Archimedes
,
Natation of bodies
,
1662
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<
archimedes
>
<
text
>
<
body
>
<
chap
>
<
p
type
="
main
">
<
s
>
<
pb
xlink:href
="
073/01/044.jpg
"
pagenum
="
374
"/>
<
emph
type
="
italics
"/>
Remaining Angle Y V I is equall to the Remaining Angle B E
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ.</
foreign
>
<
emph
type
="
italics
"/>
And therefore the
<
emph.end
type
="
italics
"/>
<
lb
/>
<
arrow.to.target
n
="
marg1263
"/>
<
lb
/>
(e)
<
emph
type
="
italics
"/>
Line V I hath to the Line I Y the ſame proportion that the Line E
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
hath to
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B: But
<
lb
/>
the
<
emph.end
type
="
italics
"/>
(f)
<
emph
type
="
italics
"/>
Line P I, which is greater than V I, hath unto I Y greater proportion than V I hath un-
<
emph.end
type
="
italics
"/>
<
lb
/>
<
arrow.to.target
n
="
marg1264
"/>
<
lb
/>
<
emph
type
="
italics
"/>
to the ſame: Therefore
<
emph.end
type
="
italics
"/>
(g)
<
emph
type
="
italics
"/>
T I ſhall have greater proportion unto I Y, than E
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
hath to
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B:
<
lb
/>
And, by the ſame reaſon, the Square T I ſhall have greater proportion to the Square I Y, than
<
emph.end
type
="
italics
"/>
<
lb
/>
<
arrow.to.target
n
="
marg1265
"/>
<
lb
/>
<
emph
type
="
italics
"/>
the Square E
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
hath to the Square
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B.
<
emph.end
type
="
italics
"/>
<
lb
/>
<
arrow.to.target
n
="
marg1266
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1262
"/>
E</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1263
"/>
(e)
<
emph
type
="
italics
"/>
By 4. of the
<
lb
/>
ſixth.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1264
"/>
(f)
<
emph
type
="
italics
"/>
By 8. of the
<
lb
/>
fifth.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1265
"/>
(g)
<
emph
type
="
italics
"/>
By 13 of the
<
lb
/>
fifth.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1266
"/>
F</
s
>
</
p
>
<
p
type
="
main
">
<
s
>But as the Square P I is to the Square Y I, ſo is the Line K R unto
<
lb
/>
the Line I Y]
<
emph
type
="
italics
"/>
For by 11. of the firſt of our
<
emph.end
type
="
italics
"/>
Conicks,
<
emph
type
="
italics
"/>
the Square P I is equall
<
lb
/>
to the Rectangle contained under the Line I O, and under the Parameter; which
<
lb
/>
we ſuppoſed to be eqnall to the Semi-parameter; that is, the double of K R
<
emph.end
type
="
italics
"/>
: </
s
>
</
p
>
<
p
type
="
main
">
<
s
>
<
arrow.to.target
n
="
marg1267
"/>
<
lb
/>
<
emph
type
="
italics
"/>
But I Y is double of I O, by 33 of the ſame: And, therefore, the
<
emph.end
type
="
italics
"/>
(h)
<
emph
type
="
italics
"/>
Rectangle made of K R
<
lb
/>
and I Y, is equall to the Rectangle contained under the Line I O, and under the Parameter;
<
emph.end
type
="
italics
"/>
<
lb
/>
<
arrow.to.target
n
="
marg1268
"/>
<
lb
/>
<
emph
type
="
italics
"/>
that is, to the Square P I: But as the
<
emph.end
type
="
italics
"/>
(i)
<
emph
type
="
italics
"/>
Rectangle compounded of K R and I Y is to the
<
lb
/>
Square I Y, ſo is the Line K R unto the Line I Y: Therefore the Line K R ſhall have unto I
<
lb
/>
Y, the ſame proportion that the Rectangle compounded of K R and I Y; that is, the Square P I
<
lb
/>
hath to the Square I Y.
