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asquad · Icarian Member

Can anyone help me to understand this one.
I tried to work it out but get to a stage where i cannot prove my answer.
I think i'm just baffling myself : )

CrystyB · Misunderstood Guy

arc AB must be 1/3 of the whole circumference, hence AOB=120 degrees, so AOX=60 and since OAX=90, OXA=30 so the opposing side (OA=R) must be half the hypothenuse (OX). Then X must be R high, which is the given answer.

What couldn't you prove?

Sofis · Beautiful and Decadent

Crysty, first off, seeing one third of the circumference is not the same as seeing one third of the surface. Second, I think you just proved that the distance between O and X is one diameter, which is not the same as the distance between the surface of the earth and X being one radius. Besides, the given answer was that one would have to reach an altitude equal to one diameter of the earth, not one radius.

[This message has been edited by Sofis (edited 04-20-2002 11:05 AM).]

asquad · Icarian Member

hmmmm. My first assumption/answer did work out to be a radius away. After re-reading the given answer it said a diameter away.
Can anyone provide the math and any suitable diagrams that could help me fully understand this ?

CrystyB · Misunderstood Guy

With the other two i agree i was wrong. BUT care to explain why "the distance between O and X is one diameter, which is not the same as the distance between the surface of the earth and X being one radius."???

K Surface of the 'spherical cone' is 3v/R so its v must be V/3. Bah i can't remember 10 y.o. spacial geometry. I'll try to compute this tomorrow.

kartelite · Daedalian Member

All right, I think you have to find the height (z coord) of the band that runs around the earth between the equator and the height where the area of that strip equals 2pi(r*r)/3. Then you'll know how high up to measure your angle from. To get the area of the strip you'd have to look up or derive the formula for a strip around a sphere, it'd probably take some 3-D calc integration. Well once you know the z coord, the job's pretty easy, and you can use geometry and trigonometry. You know that the section above the z coord contains 1/3 the SA of the sphere (earth), and to find the height above the earth you take half the chord length of the line cutting across the circle at height z. This will be the base of your triangle (also found by taking the cos of the line segment from the origin to a point on the surface at height z). The height is (1-z) + [the unknown]. I believe taking the directional derivative at height z will yield the arctan of the angle you'll need to finish up the equation. Then just plugging in the values of z should get you something like theta=arctan[(1-z+unkown)/(half chord length)], also (1-z+unkown)/(half chord length)=directional derivative at height z. I hope this is the correct approach, if anyone sees any mistakes or a simpler way please let me know.

Sofis · Beautiful and Decadent

CrystyB · Misunderstood Guy

BUMP. I'm ashamed to admit i'm making no progress...

Anyone care to remind me of some geometry obvious thing i keep overlooking?

kartelite · Daedalian Member

doesn't my explanation make sense?

CrystyB · Misunderstood Guy

It's just an explanation, it's no solution...

And it's not an obvious thing at all!!!

[This message has been edited by CrystyB (edited 05-21-2002 11:25 AM).]

kartelite · Daedalian Member

well, it is a solution in that you only have to plug numbers in...let's call it an equation where the only thing we don't know off the top is the equation for a strip around a sphere. rather than trying to find the limits for this (spherical coordinates) by hand you could probably look up the equation for it in a textbook or on the internet. however, there may be an easier way, considering most GL puzzles don't require second year college math.

CrystyB · Misunderstood Guy

mikegoo · Daedalian Member

The knowledge (which can be shown through some calculus that I don't feel like posting) that any "strip" (including a "cup" at the top) with a vertical hieght h of a sphere with radius r has the same surface area as a cylinder of radius r with the same height h... 2*pi*r*h makes this a fairly straight forward geometry problem.

If anybody really wants the calculus and/or the rest of the work I'll post it.