Propositions 6-7

Introduction to the translation
Assumptions or Postulates
Proposition 1: Weights that incline-equally from equal lengths are equal.
Proposition 2: Unequal weights from equal lengths do not incline-equally, but will incline towards the larger.
Proposition 3: Unequal weights will incline-equally from unequal lengths and the larger weight from the smaller length.
Proposition 4: If two equal magnitudes do not have the same center of weight, the center of weight of the magnitude composed from them both will be the middle of the straight-line joining the centers of weight of the magnitudes.
Proposition 5: If the centers of weight of three magnitudes are positioned on a straight-line, and the magnitudes have equal weight, and the straight-lines between the centers are equal, the center of weight of the magnitude composed from all the magnitudes will be the point which is also the same as the center of weight of the middle magnitude.
Corollaries to Proposition 5:generalizations

Eutocius

Note: Eutocius notes will be marked, e.g., (*e1*), the translator's longer notes, e.g., (*1*).

Note on the translation: It is difficult to find an adequate and intuitive translation of the Greek word, isorrhopia. The root verb, rhepô is to sink, incline downwards, or to preponderate. So the compound is to, sink-equally, incline-equally, preponderate equally, that is to balance. We could then translate our word as ‘equilibrium’ or ‘balancing’, and its other forms as, ‘to balance’. However, that would lose the contrast between one weight equally sinking (and hence not sinking at all) and one weight preponderating and so sinking. My hope is that the reader will get used to the hyphenated word, ‘to incline-equally’, which seems to capture the notion best. Nonetheless, following convention, the title will remain, “Equilibria of Planes.”

Note on the brief analyses: Let A and B be weights, a and b lengths from the fulcrum of the balance, normally of A and B, respectively, R(X,x) the inclination of weight X at length x. So R(X,x) = R(Y,y) means that X and Y equally-incline at distances x and y, and R(X,x) > R(Y,y) or R(Y,y) < R(X,x) that X inclines more than Y at distances x and y (i.e., the balance inclines on the side of X or that X is closer to the horizon than Y).

Keep in mind that the horizon, beams, and balances are not mentioned in Equilibria of Planes. Even inclination is not discussed after I Prop. 8. So these definitions are very tenuously connected to the text. However, a careful analysis of the arguments will show that the notion of center of weight is fundamental to all the assumptions and to Props. I 1-7. In particular it is important that if a weight hangs freely from a beam, one can replace the weight by one whose center of weight is on the beam, and vice versa.

Throughout the Equilibria of Planes there is an important, but unstated, basic assumption.

CW1: Every magnitude or collection of magnitudes has exactly one center of weight.

Additionally, Archimedes never gives a definition of ‘center of weight’, and the text has been criticized for this omission. He also never defines 'inclination’ or ‘equal-inclination’. Eutocius gives a definition, which seems inadequate. Here are two attempts at definitions, the first of which captures better what Archimedes does in Plane Equilibria and Quadrature of the Parabola, but keeping in mind the abstract nature of his works also fits the solid geometry of the Method and Floating Bodies. The second definition fits better with the account of Archimedes found in Heron' Mechanica.

1. The center of weight is a point from which a freely hanging body is stable, no matter how it is positioned about the point.

2. The center of weight is a point such that if a body is hung freely from any point on the body, a perpendicular from the point of suspension to the horizon will go through it.

It is obvious that either of these definitions taken together with the basic assumption are not obvious.

Centers of Equal-Inclinations of Planes or of Weights of Planes, 1st

Assumptions (*e1*)(*1*)

1. We postulate that equal weights from equal lengths incline equally and that equal weights from unequal lengths do not incline equally but incline on the side the weight of the larger length.

A1a. A = B & a = b => R(A,a) = R(B,b)

A1b. A = B & a > b => R(A,a) > R(B,b)

2. that if when weights incline-equally from certain lengths something is added to one of the weights they do not incline-equally but incline to that weight to which there was added.

A2. R(A,a) = R(B,b) => R(A+C,a) > R(B,b)

3. similarly that if something is taken away from one of the weights they will also not incline-equally, but will incline to the weight from which it was not taken away.

