**Archimedes,**__On the Equilibria of Planes__Props. 6-7- translated by Henry Mendell (Cal. State U., L.A.)

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On the Equilibria of Planes, Table of Contents

Introduction and I Props. 1-5

I Prop. 8

6. Commensurable magnitudes incline-equally when they have the
same ratio for their lengths inversely as their weights.

(general diagram)

(diagram 1) Let there be commensurable magnitudes A, B whose centers are A, B and let there be a length ED, and let it be as A is to B so is length DG to length GE. (diagram 2) One must prove that when A, B are put together on each side, the center of weight is G.

(diagram 2) For since as A is to B so is DG to GE, but A is commensurable with B, and therefore GD is commensurable with GE, i.e., a straight line with a straight line. (diagram 3) Thus there is a common measure of EG, GD. Let it be N, and (diagram 4) let each of DH, DK lie equal to EG, and EL be equal to DG. (diagram 5) Thus LE is equal to EH. Therefore, LH is double DG, and HK is double GE. Thus N measures each of LH, HK, since also it measures their halves. And since as A is to B so DG is to GE, but as DG is to GE so is LH to HK, since each is respectively double. And therefore as A is to B so is LH to HK. (diagram 6) And as many times as LH is of N, let A be so many times of Z. Therefore, as LH is to N, so is A to Z. And as KH is to LH, so is B to A. (diagram 7 = general diagram) Therefore, ex aequali, as KH is to N so is B to Z. Therefore, KH is an equal multiple of N as B is of Z. But A was also proved as being a multiple of Z. Thus Z is a common measure of A, B. And so, LH having been divided in parts equal to N, and A into parts equal to Z, while the segments in LH which are equal in magnitude to N, will be equal in number to the segments in A, which are equal to Z. (diagram 8) Thus, if a magnitude equal to Z is placed on each of the segments in LH, having the center of weight on the middle of the segment,(*1*) then the totality of magnitudes are equal to A and E will be the center of the weight composed from all of them. For the totality will be even in number and the ones on each side of E will be equal in number since LE is equal to HE.(*2*) (diagram 9) Similarly, it will be proved that even if a magnitude equal to Z is placed on each side of the segments in KH, having its center of weight on the middle segment, then the totality of the magnitudes will be equal to B, and D will be the center of the weight composed from all of them. (diagram 10) And so A placed on it will be at E, while B placed on it will be at D. (diagram 11) In fact, there will be magnitudes equal to one another lying in a straight line whose centers of weight will be equally distant from one another, while being even in multitude. And so it is clear that the bisection of the line having the centers of the magnitudes in between is the center of the magnitude composed from all of them. (I Props. 4 and its corollary, Corollary 2 after Prop. 5) Since LE is equal to GD, and EG to DK, therefore the whole LG is also equal to GK. Thus, point G is the center of the weight of the magnitude composed from all of them. (diagram 10) Therefore, when A is placed at E and B at D, they incline-equally at G. (Assumption 6) (*3*)

(*1*) It is crucial that the magnitudes have the same configuration as in one of I prop. 4, prop. 5, corollary 1, or corollary 2. Only then can one apply Assumption 6. It is a reasonable question whether the proof requires that the centers of weight of every object suspended from the beam be on the beam. Elsewhere in Archimedes, this is explicitly not the case (Quadrature of the Parabola props. 6-15). However, *Equilibria of Planes *is concerned only with findng properties of centers of weight. So one might conceive the argument as setting up the line EH and showing where the center of weight of areas A and B and then, only at the end, do we actually place the weights on the line, with the centers at E and H. Heiberg's own diagram suggests this interpretation. Note that it is assumed that we can place each weight on the line (beam) with its center of weight at any desired point.

If one wished to extend the proof to the case where the center of weight is not on the beam, one would then have to prove or assume:

It makes no difference for the assumptions 1-4, 7 whether the center of weight is on the perpendicular to the horizon going through the balance beam at

aor is on the beam at pointa.

Archimedes alludes to a proof of this in Quadrature of the Parabola 6.

Cf. E. Mach, *The Science of Mechanics* (trans. T.J. McCormack, Chicago, 1902); E.J. Dijksterhuis. *Archimedes*, trans. by C. Dijksterhuis with a forward by Wilbur Knorr (Princeton, 1987)

(*2*) The number of weights is even because N measures EG and DG, but B : Z = KH : N = 2*DG : N, so that also A : Z = LH : N = 2*GD : N.

(*3*) It is crucial for the structure of the book, and the proof of the next theorem, that Archimedes prove this theorem, which may not be possible given his assumptions:

If A and B are commensurable and placed at E and D with lengths from G, such that A : B > EG : DG, then they will incline to A.

(diagram 1) Suppose that they do not incline to A and that A : B > EG : DG. If they incline equally, then A : B = EG : DG, contrary to the hypothesis. So they incline to B. (diagram 2) If EG is commensurable with DG, add to A an amount X such that AX inclines equally with B. Hence, AX : B = EG : DG. But A : B > EG : DG, so that A : B > AX : B. Hence AX < A.

(diagram 2) Suppose that EG, DG are not commensurable (here things get not a little messy and may need to mimic some of I Prop. 7, cf. introductory note 3). Again, A : B > EG : DG. (diagram 3) So pick a point P such that A : B > EP : PD > EG : DG, but where EP is commensurable with PD. Then AB incline towards A by the previous proof. A fortiori, it will incline when the we take G as the balance point.

Notes: this proof assumes that the inclination relation corresponds to a monotonic function. There is nothing to indicate that something may incline in one direction when the distances are commensurable and in the opposite direction when incommensurable. It is a natural assumption, of course, but one that needs to be made explicit. Again, see introductory note 3.

