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   How fast is Santa flying?
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   Author  Topic: How fast is Santa flying?  (Read 5775 times)
Grimbal
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How fast is Santa flying?  
« on: Apr 7th, 2013, 8:13am »
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In another thread
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=truth;actio n=display;num=1062802494
 
on Apr 2nd, 2013, 9:01am, towr wrote:
Surely you've noticed Santa's only around in the winter, in the summer I figure he's in the antarctic, where it's dark then.

This makes me wonder how fast he needs to fly to be able to pull that trick, to go from the North Pole to the South Pole while avoiding daylight.  It is well known Santa flies faster than light to deliver all presents, but what is the minimum speed required?
« Last Edit: Apr 12th, 2013, 3:58pm by Grimbal » IP Logged
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Re: How fast is Santa flying.  
« Reply #1 on: Apr 7th, 2013, 11:13am »
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1670 km/h is enough to stay on the dark side of the earth perpetually at the equator. But less would do, since he has the whole night to pass the equator.
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Re: How fast is Santa flying.  
« Reply #2 on: Apr 7th, 2013, 11:30am »
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It depends what route and when he takes it.
 
If he wants to go straight North to South, the best time is in March. Earlier he risks to get too much sun at the South, later at the North pole. Then a daylight theoretically 12 hours, but he might be extra careful, so can use only 10 hours. The distance between the two poles is roughly 20k Km, i.e. his speed is about 2000Km/h. It is nothing compared to his speed on Xmas eve, so he can do it.
If we allow him all the 12 hours (dawn and sunset are OK), then it is 1670Km/h.
 
He can also choose to go in a spiral, i.e. hiding always on the other side of the Earth. Ideally he crosses the equator in March again, when the Sun is above for simple symmetry reasons.
If he takes a very winding spiral, he has to go abouth 40k Km in 24 hours at the equator, at about 1670Km/h to remain totally on the opposite side. North and South he has an easier task, since the circumfence of the Earth is less there.
But, here he can also make a bit of a trick, since he has more than 24 hours (strictly looking 36 hours) to go around without sunshine. That gives a lower estimate of about 1115Km/h.
The problem is that a bit North or South the Earth is still almost as big, as at the equator, so there he has to go by 1670Km/h, being already close to sunshine.
The optimal speed is therefore somewhere in between. Taking a rough guess, about 1400Km/h.
 
To find the perfect path, is a bit more complicated. First of all, it might be a good idea to look at the sun strictly (dawn and sunset is still OK), so we have 12 and not 10 hours. We can see that going in spiral his speed can be less than going straight. Therefore the journey will take more than one day, i.e starting before March 21 and arriving after that. Few days prior March 21 it is perfect darkness not only at the pole, but also a bit away from there, regardless of the hour of the day. So he can slowly and safely walk down there to a point, where he meets the sun. The he should start at a local evening his spiral with a constant speed. It is an uneven spiral. Closer to the poles, he can use more of his speed to go South and less to go West. At the equator he uses most of his speed to West and minimal to go South. He crosses the equator at midnight local time.
Then he goes in a steepier and steepier spiral again South, reaching the always dark point few days after 21 March. Form thereon he walks slowly to the pole if he wants. Actually he can even stop or start to walk backwards, and be the farest from the South pole in June, when he is still safe.
« Last Edit: Apr 7th, 2013, 11:36am by jollytall » IP Logged
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Re: How fast is Santa flying.  
« Reply #3 on: Apr 7th, 2013, 1:21pm »
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I have just finished my evening talk with my 12 year old. I told him this question, and he came up with a different solution, what was good enough to confuse me.
His solution:
Choose the starting point and time smartly (few days before 21 March, a bit south from the North pole.
Then start going with the optinmal speed in a spiral, but unlike in my solution, where you leave at sunset and gradually let the night "overtake" you, you go with the constant speed and always follow the sun at local sunset. This is also a less and less steep spiral as more and more of your speed is needed to keep up with the sun and less and less can be used to go south. At one point you cannot go further South, because your speed is not enough to follow the sun. Then you make a sharp turn and go straigh south, across the equator on 21 March, in 12 hours. On the other side, you do the same again symmetrically. You continue with your constant speed almost perfectly West, escaping from the sunset. You can go steeper and steeper until you get close enough to the South pole, where you are safe again.
My first problem with this solution was, how close you can get to the point, where you go almost parallel to the equator. In a perfect world you can do it infinite times and never get parallel. Luckily it is not true for the Earth, since the days get longer and longer, so you cannot do it on and on. There is a point, where even with a bit of southward spiral you cannot keep pace with the sun, as it gets faster and faster.
So, I am confused.
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Re: How fast is Santa flying.  
« Reply #4 on: Apr 7th, 2013, 9:48pm »
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If I am not mistaken, the lattitude, where the circumfence/24h equals to the section to cut through/12h is at lattitude of 53.5. There the section to cover is 0.59x the total distancce of Pole-to-Pole, i.e. his speed has to be about 1012Km/h. The best so-far.
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Re: How fast is Santa flying.  
« Reply #5 on: Apr 9th, 2013, 10:09am »
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I was actually considering a simplified version:
The North Pole and the South Pole are safe, the Sun stays over the Equator, circling it in exactly 24 hours.  And the earth is a perfect sphere with a circumference of 40 Mm*.
 
