A page of calculations (folio 70 recto
= f.70r) from Ms.
Gal. 72, the collection of Galileo's manuscripts in the
Biblioteca Nazionale Centrale, Florence.
Larger image. Along
with the record of measurements, this page contains examples of
all the arithmetic operations, including extractions of square roots.
Some items from this page will be analyzed in detail below. 
The manuscripts allow us very unusual, if not unique, access
to the private calculations of a great scientist; it is as if we
could look over his shoulder and watch him at work.
From a mathematical
point of view they allow us to see the very state of the art
in computation in the period 16041636 when they were written.
In fact
they show that in Galileo's hands the use of baseten
placevalue notation in arithmetic had reached its modern form. The
way he performs multiplication of large numbers, long division and
squareroot extraction is exactly the way 20th century students
were taught. One exception is that Galileo never used a decimal
point, and represented numbers smaller than unity by fractions.
Another more minor one is that Galileo performed as many steps
as possible in his head, and only wrote down the essential.
Before looking at examples of Galileo's calculations, it may
be useful to examine how arithmetic was taught in his day.
Again we are fortunate to have available online, thanks to the ECHO project,
a
mathematics textbook published in 1559,
just before Galileo was born.
Cataneo (born in Siena, died c.1569) is best known as an influential theorist of architecture. But his first published work, Le Pratiche delle due prime matematiche, ("Methods of the first two kinds of mathematics," i.e., arithmetic and geometry) is considered a prototype of the modern mathematics textbook. Cataneo starts by defining the elementary operations for integers and fractions; he begins with addition and subtraction, which are standard.
Multiplication. First comes "multiplication
from memory" (mutiplicar a la memoria). Cataneo lists
the multiplication facts one needs to know by heart:
a multiplication table up to $10\times 10$ and the
first ten numbers times the numbers $12,\dots, 20, 24, 32, 48$;
he then covers multiplication of a severaldigit number
by a single digit and multiplication by powers of 10.
Cataneo presents two methods for multiplying severaldigit
numbers. The first, per crocetta (crosswise), is the
way multiplication was explained by Fibonacci. It is lovely
but since there is no trace of it in Galileo's work it is
postponed to an Appendix. The other, multiplication
per biricuocolo, is the method we teach
children today. Cataneo's digitbydigit explanation in the text
is illustrated in the margin as shown here. 
$$\begin{array}{cccccc} &&&9&7&8\\ &&&3&6&5\\\hline &&4&8&9&0\\ &5&8&6&8&\\ 2&9&3&4&&\\\hline 3&5&6&9&7&0 \end{array}$$ 

Division. For division also, Cataneo
presents several different methods. For per testa ("in
the head") division,
by single digits for example, it is enough to know the
multiplication table. Per scapezzo works with multiples
of $10$. Per ripiego, literally
"by folding" is convenient
when the divisor can be broken into small factors, so several
consecutive per testa divisions can achieve the result.

"Galleystyle" division (per galera) most likely descends from sandtable calculations where one digit would be rubbed out and replaced by another. Here during the computation a digit is crossed out and the substitute is written directly above it. The name is inherited from an earlier format in which the divisor was written below the rightmost place of the dividend and then crossed out and rewritten one place to the left with each iteration, digits always positioned vertically so as to leave no grid square empty. The tapering top and bottom of that array had some resemblance to the prow and stern of a ship, hence "galleystyle." Caetano just keeps one end, say the prow. He illustrates the method with $87654\div 53$.
$$\begin{array}{ccccc} &&0&&\\ &0&\not{2}&4&\\ 0&\not{2}&\not{3}&\not{5}&\\ \not{3}&\not{4}&\not{8}&0&5\\ \not{8}&\not{7}&\not{6}&\not{5}&\not{4}\\ &&&&\\ &&&5&3\\ &1&6&5&3\frac{45}{53} \end{array}$$ 
Square Roots. Cataneo describes a squareroot
algorithm organized like the division a danda but distinguishing
every other digit of the radicand by a (real or imaginary) dot, starting with the units'
place. Here his how he explains the extraction of the square root of
$54756$. (This better version comes from the 1546 edition).
$$\begin{array}{ccccc} 5&4&7&5&6\\ 4&\mbox{first}&\!\!\!\!\mbox{doub}&\!\!\!\!\!\!\!\!\!\!\mbox{led}&\!\!\!\!4\\\hline 1&4&\mbox{seco}&\!\!\!\!\!\!\!\!\!\mbox{nd}&\!\!\!\!46\\ 1&2&&\\\hline &2&7&&\\ &&9&&\\\hline &1&8&5&\\ &1&8&4&\\\hline &&&1&6\\ &&&1&6\\\hline &&&&0\\ \end{array}$$  $(\!\underline{~2~~3~~4~}~$

