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>
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