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From: "Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid>
Newsgroups: comp.soft-sys.matlab
Subject: Re: volume of a pyramid
Date: Fri, 23 Nov 2007 03:47:31 +0000 (UTC)
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"Pinpress " <nospam__@yahoo.com> wrote in message <fi5g2k$37v
$1@fred.mathworks.com>...
> First, thanks for the reply.
> 
> As for the volume, it is a solid that both radial surfaces
> have identical radius toward a common origin.  All other 4
> surfaces are on a straight plane (as opposed to the two
> radial surfaces), and their angles with the vertical axis is
> known (i.e.g, vector-to-plane angles are known). 
> 
> So are your equations supposed to calculate the solid volume
> as describe above.  I will examine the equations too. 
> Thanks again. 
-------
  I'm not sure what is meant by the "vertical axis" here, but it doesn't sound as 
though that would be enough information to uniquely determine the spherical 
surface area.  You need some information about the spread of the four linear 
edges from one another, and that is different from knowing what angle they 
make with some vertical direction.  In other words what are the relative 
lengths of the four circular arcs which bound the spherical quadrilaterals?  
Are they all equal?  Are opposite pairs equal?

  The four angles, A, B, C, and D I mentioned are equal to the four dihedral 
angles between the four pairs of planar surfaces.  For example, if these were 
all equal to pi radians, the four planes would then all be part of the same 
plane and would cut the sphere precisely in half, which is consistent with the 
given formulas.  On the other hand, if they were all equal to pi/2, that would 
be an area and volume of zero.  The difference A+B+C+D-2*pi can be 
considered as the "spherical excess" over the normal value of zero for a 
planar quadrilateral.

  (By the way, there seems to be a multiplicity of copies of this last article of 
yours, Pinpress.  Something may have gone wrong in sending it.)

Roger Stafford