Santa Claus's speed

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Assume that the earth is a perfect sphere with a circumference of 40Mm. Santa needs to travel from the North Pole to the South Pole while avoiding daylight. Assuming that he can go faster than the speed of light, what path would be the best path to take and what is the slowest he can travel?

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Posted · Report post

To be clear.... the best path is the one that provides Santa the slowest speed (gives more time for gift giving)

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Posted · Report post

He can't do it -- certainly not within one day.

Whatever season of the year, one of the poles will be in 24-hour daylight.

Well, he could begin in Winter at N pole, circle the earth spiraling westward and southward at approximately 1 circumference per day to stay in the dark and arrive six months later at S pole when it would be winter there.

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Posted · Report post

Is this the slowest?

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Posted · Report post

It's hard to envision Santa being a tachyon at all, much less a slow one. And I'm not sure of the definition of best.

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Posted · Report post

To be clear.... the best path is the one that provides Santa the slowest speed (gives more time for gift giving)

@bonanova

The path/strategy that allows Santa to continiusly travel around while remaining forever at "night" going at the minimum speed possible.

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Posted · Report post

To make writing easier, I'll define the speed S = 40 Mm / day = the circumference of the Earth / day.

If Santa moves at speed S then he can constantly stay 180 degrees away from the sun, moving along the lines of latitude at speed S*cos(latitude) to stay opposite the sun while using "the rest" of his speed to move down a line of longitude toward the South pole. His speed from North to South pole would hit zero exactly at the equator, but I will blithely ignore that fact for the purposes of answering this question.

If Santa moves at a speed T which is slower than S, then he can maintain a constant angle [santa - axis of Earth - Sun] as long as he's at a latitude far enough from the equator so that

S * cos(latitude) < T

latitude > arccos(T/S)

Santa could move at speed T from the North pole to the latitude arccos(T/S) in the Northern hemisphere while staying right at the edge of sunset, then make a break straight toward the South pole while the world rotates for 12 hours to reach arccos(T/S) in the Southern hemisphere just as the sun is about to rise, and then make the rest of the trip to the South pole while staying on the brink of dawn. The distance through which he would make that mad dash through the equator would be (radius of the Earth) * (2 * arccos(T/S, in radians)) where the arccos is multiplied by 2 to account for the fact that Santa has to travel that distance in both the Northern and Southern hemispheres. Since 2 * (radius of the earth) = (circumference of the Earth) / pi, we can write that distance as 40/pi Mm * arccos(T/S, in radians). For Santa to be able to travel that distance in 12 hours:

T * 0.5 days = 40/pi Mm * arccos(T/S, in radians)

T = arccos(T / 40 Mm/d) * 80/pi Mm/d

Feeding that to a calculator says T is roughly 23.8 (with units Mm/d). But I suspect he might be able to go even slower if he doesn't go straight from North pole to South pole during the run through the equator and instead angles a bit to head away from sunrise, with that angle changing as a function of latitude in some optimal way that I cannot currently determine because the fact that lines of latitude do not all have the same length as latitude changes makes it tough.

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Posted (edited) · Report post

The sunlight can not catch him if he is faster than light.The path is a warped 4D line.

Edited by TimeSpaceLightForce
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Posted · Report post

The sunlight can not catch him if he is faster than light.The path is a warped 4D line.

True but this task is about moving as slow as possible. Though he can travel that fast, which is how he is able to deliver all those christmas gifts, we want to know what is the slowest he could go.

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Posted · Report post

To make writing easier, I'll define the speed S = 40 Mm / day = the circumference of the Earth / day.

If Santa moves at speed S then he can constantly stay 180 degrees away from the sun, moving along the lines of latitude at speed S*cos(latitude) to stay opposite the sun while using "the rest" of his speed to move down a line of longitude toward the South pole. His speed from North to South pole would hit zero exactly at the equator, but I will blithely ignore that fact for the purposes of answering this question.

If Santa moves at a speed T which is slower than S, then he can maintain a constant angle [santa - axis of Earth - Sun] as long as he's at a latitude far enough from the equator so that

S * cos(latitude) < T

latitude > arccos(T/S)

Santa could move at speed T from the North pole to the latitude arccos(T/S) in the Northern hemisphere while staying right at the edge of sunset, then make a break straight toward the South pole while the world rotates for 12 hours to reach arccos(T/S) in the Southern hemisphere just as the sun is about to rise, and then make the rest of the trip to the South pole while staying on the brink of dawn. The distance through which he would make that mad dash through the equator would be (radius of the Earth) * (2 * arccos(T/S, in radians)) where the arccos is multiplied by 2 to account for the fact that Santa has to travel that distance in both the Northern and Southern hemispheres. Since 2 * (radius of the earth) = (circumference of the Earth) / pi, we can write that distance as 40/pi Mm * arccos(T/S, in radians). For Santa to be able to travel that distance in 12 hours:

