**Geometrical Problems from the**

*Corpus Agrimensorum*:**Mollweide meets Mommsen**

A representation is made with a purpose or goal in mind, governed by criteria of adequacy pertaining to that goal, which guide its means, medium and selectivity.

--Bas Van Fraassen

The group of Roman surveying texts known as the

*Corpus Agrimensorum*have provided a great deal of information regarding the actual practices of Roman surveyors in the field and given scholars insight into how the Romans allocated and measured land. My current project in locating and mapping the surviving remains of Roman surveying in North Africa takes its starting point from this 6th century compilation of surveying manuals. The texts themselves and the illustrations attached to them have attracted the attention of many scholars in past including historians of Roman law and agrarian practices such as Theodor Mommsen and Max Weber. In this sence the historiography associated with the texts is almost as interesting as the texts themselves.

What is less known about this historiography is the attraction the texts have held for more mathematically inclined historians of cartography such as C.B. Mollweide. Mollweide is best known as map projectionist, but also did fundamental research into some of the unsolved geometrical problems found in the

*Corpus*.

One of the most interesting of these problems concerns the method for finding south. In the Corpus there are several methods given for this, but the one that interests us here is that given by the writer,

*Hyginus.*

**There is also another method of obtaining South, by marking three shadows. On level ground we shall set up a gnomon AB, and note any three of its shadows, CDE. These shadows we shall mark with the set square, to see their distances from each other. If we set them up before noon, the first shadow will be the longest; if after noon, the last. We shall then draw these shadows in proportion by a footrule... Let AB be a gnomon, B the ground. Let us take the longest shadow and mark it [i.e. its end opposite to B] on the ground as C; the second likewise D, the third E... Let us project hypotenuses from C on to A and from D on to A. Now with centre A and radius E let us draw a circle. Then let us project lines parallel to the base, i.e. ground, on to the perpendicular [AB] from the intersections of the hypotenuses and the circumference, from F on to G and from I on to K. Then we shall apply the longest line, GF, to the largest shadow, and from B we shall mark out [the length of] GF ; the second line to the second shadow, and we shall mark out [the length of] KI. Then from F and I we shall project a straight line, and likewise from C and D, the shadow ends. These two lines will meet at T. Join TE; this will be east-west.**

The above text by Hyginus is accompanied in the

*Codex Arcerianus A*by the figure shown below.

The figure as drawn by the scribe is however totally inadequate to explain the complexities of the method outlined by *Hyginus* and one has to question its purpose in the manuscript and whether it was in fact added to the text by a later copiest with less understanding of the method. Mollweide analyzes the text in an article 'Erlduterung einer in der Scriptoribus rei agrariae. . . gegebenen Vorschrift..' published in Zach's *Monatliche Correspon-denz zur Beforderung der Erd- und Himmelkunde* (Volume 28, 1813. p. 396-425).

Click on Figures to enlarge

>>>>>>

The method that is being described in the text by

*Hyginus*is extremely complicated and there are open questions concerning the level of mathematics and solid geometry involved.

Mollweide's solution is a complex affair and according to Dilke the method as described by

*Hyginus*probably goes back to Alexandrian mathematical scholarship that has been lost but that must have been dependent on Apollonius'

*Conics.*

The modern solution that Dilke adapted from Mollweide can be condensed to the following along with the figure below:

**ABC, ABD, ABE are right-angled triangles. The lines CA, DA, EA go towards the centre of the sun. The arc EIF is part of a circle forming the base of a regular cone, parallel to the sun's daily round and so to the equator, and FI is a chord of this circle. Since GF is equal and parallel to BL, FL will be equal and parallel to GB; similarly IM to KB. As FL and IM are parallel to AB and so to each other, they lie in a plane in which FI and LM lie. But FI is also in the plane ACD, and LM is in the horizontal plane BCD; so FI is the intersection of the plane FIML with the plane ACD, and LM the intersection of the plane FIML with the horizontal plane BCD. As the plane ACD is cut by the horizontal plane at CD, which when produced meets LM produced at T, it follows that T is in the plane ACD and also in the plane FIML, and so is a point on the common intersection of both planes, i.e. of FI produced. Since the latter lies entirely on the plane of the circle through FIE, T is also in this plane, but likewise in the horizontal plane BCD, and so is a point on the common intersection of both planes. Since E is also such a point, it follows that ET is the intersection of a plane parallel to the equator plane with the horizontal plane, and so parallel to the east-west line.**

One can see how different the figure which displays the actual construction as it described by Hyginus is from the original manuscript illustration.