From: "Richard Schwartz" 
To: "Thomas Janstrom" 
Cc: 
Reply-To: "Richard Schwartz" 


There are three ways to go with this, depending on your level of math skills.

1.   Spherical Trig.   The idea here is subtract the volume of a cone from
the volume under a spherical circle that is coincident with that cone.

2.  Calculus:   Approximate the curved surface with a parabola.  For molding
purposes, that is close enought to a sphere.  Then integrate.

3.  Numerical Integration: use numerical integration such as simpson's rule
to find the volume.   This is actually similar to the calculus method
because numerical integration assumes some kind of polynomial for the
integrand.   One of my favorite methods is based on this.   It is exact for
the volume of a LOT of things, and close for most everything else.

volume= height * average cross section area.

average cross section area= (area of top + 4* area of middle + area of bottom)/6

Note that this is nothing more than Simpson's rule integration.  If you
memorize this one formula, you don't have to memorize anything else from
solid geometery.  I don't know how well it works for a sector of a sphere;
it probably depends how you define "area of the middle".  But
let's play with it...    The area of the top is zero.

If you define the "middle" as half the saggita, the radius there
is sqrt(h*R-(h/2)^2).   So the area there is pi*(h*R-(h/2)^2).

The bottom has an radius of sqrt(2*h*R-h^2), so its area is pi*(2*h*R-h^2).

If you go through the algebra with the simpson rule 1-4-1 procedure, you
get that the volume is:

(pi/6)*(6*R*h^2-2*h^3).   With h=R, you have a hemisphere, and this result
is exact.   I don't know about lesser values of h.   So just memorize that
one volume formula, the 1-4-1 simpson rule.  This also works for area of
flat shapes: area= height * average width, with "average width"
calculated by the same 1-4-1 rule.  Recall that the triumph of Newton was
replacing three complex rules for orbital motion with one simple rule.  
Likewise, the
triumph of Simpson was replacing all those mensuration formulas with one
sim ple formula.

Then there is the distasteful fourth way:  look it up in a standard
handbook.  There I found that the volume is...

(2/3)*pi*h*R^2, where h is the saggita  (which you know how to compute if
your are a telescope person)

Lesseeee... sanity check:  if the saggita is R, you have a hemisphere, and
the volume is (2/3)*pi*R^3, which is half the volume of a sphere.   Yeah,
it
works.


. . . Richard

----- Original Message -----
From: "Thomas Janstrom" 
To: "atm" 
Sent: Saturday, January 18, 2003 5:21 AM Subject: ATM calculating the volumes


>
> How do I calculate the volume of a circular section of a sphere? Sort of
> like what we grind out when hogging. What I want to do is work out how big
a
> blank I can get from brick of C1 glass (see corning optical glass thread
for
> details) I'm planning on casting one of the curves into the blank so I
want
> to work out the volume of the, at this point convex side.
>
> Anyway thoughts and more importantly the formula to do this would be
great.
>
> TIA.
>
> Clear skies, Thomas
> http://www.tjanstrom.com
> "Don't make me set the laser printer to stun"
>
>
>

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