Squaring the circle: lunes, rosettes and curvilinear octagons
A page of quadrature figures worth 'the greatest square of the greatest circle'
A densely packed sheet of quadrature experiments arranged in columns, in which Leonardo repeatedly draws circles filled with squares, curvilinear octagons and star-rosettes and asserts that once the 'four maxima' are removed the remainder is worth the greatest square of the greatest circle. He works with proportional relations such as a great circle being worth four lesser circles, removes and replaces lune-like portions, and reduces mixed curved-and-straight surfaces to squarable figures. The lower-left corner carries the single word 'Abbaco'. Much of the writing is small mirror-script and partly faded.
On this page
Draw curved figures from a square, leaving a square
Leonardo sets the programme of the sheet: from a square let curved things be drawn out and let the remainder stay square, and within a circle let various curvilinear surfaces be drawn, mixed with rectilinear ones, so the remainder stays squarable. The columns that follow work this idea through many figures.
The four maxima and the greatest square
A recurring refrain across the diagrams states that where the four maxima are missing the remainder is worth the greatest square of the greatest circle. In one eight-pointed figure the labelled ab is squared in itself and cd is likewise squarable.
A great circle worth four lesser circles
In the cancelled figure the parallel made from the greater circle is worth a quarter of the whole circle, and that quarter is worth the lesser circle. Leonardo removes the middle circle, renders the same value with the greatest circular parallel outside, then removes the middle parallel worth the four maxima so the remainder stays square. Elsewhere the greatest circle is worth four least circles abcdefhnl, of which two are removed.
Removing two least circles and the half maxima
For the circle o - iah - mn - LbK - p the half is missing once the two least circles are removed, and it also lacks the half of the four maxima for want of op. Replacing them in the least circles nm and then removing the two maxima hiKL leaves the two triangles ab.
