Elliptical Tank Volume Calculator: How It Works

An elliptical tank volume calculator works out the volume of a tank with an elliptical (oval) cross-section, using the major axis, minor axis, and length you enter. It returns the volume in your chosen unit (liters, gallons, cubic feet, etc.) for a horizontal or vertical orientation. The math behind it is straightforward: the area of an ellipse times the tank length, adjusted for the liquid depth if you want partial fill.

Formula

The full volume of a horizontal elliptical tank is the cross-sectional area of the ellipse times the length. The area of an ellipse with semi-axes a/2 and b/2 is π × (a/2) × (b/2), so:

V_full = (π × a × b / 4) × L

• a = major axis (full width at the widest point), in meters (or any unit)
• b = minor axis (full height at the tallest point), in meters
• L = length of the tank, in meters
• V_full = volume in cubic meters (or the cube of your length unit)

For a partial fill (liquid depth h from the bottom), the formula uses the circular segment area of an ellipse scaled by the aspect ratio. The exact formula is:

V_partial = (a × b × L / 4) × [ arccos(1 − 2h/b) − (1 − 2h/b) × √(4h/b − 4h²/b²) ]

Here h is the liquid depth in the same units as b, and arccos is in radians. This formula assumes the tank is horizontal and the elliptical cross-section is uniform along the length. If the tank is vertical (elliptical footprint, height as length), the same full-volume formula applies, but partial fill is simply the full volume times (h / L).

Worked Example

Using the default values from the calculator: major axis a = 2 m, minor axis b = 1 m, length L = 3 m, and liquid depth h = 0.5 m.

1. Full volume: V_full = (π × 2 × 1 / 4) × 3 = (π × 0.5) × 3 = 1.5708 × 3 = 4.712 m³. Rounded to 3 decimals: 4.712 m³.
2. Partial volume: first compute the bracket. h/b = 0.5/1 = 0.5. Then 1 − 2×0.5 = 0. So arccos(0) = π/2 ≈ 1.5708. The second term: 0 × √(4×0.5 − 4×0.25) = 0. Bracket = 1.5708 − 0 = 1.5708.
3. V_partial = (2 × 1 × 3 / 4) × 1.5708 = (1.5) × 1.5708 = 2.356 m³. Rounded to 3 decimals: 2.356 m³.
4. In liters: 2.356 × 1000 = 2356 L.

The calculator shows 4.712 m³ full, 2.356 m³ partial, and 2356 L partial. The partial volume is exactly half the full volume because the liquid depth is half the minor axis – correct for a symmetric ellipse.

How to Choose the Right Dimensions

To get an accurate volume, you need the correct dimensions of your tank. The major axis is the full width at the widest point (not the radius). The minor axis is the full height at the tallest point. The length is the overall length of the cylindrical section, excluding domed or dished ends if they are not elliptical. If your tank has elliptical heads (end caps), the volume is larger than a simple cylinder; you would need to add the head volume separately, which depends on the head shape (elliptical, torispherical, etc.).

When measuring, use the inside dimensions, not the outside, because wall thickness reduces the internal volume. If you only have outside dimensions, subtract the wall thickness twice for the axes and length. For a typical steel tank with 5 mm walls, the difference is small, but for thick plastic or concrete tanks it can be significant.

Common Mistakes and How to Avoid Them

• Mixing up axes: The major axis is the longer one, but the formula works even if you swap them – the area is the same. However, the partial-fill formula assumes the minor axis is the vertical one. If you enter the major axis as the height, the partial volume will be wrong.
• Unit inconsistency: All dimensions must be in the same unit. If you measure width in meters and length in centimeters, the result is meaningless. Convert everything to one unit first.
• Forgetting to convert cubic meters to liters: 1 m³ = 1000 liters. Many people forget the factor of 1000 when reading the result.
• Using the full-volume formula for partial fill: The partial-fill formula is not linear except when the tank is vertical. For a horizontal elliptical tank, the volume vs. depth is curved; using a simple proportion underestimates or overestimates depending on depth.
• Ignoring the liquid depth: If you want the volume of liquid in the tank, you must enter the depth from the bottom, not the height of the tank. If the tank is full, set h = b.

