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Topic: hi,friends(6)
Replies: 2   Last Post: Dec 21, 2007 2:42 AM

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Posts: 12
Registered: 2/17/07
Re: hi,friends(6)
Posted: Dec 17, 2007 12:31 AM
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On 12/14/07 noble_cbc wrote:

Dear friend:

can you help me?

what is the volume of a '3-sphere_pyramid'(that is: with 3 points A,B,C
on the surface of a sphere of radius R and the vertex O is the center of
the sphere) with

And,how you deduce it out?

As we know the spherical wedge(lune) has the volume: v=2*R^3*theta/3.

One approach (not employing the lunes):

The spherical pyramid's volume, Vsp, should stand in ratio to the
sphere's volume, V, as does the spherical triangle's area, Ast, stand in
ratio to the sphere's area, A. So, the final step is to equate those two
ratios and solve for Vsp.

The obvious quantities are the V and A, and they stand in ratio of R/3.
Then Vsp will end up being this R/3 times Ast.

So, to first find Ast, one could use Girard's Theorem...if the spherical
excess, E, is known. That requires knowing the surface angles at points
A, B, and C (then E = A + B + C - pi). Since initially only the central
angles al, be, and ga are known, along with R, then all elements of the
spherical triangle ABC, including sides a, b, and c are solvable using
the appropriate Naperian Formula's.

Date Subject Author
Read hi,friends(6)
China -->hk
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Read Re: hi,friends(6)
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