A3. R(A,a) = R(B,b) => R(A-C,a) < R(B,b)

4. When equal and similar plane figures fit on one another the centers of weights also fit on one another.(*e2*)(*2*)

5. but the centers of weights of unequal but similar figures will be similarly positioned. We say that the points are similarly positioned in similar figures from which straight lines drawn to the equal angles make equal angles to the corresponding sides.(*e3*)

6. If magnitudes incline equally from certain lengths, those equal to them will also incline equally from the same lengths.

A6. R(A,a) = R(B,b) & C = A => R(C,a) = R(B,b)

7. In every figure whose perimeter is concave-on-the-same-sides, the center of weight must be within the figure.(*e4*) And when these are supposed:

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Propositions

1. Weights that incline-equally from equal lengths are equal.

For if they will be unequal, when the excess from the larger is taken away they will not incline-equally, since it is taken away from one of those that incline equally. Thus, the weights that incline equally from equal lengths are equal. This is the converse of the first part of assumption 1a.

T1. R(A,a) = R(B,b) & a = b => A = B

Suppose A > B and C = A-B. By A1a and a = b, R(A-C,a) = R(B,b). By A3 and R(A,a) = R(B,b), R(A-C,a) < R(B,b))

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2. Unequal weights from equal lengths do not incline-equally, but will incline towards the larger.

For when the excess is taken way they will incline-equally, since equals from equal lengths incline-equally. And so when what was taken away is added it will incline to the larger, since it was added to one of those inclining-equally.

T2. A > B & a = b => R(A,a) > R(B,b)

Suppose C = A-B,. By A1 and a = b, R(A-C,a) = R(B,b). By A2 and R(A-C,a) = R(B,b), R(A-C+C,a) > R(B,b), i.e., R(A,a) > R(B,b)

3. Unequal weights will incline-equally from unequal lengths and the larger weight from the smaller length.(*3*)

(diagram 1 = general diagram) Let there be unequal weights, A, B, and let A be larger, with them inclining-equally from distances AG, GB. One must prove that AG is smaller than GB.

(diagram 2) For let it not be smaller. (diagram 3) When the excess by which A exceeds B is taken away, since it is taken away from the other of the magnitudes that incline-equally, it will incline to B. But it won’t incline. (diagram 4) For if GA is equal to GB, they will incline-equally [for they are equals from equal lengths], (diagram 5) and if GA is larger than GB, it inclines to A. For equals do not incline-equally from unequal lengths, but incline to the weight from the larger length. For these reasons in fact, AG is less than GB. It is obvious that magnitudes that incline-equally from unequal distances are also unequal, and the one from the smaller length is larger.

T3a. A > B & R(A,a) = R(B,b) => a < b

Suppose A > B and R(A,a) = R(B,b) and C = A-B. Also suppose for a reductio a = b or a > b. By A3, R(A-C,a) < R(B,b). If a = b, by A1a, R(A-C,a) = R(B,b). If a > b, by A1b and A-C = B, R(A-C,a) > R(B,b).

T3b. R(A,a) = R(B,b) & a < b =>A > B

Here is a possible proof of the obvious. Suppose R(A,a) = R(B,b), and a < b, c = b - a. For the reductio, either A = B or A < B. Suppose A = B. Then, by a < b and A1b, R(A,a) > R(B,b). Suppose A < B. Then by R(A,a) = R(B,b) and T3a, b < a.

This concludes the nine basic principles for the balance. The other three assumptions, two propositions, and two corollaries are principles about centers of weight. (*3*)

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4. If two equal magnitudes do not have the same center of weight, the center of weight of the magnitude composed from them both will be the middle of the straight-line joining the centers of weight of the magnitudes.(*4*)

(diagram 1 = general diagram) Let A be the center of weight of A, and B of B, and let AB, having been joined, be bisected at G. I say that G is the center of weight of the magnitude composed from both magnitudes.

