Harriot, Thomas, Mss. 6787

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page |< < (247) of 1155 > >|
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      <text xml:lang="eng" type="free">
        <div type="section" level="1" n="1">
          <pb file="add_6787_f247" o="247" n="492"/>
          <head xml:space="preserve" xml:lang="lat">
            <lb/>
          [
            <emph style="bf">Translation: </emph>
          ]</head>
          <p>
            <s xml:space="preserve"> A progression increasing being given,
              <emph style="super">
                <emph style="st">next</emph>
              first progresionall </emph>
              <emph style="st">first</emph>
            differences
              <lb/>
            are unæquall, & the
              <emph style="st">second</emph>
              <emph style="super">next
                <emph style="st">after</emph>
              differences</emph>
            æquall: to devide the sayde
              <lb/>
            progression into a fewer nomber of progresionall </s>
            <lb/>
            <s xml:space="preserve"> The first case. as of the
              <emph style="super">progresionall</emph>
            differences decreasing,
              <lb/>
            as it is in the progression of </s>
          </p>
          <p>
            <s xml:space="preserve"> Example of the progresion
              <emph style="super">given</emph>
            </s>
          </p>
          <p>
            <s xml:space="preserve"> This example I have
              <lb/>
            so set downe as though the
              <lb/>
            nombers were answerable
              <lb/>
            to
              <lb/>
            [
              <emph style="bf">Commentary: </emph>
            'Arkes' are arcs of a fixed circle, that is, measures of angle. Each sine (or number) corresponds to, or is 'answerable to', an arc (or angle). (Hence the modern terminology ]</s>
            <s xml:space="preserve"> That by it you may
              <lb/>
            se the use of the problem
              <lb/>
            for </s>
          </p>
          <p>
            <s xml:space="preserve"> Suppose that it be required to find the nomber answerable to </s>
            <lb/>
            <s xml:space="preserve"> The number answerable
              <emph style="super">to</emph>
            30'' is 2280. that which is answerable
              <lb/>
            to 40'' is 2925. there difference is </s>
            <s xml:space="preserve"> And because the other
              <lb/>
            differences of the same
              <emph style="super">order</emph>
            ranke are unæquall, the rule of pro-
              <lb/>
            portion will not find the number desired. But it must be
              <lb/>
            found by a speciall Canon.
              <emph style="st">which followeth</emph>
            </s>
            <s xml:space="preserve"> The purpose
              <lb/>
            of
              <emph style="st">which canon</emph>
              <emph style="super">thereof</emph>
            must
              <emph style="st">for to find</emph>
            (because 30''& 40'' do differ
              <lb/>
            [by] 10''.) to find the
              <emph style="st">
                <math>
                  <mstyle>
                    <mfrac>
                      <mrow>
                        <mn>1</mn>
                      </mrow>
                      <mrow>
                        <mn>1</mn>
                        <mn>0</mn>
                      </mrow>
                    </mfrac>
                  </mstyle>
                </math>
              </emph>
              <emph style="super">first tenth</emph>
            parte progresionall, that is to say that </s>
          </p>
        </div>
      </text>
    </echo>
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