The Grand Astrologer’s platform and ramp:
Four problems in solid geometry from Wang Xiaotong’s ‘Continuation of ancient mathematics’ (7th century AD)

Forthcoming in Historia Mathematica

Third revision, 27 July 2012

Tina Su-lyn Lim 林淑

Donald B. Wagner 華道安

Click on any image to see it enlarged


Tina Su-lyn Lim is a project manager with NNIT (Novo Nordisk Information Technology) in Copenhagen, Denmark. This article formed part of her M.Sc. thesis in Mathematics at the University of Copenhagen, 2006.

 

Donald B. Wagner has taught at the University of Copenhagen, the University of Victoria (British Columbia), and the Technical University of Berlin. He has written widely on the history of science and technology in China; his most recent book is the volume on Ferrous Metallurgy in Joseph Needham’s Science and Civilisation in China.

 

We are grateful to Karine Chemla for drawing our attention to Wang Xiaotong’s interesting text, to Jesper Lützen and Ivan Tafteberg for detailed comments on an earlier version, and to Jia-ming Ying 英家銘 and three anonymous reviewers for comments on earlier drafts of this version.

Abstract

Wang Xiaotong’s Jigu suanjing is primarily concerned with problems in solid and plane geometry leading to cubic equations which are to be solved numerically by the Chinese variant of Horner’s Method. The problems in solid geometry give the volume of a solid and certain constraints on its dimensions, and the dimensions are required; we translate and analyze four of these. Three are solved using dissections, while one is solved using reasoning about calculations with very little recourse to geometrical considerations. The problems in Wang Xiaotong’s text cannot be seen as practical problems in themselves, but they introduce mathematical methods which would have been useful to administrators in organizing labour forces for public works.

王孝通的《緝古算術》主要在處理立體與平面幾何中使用到三次方程式的問題,而這 些方程式是用中國版的霍納法求出數值解。立體幾何的問題是給定立體體積與外部尺寸的某些限制, 然後再求解外部尺寸; 本文翻譯且分析四個問題. 其中三題是用分割法求解, 另一題則是用關於計算的推理求解, 並且很少觸及幾何上的考量. 王孝通著作中的問題本質上不能看成實用問題, 但這些問題介紹的數學方法對於官吏組織勞動力進行公共工程很有幫助. (英家銘譯)



Subject classifications: 01-02, 01A25.

Keywords

Wang Xiaotong, Jigu suanjing, China, Tang Dynasty, solid geometry

Introduction

Wang Xiaotong (late 6th–7th century AD) served the Sui and Tang dynasties in posts concerned with calendrical calculations, and presented his book, Jigu suanjing 緝古算經, ‘Continuation of ancient mathematics’, to the Imperial court at some time after AD 626. In 656 it was made one of ten official ‘canons’ (jing ) for mathematical education.[1] The book contains 20 problems: one astronomical problem, then 13 on solid geometry, then six on right triangles. All but the first provide extensions of the methods in the mathematical classic Jiuzhang suanshu 九章算術 (perhaps 1st century AD):[2] the solid-geometry methods in Chapter 5 and the right-triangle methods in Chapter 9,[3] thus ‘continuing’ ancient mathematics. All but the first require the extraction of a root of a cubic or (in two cases) quadratic equation. The present article is concerned with problem 2, which in fact consists of four connected but distinct geometric problems.

Printed in smaller characters in the text are comments which generally give explanations of the algorithms of the main text. The earliest extant edition states that the comments are by Wang Xiaotong himself, but we have noticed some differences in terminology between the comments and the main text,[4] and feel therefore that the question of the authorship of the comments should be left open. Perhaps most but not all of the comments are by him.

The text

We have treated the history of Wang Xiaotong’s text in detail elsewhere.[5] The oldest extant edition is from 1684,[6] and the best traditional critical edition is that of Li Huang (1832).[7] The standard modern edition has long been Qian Baocong’s [1963, 2: 487–527; important corrections, 1966], but that of Guo Shuchun and Liu Dun [1998] has much to recommend it. In the parts of the text treated in the present article there is no important difference between the two, and we generally follow Qian Baocong’s version.

The 1684 edition is marred by numerous obvious scribal errors which must be corrected by reference to the mathematical context. Guo and Liu [1998, 1: 22] note that Li Huang introduced ca. 700 emendations to the text, and that Qian Baocong followed most of these but introduced 20 new emendations. The comments, written in smaller characters than the main text and also often more difficult in their mathematical content, are even more subject to scribal errors.

The Jigu suanjing has not been much studied in modern times. The two critical editions, already mentioned, do not explicitly comment on the mathematical content. Lin Yanquan [2001] translates the text into modern Chinese, expresses the calculations in modern notation, and gives derivations of some of the formulas. Deeper studies of individual parts of the text are by Shen Kangshen [1964], Qian Baocong [1966], He Shaogeng [1989], Wang Rongbin [1990], Guo Shiying [1994], and Andrea Bréard [1999: 95–99, 333–336, 353–356]. The comments have hardly been studied at all by modern scholars: the most recent study we have found is that of Luo Tengfeng (1770–1841) [1993].[8]

Solving cubic equations by ‘Horner’s method’

In the problem solutions given by Wang Xiaotong the coefficients of a cubic equation are calculated, after which the reader is instructed to ‘extract the cube root’. Wang Xiaotong gives no indication of how this was done, but it was obviously a well-known procedure, for algorithms are described in detail in the Jiuzhang suanshu and the Zhang Qiujian suanjing 張丘 建算經 (5th century AD) for the case of extracting a cube root (the cubic equation x3 = A), and extensions of the latter algorithm to general polynomials of all orders are given in several Chinese books from after Wang Xiaotong’s time.[9]

So much has been written about Chinese methods for extracting the roots of polynomials that it is not necessary to go into detail here. The algorithm is equivalent to ‘Horner’s method’ as sometimes taught in modern schools and colleges.[10] The first digit of the root is determined, the roots of the function are reduced by the value of that digit, the next digit is determined, and these steps are repeated until the desired precision is reached. The operation is carried out on paper in the Western version, in the Chinese version with ‘calculating rods’ laid out on a table.