The passages in italics are tough sledding. We will return to them in analyzing Galileo's approach to this calculation.
Approximation: Cataneo remarks that when the radicand is not a perfect square (and therefore the algorithm gives a remainder), the remainder divided by twice the calculated root should be added to that root. He gives the example of $17=4^2+1$; the algorithm gives integer root $4$ and remainder $1$, so $4\frac{1}{8}$ should be taken as the approximate root of $17$. (Here Fibonacci had gone much farther, explaining that the approximating step could be repeated an adumbration of Newton's method: here $(4\frac{1}{8})^2 = 17 + \frac{1}{64}$; for a better approximation Fibonacci would add to $4\frac{1}{8}$ the remainder (negative in this case) divided by twice the root, i.e. $\frac{1}{528}$.)
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In the guts of the calculation, products and
subtractions are all performed mentally; digits
that have played their part are struck out; and
instead of dropping down the digits "given" by the dividend to the
successive remainders, Galileo tacks them on by reading diagonally up;
always filling all the squares in an imaginary grid. The result is
a replica of the prow of the "galley," but upsidedown (allowing
Galileo to link a division, as here, to previous operations). If
the digits
suppressed by Galileo are made explicit (here written
$\scriptstyle{\rm small}$), the resut is the configuration
used today.

Multiplication and square root extraction.
This next excerpt from f.70r shows the multiplication $1111\frac{1}{30}\times
500$ and the extraction of the square root of that product. The
multiplication is done as above, with $500 \times\frac{1}{30}$ reduced
to $50\times\frac{1}{3}$ and approximated as $17$, giving $555517$ as
the product.
The square root algorithm shows a fundamental
improvement of the a dandalike method explained by
Cataneo (and by Fibonacci and Pacioli before him). The earlier
authors understood that the odd and evenplaced digits of the radicand
had to be treated differently, and that each pair of radicand
digits contributes a single digit to the root. Nevertheless in the
operation the radicand digits were brought ("given") down one by one.
This resulted in a convoluted explanation like those shown above
in italics in Cataneo's root extraction.
To analyze the difference anachronistically, let us look at
the spot in Cataneo's operation where he has just brought down
the $5$. At that moment the current root is 23, and the next digit
of the radicand is $6$. Cataneo has
Now you
need to find a number which multiplied by the double of the
root you have found, i.e. the double of $23$ which is $46$,
that product can be subtracted from $185$ and from the remainder
joined with the $6$, the next digit of $54756$, can be subtracted
the product of that number [with itself]
with remainder not larger than the double of the root you will
have found. Suppose we call "that number" $x$. Cataneo's
sentence translates to the three inequalities
$$0\leq 18546x$$
$$0 \leq ((18546x)\cdot 10 + 6)x^2$$
$$((18546x)\cdot 10 + 6)x^2 \leq 2\cdot(230 + x)$$
which must all be satisfied by $x$. But if the $5$ and the $6$ are brought down together
the problem simplifies to
$$0\leq 1856(460+x)x$$
$$x~ \mbox{is the largest such number.}$$
I don't know if Galileo
was taught this improvement or figured it out himself. It appears,
slightly mangled, in the work of Georg von Peuerbach (14231461).
In Galileo's
operation the digits are brought down two by two to form a new
partial radicand, and the next digit of the root will be the largest single
digit
which, when appended to double the current root, and then used
to multiply that adjunction, is smaller than that partial radicand.
This is the way square root extraction was still taught in
elementary school halfway through the 20th century. Galileo continues the
condensed notation from long division (multiplications
and subtractions preformed mentally, and only essential digits
recorded) resulting in a galeralike picture.
If the operation is written as below with
the digits suppressed by Galileo reinserted $\scriptstyle{\rm small}$,
it could have been the work of a 20thcentury schoolchild.
The schoolchild might have continued the calculation by
inserting a decimal point and bringing down pairs of zeros to
get $745.33012...$;
Galileo uses the "remainder over twice the root"
fractional approximation: $\frac{492}{1490}$ which in
decimals would be $.330201...$.

