But someone needs to check it so I can sleep tonight.

Looks ok to me, but I'm just doing approximate calculation.

Here's something I think is interesting, also doing approximations, but it conveys the picture:

Consider the set-up where there's a circle with a circumference of exactly a mile around the South Pole and a larger concentric circle with a radius a mile longer than that of the smaller circle.

The radius of the smaller circle is 1 mile divided by 2pi, approximately .15 mile, making the total radius of the larger circle approximately 1.15 miles.

Multiply 1.15 by 2pi to get the circumference of the larger circle. Approximately 8.345 miles. Round down to 8 miles for calculational convenience.

Now, start at some point - Point A - on the circumference of the larger circle.

Walk 1 mile south to the circumference of the smaller circle, then half way around the smaller circle (a distance of half a mile), then 1 mile north to a point - Point B - on the circumference of the larger circle.

You've walled 2 1/2 miles, but you're now half way around the circumference of the larger circle - a distance of approximately 4 miles going around that circle's circumference.

No, that would be like objecting to Aristotle's paradox by insisting that his wheel is not a good car wheel, it's not in the spirit of the puzzle. It's just the conversion from miles to latitudes etc. We can always assume that we're talking about Jesus or St Francis of Paola, who allegedly could walk on water.

Heaven forfend we should not be "in the spirit" of the puzzle! I misunderstood what you were asking, but my question was because I wasn't sure if Jon meant literally walkable places or not. He did specify in regard to circles around the South Pole that only about 50 or so subsets were walkable-by-a-human.

You ignore the possibility that going south and going north can be done on the same line of longitude, while going west between these two displacements.

North means a 0 degree geometric heading with respect to the poles at wither end of the axis of rotation. This is slightly different from 0 degree heading wrt magnetic north. The magnetic poles wander about.

Nope. The only places where two longitudes intersect are at the poles. Heading North means moving on a line of longitude in the direction of the North geographic pole.

Longitudes intersect only at the poles. And yes, that is the meaning of walking north.

Other places besides the North Pole also work. You end on the same point where you started.

Places that allow the northerly walk to occur on the same line of longitude that the southerly one did.

Max told you yesterday:

”You ignore the possibility that going south and going north can be done on the same line of longitude, while going west between these two displacements.”

I've tried to calculate the second problem, but I get always implicit functions that cannot be solved analytically, only numerically, so I guess this problem has no explicit solution. For generality I've used a variable for the "walked" distance instead of "1 mile".

Places that allow the northerly walk to occur on the same line of longitude that the southerly one did.

Max told you yesterday:

”You ignore the possibility that going south and going north can be done on the same line of longitude, while going west between these two displacements.”

Please provide the latitude and longitude of a point other than the North Pole which satisfies the conditions.

Please provide the latitude and longitude of a point other than the North Pole which satisfies the conditions.

Watch what Michael Palin does:

He actually travels east instead of west, but no matter, just assume he walked west instead. It looks like the circle he's walking in is about 20 feet in circumference, so let's assume it's precisely 20 feet. If, to satisfy step 2, he were to do that walk 264 times (exactly one mile), his end point would be exactly the same degree of longitude as where he started. He could then walk north one mile to satisfy step 3.

Please provide the latitude and longitude of a point other than the North Pole which satisfies the conditions.

The striped pole is the south pole. The yellow sphere is your beginning point. You follow the red path south. It's exactly one mile. Then you head west on the green circle. Its circumference is exactly one mile, which brings you back to the red path, which you take north for one mile back to your yellow sphere starting point.

I've tried to calculate the second problem, but I get always implicit functions that cannot be solved analytically, only numerically, so I guess this problem has no explicit solution. For generality I've used a variable for the "walked" distance instead of "1 mile".

Once around the South Pole and enough extra to end exactly one mile west of where you started?

Isnt’t it

1 plus X miles north of the SouthPole such that:

1 minus (pi2X) equals X divided by (1 plus X) ?