Problem 2: The Grand Astrologer’s platform and ramp

The text of the problem and solution are translated at the end of this article.[11] Here we provide an overview using modern terminology.

A tamped-earth platform and associated ramp are described, to be built by workers from two counties, A and B (translation, Sections 2.1.1 and 2.1.2. Two possible interpretations of the description are shown in Figure 1. Other interpretations are also possible, but the mathematics is the same whichever interpretation is followed.

It quickly becomes clear that the problem is not intended to reflect real life: for example, the tamped-earth ramp is broader at the top than at the bottom, and building such a construction would be a heroic engineering feat. Rather, a general mathematical method is introduced, in an entirely unrealistic context, which will be used many times in the rest of the book. The method is analyzed in greater detail further below, in the ‘Concluding remarks’.

Figure 1. Two possible interpretations of the physical situation [Qian Baocong 1966; Guo Shirong 1994].

Figure 1a

Figure 1b

Figure 2. The platform and the ramp, showing the contributions of Counties A and B.

Figure 2


The basic geometric situation is shown in Figure 2. The following quantities are given:[12]

a2a1 = 20 chi
b2b1 = 40 chi
b1a1 = 30 chi
ha1 = 110 chi                                                                            (1)
c
1c2 = 12 chi
lc1 = 104 chi
gl = 40 chi

Labour data is given which allows the straightforward calculation of the volumes of the two structures, V and W, and the volumes of the contributions of the two counties, VA, VB, WA, and WB:

V = 1,740,000 chi3
VA = 531,750 chi3
VB = 1,208,250 chi3
W = 352,800 chi3
WA = 81,900 chi3
WB = 270,900 chi3

The dimensions of the two structures and of the two counties’ contributions are required.

The dimensions of the platform

The first problem is to calculate a1 in Figure 2. In describing the calculations Wang Xiaotong gives names to various intermediate quantities so that he can refer to them later. In the translation we capitalize these names and denote the quantities by K1, K2, etc. The names used give clues to how he derived his calculations.

Wang Xiaotong’s calculation proceeds as follows (translation, Section 2.3.1).

Area for the Corner Yangma,

 

Truncated Volume for the Corner Yangma,

 

Area for the Corner Heads,

 

Truncated Volume for the Corner Heads,

 

Determined Number,

 

These names refer to one of the basic solids in classical Chinese geometry, the yangma, defined in the caption of Figure 3. The ‘areas’ and ‘volumes’ here do not correspond to any area or volume in the geometric situation: these are simply arbitrary names for intermediate quantities.

Figure 3. Two basic solids of classical Chinese geometry. A yangma is a rectangular pyramid with one edge perpendicular to the base. A bienao is a right-triangular pyramid with one edge (not at the right angle of the base) perpendicular to the base.

Figure 3


The cubic equation whose root is to be extracted is then

                                           (2)



a13 + 170a12 + 7166⅔ a1 = 1,677,666⅔ chi3

This equation has one real root,

a1 = 70 chi

Once a1 has been calculated it is trivial to calculate a2, b1, b2, and h using the differences given in (1).

In Figures 1–2 we have shown the platform as a frustum of a right pyramid, but the calculation is correct for a frustum of any rectangular pyramid, and the names given to intermediate quantities in the calculation suggest that Wang Xiaotong may have derived (2) using a volume dissection similar to that shown in Figure 4, in which one edge of the frustum is perpendicular to the base.

Figure 4. Dissection of the platform to derive its dimensions.

Figure 4


The platform is divided into pieces whose volumes are sums of products of known quantities and powers of the unknown, a1. The ‘corner heads’ are nos. I and II, with volumes,

VI =  
                                                                                                                                                                        (3)

VII =  = 

VI + VII = 

The ‘corner yangma’ is no. III, with volume,

VIII =  = 

And no. IV has volume,

VIV =  = 

Summing,

V = VI + VII + VIII + VIV 

which is equivalent to (2).

The contributions of the two counties to the platform

The next problem is to calculate hB in Figure 2 (translation, Section 2.3.2). Here again Wang Xiaotong gives names to quantities which will be used more than once:

Height for the Upper Width,

 = 630 chi                                                     (4)

Height for the Upper Length,

 = 450 chi                                                    (5)

These ‘heights’ are not equal to any dimensions in the geometric situation; they are simply intermediate results in the calculation.

The height hB of county B’s contribution is then a root of the equation

 +  +  =      (6)


hB3 + 1,620hB2 + 850,500hB = 146,802,375 chi3

This equation has one real root,

hB = 135 chi

The remaining dimensions are then calculated by considering similar triangles:

 = 85 chi

 = 130 chi

A comment (translation, Section 2.3.3) gives a remarkable explanation of (6). The text of the comment is corrupt as it has come down to us. Luo Tengfeng (1770–1841) [1993, 1: 16a–18b] has proposed an interpretation and a large number of emendations to the text; we feel no real doubt that his analysis of the underlying mathematics is correct, even if his precise emendations may be less certainly correct. In the translation we follow all of his emendations.

The comment uses some geometric designations for intermediate quantities, but the reasoning concerns calculations rather than geometry, and thus might be considered almost purely algebraic.

Figure 5. Diagram for the derivation of the contribution of County B to the platform.

