Archimedes, Natation of bodies, 1662

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page |< < of 68 > >|
PROP. II. THEOR. II.
Let us ſuppoſe a Liquid that is of ſuch a conſiſtance as that it
is not moved, and that its Superficies be cut by a Plane along
by the Center of the Earth, and let the Center of the Earth
be the Point K: and let the Section of the Superficies be the Line
A B G D.
I ſay that the Line A B G D is the Circumference of a

Figure: /permanent/archimedes/archi_natat_073_en_1662/figures/073.01.008.1.jpg not scanned
[Figure 3]

Circle, and that the Center
thereof is the Point K And
if it be poſſible that it may
not be the Circumference
of a Circle, the Right-

Lines drawn ^{*} by the Point
K to the ſaid Line A B G D
ſhall not be equal.
There-
fore let a Right-Line be
taken greater than ſome of thoſe produced from the Point K unto
the ſaid Line A B G D, and leſſer than ſome other; and upon the
Point K let a Circle be deſcribed at the length of that Line,
Now the Circumference of this Circle ſhall fall part without the
ſaid Line A B G D, and part within: it having been preſuppoſed
that its Semidiameter is greater than ſome of thoſe Lines that may
be drawn from the ſaid Point K unto the ſaid Line A B G D, and
leſſer than ſome other.
Let the Circumference of the deſcribed
Circle be R B G H, and from B to K draw the Right-Line B K: and
drawn alſo the two Lines K R, and K E L which make a Right-
Angle in the Point K: and upon the Center K deſcribe the Circum-
ference X O P in the Plane and in the Liquid.
The parts, there-
fore, of the Liquid that are ^{*} according to the Circumference

X O P, for the reaſons alledged upon the firſt Suppoſition, are equi-
jacent, or equipoſited, and contiguous to each other; and both
theſe parts are preſt or thruſt, according to the ſecond part of the
Suppoſition, by the Liquor which is above them. And becauſe the
two Angles E K B and B K R are ſuppoſed equal [by the 26. of 3.
of Euclid,] the two Circumferences or Arches B E and B R ſhall
be equal (foraſmuch as R B G H was a Circle deſcribed for ſatis-
faction of the Oponent, and K its Center:) And in like manner
the whole Triangle B E K ſhall be equal to the whole Triangle
B R K.
And becauſe alſo the Triangle O P K for the ſame reaſon

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