Last year my family gave me a five litre cask for Father’s Day. This was manufactured by Noel Sullivan of the Roll Out The Barrel Cooperage in Clontarf, Queensland. It is truly a beautiful and finely crafted piece of work.

It has since been seasoned and stocked with the very finest Jameson Irish whisky.

Apart from enjoying the occasional tipple, the other item that required attention was determining the volume of liquid in the cask given the only measurement available, being the depth of liquid at the central filling hole.

This turned out to be far from trivial; the result is presented here. ## Barrels

For a good discussion on wine barrels in general, and perhaps to provide some idea of the variety of “standards” that exist, take a peek at this Wikipedia page.

The calculation of volume of a barrel depends on the assumed shape of the sides of the barrel. This reference (in French, but translatable with web browsers) provides the integral formulas for many shapes, and also provides a comparison of calculated volumes with the volume determined using the French customs rule.

In the following, the parabolic approximation is used. This generates results sufficiently accurate for this purpose and has the advantage of performing well with Gaussian-Legendre 3 point numerical integration, which is easily implemented in a spreadsheet, or other programmable calculator.

## Volume of a Cask Lying on its Side

L, D and C are external dimensions. The wall thickness T is assumed uniform for the whole cask. H is the liquid depth below the central filling hole, and (x,y) is a 2-d coordinate system with origin at the centre of the cask, and x-axis along the mid-line of the cask. The cask wall is represented as a parabolic surface y=a.x² + b rotated about the x axis, bounded by flat ends.

At position x’, the internal radius R’ is, where the prime symbols L’, D’, C’ denote the internal dimensions. The liquid depth H measured at the central hole can be used to define a constant value Y, being the level offset from the mid-line axis of the cask along the entire length of the cask, At x’ the liquid in the cask will occupy a partial segment spanning an angle 𝜃, where Note that if |Y| > R’ the liquid level is below the cask internal radius and the cross-section area will be zero. The above and following computations should be avoided in this instance.

The cross-section area A(x) of the liquid at x=x’ will be, for Y<=0, If Y>0, the liquid area will be the full circle less the above area A, so that, The total liquid volume may then be found by integrating A(x) over x from 0 to L’/2, Here is the result for my cask. Dimensions are D=20.0 cm, C=17.0 cm, L=27.0 cm, T=1.40 cm (at the fill hole.) Full volume is calculated as 5.08 litres, which accords well with the volume of sherry used to ‘season’ it. (It has never been full of whisky!) ## Calculation

### Excel

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