Salusbury, Thomas
,
Mathematical collections and translations (Tome I)
,
1667
Text
Text Image
Image
XML
Thumbnail overview
Document information
None
Concordance
Figures
Thumbnails
Page concordance
<
1 - 30
31 - 60
61 - 90
91 - 120
121 - 150
151 - 180
181 - 210
211 - 240
241 - 270
271 - 300
301 - 330
331 - 360
361 - 390
391 - 420
421 - 450
451 - 480
481 - 510
511 - 540
541 - 570
571 - 600
601 - 630
631 - 660
661 - 690
691 - 701
>
Scan
Original
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
<
1 - 30
31 - 60
61 - 90
91 - 120
121 - 150
151 - 180
181 - 210
211 - 240
241 - 270
271 - 300
301 - 330
331 - 360
361 - 390
391 - 420
421 - 450
451 - 480
481 - 510
511 - 540
541 - 570
571 - 600
601 - 630
631 - 660
661 - 690
691 - 701
>
page
|<
<
of 701
>
>|
<
archimedes
>
<
text
>
<
body
>
<
chap
>
<
p
type
="
main
">
<
s
>
<
pb
xlink:href
="
040/01/1068.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
>