(diagram 2) For otherwise, let D be the center of weight [of that from both of magnitudes A, B](*e5*), if it is possible [for it has been proved earlier that it is on AB]. And so, since point D is the center of weight of the magnitude composed from A, B, they incline-equally from lengths AD, DB, which is impossible [for equals do not incline-equally from unequal lengths]. And so it is clear that G is the center of weight of the magnitude composed from A, B.

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5. If the centers of weight of three magnitudes are positioned on a straight-line, and the magnitudes have equal weight, and the straight-lines between the centers are equal, the center of weight of the magnitude composed from all the magnitudes will be the point which is also the same as the center of weight of the middle magnitude.

Let there be three magnitudes, A, B, G, and their centers of weight, points A, B, G placed on a straight line, and let A, B, G be equal and straight-lines AG, GB be equal. I say that point G is the center of weight of the magnitude composed from all the magnitudes. For since magnitudes A, B have equal weight, point G will be the center of weight, since AG, GB are equal. But point G is also the center of weight of G. And so, it is clear that the center of weight of the magnitude composed from all the magnitudes will be the point which is also the center of weight of the middle magnitude.

The proof assumes as a principle:

k is the center of weight of X and k is the center of weight of Y => k is the center of weight of X+Y.

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Corollary 1: From these it is in fact obvious that in the case of the centers of weight of however many odd magnitudes in multitude which are positioned on a straight-line, if the magnitudes equally distant from the middle one have equal weight, and the straight-lines between their centers are equal, the center of weight of the magnitude composed from all the magnitudes will be the point which is also the center of the magnitude in the middle of them.

Corollary 2: If there are magnitudes that are even in multitude and the centers of their weight are positioned on a straight-line, and the middles one of them and those equally distant from them have equal weight, and the straight-lines between the centers are equal, the center of weight of the magnitude composed from all the magnitudes will be the middle of the line joining the centers of weight of the magnitudes, as has been inscribed below.

Eutocius

Aristotle, as well as Ptolemy, following him, most noble Petrus, says that inclination is a common genus of weight and lightness. But Timaeus, in the hands of Plato, says that every inclination comes to be from weight, since he believes that weight is a privation of lightness. It is possible for lovers of leaning to read through the opinions from the book On Inclinations composed by Ptolemy and from the natural works of Aristotle and from the Timaeus of Plato and those who composed commentaries on these things. But Archimedes in this work believes that the center of inclination of a plane figure is that hanging from which it remains parallel to the horizon, but the the center of inclination or weight of two or more planes is that suspended from which the balance is parallel to the horizon.

For example, let there be a triangle, ABG and some point, D, in the middle of it, hanging from which it remains parallel to the horizon. And so, it is clear that parts [A,] B, G will incline equally to the horizon, and neither will incline more to the horizon than the other.

Similarly when AB is also a balance and magnitudes A, B are suspended loosely from it, if the balance, suspended from G holds parts A, B, equally inclining, it remains parallel to the horizon, and G will be the center of the suspension of magnitudes A, B.

I think Geminus spoke well about Archimedes, in saying that he calls the axioms postulates. For “that equal weights from equal lengths incline-equally” is an axiom, as are the rest, and all are clear to those who examine them moderately.(*1*)

“When equal and similar plane figures fit on one another,” he says, “the centers of weights also fit on one another.” For all their parts fit onto all.

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“The centers of weights of unequal but similar figures will be similarly positioned.” (diagram 2) As in the diagram placed below, let triangles ABG, DEZ be conceived as unequal and similar, but H as the center of weight of ABG and Q of DEZ, (diagram 3) and let AH, HG, BH, DQ, QE, QZ be joined. I say that the lines joined from points H, Q divide the angles into equals. (diagram 4) For let it come as EZ to BG so is EQ to QK, and ZQ to QL and DQ to QM, and let MK, KL, LM be joined. triangle KLM will be, in fact, similar to triangle DEZ. For since QZ is to to QL as EQ to QK, EZ is parallel to KL. So too, in fact, MK is to DE and LM to DZ. Therefore, triangle DEZ is similar to triangle KLM. Therefore, as DE is to MK, EZ is to KL and DZ to ML. Because of the similarity of triangles ABG, DEZ, as DE is to AB, EZ is supposed to BG and DZ to AG. Therefore, lines ABG are equal to MKL. Thus each line fits onto each. Therefore, triangle ABG is equal and similar to triangle KML. Thus too the center of ABG will fit onto that of MKL. But since H fits onto Q, then A, B, G fit onto M, K, L, and AH, BH, GH to MQ, KQ, LQ and make equal angles at M, K, L with those in triangle ABG. Thus, too with those in DEZ. For the same straight-line are joining from Q to M, K, L, as to D, E, Z.