The striped pole is the south pole. The yellow sphere is your beginning point. You follow the red path south. It's exactly one mile. Then you head west on the green circle. Its circumference is exactly one mile, which brings you back to the red path, which you take north for one mile back to your yellow sphere starting point.

J

And the yellow circle of latitude is located one plus (1 divided by 2pi) miles north of the South Pole.

Once around the South Pole and enough extra to end exactly one mile west of where you started?

Isnt’t it

1 plus X miles north of the SouthPole such that:

1 minus (pi2X) equals X divided by (1 plus X) ?

Could you tell me the meaning of your symbols? "1" = "1 mile: , or just "1"? pi2X: 2*pi*X? X = the "overshoot" on the small circle? Is 1 + X miles an approximation? Certainly justified in the case of 1 mile trips, but I'd like to know... At least your equation can be solved if I understand your notation correctly... I tried to be too general in my calculations, resulting in quite complex equations that can't be solved analytically.

Could you tell me the meaning of your symbols? "1" = "1 mile: , or just "1"? pi2X: 2*pi*X? X = the "overshoot" on the small circle? Is 1 + X miles an approximation? Certainly justified in the case of 1 mile trips, but I'd like to know... At least your equation can be solved if I understand your notation correctly... I tried to be too general in my calculations, resulting in quite complex equations that can't be solved analytically.

1 is the number one.

1 mile is 5,280 feet

1 minus (2*pi*X) is the overshoot, in miles.

1+X is precise.

I posit the variable X, in miles. Our solution circle of latitude is 1+X miles north of the South Pole.

Then state the overshoot in miles in terms of X, namely X divided by (1+X)

When X is 1, A is half a radian and B is a quarter mile.

A is 1 divided by (1+X) radians.

B is A radians multiplied by (X divided by (1+X))

So B equals 1 – (2*X*pi) equals X divided by ((1+X) squared)

Ah, that makes it definitely less pleasant. Your previous version was a quadratic equation, easy to solve. Now you get a third degree equation. This can be solved, but doing that is as pleasant as a third degree interrogation. You'll find many methods on the Internet, but whatever method you use, the law of conservation of misery applies.

Perhaps the easiest solution in practice is to use a numerical method like Newton-Raphson. It does also have its difficulties, but these are not too bad in comparison.

I haven't looked further at your equations, as I'm now busy with my own calculations, and I'm already confused enough...

Ah, that makes it definitely less pleasant. Your previous version was a quadratic equation, easy to solve. Now you get a third degree equation. This can be solved, but doing that is as pleasant as a third degree interrogation. You'll find many methods on the Internet, but whatever method you use, the law of conservation of misery applies.

Perhaps the easiest solution in practice is to use a numerical method like Newton-Raphson. It does also have its difficulties, but these are not too bad in comparison.

I haven't looked further at your equations, as I'm now busy with my own calculations, and I'm already confused enough...

It starts out sounding so easy.

I have to correct myself one more time.

To go around the South Pole once plus extra such that you end precisely one mile west of where you started,

I now think like last night, that the circle of latitude is 1+X miles north of the South Pole such that:

1 – (2*X*pi) = X divided by (1+X)

But I confess I have forgotten the algebra to solve it.

Instead of simplicity here we get complexity. It's okay to have fun but fun in itself is of itself and a dead end. Those who are witty only don't get the girl--one night maybe excepted.

Instead of simplicity here we get complexity. It's okay to have fun but fun in itself is of itself and a dead end. Those who are witty only don't get the girl--one night maybe excepted.

--A virgin

I got the girl.

It’s Max who’s in misery.

I want to know the path he’s following on this problem, it sounds like he chose a difficult route to develop.

Watch, he will confirm my solution. And the elegant and direct path I took to get there. I have all the girls.

But I confess I have forgotten the algebra to solve it.

My estimate, X = 0.1396522

So now it's again:

1 - 2*pi*X = x/(1+x)

(X+1)*(1 - 2*pi*X) = X

X + 1 - 2*pi*X^{2} - 2*pi*X - X = 0

X^{2} + X - 1/(2*pi) = 0

X = - 1/2 + 1/2 SQRT (1 + 2/pi) (here only the positive root makes sense)

= 0.13965220479

Your estimate was not bad!