Figure 5



The platform and the contributions of the two counties (see Figure 5) are what is known in classical Chinese mathematics as chutong 芻童, frusta of a rectangular pyramid. The calculation of the volume in the Jiuzhang suanshu[13] is (using modern notation)

      

The formula is not stated in the comment but clearly is used. Substituting this expression for VB into (6) gives

 = 

           (7)

Doubling this gives

 =
                   (8)


The four summed terms in parentheses on the right side of (8) are the areas of the rectangles PUWT, PUVS, PQRS, and PQXT respectively in Figure 5. The comment relates the left side of (8) to the four areas.

The comment first states that multiplying the Height for the Upper Width by the Height for the Upper Length (K6 and K7, see (4) and (5)) ‘gives two of the small area.’ (In fact, twice this product gives this result.) The ‘small area’ is PUWT in Figure 5, with area a1b1:

                 (9)


The linear term of (8) is 6K6K7hB . Equation (9) indicates that two of these six K6K7’s correspond to the term 2a1b1 in (8); four remain to be accounted for.

The comment introduces two more named quantities,

Height for the Lower Length,

                                                                    (10)

Height for the Lower Width,

                                                                   (11)

These quantities correspond to no dimensions in the geometric situation, and at this point their numeric values are unknown; they are abstract quantities which figure in the algebraic reasoning which follows. They are not actually defined in the commentary; presumably the reader is expected to deduce their definitions by analogy with the Heights for the Upper Length and Width (see equations (4) and (5) above).

The comment states, ‘multiplying the Height for the Lower Length by the Height for the Upper Width gives one of the middle areas’. The ‘middle areas’ are PQXT (area a1b3) and PUVS (area a3b1). In this case

This is followed by a statement that

 
 

This is true because, considering similar triangles,

so that

Then ‘the middle area is composed of one small area and also an area obtained by multiplying the Height for the Upper Width by the truncated height’:


This has accounted for one more of the K6K7’s and one of the K6’s on the left side of (8).

Further, ‘multiplying [. . .] by the Height for the Upper Length gives one of the middle areas’. The other multiplicand is missing in the text, but the context indicates that it can only be the Height for the Lower Width, so that the ‘middle area’ is PUVS:


and ‘within the middle area there is one small area [PUWT, see (9)] together with the area obtained by multiplying the Height for the Upper Length by the truncated height’:


which accounts for two more of the K6K7’s and one of the K7’s on the left side of (8).

Finally, ‘multiplying the Height for the Lower Width by the Height for the Lower Length gives two of the large area’. The ‘large area’ is PQRS, with area a3b3:


and ‘within the large area there is one small area, and in addition there are the Heights for the Upper Width and the Upper Length, each multiplied by itself to make one area’:

Or,


which accounts for the rest of the terms on the left side of (8).

In conclusion, ‘Thus there are two of the area formed by multiplying the truncated height by itself [hB2] and six of the small area [K6K7]. There are also three each of the Heights for the Upper Width and the Upper Length [K6, K7].’ The quantities mentioned here are

2hB2
6K6K7
3K6hB
3K7hB

‘These, multiplied by the truncated height [hB], give six [times the sum of the] areas’:

  =
             (12)


‘When all are halved, the small area [the term K6K7] is multiplied by three. Further, there are three each of the Heights for the Upper Width and the Upper Length [K6, K7]. When [the terms 3K6hB and 3K7hB in (12)] are halved once, in each case one and one half is obtained; this is why [these terms] are multiplied by three and divided by two. All the areas are multiplied by the truncated height [hB] to make volumes in [cubic] chi.’ Halving equation (12) and multiplying by hB gives (7), which is equivalent to (6).

The dimensions of the ramp

The next problem is to calculate c2 in Figures 1 and 2 (translation, Section 2.3.4). Names given for quantities in the calculation are:

Area for the Corner Yangma,

K11 = (gc2) (lc2) = 18,096 chi2

Volume for the Bienao Corner,

K12 = K11 (c1c2) = 217,152 chi3

Area for the Vertical and Horizontal Edge Bienao,

K13  =  ([lc2]  +  [gc2]) (c1c2)  =  3,264 chi2

(Yangma and bienao are defined in the caption to Figure 3.) The cubic equation whose root is to be extracted is then

     (13)

c23  +  276 c22  +  19,184 c2  =  633,216 chi3

This equation has one real root,

c2 = 24 chi

The remaining dimensions of the ramp are then easily calculated.


Figure 6. Dissection of the ramp to derive its dimensions.

Figure 6


The names given to intermediate quantities suggest that (13) was derived using a dissection similar to that shown in Figure 6. The volume is divided into parts whose volumes are sums of products of known quantities and powers of c2. The ‘vertical and horizontal edge bienao’ is no. I, with volume

WI = 

=  

The ‘bienao corner’ is no. II, with volume

WII  =  

The ‘corner yangma’ is no. III, with volume

WIII  =  

Combining gives the volume of the ramp:

W = W+ WII + WIII  = W+ 3WII

6W = 

And from this (13) can easily be derived.

The contributions of the two counties to the ramp

The final problem is to calculate lA in Figure 2 (translation, Section 2.3.5). The cubic equation whose root is to be extracted is

                         (14)

 chi3

This equation has one positive root,

lA = 70 chi

The remaining dimensions are then easily calculated:

  =  30 chi

  =  90 chi

In this case Wang Xiaotong gives us no clue as to how he derived his calculation, but the dissection shown in Figure 7 will serve. A necessary implicit assumption is that NPQR is parallel to ABEF.

Figure 7. Dissection to derive the contribution of County A to the ramp.

Figure 7


(Figure 7 is drawn with ABDC  CDEF  BDE, but this is not a necessary condition.) The contribution of county A is NPQRCD. Dissecting this into the qiandu SPQRCD and the bienao NSRC gives for its volume

Considering similar triangles,

whence,

and multiplying by  gives (14).

