Aquarium Volume Formulas

V = L × W × H

Multiply the three interior dimensions. In inches, divide the result by 231 for US gallons. In centimetres, divide by 1,000 for litres.

Example: A 36 × 18 × 16 inch interior gives 10,368 cubic inches. Dividing by 231 gives 44.9 US gallons, or about 169.9 litres.

V = S³

Cube the interior side length. A cube with 12-inch sides holds 1,728 cubic inches, which is exactly one cubic foot or 7.48 US gallons.

Example: A 12-inch cube is 12 × 12 × 12 = 1,728 cubic inches, which is 7.48 US gallons or 28.3 litres.

V = L × W × H

A breeder is a rectangular prism, so the standard box formula applies. The extra front-to-back width is what sets its volume apart from a regular tank of the same length.

Example: A 36 × 18 × 16 inch breeder is 10,368 cubic inches, which is 44.9 US gallons or 169.9 litres, noticeably more than its 40-gallon label.

V = L × W × H

Shallow tanks are rectangular prisms. Because the height is small, every inch of substrate or air gap is a large share of the total, so the adjustment options matter more here than on any other shape.

Example: A 24 × 14 × 9 inch shallow tank holds 3,024 cubic inches, or 13.1 US gallons (49.6 litres) to the rim. With a 1-inch gap and 2 inches of soil it drops to about 9 gallons.

V = (L × D × H) + (π × (L/2) × B × H ÷ 2)

The tank is split into a rectangular box (length × flat depth) plus a half-ellipse whose semi-axes are half the length and the bow depth. Both are multiplied by height and added.

Example: Box: 36 × 12 × 20 = 8,640 in³. Bow: π × 18 × 4 × 20 ÷ 2 = 2,262 in³. Total 10,902 in³ = 47.2 US gallons (178.6 litres). The bow adds about 9.8 gallons.

V = π × r² × H

Multiply the circular floor area (π times radius squared) by the height. If you measured the diameter, halve it before entering.

Example: π × 10² × 24 = 7,540 in³, which is 32.6 US gallons or 123.6 litres.

V = π × r² × H ÷ 2

Half the volume of a full cylinder with the same radius and height. The radius is half the length of the flat back panel.

Example: π × 12² × 20 ÷ 2 = 4,524 in³, which is 19.6 US gallons or 74.1 litres.

V = π × a × b × H

The floor of an oval tank is an ellipse with area π × a × b, where a and b are half the long and short diameters. Multiply by height.

Example: π × 15 × 9 × 20 = 8,482 in³, which is 36.7 US gallons or 139.0 litres.

V = ((L − r) × W × H) + (π × r² × H ÷ 2)

The rectangular section runs from the flat end to the start of the curve (length minus radius). The rounded end is a half cylinder with the given radius. Both are multiplied by height and added.

Example: Box: (48 − 9) × 18 × 20 = 14,040 in³. Nose: π × 9² × 20 ÷ 2 = 2,545 in³. Total 16,585 in³ = 71.8 US gallons (271.8 litres).

V = ((L − W) × W × H) + (π × (W/2)² × H)

The two semicircular ends together form one full circle of diameter W. The straight middle section is length minus width. Multiply both by height and add.

Example: Straight section: (60 − 24) × 24 × 24 = 20,736 in³. Ends: π × 12² × 24 = 10,857 in³. Total 31,593 in³ = 136.8 US gallons (517.7 litres).

V = π × d³ ÷ 6

The volume of a sphere. For a bowl with a flat base, the calculator removes the spherical cap below the base. Water height is measured from the inside floor and uses the spherical segment formula.

Example: A 10-inch bowl is π × 10³ ÷ 6 = 524 in³ as a full sphere, which is 2.27 US gallons or 8.6 litres. Filled to the usual level below the opening it holds closer to 1.9 gallons.

V = (S² ÷ 4 + π × S × B ÷ 4) × H

The footprint is a right triangle between the two wall sides (area S² ÷ 4 when the front span S is the hypotenuse) plus a half-ellipse bulge with semi-axes S ÷ 2 and bow depth B.

Example: Triangle: 36² ÷ 4 = 324 in². Bulge: π × 36 × 6 ÷ 4 = 170 in². Area 494 in² × 24 = 11,852 in³ = 51.3 US gallons (194.2 litres).

V = π × r² × H ÷ 4

A quarter of a full cylinder. The two straight sides that meet in the corner are both equal to the radius.

Example: π × 24² × 24 ÷ 4 = 10,857 in³, which is 47.0 US gallons or 177.9 litres.

V = ½ × a × b × H

A right triangle has half the area of the rectangle formed by its two legs. Multiply that area by the height.

Example: ½ × 24 × 24 × 20 = 5,760 in³, which is 24.9 US gallons or 94.4 litres.

V = (3√3 ÷ 2) × s² × H

A regular hexagon has area 2.598 × s². Multiply by height. The side length is the width of one of the six glass panels.

Example: 2.598 × 12² × 24 = 8,979 in³, which is 38.9 US gallons or 147.1 litres.

V = (L × D − (L − F) × (D − S) ÷ 2) × H

Start with the full rectangle (back length × overall depth) and remove the two triangular corners cut off by the angled panels. Each triangle has legs (L − F) ÷ 2 and (D − S).

Example: Rectangle 36 × 18 = 648 in². Corners removed: (36 − 18) × (18 − 8) ÷ 2 = 90 in². Floor 558 in² × 20 = 11,160 in³ = 48.3 US gallons (182.9 litres).

V = 2 × (1 + √2) × s² × H

A regular octagon has area 4.828 × s². Multiply by height.

Example: 4.828 × 10² × 24 = 11,588 in³, which is 50.2 US gallons or 189.9 litres.

V = (√(25 + 10√5) ÷ 4) × s² × H

A regular pentagon has area 1.720 × s². Multiply by height.

Example: 1.720 × 12² × 20 = 4,954 in³, which is 21.4 US gallons or 81.2 litres.

V = (L1 × W1 + L2 × W2) × H

Split the L into two rectangles that do not overlap, add their floor areas and multiply by the common height.

Example: Section 1: 60 × 24 = 1,440 in². Section 2: 36 × 24 = 864 in². Floor 2,304 in² × 24 = 55,296 in³ = 239.4 US gallons (906.2 litres).

V = ½ × (B1 + B2) × D × H

Average the two parallel bases, multiply by the perpendicular depth for the floor area, then multiply by height.

Example: ½ × (36 + 18) × 18 = 486 in² floor. × 20 = 9,720 in³ = 42.1 US gallons (159.3 litres).