7. And then even if the magnitudes are incommensurable, they will similarly incline-equally when they have the same ratio from their lengths inversely as their magnitudes.

Let magnitudes AB, G be incommensurable as well as the lengths DE, EZ, and let AB have to G the same ratio which ED has to length EZ. I say that E is the center of weight of both AB, G. (general diagram = diagram 1)

(diagram 1) For if AB does not incline-equally when placed at Z to G placed at D, either AB is larger than it would be to incline-equally to G or not. (*e1*) (diagram 2) Let it be larger, and (diagram 3) let there be subtracted from AB a magnitude less than the excess by which AB is larger than it would be to incline-equally to G, so that the remainder A is commensurable with G (proof of this lemma, scholion to Theodosius, Sphaerica iii 9).(*e2*) And so, since magnitudes A, G are commensurable and A has a smaller ratio to G than DE to EZ, A and G do not incline-equally at distances DE, EZ, when A is placed at Z, and G at D. For the same reasons, it won't happen even if G is larger than it would be to incline-equally to AB.(*1*)

On this theorem and the history of Greek proportion theory, cf. W.R. Knorr, “Archimedes and the Pre-Euclidean Proportion Theory,” *Archives internationales d'histoire des sciences* 28 (1978), 183-244, Bernard Vitrac, Euclide, *Les Éléments: Livres V à IX*, vol. 2 (Paris, 1994), and H. Mendell, “Two Traces of Two-Step Eudoxan Proportion Theory in Aristotle: A Tale of Definitions in Aristotle, With a Moral,” *Archive for History of the Exact Sciences* 61 (2007): 3-37.

(*e1*) On the 7th:

“either AB is larger than it would be to incline-equally to G or not”

It is necessary to hear in this expression, not that G is larger in every way than AB, but that it is supposed larger than according to equal-inclination. For it is possible that even a smaller magnitude have a larger inclination because the length of the balance is much larger and makes the ratio unequal.

“let there be subtracted from AB a magnitude less than the excess by which AB is larger than it would be to incline-equally to G, so that the remainder A is commensurable with G”

It is necessary, he says, to take away some magnitude, B, from magnitude AB, that makes a remainder, A, commensurate with G and A larger than according to an equal inclination with G. It is possible to do this through what was stated at the beginning of the tenth book of Euclid’s *Elements* and in the third of the *Sphaerics* of Theodosius. It is unclear which texts in either work Eutocius refers to, unless he means in the first case that one could prove the lemma from *Elements* X 1-10 and that it is presupposed by the theorems in *Sphaerics* III or thinks that a scholion Spherics III 9 is part of the text. For the lemma, cf. scholion to Theodosius, Sphaerica iii 9.

Cf. W.R. Knorr, “Archimedes and the Pre-Euclidean Proportion Theory,” *Archives internationales d'histoire des sciences* 28 (1978), 183-244.

The proof is incomplete and problematic. Here is an outline, with R(A,*a*) the inclination of A at length *a*, R(A,*a*) = R(B,*b*) as A and B equally incline at lengths *a* and *b*, and R(A,*a*) > R(B,*b*), A and B incline towards A at lengths *a* and *b*. Commensurable(A,B) means that A and B are commensurable.

(diagram
1)

AB : G = ED : ZE

and ~Commensurable(AB,G) and ~Commensurable(ED,ZE)

Suppose R(AB,EZ)
≠ R(G,ED)

Then
R(AB,EZ)
> R(G,ED) or R(AB,EZ)
< R(G,ED)

(diagram 2)

Let
R(AB,EZ)
> R(G,ED)

(diagram
3)

Then find Commensurable(A,G) such that
R(A,EZ)
> R(G,ED)

But, A :
G < AB : G = ED : ZE

Hence, R(A,ZE) ≠ R(G,ED)

However this is not a contradiction. Since A is commensurable with G, but ZE isn't with ED isnt, we wouldn't expect them to equally incline. To get a contadiction we would need:

A :
G < ED : ZE => R(A,ZE) < R(G,ED)

Theorem I 7 does not show this. It is likely that an editor faced with a lacuna in the text drew the only correct conclusion he or she could and then didn't fill in the gap. This is not a trivial gap to fill in and does take some extra work in particular, this lemma:

Lemma: Commensurable(A,B) & R(A,a) = R(B,b) => R(A,a) < R(B,b+c)

Then istead of the last line of the proof above,

R(A,ZE) < R(G,ED)

and G was taken so that R(A,ZE)
> R(G,ED), which is impossible.

One then takes
R(AB,EZ)
< R(G,ED) and repeats the proof

Lemma: Commensurable(A,B) & R(A,a) = R(B,b) => R(A,a) < R(B,b+c)

diagram 1) Take the setup of I 6 with E the center of weight of A and D the center of weight of B, and G the center of weight of A+B, so that R(A,EG) = R(B,DG). Let the weights on GL be C. Recall that tC = the weights on GK so that R(C, LG/2) = R(weights GK, GK/2). Let PG = GK/2, so that P is the center of weight of C and let Q be the center of weight of the weights on GH = A-C, i.e., where GQ = GH/2. (diagram 2) Now, move the center of weight of B to R, so that RG > GD. Then, since C and A are commensurable with B, so are the weights on GH. Hence, the center of weight of the weights on GH+B at R will be some point S such that B : A-C = QS : SR. It is easy to see that SQ > DQ, or rather that SG > DG. Since C = A-C+B (from the original setup of I 7), it follows by I post. 1 that R(B+A-C,SG) > R(C,GP). But this arrangment is the same as R(A,EG) and R(B,GR). Hence, R(B,GR) > R(A,EG).