Concluding remarks

The first and third problems discussed here are dressed up as practical problems, but clearly are not: No practical problem in the construction of earthworks starts with a volume and the differences between dimensions. This might lead us to conclude that the book was written as a tour de force in pure mathematics, disguised as a practical textbook.[14] Nevertheless the second and fourth problems, concerned with the dimensions of the assignments of the two counties, resemble practical problems rather more. Ancient Chinese administrators were often required to calculate the dimensions of what could be accomplished by a given labour force. The first problem introduces a method which will be used many times in the rest of Wang Xiaotong’s book, including the fourth problem here.

Volume dissections

The first and third of the problems analyzed here provide clues which allow us to guess at the volume dissections used by Wang Xiaotong in deriving his results (Figures 4 and 6). The fourth provides no clues, but a dissection along the same lines seems clear enough (Figure 7).

How were the dissections communicated? There is no indication that Wang Xiaotong’s book contained diagrams of any sort, nor are diagrams of three-dimensional geometric situations known in other early Chinese mathematical works.[15] If we make the reasonable assumption that the book was intended as an adjunct to a teacher’s instruction, then an interesting question is how a teacher communicated these often rather complex geometric situations to a student. He may perhaps have possessed scale models of each problem, but study of an earlier text suggests a simpler and more flexible means.

The commentary by Liu Hui 劉徽 (3rd century AD) on the Jiuzhang suanshu, chapter 5, describes a number of volume dissections using a set of standard blocks (qi , lit., ‘chessmen’) of four kinds:

yangma 陽馬, defined in Figure 3,

bienao , defined in Figure 3,

qiandu 塹堵, right-triangular right prism,

lifang 立方, cube,

each with dimensions 1 × 1 × 1 chi (Figure 8).[16] He describes their placement to form a structure which is equivalent to the geometric structure in question except in its dimensions.


Figure 8. Standard blocks with dimensions 1 × 1 × 1 chi used in Liu Hui’s commentary on the Jiuzhang suanshu. One chi was ca. 25 cm in Liu Hui’s time.

Figure 8


If Wang Xiaotong, or a teacher using his book, possessed a set of such blocks, he could easily have demonstrated the dissection of Figure 4, for example, by building the structure shown in Figure 9. It would only be necessary for the student to abstract from the difference in dimensions between the actual geometric situation and the constructed model.


Figure 9. Example of blocks used to demonstrate the dissection of Figure 4.

Figure 9


Each of the volume dissections described here divides an object into parts whose volumes are sums of products of known quantities and powers of the dimension to be calculated. Each involves some algebraic reasoning: for example, in the first problem, equation (3),

 =
     

where (a2a1), (b1a1), and (ha1) are known and a1 is the dimension to be calculated. This may at first sight seem a difficult bit of reasoning about calculations for a pre-modern mathematician, but in fact only one non-trivial algebraic principle is required in all of the dissections described here: In modern terms, given quantities s, t, and x,

st =  (sx)(tx)  +  (sx)x  +  (tx)x  +  x2

and this is easily demonstrated by a dissection of the area of a rectangle with dimensions s × t, already demonstrated centuries earlier by Liu Hui in his explanation of the square-root algorithm.[17]

More advanced reasoning

The calculation in the second problem seems not to lend itself to a volume-dissection derivation along the same lines as the other three, and a comment gives instead a derivation which depends entirely on reasoning about calculations. This is not easy.

The commentator’s greatest problem is perhaps in the naming of abstract quantities. His usual way of naming an intermediate quantity in a calculation is to describe an actual calculation with known quantities; for example, he expresses (4) as,

Multiply the height of the platform by the upper width and divide by the difference between the widths to make the Height for the Upper Width.

But he seems unable or unwilling to give a name to a calculation with unknown quantities, for example expressing (10) as ‘let Height for the Lower Length denote the product of . . . divided by . . .’. He does not explicitly define the Heights for the Lower Width and Height ((10) and (11) above), but requires the reader to deduce the definitions of these terms by analogy with the Heights for the Upper Width and Height ((4) and (5) above). Similarly, he refers to concrete areas (in the plane PQRS in Figure 5) rather than referring to ‘the product of . . .’.

In spite of the difficulties which the author of the comment confronts, he does succeed in outlining the considerations involved, so that a student would be able to obtain an intuitive understanding of the underlying reasoning.

Problem 2: Chinese text and translation

In the Chinese text we indicate text variants with the notation {a/b/c}, where a is the 1684 version; b is Qian Baocong’s version; and c, if present, is from Luo Tengfeng’s version [1993], in the few cases in which Qian Baocong does not follow him, or from some other version, indicated in a footnote. We use Qian Baocong’s punctuation. In the translation we follow Qian Baocong’s version except in Section 2.3.3, where we follow Luo Tengfeng’s version.

Our mathematical comments are given indented in the translation, while a few philological comments are given in footnotes.

[2.1. Problem]

[2.1.1. The observatory platform]

假令太史造仰觀臺,上廣袤少,下廣袤多。上 下廣差二丈,上下 袤差四丈,上廣袤差三丈,高多上廣一十一丈。甲縣差一千四百一十八人,乙縣差三千二百二十二人,夏程人功常積七十五尺,限五日役臺 畢。

In the following see Figures 1 and 2.

Suppose that the Grand Astrologer builds a platform for observing the heavens. The upper width and length are smaller, while the lower width and length are larger. The difference between the upper and lower widths is [a2a1 =] 2 zhang, the difference between the upper and lower lengths is [b2b1 =] 4 zhang, the difference between the upper width and length is [b1a1 =] 3 zhang, and the height exceeds the upper width by [ha1 =] 11 zhang.

1 zhang = 10 chi .

County A sends 1418 corvée labourers and county B sends 3222 corvée labourers. The regulation for one man’s labour in summer is a constant volume of 75 [cubic] chi [of earth per day]. After 5 days the work on the platform is finished.

