Horizontal Cylinder Tank Volume: The Formula With a Worked Example

Short answer

In a horizontal cylinder with flat ends, liquid volume is V = L × [R² × arccos((R - h) / R) - (R - h) × √(2Rh - h²)], with R the inside radius, L the length and h the liquid height. A 2 m by 5 m tank holds 3,071 litres at 0.5 m, out of 15,708 litres full.

Key takeaways

• Liquid in a horizontal cylinder fills a circular segment, so volume is the segment’s area times the tank’s inside length.
• In a 2 m by 5 m tank, the bottom 10 cm holds about 294 litres, while the 10 cm just below the middle holds about 998 litres.
• At a quarter of the height (0.5 m) the tank holds 3,071 litres, only 19.55% of its 15,708 litre capacity.
• The top half mirrors the bottom: the volume at height h equals the full volume minus the volume at D - h, a quick check for any spreadsheet.
• The formula assumes flat ends and a perfect, level shell; real tanks with dished ends, dents or tilt need a calibration table.

01The formula, and what each letter means

A cylinder lying on its side has a round cross-section, so the liquid inside fills a shape called a circular segment: the part of a circle cut off by a straight line. The volume is that segment’s area multiplied by the tank’s length. Wolfram MathWorld gives it as V = L × [R² × arccos((R - h) / R) - (R - h) × √(2Rh - h²)]. R is the inside radius, half the inside diameter. L is the inside length of the straight shell. h is the liquid height, measured from the lowest point inside the tank. Use one unit for all three: metres give cubic metres, and NIST confirms that one cubic metre is 1,000 litres. The arccos result must be in radians, not degrees; the ACOS function in common spreadsheets already returns radians. An equivalent form uses the central angle θ = 2 × arccos((R - h) / R), with a segment area of ½ × R² × (θ - sin θ). Both forms give the same answer, so one can check the other.

02Worked example: a 2 m by 5 m diesel tank at 0.5 m

Take a common site tank with an inside diameter of 2 m, so R = 1 m, and a straight inside length of 5 m. A dipstick shows 0.5 m of diesel. Step one: R - h = 1 - 0.5 = 0.5 m. Step two: (R - h) / R = 0.5, and arccos(0.5) = 1.0472 radians; multiplied by R² (1 m²) that gives 1.0472 m². Step three: 2Rh - h² = 1 - 0.25 = 0.75, its square root is 0.8660, and 0.5 × 0.8660 = 0.4330 m². Step four: subtract, 1.0472 - 0.4330 = 0.6142 m². That is the wetted cross-section. Step five: multiply by the length, 0.6142 × 5 = 3.071 m³, or about 3,071 litres. The full tank holds π × 1² × 5 = 15.708 m³, about 15,708 litres. So the liquid reaches a quarter of the height, but the tank is only 19.55% full. We checked the area with the angle form: θ = 2.0944 radians gives the same 0.6142 m².

03The full table, and the shortcut for the top half

The table below runs the same formula at ten heights for the same 2 m by 5 m tank, and the chart shows the litres at each quarter metre. Two patterns stand out. First, the litres per centimetre change with height. The bottom 10 cm holds about 294 litres, while the 10 cm between 0.9 m and 1.0 m holds about 998 litres, more than three times as much. A drop of 5 cm near the middle is a large amount of fuel; the same drop near the bottom is small. Second, the top half mirrors the bottom. The volume at height h equals the full volume minus the volume at D - h. So 1.5 m holds 15,708 - 3,071 = 12,637 litres, which matches the direct calculation. Use this as a check on any spreadsheet you build: the values at 0.5 m and 1.5 m must add up to the full tank, and the value at exactly half the diameter must be half the capacity, here 7,854 litres.

04Where the formula stops being right

The formula describes an ideal tank: a perfect circle, flat ends, and a shell that sits exactly level. Many real tanks break at least one of these. Dished or rounded ends hold extra fuel that the straight-shell formula leaves out, so measure only the straight length for L and treat the ends as a separate allowance from the tank drawing. Use the inside diameter, not the outside; on a 2 m tank, a few millimetres of wall thickness changes the total by tens of litres. A tank that tilts along its length has a different liquid height at each end, so a reading taken near one end does not represent the whole tank. Dents, internal ladders and fittings also take up space. For these reasons the formula is the right starting point, and a measured tank calibration table is the right finish for any tank where litres matter for money. Our articles on fuel level versus fuel volume and on calibration tables cover those two steps.

05How ENTEK Fuel Tank uses the tank shape

In the ENTEK Fuel Tank solution, each tank is set up with its shape, horizontal or vertical, its dimensions and units, its capacity and its own minimum and maximum thresholds. The fill volume is then calculated from that shape, so the screen shows litres against capacity rather than a raw height. The same tank appears in the Healthy, Low (below 35%) or Critical (below 20%) state. The geometry above explains why the shape setting matters. In the 2 m by 5 m example, 35% of capacity is about 5,498 litres, which sits at only about 0.76 m of liquid, and 20% is about 3,142 litres at about 0.51 m. Read on a dipstick, both points look much lower than their percentages suggest. Rules are checked every 30 seconds and alerts reach named people by email, SMS or WhatsApp, usually within about a minute. If you need fuel level sensors for a tank, you can source them through the ENTEK marketplace or an RFQ at entek.ai.

Figures and sources

• Volume of liquid in a horizontal cylinder of radius R and length L filled to height h: V = L[R² arccos((R - h)/R) - (R - h)√(2Rh - h²)] (Source: Wolfram MathWorld, Horizontal Cylindrical Segment, as of 2026-10-19)
• Area of a circular segment in terms of the central angle θ, used as a cross-check: A = ½R²(θ - sin θ) (Source: Wolfram MathWorld, Circular Segment, as of 2026-10-19)
• Litres in one cubic metre: 1,000 L (Source: NIST Office of Weights and Measures, SI units: volume, as of 2026-10-19)
• Worked example: litres in a 2 m by 5 m horizontal tank at 0.5 m of liquid: 3,071 L (19.55% of capacity) (Source: Computed with the Wolfram MathWorld horizontal cylindrical segment formula, as of 2026-10-19)
• Worked example: full capacity of a 2 m by 5 m horizontal tank with flat ends: 15,708 L (Source: Computed with the Wolfram MathWorld horizontal cylindrical segment formula, as of 2026-10-19)
• Worked example: litres in the bottom 10 cm versus the 10 cm between 0.9 m and 1.0 m of the same tank: 294 L vs 998 L (Source: Computed with the Wolfram MathWorld horizontal cylindrical segment formula, as of 2026-10-19)

Questions

Sources

• Wolfram MathWorld, Horizontal Cylindrical Segment (volume formula)
• Wolfram MathWorld, Circular Segment (area by height and by angle)
• NIST Office of Weights and Measures, SI units: volume (1000 L = 1 m³)

Related insights

• Fuel Level vs Fuel Volume: Why a Height in Millimetres Is Not Litres
• What a Tank Calibration Table Is, and Why Every Tank Needs Its Own
• How Long Will This Tank Run the Generator? The Runtime and Days Left Formula
• Inventory Reconciliation for Fuel Tanks: Finding the Loss That No Alert Fires On
• Fuel Sensor Calibration Drift: The Correction Most Integrators Apply Twice