Horizontal Elliptical Tank Volume Calculator

This calculator works out the volume of a horizontal elliptical tank (also called a horizontal oval tank) from the length, the major axis (long diameter), the minor axis (short diameter), and the current liquid depth. You enter the tank dimensions and liquid depth in any consistent unit, and you get back the liquid volume in cubic units, plus the total tank capacity and the percentage filled.

Formula

The liquid volume in a horizontal elliptical tank is calculated by multiplying the tank length by the cross-sectional area of the liquid. The cross-section is a half-ellipse, and the area of a partially filled ellipse is given by:

V = L × (A × B) × [ arccos((B − h)/B) − (B − h) × √(2Bh − h²) / B² ]

Where:

• L = tank length (m, ft, cm, etc.)
• A = semi-major axis = major diameter / 2 (same unit as L)
• B = semi-minor axis = minor diameter / 2 (same unit as L)
• h = liquid depth measured from the bottom of the tank (same unit)
• V = liquid volume (cubic unit of L)

The formula uses the arccosine (inverse cosine) function, which must be in radians for most calculators. The term inside the brackets is the fraction of the ellipse’s area that is filled. When h = B (half full), the bracket equals π/2, so the volume is exactly half the full capacity.

Worked Example

Using the default values: L = 3 m, major diameter = 1.5 m, minor diameter = 1.0 m, liquid depth = 0.6 m.

1. Semi-major axis A = 1.5 / 2 = 0.75 m
2. Semi-minor axis B = 1.0 / 2 = 0.5 m
3. Compute (B − h) = 0.5 − 0.6 = −0.1 m
4. Compute arccos((B − h)/B) = arccos(−0.2) ≈ 1.7722 radians
5. Compute √(2Bh − h²) = √(2×0.5×0.6 − 0.6²) = √(0.6 − 0.36) = √0.24 ≈ 0.4899
6. Compute (B − h) × √(2Bh − h²) / B² = (−0.1) × 0.4899 / 0.25 = −0.19596
7. Bracket = 1.7722 − (−0.19596) = 1.96816
8. Multiply by L × A × B = 3 × 0.75 × 0.5 = 1.125
9. Volume = 1.125 × 1.96816 ≈ 2.214 m³

The calculator shows 2.214 m³, matching the manual calculation. Note that the depth (0.6 m) is greater than the semi-minor axis (0.5 m), so the tank is more than half full, which is why the bracket value exceeds π/2 (1.5708).

When to Use This Calculator

Horizontal elliptical tanks are common in fuel storage, chemical processing, water treatment, and agricultural applications. They are used when a low-profile tank is needed but a flat bottom is not ideal. This calculator is useful for:

• Estimating how much liquid remains in a partially filled tank (e.g., fuel level gauging).
• Determining the total capacity of a new tank design.
• Checking if a tank will hold a required volume before purchase.
• Converting between volume and liquid depth for inventory or process control.

The calculator assumes the tank is perfectly elliptical in cross-section and horizontal (its long axis is parallel to the ground). If the tank is tilted, the liquid surface is not flat across the entire length, and the formula becomes more complex.

How to Choose the Right Dimensions

To use the calculator, you need the tank’s major diameter (the wider horizontal dimension of the ellipse) and minor diameter (the shorter vertical dimension). These are often stamped on the tank nameplate or available from the manufacturer’s drawing. If you only have the circumference, you cannot derive the ellipse axes without additional information—the same circumference can correspond to many different ellipse shapes.

When measuring an existing tank, measure the inside dimensions to get the actual liquid volume. Outside dimensions include wall thickness, which can cause a noticeable error on small tanks. For example, a 3 mm wall on a 1 m diameter tank reduces the volume by roughly 1%.

Liquid depth is measured from the bottom of the tank interior to the liquid surface. Use a dipstick or level sensor. If the tank is buried, the depth is still measured from the bottom, not from the ground surface.

Common Mistakes and Misconceptions

One frequent error is using the full diameters instead of the semi-axes in the formula. The formula requires the semi-axes (A = D/2, B = d/2). If you plug in the full diameters, the volume will be four times too large.