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(diagram after Heiberg = general diagram) (after the alternative diagram in the Latin translation) )“In every figure whose perimeter is concave on the same sides, the center of weight must be within the figure.” What he calls “concave on the same sides” has been stated clearly by us in the introduction to On the Sphere and Cylinder (axiom 2). Since a figure having a convex perimeter on the same sides has all the parts of its plane inside and its angles, it is clear that it also has the center of the weight within the figure. For in the case of some figures the center of the figure is outside or on the perimeter. For in the case of semicircle ABG the center of the figure is H, while in the case of hyperbola DEZ the center of the figure is outside, where the diameters meet one another, as Q is. For these have been stated in the second book of the Conics of Apollonius (II 1). Nevertheless, in the case of figure ABG and in the case of DEZ, the center of weight, suspended from which the figure is clearly parallel to the horizon, is within the perimeter. For if it will be on the perimeter or outside, it will incline to one side, which is not supposed.

(diagrams with problems) The diagrams are not a little weird as is Eutocius claims. The center of weight of a semicircle is inside the semicircle, while only the center of weight of a double hyperbola may be outside.

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On the 4th

“Let D be the center of weight , if it is possible.” For it has been proved that it is on line AB. For it has been stated above that the center of two magnitudes is that suspended from which the balance holds the parts as inclining-equally, while remaining parallel to the horizon.(*e1*) Thus, accordingly, the center of magnitudes A, B is on AB.

(1) Most but not all ‘complete’ extant treatises of Archimedes have an epistolary form. Floating Bodies and the Dimension of the Circle are others. The Dimension of the Circle shows traces of heavy editing. Floating Bodies has a similar structure to that of Plane Equilibria, a first book that is largely general and foundational and a second book that is very technical and focused on a particular subject.

The book begins with some assumptions. Unlike some treatises, there is no header describing the list. Archimedes begins by saying, “We postulate...,” which Eutocius, following Geminus, takes Archimedes to label these assumptions as postulates. However, only the first three assumptions are described as postulated. The text switches from things that ‘we postulate‘ to simple statements of the assumptions. At the end, Archimedes introduces the theorems with, “When these are supposed....” We could as well take all the assumptions as hypotheses, or suppose that Archimedes means the same by ‘postulate’ and ‘hypothesis’. In other words, these are assumptions which Archimedes does not specifically label. Someone might even conclude that this shows that the list is a hodgepodge not likely to have been composed by a single author.

Cf. L. Berggren, “Spurious Theorems in Archimedes' Equilibrium of Planes: Book I,” Archive for History of Exact Sciences, 16 (1976): 87-103.

(2) It is important to emphasize that none of the assumptions in the list resemble definitions. It is a matter of great controversy that there is no definition of ‘center of weight’ in any extant work of Archimedes. Nor is there a definition of ‘inclination’ or of ‘equal-inclination’ either. The reason that the lack of definition of ‘inclination’ and of ‘equal-inclination’ causes less controversy is perhaps that the assumptions provide a fairly good, though not quite adequate, contextual definition. In any case, the postulates, hypotheses, or assumptions that open Equilibria of Planes do not include definitions, and that may be the entirety of the issue, as dissatisfying as that may be. However, the lack of definition does make it difficult to see the logical structure of this part of Archimedes' argument.

Cf. Eutocius' note.
Cf. L. Berggren, “Spurious Theorems in Archimedes' Equilibrium of Planes: Book I,” Archive for History of Exact Sciences, 16 (1976): 87-103.
E.J. Dijksterhuis. Archimedes, trans. by C. Dikshoorn with a forward by Wilbur Knorr (Princeton: Princeton University Press, 1987)

back to Assumptions

(3) Here is a list of the nine basic principles on the balance.