My best solution so far is just over 20 Mm/24 h = 833.3 km/h
 
*mega-meters.  Isn't it a bit odd to insist on speaking of 40'000 km? Wink
« Last Edit: Apr 10th, 2013, 1:00pm by Grimbal » IP Logged
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Re: How fast is Santa flying.  
« Reply #6 on: Apr 9th, 2013, 3:06pm »
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I do not get it. 20 Mm is the distance if you go straight from pole to pole. But how do you have 24h-s for it? Whereever you go, in 12h-s you have the sun.
 
P.S. Mm I like. 40k Km probably comes from Kg, what is an SI unit. So it is correct to talk about 1000 Kg instead of 1 Mg. For m you are right, Km is not an official SI unit.
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Re: How fast is Santa flying.  
« Reply #7 on: Apr 10th, 2013, 2:32am »
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Indeed, you cannot go from pole to pole in 24 hours.  I mean, in the shadow.
But the critical leg of the journey is 20 Mm in 24 hours.
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Re: How fast is Santa flying.  
« Reply #8 on: Apr 10th, 2013, 3:13am »
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I am sure, you are right as always, but still do not see it. What is the best path, where the critical you can make 20Mm in 86.4Ks?
 
Ks - isn't it even stranger? (If we want to use SI.)
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Re: How fast is Santa flying.  
« Reply #9 on: Apr 10th, 2013, 4:25am »
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Both Km and Kg are not SI units - the symbol for the kilo- prefix is a lower-case 'k'.
 
To be correct, it should me km and kg.
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Re: How fast is Santa flying.  
« Reply #10 on: Apr 10th, 2013, 5:27am »
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on Apr 10th, 2013, 2:32am, Grimbal wrote:
the critical leg of the journey is 20 Mm in 24 hours.

Aha! And somewhat surprisingly, there are whole families of solutions! Hint: great circle route.
 
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Re: How fast is Santa flying.  
« Reply #11 on: Apr 10th, 2013, 6:27am »
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Hmm.  I am not sure there are whole families of solutions.  I thought so at first but I realized it doesn't always work.
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Re: How fast is Santa flying.  
« Reply #12 on: Apr 10th, 2013, 10:13am »
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Quote:
Both Km and Kg are not SI units - the symbol for the kilo- prefix is a lower-case 'k'.

 
You are right. Kilo prefix is k not K. There is still a big difference though. kg is an SI base unit, whereby km is not. For length m is the base unit, km is 1000 base units. For mass kg is the base unit.
« Last Edit: Apr 10th, 2013, 10:13am by jollytall » IP Logged
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Re: How fast is Santa flying.  
« Reply #13 on: Apr 10th, 2013, 12:46pm »
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on Apr 10th, 2013, 6:27am, Grimbal wrote:
I thought so at first but I realized it doesn't always work.

After performing quite a bit of trig, it seems you're correct (not that I doubted it). My visualization skills have failed me once again... There is indeed a single, asymptotically-optimal solution
 
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« Last Edit: Apr 10th, 2013, 12:48pm by SMQ » IP Logged

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Re: How fast is Santa flying.  
« Reply #14 on: Apr 11th, 2013, 1:26am »
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Trig?  I only used sin(pi/6) = 0.5.
 
On the other side, I wasn't 100% sure to have the optimal solution.  I didn't check all variants.
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Re: How fast is Santa flying.  
« Reply #15 on: Apr 11th, 2013, 4:36am »
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on Apr 11th, 2013, 1:26am, Grimbal wrote:
Trig?  I only used sin(pi/6) = 0.5.

Well, technically the computer did all the trig...  I plotted a lot of routes just to discover the (obvious, in retrospect) lemma that the uinque great circle route from a point to its antipode which is symmetrical w/respect to the equator always departs and arrives bearing due east or west.  
 
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Re: How fast is Santa flying.  
« Reply #16 on: Apr 11th, 2013, 12:52pm »
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You have to help me.
From the hints, I get closer. It shows that the middle jump is not a straight south move, but going also 180 longitudes. Actually if you leave at any north lattitude at 0 longitude, and arrive to the same south lattitude at 180 longitude, it is 20Mm. Altogether you have 24 hours (12 hours night+12 hours timezone).
 
