On the Equilibrium of Planes (Ἐπιπέδων ἰσορροπιῶν αʹ βʹ)¶
De Planorum Aequilibriis, Books I & II — Archimedes of Syracuse (ed. T. L. Heath, 1897, pp. 189–220; J. L. Heiberg, Vol. II, pp. 142–241)
Note
Edition Track (English): Public-domain translation from heath:1897 (pp. 189–220), verified against 300 DPI scans of Heath (1897) and collated with Heiberg's critical Syracusan Doric text (heiberg:1881, Vol. II, pp. 142–241). Switch tracks above for the Summary or Notes.
Book I — Axiomatic Statics, the Lever & the Triangle Centroid¶
Postulates (Αἰτήματα, 1–7)¶
“I postulate the following (Αἰτούμεθα τάδε):”
Postulate 1: Equal weights at equal distances are in equilibrium, and equal weights at unequal distances are not in equilibrium but incline towards the weight which is at the greater distance.
Postulate 2: If, when weights at certain distances are in equilibrium, something be added to one of the weights, they are not in equilibrium, but incline towards that weight to which the addition is made.
Postulate 3: Similarly, if anything be taken away from one of the weights, they are not in equilibrium, but incline towards the weight from which nothing was taken.
Postulate 4: When equal and similar plane figures coincide with one another, their centres of gravity likewise coincide.
Postulate 5: In figures which are unequal, yet similar, the centres of gravity will be similarly situated. By points similarly situated in relation to similar figures I mean points such that, if straight lines be drawn from them to the equal angles, they make equal angles with the homologous sides.
Postulate 6: If magnitudes at certain distances be in equilibrium, (other) magnitudes equal to them will also be in equilibrium at the same distances.
Postulate 7: In any figure whose perimeter is concave in one and the same direction, the centre of gravity must be within the figure.
Propositions 1–5: Elementary Symmetry of the Balance¶
Proposition 1: Weights which balance at equal distances are equal (proved by Post. 3 and Post. 1).
Proposition 2: Unequal weights at equal distances will not balance, but will incline towards the greater weight (Post. 2, Post. 1).
Proposition 3: Unequal weights will balance at unequal distances, the greater weight being at the lesser distance (and conversely).
Proposition 4: If two equal magnitudes have not the same centre of gravity, the centre of gravity of both taken together is at the middle point of the straight line joining their centres of gravity.
Proposition 5 (Corollaries 1–2): If three (or any odd or even number of) equal magnitudes have their centres of gravity on a straight line at equal distances, the centre of gravity of the system coincides with that of the middle magnitude (or the midpoint between the two middle magnitudes).
Propositions 6 & 7: The Law of the Lever¶
Let A, B be commensurable magnitudes with centres of gravity A, B, and let DE be a straight line divided at C so that . Place A at E and B at D. Produce ED so that and ; then has C as its midpoint! Since A and B are commensurable, divide LH () and HK () into equal segments corresponding to a common measure N of A and B. Placing equal unit weights N at the midpoints of all equal segments along LK, the first group has its centre of gravity at E (midpoint of LH, by Prop. 5) and equals A; the second group has its centre of gravity at D (midpoint of HK) and equals B; while the whole uniform array on LK balances at its midpoint C (Post. 6, Prop. 5). Therefore A at E and B at D balance about C.
Let , where A and B are incommensurable. If Aa placed at E does not balance B placed at D, either Aa is too great to balance B or not great enough. Suppose Aa is too great. Take from Aa a magnitude less than the excess by which it fails to balance B, yet such that the remainder A is commensurable with B (Eucl. V. Def. 4). Then since A, B are commensurable and , A at E will not incline the beam at E down, by Prop. 6, rather D will incline downward—contradicting that insufficient weight was removed to stop E from inclining! Similarly B cannot incline downward. Hence incommensurable magnitudes likewise balance at reciprocally proportional distances.
Propositions 8–12: Remainders, Parallelograms & Similar Triangles¶
Proposition 8: If AB be a magnitude whose centre of gravity is C, and AD a part of it whose centre of gravity is F, then the centre of gravity G of the remaining part DE lies on FC produced so that .
