Why depth percentage is not volume percentage
In a vertical tank with a constant horizontal cross-section, volume increases directly with depth. Turn a cylinder on its side and the width of the liquid layer changes with height: narrow near the bottom, widest through the centre, then narrow again near the top.
That means a tank at 25% of its diameter is less than 25% full by volume, while a depth at exactly half the diameter is exactly 50% full for a symmetric cylinder and symmetric ends. Reading depth as a direct percentage is one of the most common horizontal-tank mistakes.
The circular segment at the tank end
Look squarely at the end of the tank. The liquid occupies a circular segment bounded by the tank wall and the horizontal liquid chord. Define:
- D = internal tank diameter
- r = D/2, the internal radius
- h = liquid depth measured vertically from the bottom
- L = straight cylindrical length
- θ = central angle of the filled segment, in radians
For 0 ≤ h ≤ 2r, the segment angle is:
θ = 2 acos((r − h) / r)The filled cross-sectional area is the sector area minus the central triangle:
Asegment = ½r²(θ − sin θ)An equivalent formula uses inverse cosine and a square-root term. The angle form matches the site’s Circular Segment Calculator and makes the geometric construction explicit.
Multiply cross-section by cylinder length
For a tank with flat ends, liquid volume is simply the filled segment area multiplied by the internal straight length:
Vcylinder = Asegment × LFull cylindrical capacity is πr²L. Fill percentage is 100 × current liquid volume / full tank capacity. The dimensions must all use the same unit before multiplication.
Worked example: a half-full flat-ended tank
Let D = 2 m, so r = 1 m; L = 3 m; and depth h = 1 m.
θ = 2 acos((1 − 1)/1) = 2 acos(0) = πAsegment = ½(1²)(π − sinπ) = π/2 m²V = (π/2)(3) = 3π/2 ≈ 4.712 m³The full cylinder is π(1²)(3) = 3π ≈ 9.425 m³, so the result is exactly 50%. Since 1 m³ = 1000 L, the liquid volume is approximately 4712 L.
A less obvious example below the centreline
Use the same 2 m diameter and 3 m length, but depth h = 0.5 m—25% of the diameter.
θ = 2 acos(0.5) = 2π/3 ≈ 2.0944Asegment = ½[2.0944 − sin(2.0944)] ≈ 0.6142 m²V ≈ 0.6142 × 3 = 1.8426 m³ ≈ 1843 LThat is about 19.55% of the 9425 L full capacity—not 25%. The drawing’s shallow filled segment makes the difference intuitive.
Hemispherical ends
The current tank tool can also include two hemispherical ends, equivalent together to one sphere. At a liquid depth h measured from the bottom, the filled volume of those two ends is the volume of a spherical segment:
Vends = πh²(3r − h) / 3The total becomes AsegmentL + Vends. Full capacity adds 4πr³/3 to the cylindrical capacity. The length input in this mode is the straight cylinder between the hemispherical tangency points; overall tank length is L + 2r.
Chord, wetted area and liquid surface
The liquid chord length in the circular end is:
c = 2√[h(2r − h)]For a flat-ended cylinder, the plan area of the liquid surface is approximately cL. For hemispherical ends, the current implementation adds the elliptical contribution from the ends. Wetted internal area is a separate surface calculation and can be useful for coatings or process estimates, but it is not the same as liquid volume.
Unit conversions
Cubic centimetres convert to litres by dividing by 1000. Cubic millimetres divide by 1,000,000. Cubic metres multiply by 1000. US gallons use 231 cubic inches, while one cubic foot is approximately 7.48052 US gallons.
Use internal dimensions for capacity. An outside diameter and outside length will overstate capacity when wall thickness or end construction is significant.
Using a dip or depth reading
Confirm what the instrument measures. A wet dip normally measures liquid depth from the internal bottom; an ullage reading measures the empty distance down from a top reference. For a simple cylinder, depth = internal diameter − ullage only when the reference corresponds to the true internal top and the tank is level.
The circular symmetry provides a useful check: at complementary depths h and D − h, the two volumes add to full capacity. At h = 0 the result should be zero; at h = r it should be half; and at h = D it should be full. These boundary cases are useful when reviewing a table or calculator result.
Measurement and modelling limitations
- The tank must be level for one depth reading to represent the full length consistently.
- The cylinder and end shapes must match the selected model. Dished, elliptical, conical or irregular ends require different geometry.
- Internal fittings, baffles, displacement, wall thickness and dead volume are not automatically included.
- Depth should be measured vertically from the internal bottom, not along the curved shell.
- Real tanks can deform, settle or differ from nominal dimensions.
- For inventory, regulated custody transfer or safety-critical operations, use the approved tank calibration and procedures.
The calculator draws the filled segment and reports liquid volume, percentage, ullage, chord and related surfaces. Formula precision and site-wide limitations are detailed in Methodology & Accuracy.