Another common mistake is mixing units. If the length is in meters and the diameters in centimeters, the result is nonsense. Always convert all dimensions to the same unit before calculating.

A common misconception is that a horizontal elliptical tank with a half-depth of liquid is half full. That is true only if the liquid depth equals the semi-minor axis (B). Because the ellipse is symmetric, the volume at h = B is exactly half the total capacity. However, the relationship between depth and volume is not linear—at 25% depth, you have much less than 25% of the volume because the bottom of the ellipse is narrow.

Rounding errors can also creep in. The arccosine function must be in radians, not degrees. If your calculator is in degree mode, the result will be wildly wrong.

Limitations of the Formula

This formula assumes the tank is a perfect elliptical cylinder—that the cross-section is the same ellipse along the entire length. Real tanks often have dished or flat ends, which add or subtract a small volume. For most practical purposes, the error is minor (a few percent) unless the tank is very short relative to its diameter.

The formula also assumes the liquid surface is flat and horizontal. In a moving vehicle or a tank that is not level, the surface will be sloped, and the volume will differ from the calculated value. For static tanks on level ground, the formula is accurate.

If the liquid depth exceeds the minor diameter (h > 2B), the tank is overfilled, and the formula will give an invalid result (arccosine of a number less than −1). The calculator clamps the depth to the tank height; if you enter a depth greater than the minor diameter, the volume will be capped at the full capacity.

Alternatives and Related Calculations

If your tank is horizontal but has a circular cross-section (a horizontal cylinder), use the horizontal cylinder volume calculator, which uses a slightly different formula based on the circle’s radius.

For vertical elliptical tanks (where the ellipse is upright), the volume calculation is simpler: the cross-sectional area is constant with depth, so volume = π × A × B × L, where L is the liquid height.

For tanks with conical, spherical, or other shaped ends, you need to add the end volumes separately. Many tank manufacturers provide calibration charts that give volume per inch or centimeter of depth, which are more accurate than a simple formula for irregular shapes.

Frequently Asked Questions

Q: How do I convert the volume to gallons or liters?
After getting the volume in cubic meters, multiply by 1000 to get liters, or by 264.172 to get US gallons. For cubic feet, multiply by 7.48052 to get US gallons. Always check which gallon you need—imperial gallons are about 20% larger.

Q: Can I use this for a tank with flat ends?
Yes, the formula assumes flat ends. If your tank has domed or dished ends, the actual capacity will be slightly higher. The error is usually small, but for precise inventory, consult the manufacturer’s calibration table.

Q: What if my tank is not exactly elliptical?
The formula is an approximation for true ellipses. Many “elliptical” tanks are actually oval with slightly different curvature. The error is typically under 2% for standard tank shapes. For exact measurements, use a strapping table or ultrasonic level sensor.

Q: Why does the volume not double when the depth doubles?
Because the cross-section is an ellipse, the width changes with height. Near the bottom, the tank is narrow, so a small increase in depth adds little volume. Near the middle, the tank is widest, so the same depth increase adds more volume. This nonlinearity is captured by the arccosine term.

Q: Can I calculate the depth from a known volume?
Yes, but it requires solving the equation iteratively because the depth appears inside the arccosine and square root. Many tank gauging systems use lookup tables or numerical methods. For a quick estimate, you can use the calculator in reverse by trying different depths until the volume matches.

Q: Does the calculator work for any liquid?
Yes, the volume calculation is independent of the liquid type. However, if you need the mass, multiply the volume by the liquid density (e.g., 1000 kg/m³ for water, 850 kg/m³ for diesel).

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in L | Tank length | m | 3 | 0
in D | Major axis (long diameter) | m | 1.5 | 0
in d | Minor axis (short diameter) | m | 1.0 | 0
in h | Liquid depth | m | 0.6 | 0
out vol | Liquid volume | m³ | L * (D/2) * (d/2) * ( acos((d/2 - h)/(d/2)) - (d/2 - h) * sqrt(2*(d/2)*h - h^2) / (d/2)^2 ) | 3
out cap | Total capacity | m³ | L * (D/2) * (d/2) * pi | 3
out pct | Percent filled | % | vol / cap * 100 | 1