[2.1.2. The ramp]

羨道從臺南面起,上廣多下廣一丈二尺,少袤 一百四尺,高多袤 四丈。甲縣一十三鄉,乙縣四十三鄉,每鄉別均賦常積六千三百尺,限一日役羨道畢。二縣差到人共造仰觀臺,二縣鄉人共造羨道,皆從先給 甲縣,以次與乙縣。臺自下基給高,道自初登給袤。問臺道廣、高、袤及縣別給高、廣、袤各幾何?

A ramp rises from the southern surface of the platform. The upper width is [c1c2 =] 1 zhang 2 chi longer than the lower width, and [lc1 =] 104 chi shorter than the length [l]. The height exceeds the length by [gl =] 4 zhang.

County A [sends labourers from] 13 districts and county B [sends labourers from] 43 districts. Each district is uniformly assessed with a constant volume of 6300 [cubic] chi [of earth]. After one day the work on the ramp is finished. The delegate corvée labourers from the two counties work together to build the platform. The men from the two counties’ districts work together to build the ramp. In both cases the work is first assigned to county A and thereafter to county B. The platform is built from the base upward, while the ramp is built from the beginning of elevation along the length.[18] What are the width, height and length of the platform and the ramp, and how much does each of the counties contribute in terms of height, width and length?

[2.2.] Answer

答曰:

臺高一十八丈,上廣七丈,下廣九丈,上袤一 十丈,下袤一十四 丈。

甲縣給高四丈五尺,上廣八丈五尺,下廣九 丈,上袤一十三丈, 下袤一十四丈。

乙縣給高一十三丈五尺,上廣七丈,下廣八丈 五尺,上袤一十 丈,下袤一十三丈。

羨道高一十八丈,上廣三丈六尺,下廣二丈四 尺,袤一十四丈。

甲縣鄉人給高九丈,上廣三丈,下廣二丈四 尺,{上 /}袤 七丈{, 下袤一十四丈/}

乙縣鄉人給高九丈,上廣三丈六尺,下廣三 丈,{下 /}袤 七丈。

The platform:
            height [h =] 18 zhang [= 180 chi];
            upper width [a1 =] 7 zhang [= 70 chi];
            lower width [a2 =] 9 zhang [= 90 chi];
            upper length [b1 =] 10 zhang [= 100 chi];
            lower length [b2 =] 14 zhang [= 140 chi].

The contribution of county A:
            height [hA =] 4 zhang 5 chi [= 45 chi];
            upper width [a3 =] 8 zhang 5 chi [= 85 chi];
            lower width [a2 =] 9 zhang [= 90 chi];
            upper length [b3 =] 13 zhang [= 130 chi];
            lower length [b2 =] 14 zhang [= 140 chi].

The contribution of county B:
            height [hB =] 13 zhang 5 chi [= 135 chi];
            upper width [b1 =] 7 zhang [= 70 chi];
            lower width [a3 =] 8 zhang 5 chi [= 85 chi];
            upper length [b1 =] 10 zhang [= 100 chi];
            lower length [b3 =] 13 zhang [= 130 chi].

The ramp:
            height [g =] 18 zhang [= 180 chi];
            upper width [c1 =] 3 zhang 6 chi
[= 36 chi];
            lower width [c2 =] 2 zhang 4 chi
[= 24 chi];
            length [l =] 14 zhang
[= 140 chi].

The contribution of the men in the district of county A:
            height [gA =] 9 zhang
[= 90 chi];
            upper width [c3 =] 3 zhang
[= 30 chi];
            lower width [c2 =] 2 zhang 4 chi
[= 24 chi];
            length [lA =] 7 zhang
[= 70 chi].

The contribution of the men in the district of county B:
            height [gB =] 9 zhang
[= 90 chi];
            upper width [c1 =] 3 zhang 6 chi
[= 36 chi];
            lower width [c3 =] 3 zhang
[= 30 chi];
            length [lB =] 7 zhang
[= 70 chi].

[2.3.] Method

[2.3.1. The dimensions of the platform]

術曰:以程功尺數乘二縣人,又以限日乘之, 為臺積。又以上下 袤差乘上下廣差,三而一為隅陽冪。以乘截高,為隅陽截積{冪/}。又半上下廣差,乘斬上袤為隅 頭冪,以乘截高為隅頭截積。{所得/}并二積,以減臺積,余為實。以上下廣差并上下袤 差,半之為正數。加截上袤,以乘截高,所得,增隅陽冪加隅頭冪,為方法。又并截高及截上袤與正數,為廉法,從。開立方除之,即得上 廣。各加差,得臺下廣及上下袤、高。

Multiply the regulation labour in [cubic] chi by the number of men of the two counties. Multiply this by the allotted number of days to make the volume [V] of the platform.

V = 75 chi3/man-day × 4640 men × 5 days = 1,740,000 chi3

Multiply the difference between the upper and lower lengths [b2b1] by the difference between the upper and lower widths [a2a1] and divide by three to make [K1 =] the Area for the Corner Yang[ma].[19] Multiply this by the truncated height[20] [ha1] to make [K2 =] the Truncated Volume for the Corner Yang[ma].

 

Halve the difference between the upper and lower widths [a2a1] and multiply by the shortened upper length [b1a1] to make [K3 =] the Area for the Corner End. Multiply by the truncated height [ha1] to obtain [K4 =] the Truncated Volume for the Corner End.

Add the two volumes [K2 and K4] and subtract from the volume of the platform [V]. The remainder is the shi [the constant term of the cubic equation].

Add the difference between the upper and lower widths [a2a1] to the difference between the upper and lower lengths [b2b1] and halve [the sum] to obtain [K5 =] the Determined Number.

Add the truncated upper length [b1a1] and multiply by the truncated height [ha1]. To the result of this, add the sum of the Area for the Corner Yang[ma] [K1] and the Area for the Corner End [K3] to make the fangfa 方法 [the coefficient of the linear term].