<
emph.end
type
="
italics
"/>
<
lb
/>
<
arrow.to.target
n
="
marg1269
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1267
"/>
(h)
<
emph
type
="
italics
"/>
By 26. of the
<
lb
/>
ſixth.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1268
"/>
(i)
<
emph
type
="
italics
"/>
By
<
emph.end
type
="
italics
"/>
Lem. </
s
>
<
s
>22
<
emph
type
="
italics
"/>
of
<
lb
/>
the tenth.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1269
"/>
G</
s
>
</
p
>
<
p
type
="
main
">
<
s
>And as the Square E
<
foreign
lang
="
grc
">Ψ</
foreign
>
is to the Square
<
foreign
lang
="
grc
">Ψ</
foreign
>
B, ſo is half of the
<
lb
/>
Line K R unto the Line
<
foreign
lang
="
grc
">ψ</
foreign
>
B.]
<
emph
type
="
italics
"/>
For the Square E
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
having been ſuppoſed equall
<
lb
/>
to half the Rectangle contained under the Line K R and
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B; that is, to that contained under
<
lb
/>
the half of K R and the Line
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B; and ſeeing that as the
<
emph.end
type
="
italics
"/>
(k)
<
emph
type
="
italics
"/>
Rectangle made of half K R
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
main
">
<
s
>
<
arrow.to.target
n
="
marg1270
"/>
<
lb
/>
<
emph
type
="
italics
"/>
and of B
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
is to the Square
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B, ſo is half K R unto the Line
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B; the half of K R ſhall have
<
lb
/>
the ſame proportion to
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B, as the Square E
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
hath to the Square
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B.
<
emph.end
type
="
italics
"/>
<
lb
/>
<
arrow.to.target
n
="
marg1271
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1270
"/>
(k)
<
emph
type
="
italics
"/>
By
<
emph.end
type
="
italics
"/>
Lem. </
s
>
<
s
>22
<
emph
type
="
italics
"/>
of
<
lb
/>
the tenth.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1271
"/>
H</
s
>
</
p
>
<
p
type
="
main
">
<
s
>And, conſequently, I Y is leſſe than the double of
<
foreign
lang
="
grc
">ψ</
foreign
>
B.]
<
lb
/>
<
emph
type
="
italics
"/>
For, as half K R is to
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B, ſo is K R to another Line: it ſhall be
<
emph.end
type
="
italics
"/>
(1)
<
emph
type
="
italics
"/>
greater than I Y; that
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
main
">
<
s
>
<
arrow.to.target
n
="
marg1272
"/>
<
lb
/>
<
emph
type
="
italics
"/>
is, than that to which K R hath leſſer proportion; and it ſhall be double of
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B: Therefore
<
lb
/>
I Y is leſſe than the double of
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B.
<
emph.end
type
="
italics
"/>
<
lb
/>
<
arrow.to.target
n
="
marg1273
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1272
"/>
(l)
<
emph
type
="
italics
"/>
By 10 of the
<
lb
/>
fifth.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1273
"/>
K</
s
>
</
p
>
<
p
type
="
main
">
<
s
>And I
<
foreign
lang
="
grc
">ω</
foreign
>
greater than
<
foreign
lang
="
grc
">ψ</
foreign
>
R.]
<
emph
type
="
italics
"/>
For O having been ſuppoſed equall to B R,
<
lb
/>
if from B R,
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ψ</
foreign
>
<
emph
type
="
italics
"/>
B be taken, and from O
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ω,</
foreign
>
<
emph
type
="
italics
"/>
O I, which is leſſer than B, be taken; the
<
lb
/>
Remainder I
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ω</
foreign
>
<
emph
type
="
italics
"/>
ſhall be greater than the Remainder
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">Ψ</
foreign
>
<
emph
type
="
italics
"/>
R.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
main
">
<
s
>
<
arrow.to.target
n
="
marg1274
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1274
"/>
L</
s
>
</
p
>
<
p
type
="
main
">
<
s
>And, therefore, F Q is equall to P M.]
