Tank Volume Calculator
How much is in a horizontal tank that's only partly full? Enter the diameter, the length and the liquid depth — a dipstick reading — and the fill volume, litres and percentage full are worked out and drawn to scale, with flat or hemispherical ends. Click any amber value on the drawing, including the liquid depth, to change it. The cross-section is a circular segment; the rounded-end tank is a capsule on its side.
Tank & liquid
horizontalDepth is measured from the bottom of the tank up to the liquid surface.
Results
How the fill volume is worked out
Looking at the end of a horizontal tank, the liquid is a circular segment of height d. That segment area, swept along the length, is the volume in the barrel.
θ = 2·acos(1 − d⁄r)
A = ½r²(θ − sin θ)
V = A × L
Hemispherical ends
Rounded ends add the liquid in the two half-domes, which together behave as one partly filled sphere: Vends = πd²(3r − d)⁄3. The calculator adds it automatically when you pick hemispherical ends.
Tips
Work in cm and the litres appear directly (1,000 cm³ = 1 L); in inches you get US gallons. Click the depth on the drawing and nudge it until the litres match a target to answer the reverse question.
How to calculate the volume in a horizontal tank
Fuel tanks, water tanks, LPG vessels and drums usually lie on their side, and the awkward thing about a horizontal tank is that the depth-to-volume relationship isn't linear: the first few centimetres at the bottom hold very little, the middle holds a lot per centimetre, and it thins out again near the top. The reason is visible on the drawing — the cross-section of the liquid is a circular segment, and its area grows with the chord width, which is widest at half depth.
V = ½r²(θ − sin θ) × L · with θ = 2·acos(1 − d⁄r)
Here r is the tank radius, L its cylindrical length, d the liquid depth from the bottom, and θ the segment's central angle in radians. If the tank has hemispherical (rounded) ends, the two half-domes together hold one partly filled sphere's worth on top of that:
Vends = πd²(3r − d)⁄3
Worked example
A flat-ended tank 120 cm in diameter and 200 cm long holds diesel to a dipstick depth of 30 cm.
Radius r = 60. The half-depth ratio sets the angle.
θ = 2·acos(1 − 30⁄60) = 2·acos(0.5) = 2.0944 rad (120°)
Half r squared times (θ − sin θ).
A = ½ × 3,600 × (2.0944 − 0.8660) = 1,800 × 1.2284 ≈ 2,211.1 cm²
Area times length, then cubic centimetres to litres.
V = 2,211.1 × 200 = 442,213 cm³ ≈ 442.2 L (19.6% of the 2,262 L capacity)
Had the same tank hemispherical ends, the domes would add πd²(3r − d)⁄3 = π × 900 × 150⁄3 ≈ 141,372 cm³ ≈ 141.4 L, and the capacity would rise by the full sphere, 4⁄3πr³ ≈ 904.8 L.
Why you can't just use depth ÷ diameter
A tank 25% deep is not 25% full — it's about 19.6% full, as the example shows. Scaling the capacity by the depth fraction overstates low readings and understates high ones. The segment formula, or a dipstick chart built from it, is the honest answer; every dipstick chart taped to a fuel tank was produced with exactly this calculation.
Where it comes up
Fuel storage and farm diesel tanks, rainwater and molasses tanks, LPG vessels (almost always with rounded ends), transport tankers, part-full pipes and culverts running below capacity, and drums on their side. Standing upright, a tank is just a cylinder and the volume really is depth times the circular base — the horizontal case is the one that needs this tool.
Frequently asked questions
How do I get litres from the tank measurements?
Measure in centimetres and divide the cubic-centimetre volume by 1,000 — the calculator shows this litres figure automatically when the unit is set to cm (and converts sensibly for mm and m, or to US gallons for inches and feet).
Is a tank at half depth half full?
At exactly half depth, yes — the segment is a semicircle, so 50% depth is 50% volume, whatever the ends. Everywhere else the relationship is non-linear: 25% depth is about 19.6% full and 75% depth about 80.4%.
What difference do rounded ends make?
Hemispherical ends add the liquid held in the two half-domes, which together equal one partly filled sphere: πd²(3r − d)⁄3. They also add ⁴⁄₃πr³ to the total capacity. Pick "hemispherical" and both are included.
I need the depth for a target volume — can it work backwards?
There's no simple closed formula in that direction, but the drawing makes it quick: click the depth value on the figure and adjust it until the litres read what you need. A couple of nudges usually lands it.
What about a vertical cylindrical tank?
Standing on end, the problem disappears: every centimetre of depth holds the same amount, so V = πr²d. The cylinder calculator covers it — this tool is for tanks on their side, where the segment maths is needed.