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of Infinitary Arguments
We go through the theorem, with the additions due to the new reading of the palimpsest the Method, by Reviel Netz and Ken Saito (Sciamus, 2001). I shall then make some observations.
Let there be a cylinder enclosed in a square based prism. The base of the prism is a, b, g, d. 
Take a plane from the top of the cylinder tangent to the prism and cut the prism and cylinder so that it cuts through the base of the cylinder at its tangents to the prism, as shown, ae = ed. 
The result is a segment of the cylinder, which Archimedes calls a hoof. 
On the base of the hoof, draw a parabola hze, where z is on the tangent of the cylinder and prism. 
We will now take an arbitrary ordinate of the parabola, ls, and parallel to the diameter of the parabola kz, we draw mlxn with perpendiculars at to the top of the hoof (cylindrical section) and at n to the top of the prism. 

It is fairly easy to see that (since they are similar triangles):
the triangle in the prism on mn : the triangle in the cylindrical section on mxBut, by the basic property of a parabola (note hk = kz) mn : nl ≈ hk^{2} : ls^{2}Hence, mn : nl ≈ hk^{2} : ls^{2} ≈ mn^{2} : ls^{2}It follows that, mn : mlHence, mn : ml ≈ mn^{2} : mx^{2}That is, where all are distance ls from kz: the straight line parallel to kz : the line in the parabola ≈ the triangle in the prism : the triangle in the cylindrical segment 
Let the triangles in the prism be A_{1}, ..., A_{n}, ... Let the lines in the rectangle dh be B_{1}, ..., B_{n}, ... Let the lines in the parabola be C_{1}, ..., C_{n}, ... Let the triangles in the cylindrical segment be D_{1}, ..., D_{n}, ... Since A_{i} = A_{j} and B_{i} = B_{j}, it trivially follows that A_{i} : A_{j} ≈ B_{i} : B_{j}Archimedes has also shown that: B_{i} : C_{i} ≈ A_{i} : D_{i} We now know that Archimedes claims:
From these it follows by the Conoids and Spheroids 1, (B_{1} +...+ B_{n}, ...) : (C_{1} +...+ C_{n}, ...)
all the lines in rectangle dh parallel to kz : all the lines in the parabola ≈ all the triangles in the prism : all triangles in the section of the cylinderOr, if we think of the magnitudes as 'filled up' by the slices', the triangle based prism : the cylindrical section or hoofHence, since the prism on the square is quadruple the triangle based prism, the whole prism on the square : the hoof ≈ 12 : 2 = 6 : 1. 
Several points about the argument:
Archimedes has no difficulty in saying that