How to Read the Results

The calculator gives you three outputs: full volume in cubic meters, partial volume in cubic meters, and partial volume in liters. The first output (full volume) is the headline answer – the total capacity of the tank when completely full. The partial volume is what you actually have when the liquid depth is h. The liters output is convenient for everyday use (e.g., water tanks, fuel tanks).

If you need gallons, multiply cubic meters by 264.172 (US gallons) or 219.969 (imperial gallons). For cubic feet, multiply by 35.3147. The calculator does not include these conversions because the unit system is up to you – just keep all inputs in the same unit.

Limitations and Edge Cases

The formulas assume a perfect ellipse and a uniform cross-section along the entire length. Real tanks often have rounded ends, baffles, internal pipes, or a slight taper, which change the volume. For a tank with elliptical heads, add the head volume: for a 2:1 elliptical head, the volume is (π/6) × a × b × (L/2) per head, but this is an approximation. The partial-fill formula also assumes the tank is level. If the tank is tilted, the liquid surface is not parallel to the bottom, and the volume is different.

For very shallow depths (h close to 0), the formula still works, but numerical precision may be an issue. For h = 0, the bracket becomes arccos(1) − 1×√(0) = 0, so volume is 0. For h = b, the bracket becomes arccos(−1) − (−1)×√(4−4) = π, so volume is (a×b×L/4)×π, which equals the full volume. The formula is continuous and correct at the endpoints.

Alternatives and When to Use Them

If your tank is a perfect cylinder, use a cylinder volume calculator (π × r² × L) – it is simpler and more accurate for that shape. For rectangular tanks, length × width × height. For spherical tanks, (4/3)πr³. The elliptical tank calculator is specifically for tanks with an oval cross-section, which are common in truck trailers, fuel storage, and some water tanks. If you have a horizontal tank with flat ends, the formula is exact. If the ends are dished, you need a different calculator that accounts for the head shape.

For a vertical elliptical tank (elliptical base, height as the length), the full volume is the same as above, but partial fill is simply V_full × (h / L). The calculator above does not handle that case directly – you would use the full volume and multiply by the fill fraction.

Frequently Asked Questions

How do I measure the major and minor axes of an elliptical tank?

Measure the full width at the widest point (major axis) and the full height at the tallest point (minor axis), using the inside dimensions. If the tank is horizontal, the minor axis is the vertical height. Use a tape measure or laser measure, and take the measurement at the center of the tank to avoid distortion from the ends.

Can this calculator work for a vertical elliptical tank?

Yes, but only for full volume. For partial fill in a vertical tank, the cross-section is constant, so the volume is proportional to the liquid depth. You can compute the full volume with this calculator, then multiply by (h / L) where L is the tank height. The partial-fill formula in this calculator is for horizontal tanks only.

What if my tank has elliptical heads (dished ends)?

This calculator ignores the head volume. For a tank with 2:1 elliptical heads, add the head volume: each head adds approximately (Ï€/6) × a × b × (L/2) where L is the head’s straight length. For a more accurate result, use a tank volume calculator that includes head shapes, or measure the actual water capacity by filling the tank.

Why is the partial volume not proportional to the liquid depth?

Because the cross-section is an ellipse, not a rectangle. When the liquid is low, the width of the liquid surface is narrower, so adding a small amount of depth adds less volume than when the tank is half full. The formula accounts for this curvature, which is why the volume vs. depth is a nonlinear curve.

How do I convert the result to gallons?

If you used meters, the volume is in cubic meters. Multiply by 264.172 for US gallons or 219.969 for imperial gallons. If you used feet, the volume is in cubic feet; multiply by 7.48052 for US gallons or 6.22884 for imperial gallons. Always use the same unit system throughout.

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in a | Major axis (full width) | m | 2 | 0
in b | Minor axis (full height) | m | 1 | 0
in L | Length | m | 3 | 0
in h | Liquid depth | m | 0.5 | 0
out vol_full | Full volume | m³ | (pi * a * b / 4) * L | 3
out vol_part | Partial volume | m³ | (a * b * L / 4) * (acos(1 - 2*h/b) - (1 - 2*h/b) * sqrt(4*h/b - 4*h²/b²)) | 3
out vol_part_L | Partial volume | L | ((a * b * L / 4) * (acos(1 - 2*h/b) - (1 - 2*h/b) * sqrt(4*h/b - 4*h²/b²))) * 1000 | 0