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This page is an extension of my original tank volume page, a page that analyzes a storage tank with elliptical end caps, lying horizontally. Since I published the original article, I have received a number of inquiries about tanks in other orientations, most often a vertically mounted tank. This page analyzes that tank orientation.

| L | The length of the cylindrical section. |
| R | The radius of the cylinder and the major radius of the elliptical end caps. |
| r | The minor radius of the elliptical end caps. |
| y | The height of the tank's contents. |
Let's start with the half-ellipses that lie at the bottom and top of the tank. First, we need to get the area of a circle in the horizontal plane that slices through an ellipse at a vertical position given by y. For a spheroid with major radius R and minor radius r, such a function is:Center Cylindrical Section (applied to zone 2)Now let's integrate this circle area to get a partial ellipse volume:
(1) ![]()
Equation (2) completes the volume of an ellipse with major radius R and minor radius r, between the bottom of the tank and any chosen y value within the range 0 <= y <= 2r. With range adjustments, this derivation will be used during volume computation of the bottom and top sections of the tank (zones 1 and 3 in Figure 1).
(2) ![]()
The volume of a cylinder is trivially computed using:
(3) ![]()
For our multi-part, zone-oriented solution, in the above equation, H = y-r (see Figure 1).
It must be emphasized that, although the equations for this solution are not particularly complex, the solution is algorithmic, not purely algebraic. Unlike the horizontal tank problem, there is no single equation that can be applied to the entire tank, instead a three-part solution is required, and the selected zone depends on the measured height of the tank's contents.
Here is a table that describes the zones and their corresponding equations:
Zone (see Figure 1) Range of y values Solution (volume in units of y cubed) 1 (bottom ellipse) 0 <= y < r ![]()
2 (cylindrical section) r <= y < r+L ![]()
3 (top ellipse) r+L <= y <= 2r+L ![]()
It is important to emphasize that each of the zone solutions listed above provides an independent tank volume for the corresponding value of y, e.g. one does not add individual zone volumes to arrive at a solution.
Sanity Check
As readers apply this zone solution, to avoid errors they should apply common-sense tests to their methods. Zones 1 and 3 should each have volumes of 2/3 π R2 r, and Zone 2 should have a volume of π R2 L. The volume of the entire tank should equal 4/3 π R2 r + π R2 L.
Here is an online computer based on the foregoing derivation.
NOTE: This computer is meant for a vertical tank. For a horizontal tank, click here.
Click a link to access these additional resources (or use the drop-down lists at the top and bottom of this page):
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