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revolution
Okay, for the north/south reading we are using the rotational north/south, not the magnetic north/south.


28 May 2013, 11:21 

revolution
YONG wrote: Define "starting point" please. Other than using the Earth's rotation to determine north and south I don't think that the rotation would be an important consideration. But I know you guys and gals here are smart and could possibly prove me wrong here, but we shall see. 

28 May 2013, 12:25 

bitRAKE
I'm concerned about the sensitivity of measurement required to accurately traverse that distance. Not that they couldn't get close, but the exact spot might be impossible. It's easier to get close and look around for the stake, monolith, or whatever. Anyone who travels is constantly adjusting their heading as better information becomes available.


28 May 2013, 12:39 

revolution
For the purposes here the compass is perfect. So that means even a microfraction of a degree off course would be detected instantly. And the man is able to follow the compass direction perfectly also.


28 May 2013, 12:47 

YONG
I guess the trick of this problem lies here:
"Upon discovering that he can no longer walk any further south, ..." By that time, the man is situated exactly at the south pole! References: http://www.polartrec.com/forum/icecubeiniceantarctictelescope/compassthesouthpole http://wiki.answers.com/Q/Will_a_compass_work_at_the_south_pole Am I right? 

28 May 2013, 13:04 

revolution
YONG wrote: I guess the trick of this problem lies here: 

28 May 2013, 13:20 

tthsqe
revolution,
if the answer you are expecting is not the obvious one of zero, then the problem is not defined well enough, or you have some strange notion of probabilities. The man essentially starts on the latitude line 10km away from the south pole, and returns to this same latitude only now with an arbitrary longitude. 

28 May 2013, 15:34 

revolution
tthsqe wrote: revolution, tthsqe wrote: The man essentially starts on the latitude line 10km away from the south pole, and returns to this same latitude only now with an arbitrary longitude. Assume the starting point is a circle of one metre in diameter. Compute the probability that the man will return to some point that lies within the one metre circle. Hint: This is actually trickier to work out than it might first appear. But there is a precise answer. 

28 May 2013, 19:21 

bitRAKE
We call the solution "hole in the arc of revolution".


28 May 2013, 21:06 

tthsqe
So just to clarify, we want to calculation the probability that the man will return to within 1 meter of the point where he started given that he return point is uniformly distributed on the latitude where he started?
Someone should definitely try it and see if they got the same answer of Code: 0.00318310017018888629% chance EDIT: corrected answer reposted below Last edited by tthsqe on 31 May 2013, 15:42; edited 3 times in total 

28 May 2013, 21:45 

revolution
tthsqe wrote: So just to clarify, we want to calculation the probability that the man will return to within 1 meter of the point where he started ... 

29 May 2013, 00:49 

bitRAKE
I got 0.0124602 % chance. (which seems too large)
Given an arc length L on a circle of radius R, the distance from the center of the circle to the midpoint of line (connecting the two arc endpoints) is: d = R cos(L / 2R) The area of a spherical cap is (h=Rd): 2 pi R h The traveler can return to an area between 10000.5 and 9999.5 distance from the pole  which creates an arc length of 20001 and 19999 respectively, across the pole. First we calculate the height to the arc, subtract the smaller cap from the larger to get our destination area. Next we have the starting circle of 1 meter diameter. We use the same equations  with L=1. Divide to get probability. Edit: oh, it was 10k, 0.00124603 % chance. 

29 May 2013, 07:44 

revolution
bitRAKE wrote: The traveler can return to an area between 10000.5 and 9999.5 distance from the pole 

29 May 2013, 08:40 

bitRAKE
It was so much easier the way I chose to interpret it. (...and you can see why I called it the "hole in the arc of revolution"  triple entendre if you will, lol.)
So there is an arc destination within the original 1 meter starting position. This arc is on the earth sphere and 10km from the pole? Making the probability the ratio of that arc and the circumference of the 10km circle? 

29 May 2013, 09:57 

tthsqe
I think that revolution intends for the man to start and return to the same latitude. The problem is then: which points on that latitude are less than 0.5 meters away. In my answer, I calculated it for 1.0 meters away because I confused radius and diameter. So, my corrected solution would be the same except that the equation for phi2 should be phi2 = 0.5/r. This gives
Code: 0.0015915500843919805897% We have good reasons to believe that the answer should be close to Code: 0.5/(10^4*Pi) 

29 May 2013, 10:56 

revolution
tthsqe wrote: I think that revolution intends for the man to start and return to the same latitude. The problem is then: which points on that latitude are less than 0.5 meters away. In my answer, I calculated it for 1.0 meters away because I confused radius and diameter. So, my corrected solution would be the same except that the equation for phi2 should be phi2 = 0.5/r. This gives 

29 May 2013, 11:05 

tthsqe
Quote: Did you know that the Earth is not perfectly round but it more like an oblate spheroid? Yes, but I was hoping that it would not come up, as geodesics on ellipsoids are complicated enough that I would rather not think about them. Do you agree with my corrected solution if we assume that the earth is a sphere with that radius? 

31 May 2013, 00:27 

revolution
tthsqe wrote: ... geodesics on ellipsoids are complicated enough that I would rather not think about them. tthsqe wrote: Do you agree with my corrected solution if we assume that the earth is a sphere with that radius? 

31 May 2013, 01:45 

bitRAKE
I used numerical integration method.
0.00159155007810710868 % Code: fstcw word[rsp] pop rax push rax or word[rsp], 11'0000'0000b ; extended precision and word[rsp],not 1100'0000'0000b ; round closest fldcw word[rsp] push rax .integrate: fld [..L] ; L = length desired fld [..dt] ;dt = change between measures fldz ; S = sum of partial lengths fld [..b] ; v = previous (y), b cos 0 fldz ; u = previous (x), a sin 0 fldz ; t = current measure index (time) .slice: fadd st0,st4 ; t u v S dt L ; Ellipse: ; x = a sin t, major axis, start in center (0) ; y = b cos t, minor axis, start at top pole fld st0 ; t t u v S dt L fsincos ; y x t u v S dt L fmul [..b] ; Y x t u v S dt L fxch st1 ; x Y t u v S dt L fmul [..a] ; X Y t u v S dt L ; Extremely general numerical integration to approximate length of arc: ; S = lim Sum[sqrt( dx^2 + dy^2 )] fsubr st3,st0 ; u Y t dx v S dt L fxch st1 ; Y u t dx v S dt L fsubr st4,st0 ; v u t dx dy S dt L fxch st4 ; dy u t dx v S dt L fmul st0,st0 ; DY u t dx v S dt L fxch st1 ; u DY t dx v S dt L fxch st3 ; dx DY t u v S dt L fmul st0,st0 ; DX DY t u v S dt L faddp st1,st0 ; dl t u v S dt L fsqrt ; dl t u v S dt L fadd st0,st4 ; S t u v # dt L fcomi st6 ; S t u v # dt L fxch st4 ; # t u v S dt L fstp st0 ; t u v S dt L jc .slice ; stop if sum over L fstp st0 fldpi fmulp fadd st0,st0 fld1 fdivrp st1,st0 ; < ST0 is lost traveler fldcw word[rsp] int3 ..dt dq 1e12 ..L dq 10000.0 ..a dq 6378136.6 ; major (equatorial) and ..b dq 6356751.9 ; minor (polar) radii of approximate Earth oblique spheroid 

31 May 2013, 09:14 

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