Further add together the truncated height [ha1], the truncated upper length [b1a1], and the Determined Number [K5] to make the lianfa 廉法 [the coefficient of the quadratic term].

Extract the cube root to obtain the the upper width [a1].[21]

a13  +  170a12  +  7166⅔ a  =  1,677,666⅔ chi3

a1 = 70 chi

Add to each of the differences to obtain the lower width, the upper and lower lengths, and the height of the platform.

a2 = (a2a1) + a1 = 90 chi

b1 = (b1a1) + a1 = 100 chi

b2 = (b2b1) + b1 = 140 chi

h = (ha1) + a1 = 180 chi

[2.3.2. The contributions of the two counties to the platform]

求均給積尺受廣袤術曰:以程功尺數乘乙縣 人,又以限日乘之, 為乙積。三因之,又以高冪乘之,以上下廣差乘袤差而一,為實。又以臺高乘上廣,廣差而 一,為上廣之高。又以臺高乘上袤,袤差而一,為上袤之高。又以上廣之高乘上袤之高,三之,為方法。又并兩高,三之,二而一,為廉法, 從。開立方除之,即乙高。以減本高,餘,即甲高。此是從下給臺甲高。又以廣差乘{之/乙}高,{以/如/[22]}本高而一,所得,加上廣,即甲 上廣。又以袤差乘 乙高,如本高而一,所得,加上袤,即甲上袤。其甲上廣、袤即乙下廣、袤。臺上廣、袤即乙上廣、袤。其後求廣、袤,有增損者,皆放此。

In the following see Figure 2.

The method to calculate the volume [of earth] supplied in [cubic] chi and the width and length [of the part of the platform built by county B] is: Multiply the regulation volume in [cubic] chi by the [number of] men from county B. Multiply this by the allotted number of days to obtain the volume [of the part of the platform built by county] B.

VB = 75 chi3/man-day × 3222 men × 5 days = 1,208,250 chi3

Multiply it by three and multiply by the area of [the square on] the height [h]. Divide by the difference between the upper and lower width [a2a1] multiplied by the difference between the lengths [b2b1]. This is the shi [the constant term in the cubic equation].

 = 146,802,375 chi3

Multiply the height of the platform [h] by the upper width [b1] and divide by the difference between the widths [a2a1] to make [K6 =] the Height for the Upper Width.

 = 630 chi                                                   (15)

Multiply the height of the platform [h] by the upper length [b1] and divide by the difference between the lengths to make [K7 =] the Height for the Upper Length.

 = 450 chi                                                  (16)

Multiply the Height for the Upper Width [K6] by the Height for the Upper Length [K6] and multiply by 3 to make the fangfa [the coefficient of the linear term].

  =  850,500 chi2

Add the two heights [K6 and K7], multiply by three, and divide by two to make the lianfa [the coefficient of the quadratic term].

 = 1620 chi

Extract the cube root to obtain [hB =] the height [of the part of the platform built by county] B.

 +  +  = 

hB3 + 1,620hB2 + 850,500hB = 146,802,375 chi3

hB = 135 chi

Subtract this from the original height [h]. The remainder is then [hA =] the height [of the part of the platform built by county] A. This is the height [of the part built by county] A from the base upwards.[23]

 = 45 chi

Multiply the difference between the widths [a2a1] by the height [of the part of the platform built by county] B [hB] and divide by the original height [h]. To the result of this, add the upper width [a1]. This is then [a3 =] the upper width [of the part of the platform built by county] A.

  =  85 chi

Multiply the difference between the lengths [b2b1] by the height [of the part of the platform built by county] B [hB] and divide by the original height [h]. To the result of this, add the upper length [b1]. This is then [b3 =] the upper length [of the part of the platform built by county] A.

  =  130 chi

The upper width and length of [the part built by] county A are the lower width and length of [the part built by] county B [a3 and b3]. The platform’s upper width and length [a1 and b1] are the upper width and length [of the part built by] county B.

If later [one wants to] calculate the width and length, [when] there are more or fewer [counties], in all cases [the procedure] is the same.

[2.3.3. Comment: Derivation of county B’s contribution to the platform]

(此 應{三/六}因乙積,臺高再乘,上下廣差乘袤差而一。又 以臺高乘上廣,{/ 廣差而一,}為上 廣之高。又以臺高乘上袤,{/袤差而一,}為上袤之高。{以上廣之高乘上袤之高/相乘}為小冪二。{因/因/又}下袤之高{/乘上廣之高},為中冪一。凡下袤、下廣之高即是截高與上袤、{與//與[24]}上廣之高相連并數。然{此/此/則}有中冪定有小冪一,又有上 廣之高乘截高為冪{各/}一。又下廣之高乘下袤之高為大冪二。乘上袤之高為中冪一。其大冪之中{又/又有/有}小冪一,復有上廣、上袤之 高{為中冪/}各乘截高為{中/中/}冪各一。又截高自乘,為冪一。其中冪之內有 小冪一。又上袤之高乘截高為冪一。然則截高自相乘為冪二,小冪六。又上廣上袤之高各三,以乘截高為冪六。令皆半之,故以三乘小 冪。又上廣上袤之高各三,今但半之,各得一又二分之一,故三之二而一。諸冪{/乘}{/高}為積尺。)

This reflects multiplying the volume [VB] [of the part of the platform built by county] B by six, then multiplying twice by the height of the platform [h] and dividing by the difference between the upper and lower widths [a2a1] multiplied by the difference between the lengths [b2b1].

2  ×   shi  =  

Further, multiplying the height of the platform [h] by the upper width [a1] and dividing by the difference between the widths [a2a1] gives [K6 =] the Height for the Upper Width. Multiplying the height of the platform [h] by the upper length and dividing by the difference between the lengths [b2b1] gives the Height for the Upper Length.

These are the same calculations as above, equations (15) and (16).