Nonetheless it is still not that easy.
To have the 12 hours night used, it means that you have to leave exactly at sunset and arrive at sunrise. It means that you do not have any "night" left for the upper and lower part of the journey, i.e. you have to use only the timezone.
If the lattitude is too small (too close to the equator), then the circumfence remains too big, and so with the same speed you cannot hide from the sun. With the 833.3km/h, the circumfence is 20Mm, i.e. at lattitude 60 you can just keep up with the sun. So you need to start the middle section from north at least 60N and arrive to south more than 60S. This is the sin(pi/3)=0.5 logic. So far easy.
 
My problem is the other limit, i.e. how North you can leave from. Following the logic of going from any random point to its antipode gives 24 hours, it could also work for the Poles (or 1 mm from it). I do not see, how it could give more than 12 hours.
 
I also do not see SMQs point of going first West. Unless if Santa leaves exactly at 60N, going West at Sunset with max speed would mean that Santa catches up with the sun and runs into daylight. Even is we say, there is only one lattitude (60N) to start this special spiral from, I do not see how the route would start West.
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Re: How fast is Santa flying.  
« Reply #17 on: Apr 11th, 2013, 1:33pm »
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You start somewhere on the sunset line, and end at the antipode on the sunrise line.  What does it tell you about where the Sun is at the start and at the end of the journey?
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Re: How fast is Santa flying.  
« Reply #18 on: Apr 11th, 2013, 4:28pm »
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on Apr 11th, 2013, 12:52pm, jollytall wrote:
My problem is the other limit, i.e. how North you can leave from. Following the logic of going from any random point to its antipode gives 24 hours, it could also work for the Poles (or 1 mm from it).

That's exactly the issue I was struggling with that led me to graphing spirals and great circles. The upper limit is set by not outpacing the sun and crossing the sunset line into evening. As Grimbal implied in response to my first post, that leaves only a single ideal solution.
 
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Re: How fast is Santa flying.  
« Reply #19 on: Apr 12th, 2013, 2:50am »
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Thanks, now I have it. If you have to depart Westwards, then indeed 60N is also an upper boundary. What was not clear to me, why you depart Westward. Now thinking it over, it is clear, that because of symmetry reasons, the "top" of the circle is indeed E-W.
Very strange to imagine though, that going from any point on the Northen hemishere to any point on the Southern hemishpere, the shortest route is to leave West... Obviously because these are now antipodes, any direction has the same lenght...
 
So now we know the middle segment. There are still the upper (and lower) spiral segments. Obviously close to 60N and 60S, most of the speed has to be used to go West to keep up with the sunset or sunrise. The further to the North (and South) more and more of the speed can be used to N-S move. It is still unclear to me how much time (how long spiral) it takes to get close enough to the poles to be in the ever-dark segments. This also depends on the spiral. If it takes too long (many rounds, i.e. many days) then the "safe" zone is larger, but if the spiral can be steep, then it is very small.
In the ever dark segment he can go safely directly South.
 
So, what it the total time needed pole-to-pole, covering all the 5 segments?
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Re: How fast is Santa flying.  
« Reply #20 on: Apr 12th, 2013, 7:19am »
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on Apr 12th, 2013, 2:50am, jollytall wrote:
So, what it the total time needed pole-to-pole, covering all the 5 segments?

Sticking with Grimbal's ideal Earth (no axial tilt; every day is a perfect equniox) it's only possible to asymptotically approach the minimum speed, never quite reach it. That's because the spiral from the pole along the terminator never perfectly flattens out (or alternatively never quite reaches 60N).
 
From a practical standpoint, however, after following the terminator spiral for only 6.5 hours, the difference between the true position and ideal position is smaller than the floating-point accuracy of my simulation (approximately 16 decimal digits). So as a practical matter, a 37 hour journey can be accomplished with a speed immeasurably-close to 833.3... km/h.
 
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Re: How fast is Santa flying?  
« Reply #21 on: Oct 21st, 2013, 5:03pm »
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I had a funny dream.
 
Not only was Santa Claus a vampire avoiding daylight, but the reindeer was a lion.  And there was a lion tamer running on the 60th parallel at an incredible speed in an effort to witness a never-ending sunset.
 
I wonder what that is supposed to mean?
« Last Edit: Oct 21st, 2013, 11:23pm by Grimbal » IP Logged
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Re: How fast is Santa flying?  
« Reply #22 on: Nov 6th, 2013, 4:55am »
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I actually was pointing at the fact that there is a solution at exactly 20000 km/24h in a finite time, not only asymptotically close.
 
At least if you assume that the shadow border is still safe.  (Which, to be honest, I can't see how it could be).
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