T * 0.5 days = 40/pi Mm * arccos(T/S, in radians)

T = arccos(T / 40 Mm/d) * 80/pi Mm/d

Feeding that to a calculator says T is roughly 23.8 (with units Mm/d). But I suspect he might be able to go even slower if he doesn't go straight from North pole to South pole during the run through the equator and instead angles a bit to head away from sunrise, with that angle changing as a function of latitude in some optimal way that I cannot currently determine because the fact that lines of latitude do not all have the same length as latitude changes makes it tough.

on the right track

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Posted · Report post

The sunlight can not catch him if he is faster than light.The path is a warped 4D line.

True but this task is about moving as slow as possible. Though he can travel that fast, which is how he is able to deliver all those christmas gifts, we want to know what is the slowest he could go.

Well i don't know.. i read about the twin brothers where one travel at relativistic speed.. and when he returned the brother that stayed got much older.. What i do not get or missing is "why are we assuming he can travel faster than light?"

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Posted (edited) · Report post

The sunlight can not catch him if he is faster than light.The path is a warped 4D line.

True but this task is about moving as slow as possible. Though he can travel that fast, which is how he is able to deliver all those christmas gifts, we want to know what is the slowest he could go.

Well i don't know.. i read about the twin brothers where one travel at relativistic speed.. and when he returned the brother that stayed got much older.. What i do not get or missing is "why are we assuming he can travel faster than light?"

I believe the American term is that it is a Red Herring. Just because my car can travel 90mph doesn't mean that I will or should. So just because Santa Clause could travel faster than the speed of light doesn't mean he will or should. He may need to on December 25th in order to deliver the presents but otherwise he would probably prefer a leisurely trip as he goes south for his vacation.

Edited by BMAD
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Posted · Report post

The poles has semi year of sunshine and semi year of shadow ..If he depart from North pole on its last day of sunshine and fly in shadow on semi circular path (shortest) with respect to earth axis toward South pole, he will be above Equator when it is last day of sunshine for the South pole. Continue to travel for another semi year to arrive at South pole on last day of shadow..he would have an annual trip for one holiday/ year .

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Posted · Report post

The poles has semi year of sunshine and semi year of shadow ..If he depart from North pole on its last day of sunshine and fly in shadow on semi circular path (shortest) with respect to earth axis toward South pole, he will be above Equator when it is last day of sunshine for the South pole. Continue to travel for another semi year to arrive at South pole on last day of shadow..he would have an annual trip for one holiday/ year .

Can he go slower?

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Posted · Report post

Define "going slower." What does it mean?

  1. Longer elapsed time before reaching South Pole
    We could find an arbitrarily long path and define speed = (distance between poles)/(elapsed time.)
    This definition gives one value of speed for the entire trip and does not depend on the layout of the path.
    We could go slower under this definition by flying endlessly at night at a constant latitude.

  2. Lower path speed.
    This means speed = ds/dt where s is path length.
    Here speed is an instantaneous value.
    With this definition we could go slower just by stopping to rest for a while.
    That is, we could make our (instantaneous) speed zero. That is very slow.

Also define "best path."

I think of best path as most efficient - shortest path or earliest arrival.

Going slower seems to ask for the path that will tire him the least.

So, instantaneous speed, average speed, longest transit time, shortest path, rest stops for the reindeer to regain strength, least time between houses so children don't have to wait for their toys, lowest to the ground so everyone can hear HO HO HO clearly ... there are a lot of things that might make a path "best."

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Posted · Report post

Assuming that Santa must always get progressively closer to the south pole. What is the slowest (longest duration of travel) he can travel and remain forever at night? So far both you (bonanova and TSLF) claim it is 1 year. I am asking if this is correct.

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Posted · Report post

I did not really get why the 12 hours...

- by symmetry, we can calculate only the trajectory to the Equator

- if Santa moves faster then the speed of the shadow line, he should move West as much as possible (I can develop if necessary)

- his move to the West is limited by the shadow line

(- when his speed equals the speed of the shadow line, he should make some turns around the Earth until the South pole gets into the shadow - solves Bonanova's remark, but we can omit this in the calculation)

- when his speed is smaller than the speed of the shadow line, he should move plain South

When he moves to the West, he trades somehow time against distance, but I cannot find the equivalence.

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Posted (edited) · Report post

Assuming that Santa must always get progressively closer to the south pole. What is the slowest (longest duration of travel) he can travel and remain forever at night? So far both you (bonanova and TSLF) claim it is 1 year. I am asking if this is correct.

I am assuming we are looking for the smallest CONSTANT speed.

