Lecture 4 A Chinese surveying problem 


Chinese mathematics 
The history of Chinese mathematics is complex. Early works were incorporated with commentary into
later works, and it is often difficult to determine who did what and when. For a brief outline, check this site. For a more complete and reliable reference, see J.C. Martzloff: "A History of Chinese Mathematics". Surveying and mapmaking have always been important in Chinese maths, for both military and civil use. The most important early work was the Jiuzhang Suanshu (Nine Chapters of the Mathematical Art), compiled about +220, a compilation of works dating back to c1100. It is a Problem Text, divided into chapters dealing with different types of field measurements, constructions and rightangled triangle problems. In 263, the scholar Liu Hui wrote a commentary on the Nine Chapters in which he verified theoretically the solution procedures, and added some problems of his own. About 600 his work was separated out and published as the Huidau Suanjing (Sea Island Mathematical Manual), and various versions have been republished till modern times. It consists of a series of problems about a mythical Sea Island which describe a range of surveying and mapmaking techniques which are a precursor of trigonometry, but using only properties of similar triangles, the area formula and so on. The text is a modern free translation of a Chinese edition of 1960. Typically for Chinese works, a problem is presented, a solution given and the solution is explained. In early China, all computations are done using computing rods on a counting board and the results transcribed into the text.

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The text

Now for [the purpose of] looking at a sea island, erect two poles of the same height, 3 zhang [on the ground], the distance between the front and rear [pole] being a thousand bu. ,p. Assume that the rear pole is aligned with the front pole. Move away 123 bu from the front pole and observe the peak of the island from ground level; it is seen that the tip of the front pole coincides with the peak. Move backward 127 bu from the rear pole and observe the peak of the island from ground level again; the tip of the back pole also coincides with the peak. What is the height of the island and how far is it from the pole? [Answer:] The height of the island is 4 Ii 55 bu. It is 102 li 150 bu from the pole. [Method:l Multiply the distance between poles by the height of the pole, giving the shi. Take the difference in distance from the points of observations as the fa to divide [the shil, and add what is thus obtained to the height of the pole. The result is the height of the island. To find the distance of the island from the front pole, multiply [the distance of the] backward movement from the front pole by the distance between the poles, giving the shi. Take the difference in distance at the points of observation as the fa to divide the shi. 'Re result is the distance of the island from the pole in li.

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Explanation 
To find the distance and height of an inaccessible mountain, plant two poles of equal height in line with the mountain top so that the sightlines from ground level through the tops of the poles converge at the top of the mountain. From the lengths of the poles, the distance between them, and the distance from the poles back to the sighting points, find the unknown distances by similar triangles. The technical terms: chi ~ 1 cm, bu = 6 chi, zhang = 10 chi, li = 1800 chi ~ 50 m. fa is the divisor, shi the dividend, represented by certain positions on the counting board. The solution states that the shi is the distance between the poles (SN) times the height of the poles (AS), and the fa is the difference in the distances of the observing positions (NDSB). Then the height of the island is the height of the pole + shi/fa. Thus we have an algorithm, not a calculation. An important theorem used in Chinese mathematics often used in situations where we would employ similar triangles or trigonometry concerns the figure obtained by dropping perpendiculars from points on two adjacent sides of a rectangle to meet a diagonal at a common point. In the diagram above for example, one perpendicular is HN and the other is RE, meeting the diagonal PD at C. The reult is that area HCEF = area RCNQ.

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Exercises 

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From L. L. Yong and A. T. Se `Fleeting Footsteps' p. 112

Now there are 3 bundles of top grade cereal, 2 bundles of medium grade cereal and 1 bundle of low grade cereal, which yield 39 dou of grains as shi; 2 bundles of top grade cereal, 3 bundles of medium grade cereal and 1
bundle of low grade cereal yield 34 dou as shi; 1 bundle of top grade cereal, 2 bundles of medium grade cereal
and 3 bundles of low grade cereal yield 26 dou as shi. Find the measure of grains in each bundle of the top,
medium and low grade cereal. Answer: One bundle of top grade cereal has 9 1/4 dou, one bundle of medium grade cereal has 4 1/4 dou, and one bundle of low grade cereal has 2 3/4 dou.
Method: Put down 3 bundles of top grade cereal, 2 bundles of medium grade cereal, 1 bundle of low grade cereal and 39 dou as shi in a column on the right. Set up the columns in the centre and on the left in the same way as the column on the right ([i] in diagram above). Take the number representing top grade cereal in the right column to multiply all numbers in the central column [ii], and then use the method of direct subtractions [iii]. Once again multiply the numbers in the next column, that is, the left column, by the number representing top grade cereal in the right column [ivl, and then use the method of direct subtractions [v]. Next multiply all the numbers in the left column by the remaining number representing medium grade cereal in the central column [vi], and then use the method of direct subtractions [vii]. The left column has the remaining number representing low grade cereal. The fa (divisor) is above and the shi (dividend) below; the shi here is the shi for low grade cereal. To find the measure for medium grade cereal, multiply the shi in the central column by the fa of the left column and subtract the shi for low grade cereal [viii]. The remainder is divided by the number of bundles of medium grade cereal in the central column, yielding the shi for medium grade cereal [ix]. To find the measure for top grade cereal, once again multiply the shi in the right column by the fa of the left column, and subtract the respective shi for low and medium grades [xl. The remainder is divided by the number of bundles of top grade cereal in the right column, yielding the shi for top grade cereal [xi]. The shi for all grades are each divided by the fa to yield the measures per bundle of the respective grades [xii]. 
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Last update: 3 Nov., 2000
Author: Phill Schultz, schultz@maths.uwa.edu.au