Multiplying these together gives two of the small area [PUWT].

 

Further, multiplying the Height for the Lower Length [K8] by the Height for the Upper Width [K9] gives one of the middle areas [PQXT].

The comment does not define the Heights for the Lower Length and Lower Width, but they turn out to be

Further, multiplying the Height for the Lower Length [K8] by the Height for the Upper Width [K9] gives one of the middle areas [PQXT].

The Heights for the Lower Length and the Lower Width [K8 and K9] are in fact the sum of the truncated height [hB] together with the Heights for the Upper Length and the Upper Width [K6 and K7] lined up.


Thus it has been determined that the middle area [KLPQ] is composed of one small area [PUWT] and also an area obtained by multiplying the Height for the Upper Width [K6] by the truncated height [hB].[25]

Further, multiplying [twice] the Height for the Lower Width [K8] by the Height for the Lower Length [K9] gives two of the large area [PQRS].

Multiplying . . .[26] [K7] by the Height for the Upper Length [K9] gives one of the middle areas [PUVS].

Within the large area [PQRS] there is one small area [PUWT], and in addition there are the Heights for the Upper Width and the Upper Length [K6, K7], each multiplied by the truncated height [hB] to make one middle area, and the truncated height [hB] multiplied by itself to make one area.

Within the middle area [PUVS] there is one small area [PUWT] together with the area obtained by multiplying the Height for the Upper Length [K7] by the truncated height [hB].

Thus there are two of the area formed by multiplying the truncated height by itself [hB2] and six of the small area [K6K7]. There are also three each of the Heights for the Upper Width and the Upper Length [K6, K7].

2hB2  +  6K6K7  +  3K6hB  +  3K7hB

These, multiplied by the truncated height [hB], give six [times the sum of the] areas. When all are halved, the small area [the term K6K7] is multiplied by three. Further, there are three each of the Heights for the Upper Width and the Upper Length [K6, K7]. When [the terms K6hB and K7hB] are halved once, in each case one and one half is obtained; this is why [these terms] are multiplied by three and divided by two. All the areas are multiplied by the truncated height [hB] to make the volumes in [cubic] chi.

 = 

     

[2.3.4. The dimensions of the ramp]

求羨道廣袤高術曰:以均賦常積乘二縣五十六 鄉,又六因為積。 又以道上廣多下廣數加上廣少袤為下廣少袤。又以高多袤加下廣少袤為下廣少高。以乘下廣少袤為隅陽冪。又以下廣少上廣乘之,為鼈隅{/積}。以減積,餘,三而一,為實。 并下廣少袤與下廣 少高,以下廣少上廣乘之,為鼈從橫廉冪。三而一,加隅{//陽[27]}冪,為方法。又以三除上廣多下 廣,以下廣少袤、 下廣少高加之,為廉法,從。開立方除之,即下廣。加廣差即上廣,加袤多上廣於上廣即袤,加{廣/高}多袤即道高。

The method for calculating the width, length and height of the ramp is: Multiply the constant volume of the uniform assessment by the two counties’ 56 districts and multiply by six to make [6W =] the ‘Volume’.

W  =  6300 chi3/district  ×  56 districts  =  352,800 chi3

6W  =  2,116,800 chi3

In the following see Figure 2.

Add the difference between the upper width and the lower width of the ramp [c1c2] to the difference between the upper width and the length [lc1] to obtain the difference between the length and lower width [lc2].

lc2  =  (c1c2)  +  (lc1)  =  116 chi

Add the difference between the height and the length [gl] to the difference between the length and the lower width [lc1] to make the difference between the height and the lower width [gc2].

gc2  =  (gl)  +  (lc1)  =  144 chi

Multiply this by the difference between the length and the lower width [lc2] to make [K11 =] the Area for the Corner Yang[ma].

K11  =  (gc2) (lc2)  =  18,096 chi2

Multiply by the difference between the upper width and lower width [c1c2] to make [K12 =] the Volume for the Bie[nao] Corner.

K12  =  K11 (c1c2)  =  217,152 chi3

Subtract this from the ‘Volume’ [6W] and divide the difference by three to make the shi [the constant term of the cubic equation].

  =  633,216 chi3

Add the difference between the length and the lower width [lc2] and the difference between the height and the lower width [gc2] and multiply by the difference between the upper width and the lower width [c1c2] to make [K13 =] the Area for the Vertical and Horizontal Edge Bie[nao] .

K13  =  ([lc2] + [gc2]) (c1c2)  =  3,264 chi2

Divide by three and add the Area for the Corner [Yangma] [K11] to make the fangfa [the coefficient of the linear term].

fangfa  =    =  19,184 chi2

Divide the difference between the upper width and the lower width [c1c2] by three and add to this the difference between the length and the lower width [lc2] and the difference between the height and the lower width [gc2] to make the lianfa [the coefficient of the quadratic term].

lianfa  =   = 276 chi

Extract the cube root to obtain the lower width [c2].

c23  +  276 c2 +  19,184 c =  633,216 chi3

c2 = 24 chi.

Add this [c2] to the difference in the widths [c1c2] to make the upper width [c1]. Add the difference between the length and the upper width [lc1] to the upper width [c1] to make the length [l]. Add the difference between the height and the length [gl] to make the the height [g] of the ramp.

c1 = c2 + (c1c2) = 36 chi
l = c1 + (lc1) = 140 chi
g = l + (gl) = 180 chi

[2.3.5. The contributions of the two counties to the ramp]

求羨道均給積尺,甲縣受廣、袤,術曰:以均 賦常積乘甲縣一十 三鄉,又六因為積。以袤再乘之,以道上下廣差乘臺高為法而一,為實。又三因下廣,以袤乘之,如上下廣差而一,為都廉,從。開立方除 之,即甲袤。以廣差乘甲袤,本袤而一,以下廣加之,即甲上廣。又以臺高乘甲袤,本袤除之,即甲高。

The method to calculate the volume [of earth] supplied in [cubic] chi and the width and length of [the part of the ramp supplied by] county A is: Multiply the constant volume of the uniform assessment by county A’s 13 districts and multiply this by six to make [6WA =] the ‘Volume’.