A1a. A = B & a = b => R(A,a) = R(B,b)

A1b. A = B & a > b => R(A,a) > R(B,b)

A2. R(A,a) = R(B,b) => R(A+C,a) > R(B,b)

A3. R(A,a) = R(B,b) => R(A-C,a) < R(B,b)

T1. R(A,a) = R(B,b) & a = b => A = B

T2. A > B & a = b => R(A,a) > R(B,b)

T3a. A > B & R(A,a) = R(B,b) => a < b

T3b. R(A,a) = R(B,b) & a < b =>A > B

There are 15 possible, distinct hypothesized situations. The ones marked have a condition that is covered by one of the principles.

Case Conditions Principle Result
1 A = B & a = b A1a R(A,a) = R(B,b)
2 A = B & a > b A1b R(A,a) > R(B,b)
3 A > B & a = b T2 R(A,a) > R(B,b)
4 A > B & a > b trivial 13, non-trivial R(A,a) > R(B,b)
5 A > B & a < b   indeterminate
6 A = B & R(A,a) = R(B,b) trivial 2 (A1b) a = b
7 A = B & R(A,a) > R(B,b) trivial 1 (A1a), 2 (A1b) a > b
8 A > B & R(A,a) = R(B,b) T3a a < b
9 A > B & R(A,a) > R(B,b)   indeterminate
10 A > B & R(A,a) < R(B,b) trivial 3, non-trivial a < b
11 a = b & R(A,a) = R(B,b) T1 A = B
12 a = b & R(A,a) > R(B,b) trivial 1 (A1a), 3 (T2) A > B
13 a > b & R(A,a) = R(B,b) T3b A < B
14 a > b & R(A,a) > R(B,b)   indeterminate
15 a > b & R(A,a) < R(B,b) trivial 2 (A1b), non-trivial A < B

The indeterminate cases will be handled by the general Theorems 6 and 7. The cases that are marked as trivial/non-trivial involve two case, where the consequent of the theorem is an inequality, x > y, and the proof would involve considering two cases, where x = y and where x < y. The first is trivial, the second non-trivial. One might think that these are trivial extensions of the other principles, but that would only be the case if the R(A,a) < R(B,b) were somehow monotonic (ignoring our initial description of R, the value of R(A,a) increases with respect to R(B,b) as a increases). This seems obvious and is very natural, but it is not an explicit assumption of the theory. To show this, one might want to assume an analogous principle to A2 or A3 that gives monotonicity.

P4. a = b & R(A,a) > R(B,b) => R(A,a+c) > R(B,b)

If so then to show (4) A > B & a > b => R(A,a) > R(B,b), one does the following.

Let C = A-B and d = a-b
R(A-C,a-d) = R(B,b), by A1a
R(A,a-d) > R(B,b), by A2
R(A,a-d+d) > R(B,b), by P4

Since the other two propositions, (10) and (15), are equivalent to (4), they may now be trivially proved. There are, of course, a great variety of ways to ground the three propositions.

(*4*)

There is a deep logical gap in the proof of this proposition. The proof shows that the center of weight of AB cannot be anywhere other than G, IF the center of weight is on AB. However, it does not show that the center of weight must be on AB. Eutocius sees the problem (*e5*) and thinks he has proved this in his introductory note.(*e1*) However, nothing shows that the center of weight of AB couldn't be on line DG. Of course we could rotate the plane about AB or about point G (e.g., clockwise) and then treat GD as the balance. We would then see that D could not be the center of weight, as Eutocius says. To prove this, however, we would still need an argument to say where the center of weight is of magnitude AB, or rather parts of AB on each side of D, to show the two parts (each from A and B) to show that they do not balance at D. And that requires us to be able to determine where the center of two equal magnitudes is, what the theorem sets out to prove. A much more complicated argument might do the job, but we need to assume that the center of weight is the intersection of any two balance points..