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© Veneranda Biblioteca Ambrosiana, Milan

In alto dal centro del margine verso sinistra:

* Io leverò a, che val tanto quanto li ripieni delle 2 curve opposite.

Figura colle lettere:

a -b -c

a b è pari al b c. Leva b all'una e all'altra; resta a di 3 lati curvi equale al c d'un sol lato curvo.

Figura colle lettere:

a — n — m

a n è pari al n m. Leva n all'uno e all'altro; resta m equale al a1.

A destra due figure con:

K – h a – b g –d c - e-f; n-tS-m

* Io ho la superfizie a g b equale alla superfizie e f, onde io presterrò2 un altro conio al lato retto K h, e l'apirrò con 3 lato retto; e a[l]trettanto darò al conio e f. Di poi sopra porrò la lu[nola] ...

a b d è pari al n di sotto, c d e è pari al m di sotto, n m infra lor sono equali. Ora io apro n 'n un de' sua lati retti e lo riempio col conio g c, e questo dimando parte prestata, e questo sol fo per allargare la fronte a esso n tanto che di punto riceva la curva ovver portione m; e così presto altrettanta superfizie al m, che di sopra si dimanda f e così ho fatto due superfizie equali, e tolgo la porzione c d all'una superfizie e l'altra. Resta a g b di 3 lati curvi equale al e f di 3 lati rettilinii; e per essere la porzione a doppia alle 2 porzioni delle 2 concavità opposite, leva a e riempi poi le 2 concavità di tanta superfizie che basti, e arai fatto una superfizie di 4 lati retti.

Metà inferiore, a destra. Triangolo, e quindi lo stesso con tre cerchi generati dai suoi lati, con:

a -cb

* Se farai l'ortogonio di 2 lati equali, che non possano essere se non quegli che contengan l'angol retto, li cerchi che si faran diamitri di tali lati, il maggiore sarà equale a li 2 minori.

* È bon per fare un circolo duplo a un altro.

A sinistra, due figure con:

t; K—h—ba — c—d

Io ho qui la superfizie a b equale al triangolo d, e la porzione a è equale alli 2 ripieni delle curve opposite4. Ora io apirrò la fronte retta K h e presterrolle il conio rettilinio t e altrettanto conio presterrò al d; e così non ho isviato la equalità delle 2 superfizie e tal conio apirrà tan[to] la fronte Kh ch'ella sarà capace di ricevere la mezza lunola quadrabile; la quale lunola, essendo intera, sarà equale al triangolo d co' sua prestamenti5 . (Nel margine figura di lunola colle lettere e-f S).

E farò che tal mezza lunola arà una porzione di fori in S equale al tutto o alla metà della porzione a (Figura colle lettere: q-p-r-om-n) e leverò essa tutta o essa mezza porzione a del suo sito e riempironne il lato della concavità fatto della gibbosità d'essa mezza lunola; e così arò fatto una superfizie n equale al d, e leverò m e metterollo in o e resterà medesimamente r n o equale al d. Restami la concavità p alla quale aggiungo altrettanta aria 6 e sarà q equale al p: e se …

Quattro figure colle lettere:

t; S; o b; n - r

Levo n e la metto in b e levo r e lo metto in o, e ho una superfizie di 4 lati retti e una curva equale alla mezza lunola quadrabile.

Andando verso destra. Figura colle lettere:

o - e - f n r a – g c b

Due figure colle lettere:

q – p – t o r - S; x

Figura colle lettere:

e d b c a

E trarrò7 il rimanente della mezza lunola quadrata e 'l rimanente di tale quadrato sarà equale a essa porzione dell'ottava parte d'esso cerchio8, onde spiccherò col lato curvo una porzione equale alla ...

At the top, from the centre of the margin toward the left:

*I will remove a, which is worth as much as the fillings of the 2 opposite curves.

Figure with the letters:

a -b -c. ab is equal to bc. Remove b from the one and the other; there remains a, of 3 curved sides, equal to c, of a single curved side.

Figure with the letters:

a - n - m. an is equal to nm. Remove n from the one and the other; there remains m equal to a.

At the right, two figures with:

K - h a - b g -d c - e-f; n-tS-m. *I have the surface agb equal to the surface ef, whence I will lend another wedge to the right side Kh, and I will open it with a straight side; and I will give as much to the wedge ef. Then over it I will place the lune ... abd is equal to the n below, cde is equal to the m below, and n and m are equal among themselves. Now I open n on one of its straight sides and I fill it with the wedge gc, and this I call the lent part, and I do this only to widen the front of that n so much that it may exactly receive the curve or portion m; and thus I lend as much surface to the m, which above is called f, and so I have made two equal surfaces, and I remove the portion cd from the one surface and the other. There remains agb, of 3 curved sides, equal to ef, of 3 rectilinear sides; and since the portion a is double the 2 portions of the 2 opposite concavities, remove a and then fill the 2 concavities with enough surface, and you will have made a surface of 4 straight sides.

