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## Visiting Santa

Posted by trickofmind on June 12, 2010 – 7:31 pm

How far do you have to travel to reach the North Pole, if you start from the equator and always head North West?

Karl

This post is under “SharedPuzzle” and has 27 respond so far.

### 27 Responds so far- Add one»

1. 1. James Said：

you’d never reach the north pole.

2. 2. Random Guy Said：

I agree with James. Youll eventually end up halfway between the West side of the equator and the North Pole, then continue downwards, going in a wave-like style.

3. 3. Tiana Said：

Forever… And you’d still never reach it…

4. 4. karl Sharman Said：

James…. where would you get to then?

5. 5. Karys Said：

Mathematically, assuming earth is a sphere and WE are a point, we will never get on the north pole, just as near as it is possible, but THE point which would be north pole won’t be reached.

A related case is with a circle :
Draw a disc (a “full circle”), centered on N, North pole, that’s Earth’s north hemisphere (ugly, I know…). Equator is the extern circle.
From here, East means “tangential to the circle, counterclockwise directed”, or “orthogonal to ((We)N), counterclockwise directed”. At any point except N on the disc, there is an East, as on a spherical Earth.

North obviously is in direction (We)>N.

So North West is the direction bissecting North and West (opposite to East). and I assume you cannot get on a point as long as the direction is never oriented to this point. Then we are never directly directed (:D ) to North, you cannot reach North pole that way.

BUT we are on ToM, so all assumptions are good as long as they can. James, your intuition was probably guided by the fact that many problems of this type have no solution, because the aim is to trick people into doing useless calculus.
But we are not in an all mathematical world (at least not in what most people think of maths geometry) and WE are NON-ponctual people. We are not points, so if you say “NW” is from our gravity center, then you could say we will reach North pole when G is in a certain area around NP…

6. 6. Karl Sharman Said：

There’s also the tricky part to consider about magnetic north and true north, and which one a compass or GPS would lead you to….

7. 7. Chris Said：

If the distance between the equator and the north pole is d, then the total distance is d*sqrt(2).

This is because lines of latitude and longitude are orthogonal.

8. 8. Chris Said：

Opps! I submitted that by accident. I meant that lines of latitude and longitude are orthogonal everywhere, except for a co-ordinate singularity at the pole.

The time taken is finite.

9. 9. Karl Sharman Said：

Working backwards, if you were to leave the North Pole, the only direction you can go is South. Conversely getting to the North Pole you must travel North.
This means you cannot get to the North Pole by travelling North West from anywhere on the globe.
In reality you would get to a point where you could stop going round in ever decreasing circles and step onto the North Pole.

Karl

10. 10. Gary Holmes Said：

Karl,

I mentioned the difference between Magnetic North Pole and Geographical North Pole. (That article disappeared…)

Compass will take you to a point nearby Mag N Pole.

Santa lives at Geogrph N Pole.

Ur Title said, “Visiting Santa”. Yet, they’re orange and apple.

Yes, Karl,
When you follow the compass, going around and around, You WILL Pass Where Santa Lives.

Gary Holmes

11. 11. Chris Said：

Hi Karl, your post doesn’t quite do it for me. Start at the North pole facing in a given direction, then start the backwards journey. The moment you start forwards, West and East are defined, the rest of the journey doesn’t require any such trickery

12. 12. Euclid's Brother Said：

I’m inclined to believe that you would never get there, because the instance that you stepped on the north pole, you would not be traveling NW.. only N.

But… what about looking at it this way. Say you are 50,000 meters from the north pole. You take a step going NW that of some specific distance, which leaves you exactly 1 meter north and 1 meter west of where you started. That specific distance would be slightly more than sqrt(2) when accounting for the curved surface of the planet.

So every step you take, would take you 1 meter further north and 1 meter further west. After 50,000 of such steps, would you not be 50,000 meters further north than you were (lattitudial speaking).

I’m not sure how to calculate the arc-distance over a sphere, but if you could come up with a formula, you should be able to calculate it.

If the Theroem Pytagoris worked in this case (which it doesn’t because it’s on a sphere), it would be something like sqrt(2 * (D^2)) where D is the distance between the equater and the north pole. AGAIN: i know this formula doesn’t work for a sphere.

Yep.. this is definately ToM calibur…

13. 13. Karl Sharman Said：

Sorry, Euclid’s Brother, but even taking a step north at any time will not get you to the North Pole. The only time that step would work is when you can step directly onto the North Pole, which as you have already surmised would be travelling North, breaking the North West caveat.

Karl

14. 14. Unknown Said：

I think that (not too sure) but you would go around in almost a swirl around and around and you would eventually see santa!

15. 15. Chris Said：

The path goes round the North pole infinitely often.

Near the pole, the curvature of the Earth can be neglected, reducing the problem to a 2D one. Let r and A represent the polar co-ordinates of a point on the path. The path is such that dr = r dA => ln(kr) = A => r = c * e^A, where c and k are arbitrary constants, ln is the natural logarithm and e is Euler’s number (approx 2.7183). For r to → 0, A must → -∞, so the path winds round the North pole infinitely often.

I still believe that the path length is finite though.

16. 16. Chris Said：

Each time the path does one revolution, it becomes e^(2*pi) ≈ 535.49 times close to the pole.

17. 17. Chris Said：

More generally, if you start at a distance r1 from the pole (measured over the Earth’s surface), and stop at a distance r2 from the pole, always travelling at an angle A (=45 degrees in the original problem, A = 0 => directly towards the pole), the distance you travel will be (r2 – r1)/cos(A).

