Harriot, Thomas, Mss. 6789

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              <s xml:space="preserve">[
                <emph style="bf">Commentary:</emph>
              </s>
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            <p>
              <s xml:space="preserve"> The page is part of Harriot’s
                <ref target="http://echo.mpiwg-berlin.mpg.de/content/scientific_revolution/harriot/harriot-bl/maps/8.1.1_Compound.pt">introduction of compound diagrams of motion</ref>
              , a group of folios on which Harriot eventually succeeds in deriving, by means of such diagrams, a general expression for the time of flight of a projectile
                <ref id="Schemmel_2008">(Schemmel 2008, Section 8.2)</ref>
              . </s>
              <s xml:space="preserve">]</s>
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            <p>
              <s xml:space="preserve"> On this page, Harriot first formulates the problem to find the range of a shot depending on the elevation angle and describes the first step to solve it, the determination of the time of flight of a projectile. The compound diagram of motion he puts forward (lower drawing on this page) is inadequate for solving this problem. The calculations based on this erroneous diagram are found on
                <ref target="http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/permanent/library/0VGM2B80/&start=50&viewMode=image&pn=49">f. 25</ref>
              . The problem is adequately formulated and solved on
                <ref target="http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/permanent/library/0VGM2B80/&start=50&viewMode=image&pn=45">f. H-23</ref>
              . </s>
              <s xml:space="preserve">]</s>
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          <head xml:space="preserve"> 3) For oblique motions. </head>
          <p>
            <s xml:space="preserve"> To find where a motion at
              <lb/>
            randon will cut the horizon. </s>
          </p>
          <p>
            <s xml:space="preserve"> Suppose it cut in the poynt ι.
              <lb/>
            & let ιδ be a perpendicular.
              <lb/>
            the time of δι is aequall to
              <lb/>
            the time of δα; for γθ is aequall
              <lb/>
            to γα & βH toβα .&c. </s>
          </p>
          <p>
            <s xml:space="preserve"> Now the space of αε is geuen & the time:
              <lb/>
            the time of δα or δι is required. </s>
          </p>
          <p>
            <s xml:space="preserve"> It is performed by the
              <lb/>
            probleme following: </s>
          </p>
          <p>
            <s xml:space="preserve"> Data. ∆.
              <math>
                <mstyle>
                  <mi>m</mi>
                  <mi>a</mi>
                  <mi>n</mi>
                </mstyle>
              </math>
            .
              <lb/>
            Λ.
              <math>
                <mstyle>
                  <mi>m</mi>
                  <mi>a</mi>
                  <mi>k</mi>
                </mstyle>
              </math>
            . </s>
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          <p>
            <s xml:space="preserve"> Quaeritur: to draw the line
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            in
              <lb/>
            such sort, as that:
              <lb/>
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>f</mi>
                  <mi>n</mi>
                  <mi>m</mi>
                </mstyle>
              </math>
            ,
              <math>
                <mstyle>
                  <mi>b</mi>
                  <mi>a</mi>
                  <mi>c</mi>
                </mstyle>
              </math>
            : αδ, δι. […] </s>
          </p>
          <p>
            <s xml:space="preserve"> This probleme is answered in the page following.
              <lb/>
            But that which in deed answereth the question is in page .5.) </s>
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