MATLAB Central Newsreader - volume of a pyramid

volume of a pyramid

Pinpress — Fri, 23 Nov 2007 01:35:11 -0500

Hi,

Anyone knows how to calculate the volume of a 3D ultrasound
imaging volume, like the one in the following figure:

http://www.gehealthcare.com/usen/ultrasound/education/images/u3d4d/fig2.jpg

Or any pointer to useful links would be much appreciated.
thanks and happy Thanksgiving!

Re: volume of a pyramid

Roger Stafford — Fri, 23 Nov 2007 02:52:51 -0500

"Pinpress " <nospam__@yahoo.com> wrote in message <fi5aof$gp
$1@fred.mathworks.com>...
> Hi,
>
> Anyone knows how to calculate the volume of a 3D ultrasound
> imaging volume, like the one in the following figure:
>
> http://www.gehealthcare.com/usen/ultrasound/education/images/u3d4d/
fig2.jpg
>
> Or any pointer to useful links would be much appreciated.
> thanks and happy Thanksgiving!
--------
  You haven't stated explicitly what shape your image has. It looks like a solid
defined by projecting a spherical quadrilateral surface inwards along radial
lines to a smaller spherical quadrilateral. Is that correct?

  If so, then its volume can be calculated in terms of the area of the outer
quadrilateral:

V = a*R/3*(1 – (r/R)^3)

Where R is the outer radius, r the inner radius, and a the outer area.

  As for computing a, it is equal to

a = R^2*(A + B + C + D – 2*pi)

where A, B, C, and D are the four angles in radians at the four vertices of the
outer quadrilateral.

  So your problem becomes that of determined what those four angles are.
There is no way of determining them without additional information about
that quadrilateral.

Roger Stafford

Re: volume of a pyramid

Pinpress — Fri, 23 Nov 2007 03:05:56 -0500

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.

> --------
> You haven't stated explicitly what shape your image has.
It looks like a solid
> defined by projecting a spherical quadrilateral surface
inwards along radial
> lines to a smaller spherical quadrilateral. Is that correct?
>
> If so, then its volume can be calculated in terms of the
area of the outer
> quadrilateral:
>
> V = a*R/3*(1 – (r/R)^3)
>
> Where R is the outer radius, r the inner radius, and a the
outer area.
>
> As for computing a, it is equal to
>
> a = R^2*(A + B + C + D – 2*pi)
>
> where A, B, C, and D are the four angles in radians at the
four vertices of the
> outer quadrilateral.
>
> So your problem becomes that of determined what those
four angles are.
> There is no way of determining them without additional
information about
> that quadrilateral.
>
> Roger Stafford
>

Re: volume of a pyramid

Pinpress — Fri, 23 Nov 2007 03:06:01 -0500

Re: volume of a pyramid

Pinpress — Fri, 23 Nov 2007 03:06:13 -0500

Re: volume of a pyramid

Pinpress — Fri, 23 Nov 2007 03:06:13 -0500

Re: volume of a pyramid

Pinpress — Fri, 23 Nov 2007 03:06:14 -0500

Re: volume of a pyramid

Pinpress — Fri, 23 Nov 2007 03:08:14 -0500

Re: volume of a pyramid

Pinpress — Fri, 23 Nov 2007 03:13:44 -0500

Didn't mean to send so many identical reply -- looks like
mathworks server problem.

Re: volume of a pyramid

Roger Stafford — Fri, 23 Nov 2007 03:47:31 -0500

"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