WA = 6,300 chi3/district × 13 districts = 81,900 chi3

6WA = 491,400 chi3

Multiply this twice by the length [l]. Multiply the difference between the upper and lower width [c1c2] of the ramp by the height of the platform [h = g][28] to make the denominator, and divide to make the shi [the constant term of the cubic equation].

shi  =    =  4,459,000 chi3

Multiply the lower width [c2] by three, multiply by the length [l], and divide by the difference between the upper and lower width [c1] to make the dulian 都廉 [the coefficient of the quadratic term].[29]

dulian  = 840 chi

Extract the cube root to obtain [lA =] the length [of the part of the ramp built by county] A.

 chi3

lA = 70 chi

Multiply the difference in widths [c1c2] by the length [lA] [of the part built by county] A. Divide by the original length [l] and add the lower width [c2] to make [c3 =] the upper width [of the part built by county] A.

  =  30 chi

Multiply the height of the platform [h = g] by the length [lA] [of the part built by county] A and divide by the original length [l]. This is then [gA =] the height [of the part built by county] A.

  =  90 chi

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[1] Jiu Tang shu 1975, 32: 1168, 47: 2039, 4: 76, 44: 1892; Qian Baocong 1963: 487.

[2] ‘Nine chapters of the mathematical art’ is a common translation of this title.

[3] Chemla and Guo [2004: 387–457, 661–745]; see also Cullen [1993]; Shen Kangshen et al. [1999: 250–306, 439–517].

[4] E.g. in the commentary to problem 3, Qian Baocong [1963: 506].

[5] Lim and Wagner [forthcoming]; see also e.g. Qian Baocong [1963: 487–491]; Guo Shucun and Liu Dun [1998, 1: 21–22]; He Shaogeng [1989].

[6] This is the Tianlu Linlang congshu 天祿琳瑯叢書 edition, a facsimile printed in 1931, available at http://www.scribd.com/doc/78515037.

[7] On Li Huang’s edition see Huang Juncai [2009].

[8] The only exceptions to this statement appear to be the highly speculative reconstructions of the fragmentary commentary to Problem 17 by He Shaogeng [1989] and Wang Rongbin [1990].

[9] Chemla & Guo [2004: 322–335, 363–379]; Shen et al. [1999: 175–195, 204–226]; Wang & Needham [1955]; Lam Lay Yong [1970; 1977: 195–196, 251–285; 1986]; Libbrecht [1973: 175–191]; Chemla [1994]; Martzloff [1997: 221–249].

[10] See e.g. Rees & Sparks [1967: 294–297] as well as numerous pages on the World Wide Web. Horner [1819] presented a procedure for approximating roots of any infinitely differentiable function, but modern descriptions of ‘Horner’s method’ consider only the special case of polynomial functions.

[11] Andrea Bréard [1999: 95–99, 353–356] gives an abridged translation of the text and an analysis of the first of the four problems along the same lines as ours. Her analysis uses a dissection into a larger number of pieces and consequently a smaller amount of algebraic reasoning. She seems to have been the first to note that the names given by Wang Xiaotong to intermediate quantities give clues to how the calculations were derived.

[12] Chi is a unit of linear measure, at this time equal to about 30 cm.

[13] See e.g. Chemla & Guo [2004: 390, 434, 436–439, 826–827, 912].

[14] And indeed, Wang Xiaotong’s six problems on right triangles [Lim & Wagner forthcoming] seem to be entirely without practical applications.

[15] A possible exception is in Liu Hui’s explanation of the cube-root algorithm in Jiuzhang suanshu, which may have used a diagram [Lam Lay Yong 1970]; but see Volkov [2007]; Chemla [2010].

[16] Wagner [1979].

[17] Lam Lay Yong [1970].

[18] Tai zi xia ji ji gao, dao zi chu deng ji gao 臺 自下基給高,道自 初登給袤. The word ji would seem to have some specific meaning in the administrative terminology of the time.

[19] In keeping with the writing fashion of his time, in which multisyllabic words were considered inelegant, Wang Xiaotong abbreviates the terms yangma and bienao to yang and bie respectively.

[20] Jiegao 截 高. The same name is given to a different quantity in Section 2.3.3.

[21] Commentators are divided on the function of the word cong / zong in the phrase 從開立方除. Qian Baocong [1966: 46–47] reviews the various attempts to explain it, and comes to the conclusion that in each case it belongs at the end of the previous sentence, and means ‘follow, accompany’ (gensui 跟随). This is difficult for us to understand, and we omit the word in our translation.

[22] Guo Shuchun & Liu Dun [1998, 2: 4, n. 14].

[23] The point of this redundant sentence is not apparent. Perhaps it is a comment by another hand.

[24] Guo Shuchun & Liu Dun [1998, 2: 4, n. 19].

[25] Jiegao 截高. This term is used for a different quantity in Section 2.3.1.

[26] The multiplicand is missing in the text here. The mathematical context indicates that it is K7.

[27] Guo Shuchun & Liu Dun 1998, 2:4, n. 26.

[28] The height of the ramp, g, was calculated in Section 2.3.4 above (equation (2.31)), and it turned out to be the same as the height of the platform, g = h = 180 chi. It is therefore not important, but something of a surprise, that the text refers here to the height of the platform rather than that of the ramp.

[29] The term dulian 都廉 is occasionally used by Wang Xiaotong for the quadratic coefficient instead of the more usual lianfa 廉法. We have been unable to determine whether the two terms are exact synonyms, or differ in meaning in some way.