Harriot, Thomas, Mss. 6785

List of thumbnails

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341
341 (171)
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342 (171v)
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page |< < (216) of 882 > >|
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            <p>
              <s xml:space="preserve">[
                <emph style="bf">Commentary:</emph>
              </s>
            </p>
            <p>
              <s xml:space="preserve"> The reference on this page is to
                <emph style="it">Variorum responsorum liber VIII</emph>
              , Chapter XVIII, Proposition 2, Corollary
                <ref id="Viete_1593d" target="http://www.e-rara.ch/zut/content/pageview/2684269"> (Viete 1593d, Chapter 18, Prop </ref>
              . </s>
              <lb/>
              <quote xml:lang="lat">
                <s xml:space="preserve"> Corollarium.
                  <lb/>
                Itaque quadratum circulo inscriptum erit ad circulum, sicut latus illius quadrati ad potestatem diametri altissimam adplicatam ad id quod fit continue sub apotomis laterum octogoni, hexdecagoni, polygoni triginta duorum laterum, sexagintao quatuor, centum viginti octo, ducentorum quinquaginta sex,& reliquorum omnium in ea ratione angulorum laterumve </s>
              </quote>
              <lb/>
              <quote>
                <s xml:space="preserve"> Thus a square inscribed in a circle will be to the circle as the side of the square to the greatest power of the diameter applied to that which is successively under the apotome of the sides of octagons, hexdecagons, polygons with thirty-two sides, sixty-four, one hundred and twenty eight, two hundred and fifty six, and so on, all in the ratio of halved angles and </s>
              </quote>
              <s xml:space="preserve">]</s>
            </p>
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          <head xml:space="preserve" xml:lang="lat"> Responsorum. pag. 30.
            <lb/>
          in
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          Responsorum, page 30, on the ]</head>
          <p xml:lang="lat">
            <s xml:space="preserve"> per propositione
              <lb/>
            [
              <emph style="bf">Translation: </emph>
            by the preceding ]</s>
          </p>
          <p>
            <s xml:space="preserve"> As
              <math>
                <mstyle>
                  <mi>B</mi>
                </mstyle>
              </math>
            to […] so let
              <math>
                <mstyle>
                  <mi>α</mi>
                  <mi>β</mi>
                </mstyle>
              </math>
            be to
              <math>
                <mstyle>
                  <mi>β</mi>
                  <mi>γ</mi>
                </mstyle>
              </math>
            .
              <lb/>
            So will the square
              <math>
                <mstyle>
                  <mi>α</mi>
                  <mi>ɛ</mi>
                </mstyle>
              </math>
            be to the oblong
              <math>
                <mstyle>
                  <mi>β</mi>
                  <mi>θ</mi>
                </mstyle>
              </math>
            .
              <lb/>
            And as ... to ...
              <math>
                <mstyle>
                  <mi>O</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Therefore if
              <math>
                <mstyle>
                  <mi>α</mi>
                  <mi>β</mi>
                </mstyle>
              </math>
            be æquall to
              <math>
                <mstyle>
                  <mi>B</mi>
                </mstyle>
              </math>
              <lb/>
            these æquations will also follow:
              <lb/>
            And therfore the oblong made of
              <math>
                <mstyle>
                  <mi>B</mi>
                </mstyle>
              </math>
            and […]
              <lb/>
            is æquall to the </s>
          </p>
          <p>
            <s xml:space="preserve"> The oblong or square therefore æquall to the circle is
              <lb/>
            Devide it by the semidiameter
              <lb/>
            And the Quotient wilbe the semiperimeter </s>
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