My attempt was more difficult because I used a general distance "walked" Z instead of 1 mile, and the fact that I made a correction for the difference between the radius of a circle in a plane and measured along a meridian. For circles with radius about 1 mile on Earth the difference is negligible, but not for larger circles. Perhaps I was just too ambitious...

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## Popular Posts

## Jonathan

Thanks for playing with the math, Jon and Max. It's fun to see. And I'm enjoying witnessing the spectrum that people fall on: mechies to mathies. I'm way over on the mechie side lacking in math, Merli

## Ellen Stuttle

Imagine a circle surrounding the South Pole as its center and exactly a mile larger in radius than a concentric smaller circle with circumference of exactly a mile around the South Pole. Then the

## Brant Gaede

The true north pole? This can't be the answer; it's too easy. Des Moines? --Brant

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## Ellen Stuttle

Looks ok to me, but I'm just doing approximate calculation.

Here's something I think is interesting, also doing approximations, but it conveys the picture:

Consider the set-up where there's a circle with a circumference of exactly a mile around the South Pole and a larger concentric circle with a radius a mile longer than that of the smaller circle.

The radius of the smaller circle is 1 mile divided by 2pi, approximately .15 mile, making the total radius of the larger circle approximately 1.15 miles.

Multiply 1.15 by 2pi to get the circumference of the larger circle. Approximately 8.345 miles. Round down to 8 miles for calculational convenience.

Now, start at some point - Point A - on the circumference of the larger circle.

Walk 1 mile south to the circumference of the smaller circle, then half way around the smaller circle (a distance of half a mile), then 1 mile north to a point - Point B - on the circumference of the larger circle.

You've walled 2 1/2 miles, but you're now half way around the circumference of the larger circle - a distance of approximately 4 miles going around that circle's circumference.

Ellen

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## Ellen Stuttle

Heaven forfend we should not be "in the spirit" of the puzzle! I misunderstood what you were asking, but my question was because I wasn't sure if Jon meant literally walkable places or not. He did specify in regard to circles around the South Pole that only about 50 or so subsets were walkable-by-a-human.

Ellen

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## BaalChatzaf

North means a 0 degree geometric heading with respect to the poles at wither end of the axis of rotation. This is slightly different from 0 degree heading wrt magnetic north. The magnetic poles wander about.

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## Max

We're talking here about the geometric north, ignore the magnetic poles.

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## Jon Letendre

AuthorLongitudes intersect only at the poles. And yes, that is the meaning of walking north.

Other places besides the North Pole also work. You end on the same point where you started.

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## BaalChatzaf

what other places?

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## Jon Letendre

AuthorPlaces that allow the northerly walk to occur on the same line of longitude that the southerly one did.

Max told you yesterday:

”You ignore the possibility that going south and going north can be done on the same line of longitude, while going west between these two displacements.”

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## Max

I've tried to calculate the second problem, but I get always implicit functions that cannot be solved analytically, only numerically, so I guess this problem has no explicit solution. For generality I've used a variable for the "walked" distance instead of "1 mile".

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## BaalChatzaf

Please provide the latitude and longitude of a point other than the North Pole which satisfies the conditions.

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## 9thdoctor

Watch what Michael Palin does:

He actually travels east instead of west, but no matter, just assume he walked west instead. It looks like the circle he's walking in is about 20 feet in circumference, so let's assume it's precisely 20 feet. If, to satisfy step 2, he were to do that walk 264 times (exactly one mile), his end point would be exactly the same degree of longitude as where he started. He could then walk north one mile to satisfy step 3.

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## Jonathan

The striped pole is the south pole. The yellow sphere is your beginning point. You follow the red path south. It's exactly one mile. Then you head west on the green circle. Its circumference is exactly one mile, which brings you back to the red path, which you take north for one mile back to your yellow sphere starting point.

