falls, as the local 'circumference' measured on a latitude drops as cos(latitude). Over the ~13 hour trip to latitude 63N the 'integrated' horizontal distance [scaled by cos(latitude)] wins him ~7 hours (not 6) so in total he has lost 6 hours. – Penguino Sep 16 '14 at 23:28
  • It is still relevant. If you start anywhere on the Greenwich meridian and travel 10000 km, you cannot reach the equator further than 90°W. At that point you have advanced 6h on local time. – Florian F Sep 17 '14 at 04:46
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    @Penguino Take a look at this image: http://www.bcs.org/upload/img_400/diagram-a.jpg If z is the north pole, and Santa start at the point marked y, then he is exactly 10000 km from any point on the circle that goes through x and z, that circle also happens to mark the longitude that is exactly 6 hours away. So there is absolutely no way for Santa to gain more than 6 hours in 10000 km starting from the equator. – aaaaaaaaaaaa Sep 17 '14 at 06:50
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This actually has a fair amount of similarity with The lake monster. But for this puzzle I'll skip all the advanced maths and find a solution based on simple logic.

  1. Obviously, the first and last legs of the travel can be made without losing any ground to the sun, by following a path that ends in the latitude circle where Santa can only just keep up with the sun. At the end point this path is tangential to the latitude circle.

  2. There is no kinks in the optimal path. As the night area doesn't have any corners any path that includes a kink can trivially be made shorter by cutting some of that corner off.

  3. During the part where Santa can't keep up with the sun there is no reason for him to go along anything but a straight line, a line that turns means he could have made the same trip shorter by following a straight line between the same points. As he moves slower than the sun such a correction will not put him in daylight.

  4. Given these three observations his middle leg must consist of leaving the circle where he can keep up with the sun in a straight line that start out going west and follow this line until he reach the circle where he can keep up with the sun on the other side of the planet.

  5. No matter how large or small the circles where Santa can keep up with the sun are, the line described in 4. is 20000 km long and moves Santa 12 hours forward relative to the sun. This means Santa has 24 hours to complete those 20000 km, so he must move at a speed of $833\frac{1}{3}\mathrm{km/h}$.

  6. This in turn means that the circles where Santa can keep up with the sun have a circumference of 20000 km, so they are located at 60°N and 60°S.

I guess the optimal path to and from these circles are approximately half circles. I'll use some differential maths to figure that part some day when I feel like my head doesn't hurt enough.

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  • Well, this is not only the expected answer, but it is also a perfectly clear explanation why it is correct. And indeed, it is my remark on the Lake Monster puzzle that reminded me of this problem. – Florian F Sep 15 '14 at 21:10
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    @eBusiness - I agree with everything down to point 3. But I disagree that "there is no reason for him to go along anything but a straight line". If he follows an initially gentle curve away from the sunset line and steepens this as he approaches the equator, then he can travel slower. – Penguino Sep 16 '14 at 00:23
  • If by "a straight line" you mean a geodesic on the sphere (i.e. a great circle), how can he follow a straight line which starts out going west in step 4 unless he starts out at the equator? – Peter Taylor Sep 16 '14 at 08:52
  • @PeterTaylor You start heading west at 60°N, then follow a great circle. You don't head west all the way. You naturally fly down to the antipode at 60°S. – Florian F Sep 16 '14 at 09:17
  • @FlorianF, to get from the Shetland Islands to the Galapagos Islands along a great circle you at no point head W. At best you might go WSW. – Peter Taylor Sep 16 '14 at 11:21
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    @PeterTaylor Straight line, by Google Maps: http://i.imgur.com/mj6uvqd.png – aaaaaaaaaaaa Sep 16 '14 at 13:11
  • Even though I accepted the answer years ago, I now realize that your explanation gives necessary conditions on the path, but doesn't prove they are sufficient. – Florian F Dec 15 '20 at 07:22