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Reproduction forbidden in any medium. See copyright notice above. 
Galileo does not always use the fractional approximation; of over 300 square root extractions in this archive, fewer than 80 are completed in this way. Many of the numbers going into the computations come from measurements in his laboratory (there is a list of such data at the top of f.70r); it is unlikely that there were ever more than 3 or 4 significant digits at stake, so the extra precision would not have made sense. This item from f.156r has an example of the calculation of the geometric mean of two numbers, one among many in the collection. Here Galileo calculates $\sqrt{184776\times153072}$. The product gives a nice example of our biricuocolo algorithm; the root is calculated as $168178$. Note that Galileo did not write the last $8$ in the lefthand column of numbers, but only at the end of the root. Note also the prow of the "galley" taking shape in the center of the calculation. 
On five of the folios (117r, 114v, 161r, 132r, 132v), all dated
by Stillman Drake to 16081609, Galileo uses twice the root plus
one in the denominator of the fractional part (there are 8
such extractions).
This image shows an item from f.132r, where $\sqrt{7200}$ is calculated as $84\frac{144}{169}$. In fact this is the better approximation when (as in this case) the integer part plus one is closer to the root than the integer part. But on the same page Galileo calculates $\sqrt{1625}$ as $40\frac{25}{81}$ when $40\frac{25}{80}$ is better. Writing $N$ for the radicand, $a$ for the remainder and $r$ for the integer part of the root, it had been known since the Middle Ages that $$ (r+\frac{a}{2r+1})^2 \leq N \leq (r+\frac{a}{2r})^2, $$ so it is possible that on these folios Galileo needed an underestimate of the roots he was calculating. 
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Galileo published Discorsi e dimostrazioni matematiche intorno à due nuove scienze, usually called "Two New Sciences" in English, in 1638. In his discussion of trajectories, he prints a table of ranges ("amplitudes") and maximum heights ("altitudes") of projectiles all fired with the same initial speed, at angles from the horizontal varying from $1^{\circ}$ to $90^{\circ}$. The distances are normalized so that a projectile fired at $45^{\circ}$ will travel 20,000 units, and are described in terms of the semiparabola representing half the trajectory. Proposition XII explains how to compute the amplitudes, and Proposition XIII how to compute the altitudes from the amplitudes. His calculations for the second proposition are part of the Biblioteca Nazionale archive; all fortyfour (see below) are written in columns on three folios, f.103v, f.104r and f.104v. We can (almost) match his work with the sample calculation he gives in "Two New sciences," here in Stillman Drake's translation.
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To be found is the altitude of the semiparabola described at elevation $55^{\circ}$. The amplitude, from the preceding tabulation, is $9396$; half of this is $4698$, of which the square is $22071204$. Subtract this from the square of half $~$BO, which is $25000000$ and is always the same; the remainder is $2928796$, of which the square root is approximately $1710$. This, added to half$~$ BO (that is, $5000$), gives $6710$, which is the altitude BF. Galileo does not have a calculation for $55^{\circ}$ in his notes: he only treats angles from $1^{\circ}$ through $44^{\circ}$. He derives the others using the principle: if two semiparabolas have angles symmetrically spaced about $45^{\circ}$, their altitudes will be symmetric about the altitude corresponding to $45^{\circ}$ exactly, which is $5000$. The calculation for $35^{\circ}$ matches his description for $55^{\circ}$ except that the root is subtracted from $5000$ instead of being added. (There is an additional curious discrepancy: on the folio the root is calculated correctly as $1711$, whereas in the text it is given as "approximately $1710$"). 
$$\begin{array}{ccc} 5&&9\\ (& \Large\mbox{X} &)\\ 4&&7\\ 27&7&3 \end{array}$$  In this first example Cataneo applies the crocetta method to $59\times 47$. "Multiply $7$ by $9$, which makes $63$, of which you will mark $3$ in the first place & you save $6$; then multiply crosswise saying $4$ times $9$ makes $36$ & $5$ times $7$ makes $35$, which two products added together & to that sum added the saved $6$ makes $77$, of which you will write $7$ and save $7$; then multiply $4$ by $5$ & add to them the saved $7$, & that makes $27$, which you write down as shown; & you will have $2773$ for the said multiplication." 
$$\begin{array}{ccccc} 4&&5&&6\\ &&& \Large\mbox{X} &)\\ 3&&7&&5\\\hline &&&0&0 \end{array}$$  "To multiply $456$ by $375$ you write one quantity under the other as shown in the margin and you multiply $5$ by $6$ which makes $30$, of which you put the $0$ in the first place & save $3$; then you multiply $5$ by $5$ & $6$ by $7$ & form the sum of those products & add the $3$ you saved, & that will make $70$, of which you will mark $0$ in the second place & save $7$;" 
$$\begin{array}{ccccc} 4&&5&&6\\ && \Large{\times}\!\!\!\!\!\!\!\! &&\\ 3&&7&&5\\\hline &&0&0&0 \end{array}$$  "then you multiply $4$ by $5$ & $3$ by $6$ & sum those $3$ products & add to them the $7$ you saved, & that will make $80$, of which you will mark down $0$ & save $8$. " 
$$\begin{array}{ccccc} 4&&5&&6\\ (&\Large\mbox{X}&& &\\ 3&&7&&5\\\hline 17&1&0&0&0 \end{array}$$  "then you multiply $4$ by $7$ & $3$ by $8$ & sum those two products together & add to them the $8$ you saved which will make $51$ of which you will mark down one & save $5$; then you multiply $3$ by $4$ & add to them the $5$ you saved & that will be $17$ which when marked down as shown you will have for the said multiplication $171000$." 
The translations in this column, unless otherwise noted, are mine.
References