If we want to maximize the travel time (at varying speed), he can travel almost forever: he follows the shadow line moving imperceptibly to the south.

Edited by harey
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Posted · Report post

Define "going slower." What does it mean?

  • Longer elapsed time before reaching South Pole

    We could find an arbitrarily long path and define speed = (distance between poles)/(elapsed time.)

    This definition gives one value of speed for the entire trip and does not depend on the layout of the path.

    We could go slower under this definition by flying endlessly at night at a constant latitude.

  • Lower path speed.

    This means speed = ds/dt where s is path length.

    Here speed is an instantaneous value.

    With this definition we could go slower just by stopping to rest for a while.

    That is, we could make our (instantaneous) speed zero. That is very slow.

Also define "best path."

I think of best path as most efficient - shortest path or earliest arrival.

Going slower seems to ask for the path that will tire him the least.

So, instantaneous speed, average speed, longest transit time, shortest path, rest stops for the reindeer to regain strength, least time between houses so children don't have to wait for their toys, lowest to the ground so everyone can hear HO HO HO clearly ... there are a lot of things that might make a path "best."

I do not know if you recognize it but this is a variation of one of your puzzles (I believe a year old now?). You asked us to try and solve a travel salesman problem that maximized travel as one moved progressively closer to their target. Mine is similar to yours except yours was movement on a grid.

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Posted (edited) · Report post

I chose to define "best path" as the path which gets Santa from the North pole to the South pole with the lowest possible top speed at any point in the path, neglecting the tilt of the Earth (and hence possibility of one pole being in constant daylight). This seemed to be the definition that makes the problem non-trivial.

If you consider the way it's phrased in post #16, asking what the longest duration of travel Santa could make while staying in darkness, then I agree with harey's answer, but I suspect that BMAD meant something else.

I did not really get why the 12 hours...

12 hours is necessary because I'm asking Santa to start from the Northern hemisphere at the edge of evening-night, travel South along a line of longitude as the world is turning, and make it to a point in the Southern hemisphere just before sunrise where he can outrace the sun.

- by symmetry, we can calculate only the trajectory to the Equator

You're correct that you can, by symmetry, calculate only the trajectory to the Equator. As long as he makes it to the Equator when he's on the direct opposite side of the Earth as the Sun. I just didn't happen to invoke that symmetry in my explanation.

- if Santa moves faster then the speed of the shadow line, he should move West as much as possible (I can develop if necessary)

I agree that when he's in the Northern hemisphere (and before he starts heading straight South) he should move West as much as possible to stay on the brink of the evening-night transition, but in the Southern hemisphere he should travel West only as much as necessary to avoid sunrise while using the rest of his speed to travel Southward.

- his move to the West is limited by the shadow line

You are correct that his movement Westward is limited by the shadow line. When he's in the parts of the Northern and Southern hemisphere closest to the poles, he travels Westward only fast enough to keep up with the shadow line, while using "the rest" of his speed to travel Southward.

(- when his speed equals the speed of the shadow line, he should make some turns around the Earth until the South pole gets into the shadow - solves Bonanova's remark, but we can omit this in the calculation)

If you want to deal with the Earth's tilt, then sure. I just chose to ignore the Earth's tilt, or suppose that Santa travels at an equinox.

- when his speed is smaller than the speed of the shadow line, he should move plain South

That's what I had him doing in my answer, but that's not actually optimal. For the sake of making a simplifying example, suppose the Earth is a cylinder and he must travel from the top to the bottom in darkness as it's rotating. Define H = height of cylinder, C = circumference of cylinder. If he travels straight down, he must travel a distance H in 12 hours. If he travels at a small angle theta westward, then he will travel a total distance of H / cos(theta) and will travel a Westward distance of H tan(theta). Because he travels a Westward distance, he can take more time before the sun catches him, and you can calculate that extra time as [24 hours * westward distance traveled / C] = [24 * H tan(theta) / C]. That's time in addition to the 12 hours that he would have if he were to just head straight South. So total distance traveled would be

H / cos(theta)

And total time spent would be

12 + (24 H tan(theta) / C)

And the speed would be distance / time.

If you were to increase theta by a miniscule amount from 0 to delta, then the derivative of H / cos(theta) is zero and the derivative of 12 + (24 H tan(theta) / C) is positive, so you would be decreasing his necessary speed by making a small angle.

As you can see, that formula for speed as you make a slight angle is already sort of nasty, and I dare not try to take it on if we need to account for a circular Earth where lines of latitude do not have constant length.

When he moves to the West, he trades somehow time against distance, but I cannot find the equivalence.

Comments are within the spoiler, although I somehow get the feeling that maybe you were trying to make a point that I'm not picking up on? Edited by plasmid
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