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Codex Atlanticus — 0395 r

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© Veneranda Biblioteca Ambrosiana, Milan

(a) Figura colle lettere e i numeri:

z-2-a1b-88 h -8 1g4e
... [a]nchora altrimenti chome e in che modo si truovi infra 2 linie 2 altre linee in tal modo ch'elle si chontinuino infralloro tutte sechondo la proporzione chontinua. Porrò adunque 2 linee a b e b g. L'angholo loro sia recto. Di poi chontinuirò a g e farò sopra la linea a g del triangholo a b g il cerchio a b g. Di poi menerò dal punto a la perpendiculare ad e di poi la linea ab e menerolle amendun sechondo la rettitudine infino al punto z, e di poi farò passare la reghola1, che si muova sopra il punto b e non si separi2 da quello e chosì taglierò 2 linee, cioè d z e d e, e non rimanghi di muoversi infino a tanto che quello che chade di quelle infra 'l z b3 sta iguale a quello che chade infra l'e h. Adunche la linea z h b è iguale alla linea e h b. Adunche la multiplichatione del z b in z h è iguale alla multiplicatione del b e in h e; ma e b in h e è iguale al d e in g e, e b z in z h4 è iguale a quello ch'è fatto del d z in z a. Adunque la multiplichatione del d z in d e5 è chome6 la proportione del a z al g e; ma lle proportione fatte del e d del d z è chome la proportione del a z del g e7; e anchora chome la proportione del g e del g b è quello che volemo mostrare. Provasi questa fighura per la penultima del t8 d'Euclide per la similitudine de' triangoli.
In questo modo se il g b si parte alchun numero e a b sia posto uno, sarà a z radice chubicha del g b, in però che, chome egl'è provato, chosì è in proportione b a al a z, chome a z al g e, e chosì chome g e al g b. Adunque infra 'l g b e g a già sono chadute 2 linee nella chontinua proportionalità, cioè la linea g e e la linea a z, e a z è due9. Adunque seghuita essere radice chubicha, chome volavamo.

(b)

12000 ¼ 16384
4
1
04384
16384
12000

(a) Figure with the letters and numbers:

z-2-a 1 b-8 8 h -8 1 g 4 e
... again, otherwise, how and in what manner one may find between 2 lines 2 other lines in such a way that they all continue among themselves according to continuous proportion. I shall therefore set down 2 lines, a b and b g. Let their angle be right. Then I shall continue a g, and I shall make upon the line a g of the triangle a b g the circle a b g. Then I shall draw from the point a the perpendicular a d, and then the line a b, and I shall extend them both in a straight line as far as the point z; and then I shall pass the ruler, which moves upon the point b and does not part from it, and thus I shall cut 2 lines, namely d z and d e, and let it not cease to move until that which falls of them within z b is equal to that which falls within e h. Therefore the line z h b is equal to the line e h b. Therefore the multiplication of z b by z h is equal to the multiplication of b e by h e; but e b by h e is equal to d e by g e, and b z by z h is equal to that which is made of d z by z a. Therefore the multiplication of d z by d e is as the proportion of a z to g e; but the proportions made of e d and d z are as the proportion of a z to g e; and also as the proportion of g e to g b, which is what we wished to show. This figure is proved by the penultimate [proposition] of the [book] t of Euclid, by the similitude of the triangles.
In this way, if g b be divided [into] some number and a b be set as one, a z will be the cube root of g b; because, as is proved, so it is in proportion, b a to a z as a z to g e, and so as g e to g b. Therefore between g b and g a there have already fallen 2 lines in continuous proportionality, namely the line g e and the line a z, and a z is two [?]. Therefore it follows that it is the cube root, as we wished.

(b)

12000 ¼ 16384
4
1
04384
16384
12000

Two Mean Proportionals and the Cube Root

A sliding-ruler (neusis) construction with a triangle and circle to find the cube root

Geometry Arithmetic & Proportion

The upper fragment sets out a construction for inserting two mean proportionals between two lines in continuous proportion: on a right triangle a b g Leonardo circumscribes the circle a b g, draws the perpendicular a d, and slides a ruler pivoting on point b until equal segments are cut, so that a z becomes the cube root of g b. He justifies the result by the similitude of triangles and the penultimate proposition of a book of Euclid, then gives a worked numerical case. The lower fragment carries related triangle diagrams and a calculation (12000, 16384, one quarter); a cut-out drawing of a pointing hand and a small dark disk are mounted upon it, features the transcription does not cover.

On this page

Inserting two mean proportionals by a sliding ruler

Setting two lines a b and b g at a right angle, Leonardo joins a g and circumscribes the circle a b g, drops the perpendicular a d, and extends the lines to a point z. A ruler pivoting on b is slid until the segment falling within z b equals that within e h, cutting the two mean proportionals; by the similitude of the triangles a z is the cube root of g b.

The cube root through continuous proportion

With a b set as one, a z is the cube root of g b, because in proportion b a is to a z as a z is to g e and as g e is to g b, so two lines have fallen between g b and g a in continuous proportionality. The lower fragment adds a worked calculation involving 12000, 16384 and a quarter.

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