Chronology |
Current Month |
Current Thread |
Current Date |

[Year List] [Month List (current year)] | [Date Index] [Thread Index] | [Thread Prev] [Thread Next] | [Date Prev] [Date Next] |

*From*: David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>*Date*: Fri, 13 Aug 2004 11:42:24 -0400

Regarding Joel R's comment:

For me, the impediment is partially that; but more so that I've

heard this one since I was a kid and was always told the answer was

the North Pole (combined probably with a northern hemisphere

prejudice) all of this combines in my head to provide a powerful

disinclination to search for other answers beyond the one given.

As an aside, both answers provide a nice example of how geometry on

S2 (surface of a 3dim sphere) is different from geometry on R2

(Euclidean plane).

Joel R

This discussion reminds me of a somewhat related problem. The

Northern Hemisphere solution can be considered as a closed path that

is, to a good approximation, a closed circular sector whose wedge

angle is 1 radian. The 2nd leg of the path is a circular arc of

approximately 1 radian in turning angle and the 1st and 3rd legs of

the path are two (geodesically straight) radii connecting the arc

ends to the center point of the arc.

Suppose we straightened out the 2nd leg of the path so *all three*

legs of the path are geodesically straight and the length of the 2nd

leg is the same as the length of the 1st and 3rd legs. The whole

closed path is now an equilateral triangle as inscribed onto the

spherical surface. The problem is to find a formula for the measure

of the interior angle of such an equilateral triangle as a function

of the length s of the sides of the triangle (conveniently in units

of the sphere's radius). A few hints are that 1) the value of the

formula must boil down to 60 deg in the limit of s becoming a

zeroth fraction of the sphere's radius, 2) the value of the formula

becomes 90 deg when s is 1/4 of the circumference of the sphere,

3) the maximum size triangle occurs for a great circle with 3

equally-spaced vertices (120 deg apart from each other) on it with

the interior angle at each vertex being 180 deg across the vertex and

each side having a length s of 1/3 of the sphere's circumference, and

4) the messy intermediate math eventually simplifies to a relatively

simplified formula in the general case.

For a lot of extra credit points you can also find the proper formula

for the *area* of this spherical equilateral triangle in terms of the

length s of the sides of the triangle (making sure that the formula

boils down to all the correct values for the variously known special

cases).

David Bowman

- Prev by Date:
**Re: navigation riddle** - Next by Date:
**Re: navigation riddle** - Previous by thread:
**trove of physics questions** - Next by thread:
**Re: spherical geometry (was Re: navigation riddle)** - Index(es):