The angle you will travel through will be ln(r1/r2)/tan(A) radians.

If r2 = 0 and A 0, then the journey is not physically realisable, it is only mathematically realisable as infinitely large rates of rotation will be required. Or if you constrain the rate of rotation to be finite, then the journey time will become infinite.

18. 18. The Winner Said：

It all depends on how west, or how north you go. You will get there, it just depends on how tedious you want it to be, because if you go a half a degree north of going west, you will travel the earth 181 times before you accually reach the north pole. However if you go a half a degree west of north, you’ll only go around the earth once or twice. On either of those, you could skip alot of humdrum if you call it quits, at about 1* variance from the accual 180* north.

But it all depends on which north you locate as the north pole, or if you are talking about the same planet, or if you imply the name of this clue or not.

I was lazy and only put aproxxamates, rather than the accual distance, but i still win

19. 19. Chris Said：

Hi The Winner. I doubt than many rational people would agree with your conclusion.

20. 20. Chris Said：

The Winner – you didn’t mention the distance at all.

21. 21. Chris Said：

See http://mathworld.wolfram.com/SphericalCoordinates.html for the polar co-ordinates that I’m using.

Assume that we start at (r,θ,φ) = (r,0,π/2). The North pole is at (r,θ,φ) = (r,?,0), where ? is undefined. The distance to the pole is rφ (measured over the Earth’s surface). At any position, an incremental movement North is -r dφ, and an incremental movement West is -r sin(φ) dθ. As we are moving NW at every position, we have r dφ = r sin(φ) dθ, so dθ = dφ/sin(φ) = cosec(φ) dφ.
Integration => θ = ln|k tan(φ/2)| for some constant k.
Hence, e^θ = k tan(φ/2). At the start, θ = 0, φ = π/2 => k = 1, so e^θ = tan(φ/2).

As we approach the pole, φ → 0, so θ must → -∞ and we go round the pole an infinite number of times.

It should be obvious that the incremental distance travelled NW = √2 * the incremental distance travelled N EVERYWHERE i.e. the total distance travelled is √2 * the distance travelled North.

I’m completely confident that my analysis is correct.

22. 22. Chris Said：

The question seems to suggest that a living being (oneself) is doing the journey. As you have a non-zero size, at least a part of you can get to the pole, having only gone around it a few times.

23. 23. Chris Said：

Starting with e^θ = tan(φ/2). If we want to get to 0.1 nm (approx diameter of an atom) of the pole, then φ ≈ 0.1*10^(-9)/(6400*10^3) = 1.5625*10^(-17), so θ = ln(1.5625*10^(-17)/2) ≈ -39.4. So we’d go round the pole 39.4/(2π) ≈ 6.27 times.

The carbon nucleus has a diameter of about 5 fm (this is a guess on my side), so we’d end up going round the pole about 7.85 times to get that close. These are very far from infinite numers.

Heisenberg’s uncertainty principle will definitely have cut in, so you’ll get there

24. 24. Chris Said：

@slavy. Does the assumed point get to the North pole? I’m confused about it. I believe it is to do with the real and the superreal numbers. I’m pretty sure that 1 + 1/2 + 1/4 + 1/8 … = 2 exactly. But I assume that is only true for the reals, but not true for the superreals. Can you explain how to deal with this? A link would be fine. Thank you (in anticipation).

25. 25. slavy Said：

I don’t know how to help here I, myself, am a terrible geometer, but what I can say is that Chris’ computations are correct and the length of the path is indeed sqrt(2) times half the length of a meridian. However the curve (trajectory/path) is not defined at its endpoint so mathematically one will never exactly reach the north pole. But in the physical world this will happen, since we are not dots with infinite sizes that can be assumed to equal zero… And it is not true that we have to walk infinite time, since the trajectory at the end is so spinning that one can never be able to follow it in practice. It is like a whirligig changing its center by very very small quantities…

In conclusion – the problem here is that we try to unite a pure theoretical (mathematical) model with a pure phisical situation and usually this is a lost cause…

26. 26. Chris Said：

@slavy. Thank you. I feel a lot better now.

27. 27. Chris Said：

Santa resolved. I think I’ve solved the dilemma. As the numbers in the problem are awkward, I’ll first do it by analogy.

Let S = 1 + 1/2 + 1/4 + 1/8 + …
The 2S = 2 + 1 + 1/2 + 1/4 + …
So S = 2S – S = (2 + 1 + 1/2 + …) – (1 + 1/2 + …) = 2 exactly.

You may initially have been tempted to think that S = 2 – d where d is an infinitesimal – that would have been wrong.

The point moving to the pole has a similar series (to the above). Each time you go round the pole, the distance travelled is approx 1/535 of the previous term (once you are so near to the pole the the curvature of the Earth may be neglected). So the series is more like D = X + 1/535 + 1/(535^2) + 1/(535^3) + …, where X is some fixed and definite value representing the distance travelled over the region where the curvature of the Earth can’t be neglected (e.g. 1 picometre from the pole should do the trick ). Then use the same logic as above. i.e. 535D – D = 534D = 534X + 1, so D = X + 1/534, and D is thus well defined and has no infinitesimal part. I’ve worked with units of length that allow 1/…, to avoid introducing more letters.

If you think I’m doing some smoke and mirror tricks, forget that it’s a spherical problem, and think about the corresponding problem on a flat disc (where the same objections about never getting there would apply), then the above argument is exact (in principle).

These sort of considerations, boil down to the fact that real number system does not include the infinitesimals (or the infinites). The real numbers possess the “Archimedean property”.

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