J

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## Jon Letendre

AuthorOnce around the South Pole and enough extra to end exactly one mile west of where you started?

Isnt’t it

1 plus X miles north of the South Pole such that:

1 minus (pi2X) equals X divided by (1 plus X) ?

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## Jon Letendre

AuthorAnd the yellow circle of latitude is located one plus (1 divided by 2pi) miles north of the South Pole.

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## Max

Could you tell me the meaning of your symbols? "1" = "1 mile: , or just "1"? pi2X: 2*pi*X? X = the "overshoot" on the small circle? Is 1 + X miles an approximation? Certainly justified in the case of 1 mile trips, but I'd like to know... At least your equation can be solved if I understand your notation correctly... I tried to be too general in my calculations, resulting in quite complex equations that can't be solved analytically.

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## Jon Letendre

Author1 is the number one.

1 mile is 5,280 feet

1 minus (2*pi*X) is the overshoot, in miles.

1+X is precise.

I posit the variable X, in miles. Our solution circle of latitude is 1+X miles north of the South Pole.

Then state the overshoot in miles in terms of X, namely X divided by (1+X)

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## Jonathan

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## Jon Letendre

AuthorX is the radius of the walked circle.

B is the overshoot.

A is the angle of the overshoot. X and B are directly related,

when X is 1, A is half a radian and B is half a mile. When X is half, or 0.5, A is two thirds of a radian and B is two thirds of a mile.

So, B is X divided by (1+X)

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## Jon Letendre

AuthorSorry, I was mistaken above.

When X is 1, A is half a radian and B is a quarter mile.

A is 1 divided by (1+X) radians.

B is A radians multiplied by (X divided by (1+X))

So B equals 1 – (2*X*pi) equals X divided by ((1+X) squared)

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## Max

Ah, that makes it definitely less pleasant. Your previous version was a quadratic equation, easy to solve. Now you get a third degree equation. This

canbe solved, but doing that is as pleasant as a third degree interrogation. You'll find many methods on the Internet, but whatever method you use, the law of conservation of misery applies.Perhaps the easiest solution in practice is to use a numerical method like Newton-Raphson. It does also have its difficulties, but these are not too bad in comparison.

I haven't looked further at your equations, as I'm now busy with my own calculations, and I'm already confused enough...

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## Jon Letendre

AuthorIt starts out sounding so easy.

I have to correct myself one more time.

To go around the South Pole once plus extra such that you end precisely one mile west of where you started,

I now think like last night, that the circle of latitude is 1+X miles north of the South Pole such that:

1 – (2*X*pi) = X divided by (1+X)

But I confess I have forgotten the algebra to solve it.

My estimate, X = 0.1396522

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## Jon Letendre

AuthorX = 0.1396522 miles.

B = 0.122539 miles

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## Brant Gaede

Instead of simplicity here we get complexity. It's okay to have fun but fun in itself is of itself and a dead end. Those who are witty only don't get the girl--one night maybe excepted.

--A Virgin

if only

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## Jon Letendre

AuthorI got the girl.

It’s Max who’s in misery.

I want to know the path he’s following on this problem, it sounds like he chose a difficult route to develop.

Watch, he will confirm my solution. And the elegant and direct path I took to get there. I have

allthe girls.## Link to comment

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## Max

So now it's again:

1 - 2*pi*X = x/(1+x)

(X+1)*(1 - 2*pi*X) = X

X + 1 - 2*pi*X

^{2}- 2*pi*X - X = 0X

^{2}+ X - 1/(2*pi) = 0X = - 1/2 + 1/2 SQRT (1 + 2/pi) (here only the positive root makes sense)

= 0.13965220479

Your estimate was not bad!

My attempt was more difficult because I used a general distance "walked" Z instead of 1 mile, and the fact that I made a correction for the difference between the radius of a circle in a plane and measured along a meridian. For circles with radius about 1 mile on Earth the difference is negligible, but not for larger circles. Perhaps I was just too ambitious...

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