Since writing my original article on this problem, I have received a number of requests for variations on the basic tank type. This page covers the most often requested variation (a tank mounted vertically), but there have been a number of other requests. Here are notes on some of these:

• Oval cross-section. This very common horizontal-mounted tank type is wider than it is tall. The solution for this case is quite simple:

  • Measure the tank's overall height (temporarily disregard the tank's width) and use the horizontal-tank solution as though the tank has a cylindrical cross-section.
  • Measure the tank's width, call this value W.
  • Using the R value measured for the tank during the horizontal solution, multiply the volume result by W/2R (more on this shape below).

• Horizontal tank installed tilted, e.g. not perfectly horizontal. For maintenance reasons, some horizontal tanks are not installed horizontally, but at a small angle to facilitate cleaning. I have looked into computing this case, but it is very complex and the number of measurements and the specifics of the tank content height measurement are so daunting, that I have decided not to try to present a method for this case. In most cases, the provided horizontal solution will provide reasonably accurate results.

Thanks!

Hey - Just wanted to thank you for one of the best Horizontal Tank volume calc pages I've ever seen. (The open source was nice as well.) Not only was the page informative it was well written and easy to understand.

Thanks! I have recently updated the page and added a vertical tank solver for the same general shape tank, and I have made a number of other improvements, particularly in the provided Java program.

Horizontal: http://arachnoid.com/tank_volume

Vertical: http://arachnoid.com/tank_volume/variations.html

Thanks again, and please let me know if any difficulties come up with the equations or code.

What about an ellipsoidal cylinder?

I wondered if your formula could be easily adapted to cover the case of the parallel section being ellipsoidal instead of cylindrical as is the case with modern fuel bowsers.

Do you mean a cylinder lying on its side, whose width is greater than its height? Yes, that case can be handled easily.
Just take all the tank's dimensions, but proceed as though the tank was a simple cylinder, using the cylinder's height as the diameter in my equations. Then, when you have gotten a result volume from my equations, multiply by the ratio of the width to the height.
Example. Let's say you have a tank that has a height of 48 inches, but a width in cross-section of 56 inches. Go through all the computations using 48 inches as the tank diameter. Then, for each result volume, multiply by the ratio of 56/48. This works for any partial volume result.
This idea works because the volume of an ellipsoidal cylinder of the kind we're speaking of is equal to π * r1 * r2 * h, where r1 and r2 are the two differing radii of the tank (in the above example, they would equal 24 and 28 inches).
Be sure to test your implementation of this procedure, comparing the results with some other source of information, to make sure you are carrying it out correctly.