Jon Letendre 443 Report post Posted January 2, 2019 Now I think it is 1 plus X miles north of the South Pole such that: 1 minus (pi2X) equals X divided by (1 plus X) But I forgot the algebra to solve that equation! By using the above equation I found by trial and error, 1.1396522 miles north of South Pole. That circle of latitude should make one lap around the Pole plus enough extra that translates to one mile west of start point when you go back north. But someone needs to check it so I can sleep tonight. Quote Share this post Link to post Share on other sites

Ellen Stuttle 177 Report post Posted January 2, 2019 5 hours ago, Jon Letendre said: 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. Ellen 1 Quote Share this post Link to post Share on other sites

Ellen Stuttle 177 Report post Posted January 2, 2019 8 hours ago, Max said: 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. Ellen Quote Share this post Link to post Share on other sites

BaalChatzaf 60 Report post Posted January 2, 2019 23 hours ago, Max said: 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. Quote Share this post Link to post Share on other sites

Max 85 Report post Posted January 2, 2019 We're talking here about the geometric north, ignore the magnetic poles. Quote Share this post Link to post Share on other sites

Jon Letendre 443 Report post Posted January 2, 2019 On 1/1/2019 at 7:02 AM, BaalChatzaf said: 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. Quote Share this post Link to post Share on other sites

BaalChatzaf 60 Report post Posted January 3, 2019 7 hours ago, Jon Letendre said: 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. what other places? Quote Share this post Link to post Share on other sites

Jon Letendre 443 Report post Posted January 3, 2019 4 minutes ago, BaalChatzaf said: what other places? 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.” Quote Share this post Link to post Share on other sites

Max 85 Report post Posted January 3, 2019 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". Quote Share this post Link to post Share on other sites

BaalChatzaf 60 Report post Posted January 3, 2019 13 hours ago, Jon Letendre said: 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. Quote Share this post Link to post Share on other sites

9thdoctor 57 Report post Posted January 3, 2019 1 hour ago, BaalChatzaf said: 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. 1 Quote Share this post Link to post Share on other sites

Jonathan 295 Report post Posted January 3, 2019 4 hours ago, BaalChatzaf said: 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. J 1 Quote Share this post Link to post Share on other sites

Jon Letendre 443 Report post Posted January 3, 2019 8 hours ago, Max said: 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 South Pole such that: 1 minus (pi2X) equals X divided by (1 plus X) ? Quote Share this post Link to post Share on other sites

Jon Letendre 443 Report post Posted January 3, 2019 10 minutes ago, Jonathan said: 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. Quote Share this post Link to post Share on other sites

Max 85 Report post Posted January 3, 2019 9 minutes ago, Jon Letendre said: 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 South Pole 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. Quote Share this post Link to post Share on other sites

Jon Letendre 443 Report post Posted January 3, 2019 26 minutes ago, Max said: 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) Quote Share this post Link to post Share on other sites

Jonathan 295 Report post Posted January 3, 2019 2 hours ago, Jon Letendre said: And the yellow circle of latitude is located one plus (1 divided by 2pi) miles north of the South Pole. 1 Quote Share this post Link to post Share on other sites

Jon Letendre 443 Report post Posted January 3, 2019 X 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) Quote Share this post Link to post Share on other sites

Jon Letendre 443 Report post Posted January 3, 2019 Sorry, 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) Quote Share this post Link to post Share on other sites

Max 85 Report post Posted January 4, 2019 1 hour ago, Jon Letendre said: Sorry, 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) 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... Quote Share this post Link to post Share on other sites

Jon Letendre 443 Report post Posted January 4, 2019 1 hour ago, Max said: 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. My estimate, X = 0.1396522 Quote Share this post Link to post Share on other sites

Jon Letendre 443 Report post Posted January 4, 2019 X = 0.1396522 miles. B = 0.122539 miles Quote Share this post Link to post Share on other sites

Brant Gaede 179 Report post Posted January 4, 2019 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 Quote Share this post Link to post Share on other sites

Jon Letendre 443 Report post Posted January 4, 2019 6 minutes ago, Brant Gaede said: 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. Quote Share this post Link to post Share on other sites

Max 85 Report post Posted January 4, 2019 8 hours ago, Jon Letendre said: 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 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... 1 1 Quote Share this post Link to post Share on other sites