Propositions 9 & 10 (with Alternative Proof): In any parallelogram (Παντὸς παραλληλογράμμου τὸ κέντρον τοῦ βάρεός ἐστιν…, Heiberg II, 162), the centre of gravity lies on the straight line joining the middle points of opposite sides (Prop. 9), and coincides with the point of intersection of its diagonals (Prop. 10).
Propositions 11 & 12: If two triangles be similar, and points within them be similarly situated with respect to them, if one point is the centre of gravity of the first triangle, the other point is likewise the centre of gravity of the second triangle.
Propositions 13 & 14: The Triangle Centroid X(2)¶
Proposition 13. In any triangle, the centre of gravity lies on the straight line joining any angle to the middle point of the opposite side.
Proposition 14. It follows at once that the centre of gravity of any triangle is at the intersection of the lines drawn from any two angles to the middle points of the opposite sides respectively.
Let ABC be a triangle, and D the middle point of BC. Join AD. If the centre of gravity H does not lie on AD, draw HIparallelBC meeting AD in I. Bisect DC, BD repeatedly and draw parallels to AD from the division points, inscribing in triangleABC a union of parallelograms symmetrical about AD such that the remaining small corner triangles (all similar to triangleABC) have a combined area having to the inscribed parallelograms a ratio less than (Eucl. X. 1, Eucl. VI. 19).
By Prop. 9, every inscribed parallelogram has its centre of gravity on AD; therefore the centre of gravity P of all the parallelograms taken together lies on AD. Let Q be the centre of gravity of the remaining corner triangles taken together; by Prop. 8, P, H, Q are collinear and:
Draw KLparallelAD through Q meeting BC, AC; since all corner triangles lie between AD and its bounding tangents, Q cannot lie beyond the boundary parallel through C, implying , contradicting our construction! (Archimedes also supplies a pure similarity proof decomposing :g:`triangle ABC` into a central parallelogram :g:`FKEC` and two half-scale triangles :g:`triangle AFE`, :g:`triangle EBD` via :just:`Prop. 10` and :just:`Post. 5`.)
Applying Proposition 13 to two medians AD and BE yields Proposition 14: the three medians concur at the unique barycenter G dividing each median in the ratio (geometor:etc, Kimberling center X(2) with homogeneous barycentric coordinates ).
Proposition 15: Centre of Gravity of a Trapezium¶
Proposition 15. In any trapezium having two sides parallel to one another, the centre of gravity lies on the straight line joining the middle points of the parallel sides, at the point dividing it so that the segment adjacent to the smaller parallel side is to the remaining segment as twice the greater parallel side together with the smaller is to twice the smaller together with the greater:
Proved by producing the non-parallel sides to meet at H, decomposing the trapezium as the difference of two triangles triangleHBCtriangleHAD whose centroids trisect HF and HE (Prop. 14), and applying the remainder formula (Prop. 8).
Book II — Centres of Gravity of Parabolic Segments¶
In Book II (10 Propositions), Archimedes combines Book I with the Quadrature of the Parabola (archimedes:quadrature-of-parabola) to determine the exact centre of gravity of any parabolic segment and parabolic frustum:
Proposition 1: Two parabolic segments placed on a balance are treated as plane laminae via Book I, Props. 6–7.
Propositions 2–7: Recursive inscription of triangles (“known manner” triangleABC + triangleADB + triangleBEC + …) proves that the centre of gravity of any parabolic segment lies on its diameter AO (Prop. 2), that similar-like exhaustion polygons have centres of gravity approaching that of the segment (Props. 3–6), and that the ratio in which the centre of gravity divides the diameter is constant across all parabolic segments (Prop. 7).
Proposition 8 (Fundamental Centroid of the Parabola):
If AO be the diameter of a parabolic segment, and G its centre of gravity, then (that is, from the vertex A and from the base midpoint O).Let identically for every parabolic segment (Prop. 7). Decompose the main segment BAb (area ) into its inscribed triangle triangleBAb (centroid at from A, Book I, Prop. 14) and two flanking sub-segments on AB, Ab (combined area ), whose diameters have length measured from the line through A and whose centroids also divide their diameters in the identical unknown ratio ! Taking moments along AO via Book I, Props. 4, 6–8 yields a linear equation in whose unique solution is , i.e. .
Proposition 9: Arithmetic lemma on six terms in continued proportion used to reduce the frustum moment equation.
Proposition 10: Exact division ratio for the centre of gravity of the parabolic frustum cut off between any two parallel chords.