<
emph
type
="
italics
"/>
By the fourteenth of the fifth of
<
emph.end
type
="
italics
"/>
<
lb
/>
Euclids
<
emph
type
="
italics
"/>
Elements: For the Line O N is equall to B D.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
main
">
<
s
>
<
arrow.to.target
n
="
marg1275
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1275
"/>
M</
s
>
</
p
>
<
p
type
="
main
">
<
s
>But it hath been demonſtrated that P H is greater than F.]
<
lb
/>
<
emph
type
="
italics
"/>
For it was demonſtrated that I
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ω</
foreign
>
<
emph
type
="
italics
"/>
is greater than F: And P H is equall to I
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">ω.</
foreign
>
</
s
>
</
p
>
<
p
type
="
main
">
<
s
>
<
arrow.to.target
n
="
marg1276
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1276
"/>
N</
s
>
</
p
>
<
p
type
="
main
">
<
s
>In the ſame manner we might demonſtrate the Line T H
<
lb
/>
to be Perpendicular unto the Surface of the Liquid.]
<
emph
type
="
italics
"/>
For T
<
emph.end
type
="
italics
"/>
<
foreign
lang
="
grc
">α</
foreign
>
<
emph
type
="
italics
"/>
is equall
<
lb
/>
to K R; that is, to the Semi-parameter: And, therefore, by the things above demonstrated,
<
lb
/>
the Line T H ſhall be drawn Perpendicular unto the Liquids Surface.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
main
">
<
s
>
<
arrow.to.target
n
="
marg1277
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1277
"/>
O</
s
>
</
p
>
<
p
type
="
main
">
<
s
>Therefore, the Square P I hath leſſer proportion unto the
<
lb
/>
Square I Y, than the Square E
<
foreign
lang
="
grc
">
<
gap
/>
</
foreign
>
hath to the Square
<
foreign
lang
="
grc
">ψ</
foreign
>
B.]
<
lb
/>
<
emph
type
="
italics
"/>
Theſe, and other particulars of the like nature, that follow both in this and the following
<
lb
/>
Propoſitions, ſhall be demonſtrated by us no otherwiſe than we have done above.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
main
">
<
s
>
<
arrow.to.target
n
="
marg1278
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1278
"/>
P</
s
>
</
p
>
<
p
type
="
main
">
<
s
>Therefore Perpendiculars being drawn thorow Z and G, unto
<
lb
/>
the Surface of the Liquid, that are parallel to T H, it followeth
<
lb
/>
that the ſaid Portion ſhall not ſtay, but ſhall turn about till that its
<
lb
/>
Axis do make an Angle with the Waters Surface greater than that
<
lb
/>
which it now maketh.]
<
emph
type
="
italics
"/>
For in that the Line drawn thorow G, doth fall perpendicu
<
lb
/>
larly towards thoſe parts which are next to L; but that thorow Z, towards thoſe next to A;
<
lb
/>
It is neceſſary that the Centre G do move downwards, and Z upwards: and, therefore, the
<
lb
/>
parts of the Solid next to L ſhall move downwards, and thoſe towards A upwards, that the
<
lb
/>
Axis may makea greater Angle with the Surface of the Liquid.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
<
p
type
="
main
">
<
s
>
<
arrow.to.target
n
="
marg1279
"/>
</
s
>
</
p
>
<
p
type
="
margin
">
<
s
>
<
margin.target
id
="
marg1279
"/>
Q</
s
>
</
p
>
<
p
type
="
main
">
<
s
>For ſo ſhall I O be equall to
<
foreign
lang
="
grc
">ψ</
foreign
>
B; and
<
foreign
lang
="
grc
">ω</
foreign
>
I equall to I R; and
<
lb
/>
P H equall to F.]
<
emph
type
="
italics
"/>
This plainly appeareth in the third Figure, which is added by us.
<
emph.end
type
="
italics
"/>
</
s
>
</
p
>
</
chap
>
</
body
>
</
text
>
</
archimedes
>