Quadrature studies: transforming curved and straight surfaces
Dense sheet of lunes, bi-angles, octagons and star figures on equating curvilinear and rectilinear areas
One of Leonardo's most crowded geometry sheets, split across a horizontal fold and covered with small figures: circles with inscribed octagons and squares, crosses, sectors, star and rosette figures, and lettered lune constructions. The heading declares the theme: transforming curvilinear surfaces into rectilinear ones and back, among surfaces of known proportion. Short lettered propositions repeatedly argue by equal areas that a figure is 'squarable' (quadrabile) or 'unsquarable', and that removing equal portions leaves equal remainders. A general rule of proportion runs through it: half the double equals the whole sub-double, a third of the triple the whole sub-triple, and so on to infinity, closed by small arithmetic figures.
On this page
On transforming curvilinear into rectilinear surfaces
The heading sets the whole program of the sheet: the transformation of curvilinear surfaces into rectilinear ones, and of rectilinear back into curvilinear, that is, of surfaces conditioned by known proportions.
Octagon equal to the parallel circle and to triangle m
This octagon is worth the parallel (ring-shaped) circle and is worth the triangle m. The adjoining sector is the eighth part of a circle eightfold the smallest.
The eighth of an eightfold circle equals the whole other
Of two circles eightfold one to the other, the eighth part of the one is worth the whole of the other. The tailed-circle figures with added sectors make the equality visible.
Halving similar surfaces by proportion, to infinity
Of similar surfaces, half of the double is worth the whole sub-double, a third of the triple the whole sub-triple, a fourth of the quadruple the whole sub-quadruple, and so on to infinity. Rules follow on when removing a squarable or unsquarable part leaves a squarable or unsquarable remainder.
Removing the greatest bi-angle leaves the greatest square
If from an unsquarable figure you take an unsquarable part, the remainder is squarable. This is proved in circle a b c: taking away the greatest bi-angle b leaves a c equal to the greatest square that can fit in that circle, as the second figure shows.
The four greatest portions and the adversary's concession
Of the four greatest portions of a circle, four others equal to them can be found, as with portions a b in square c d e f. If an opponent concedes only portion a in the half-square e c d, he is forced to admit that if triangle e c d contains portion e a d, then triangle e d f contains portion e b d.
