《孫子算經卷中》 "Sun Tzŭ's Computational Classic: Volume II"
§11. Volume of a rectangular prism (1)

This section gives a worked example of computing the volume of a rectangular prism.

The relevant unit conversions for length are

\begin{aligned} 1 \unit{rod~(丈)} &= 10 \unit{rules~(尺)} \\ 1 \unit{rule~(尺)} &= 10 \unit{inches~(寸)} \\ 1 \unit{inch~(寸)} &= 10 \unit{tenths~(分)}, \end{aligned}

and, for capacity,

\begin{aligned} 1 \unit{barrel~(斛)} &= 10 \unit{pecks~(斗)} \\ 1 \unit{peck~(斗)} &= 10 \unit{quarts~(升)}. \end{aligned}

The conversion between capacity units and volume units is given by

1 \unit{barrel~(斛)} = 1.62 \unit{rules~(尺)}^3.

See Vol. I §1 (Units of length), Vol. I §3 (Units of capacity), and Vol. II §10 Extended commentary.

Translation

Chinese source text: Version A, Version B, Version C, Version D.
Unless noted otherwise, I follow the text from Version D, 《知不足齋叢書》本.

Source text Target text Notes
今有方窖、廣四丈六尺、長五丈四尺、深三丈五尺。問受粟幾何。 Suppose there be [a] rectangular cellar, of breadth four rods [and] six rules, length five rods [and] four rules, [and] depth three rods [and] five rules. [We] ask, how much grain beareth [it]?
答曰、五萬三千六百六十六斛六斗六升三分升之二。 Answer saith: five myriad three thousand six hundred [and] sixty-six barrels, six pecks, six quarts, [and] two thirds of [a] quart.
術曰、置廣四丈六尺、長五丈四尺、相乘、得二千四百八十四尺。 Method saith: put [down the] breadth four rods [and] six rules, [and the] length five rods [and] four rules, [which], multiplied with each other, result in two thousand four hundred [and] eighty-four rules.
以深三丈五尺乘之、得八萬六千九百四十尺。 Multiplying it by [the] depth three rods [and] five rules, resulteth in eight myriad six thousand nine hundred [and] forty rules.
以斛法一尺六寸二分除之、即得。 Dividing it by [the] barrel divisor, one rule, six inches, [and] two tenths, [we] are done.
  • 斛法: [the] barrel divisor

    This is the conversion between the capacity unit "barrel", huk (), and the volume unit "cubic rule", chʻek (); see §10 Extended commentary.

  • In modern notation, the volume (or capacity) of a rectangular prism of breadth B = 46 \unit{rules}, length L = 54 \unit{rules}, and depth H = 35 \unit{rules} is
    \begin{aligned} V &= B L H \\ &= 46 \unit{rules} \times 54 \unit{rules} \times 35 \unit{rules} \div \frac{1.62 \unit{rules}^3}{\unit{barrel}} \\[\tallspace] &= 53666 \tfrac{2}{3} \unit{barrels} \\ &= 53666 \unit{barrels} + 6 \unit{pecks} + 6 \tfrac{2}{3} \unit{quarts}. \end{aligned}

Cite this page

Conway (2023). "Sun Tzŭ's Computational Classic: Volume II §11". <https://yawnoc.github.io/sun-tzu/ii/11> Accessed yyyy-mm-dd.