Volume of a Rounded Horizontal Tank

Date: 03/26/2004 at 17:01:04
From: Davia
Subject: Volume of a Rounded Horizontal Tank (capsule-shaped)

What is the formula for finding the volume of a rounded horizontal 
tank (capsule-shaped)?  I found the discussion "Finding the Volume of
a Horizontal Tank" very helpful but it doesn't take into account 
rounded ends.  I think I need a cylinder + 2 lenses/domes.

Date: 03/26/2004 at 17:30:54
From: Doctor Jeremiah
Subject: Re: Volume of a Rounded Horizontal Tank (capsule-shaped)

Hi Davia,

For this type of tank you need the length of the cylindrical main body
(which I will call L), the radius of the cylindrical main body (which
I will call R) and the distance that the curved ends stick out from
the cylindrical main body (which I will call a):

           +--------------------+
       +                            +
     +                                +
    +                                  +
    +                                  +  ---
    +                                  +   |
     +                                +    R
       +                            +      |
           +--------------------+         ---

    |--a---|---------L----------|---a--|

The ends are ellipsoidal.  Half the ellipsoid is on one end and half
is on the other.  Together they make a whole ellipsoid.  So the
problem boils down to a partially filled cylinder and a partially
filled ellipsoid.

The ellipsoid has radii in three different directions: in the vertical
direction the radius is z = R, in the horizontal direction along the
tank the radius is x = a, and in the other horizontal direction the
radius is y = R.  So we can use the formula for a partially filled
sphere and modify the result by a factor of a/R to compensate for the
one shorter radii.

A spherical cap has a volume of:

   SphereVolume = (Pi/3)(3hR - h^2)h  where h is the depth of fluid

But when we modify it to be an ellipsoid we get:

   EllipsoidVolume = (a/R)*SphereVolume

So all we need to do to get the volume of the tank at any level is add
the volume of the cylindrical part and the ellipsoid volume that
compensates for the ends.

V = L(R^2*acos((R - h)/R) - (R - h)*sqrt(2Rh - h^2))
  + (a/R)(Pi/3)(3hR - h^2)h

So now let's check.  When the tank is full, then h = 2R and:

V @ h = 2R = L(R^2*acos((R - 2R)/R) - (R - 2R)*sqrt(2R(2R) - (2R)^2))
         + (a/R)(Pi/3)(3(2R)R - (2R)^2)(2R)

V @ h = 2R = L(R^2*acos(-1) + R*sqrt(4R^2 - 4R^2))
         + (a/R)(Pi/3)(6R^2 - 4R^2)(2R)

V @ h = 2R = L(R^2*Pi) + (a/R)(Pi/3)(4R^3)

At this point it's clear that the first term is the volume of a
cylinder and the second is the volume of a sphere with a modifier to
make it into an ellipse, so it should work.

If your dimensions are in inches, then V will be cubic inches and you
will have to divide by 231 to get American gallons.

Please let me know if I can be of further help.

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/