Lower half, at the right. A triangle, then the same with three circles generated by its sides, with:

a -cb. *If you make the right-angled [triangle] of 2 equal sides, which can only be those that contain the right angle, then of the circles made with such sides as diameters, the larger will be equal to the 2 smaller. *It is good for making one circle double another.

At the left, two figures with:

t; K-h-ba - c-d. Here I have the surface ab equal to the triangle d, and the portion a is equal to the 2 fillings of the opposite curves. Now I will open the straight front Kh and lend it the rectilinear wedge t, and I will lend as much of a wedge to d; and thus I have not deviated from the equality of the 2 surfaces, and such a wedge will open the front Kh so much that it will be able to receive the squarable half-lune; which lune, being whole, will be equal to the triangle d with its lendings. (In the margin a figure of a lune with the letters e-f S.) And I will make that half-lune have a portion of holes at S equal to all or to half of the portion a (figure with the letters: q-p-r-om-n), and I will remove all or that half of the portion a from its place and fill with it the side of the concavity made by the convexity of that half-lune; and thus I will have made a surface n equal to d, and I will remove m and place it in o, and likewise rno will remain equal to d. There remains to me the concavity p, to which I add as much air [void], and q will be equal to p: and if ...

Four figures with the letters:

t; S; o b; n - r. I remove n and place it in b, and I remove r and place it in o, and I have a surface of 4 straight sides and one curved side, equal to the squarable half-lune.

Going toward the right. Figure with the letters:

o - e - f n r a - g c b

Two figures with the letters:

q - p - t o r - S; x

Figure with the letters:

e d b c a. And I will draw out the remainder of the squared half-lune, and the remainder of such a square will be equal to that portion of the eighth part of that circle, whence with the curved side I will detach a portion equal to the ...

Lunes, Wedges and the Circle on the Hypotenuse

Turning three-curved-sided figures into four straight-sided ones, and a circle equal to two smaller circles

Geometry

A crowded working leaf on squaring lunes, where Leonardo introduces 'lent' wedges (conii) that he inserts to widen a figure's front so it can receive a curved portion, always preserving the equality of two surfaces. By this bookkeeping of added and removed parts he turns figures of three curved sides into ones of four straight sides. Lower right, a right-angled triangle with circles built on its sides carries the theorem that the circle on the larger side equals the two on the smaller.

Having surface agb equal to ef, Leonardo lends a wedge (conio) to the right side Kh to widen it enough to receive the curve or portion m, and lends an equal wedge to ef so the two surfaces stay equal. Removing portion cd from each leaves agb of 3 curved sides equal to ef of 3 straight sides; since portion a is double the two opposite concavities, removing a and filling the concavities yields a surface of 4 straight sides.

Circle on the hypotenuse equals the two on the legs

For a right triangle of two equal sides containing the right angle, the circles drawn on those sides as diameters obey a rule: the larger will be equal to the two smaller. Leonardo notes this is good for making one circle double another.

A half-lune made equal to a triangle

Surface ab is taken equal to triangle d, and portion a equal to the two fillings of the opposite curves. By opening the front Kh with a rectilinear wedge t (and lending an equal wedge to d), Leonardo makes Kh wide enough to hold the squarable half-lune, which whole would equal triangle d with its lendings; portions are then shifted (m into o, and so on) to keep the areas equal.

A four-straight-sided surface from a half-lune

Moving n into b and r into o, Leonardo obtains a surface of four straight sides and one curved side, equal to the squarable half-lune. It is one of several rearrangements that reduce the curved figure toward a square.

Reader's Edition — described with AI from the e-Leo transcription.

In alto dal centro del margine verso sinistra:

Figura colle lettere:

Figura colle lettere:

A destra due figure con:

Metà inferiore, a destra. Triangolo, e quindi lo stesso con tre cerchi generati dai suoi lati, con:

A sinistra, due figure con:

Quattro figure colle lettere:

Andando verso destra. Figura colle lettere:

Due figure colle lettere:

Figura colle lettere:

At the top, from the centre of the margin toward the left:

Figure with the letters:

Figure with the letters:

At the right, two figures with:

Lower half, at the right. A triangle, then the same with three circles generated by its sides, with:

At the left, two figures with:

Four figures with the letters:

Going toward the right. Figure with the letters:

Two figures with the letters:

Figure with the letters:

Lunes, Wedges and the Circle on the Hypotenuse

On this page

Squaring a lune with 'lent' wedges

Circle on the hypotenuse equals the two on the legs

A half-lune made equal to a triangle

A four-straight-sided surface from a half-lune