Harriot, Thomas, Mss. 6784

List of thumbnails

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301
301 (151)
302
302 (151v)
303
303 (152)
304
304 (152v)
305
305 (153)
306
306 (153v)
307
307 (154)
308
308 (154v)
309
309 (155)
310
310 (155v)
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page |< < (203) of 862 > >|
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              <s xml:space="preserve">[
                <emph style="bf">Commentary:</emph>
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              <s xml:space="preserve"> The reference to Pappus is to Commandino's edition of Books III to
                <emph style="it">Mathematicae collectiones</emph>
                <ref id="pappus_1588"> (Pappus </ref>
              . The proposition on page 47 is Proposition IV.14. Harriot's diagram is the same as the one given by Commandino except for his use of lower case letters. A second diagram for the same proposition appears on Add MS 6784
                <ref target="http://echo.mpiwg-berlin.mpg.de/ECHOdocuView?url=/permanent/library/XT0KZ8QC/&start=400&viewMode=image&pn=407"> f. </ref>
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                <s xml:space="preserve"> Theorema XIIII. Propositio XIIII.
                  <lb/>
                Sint duo semicirculi BGC BED: & ipsos contingat circulus EFGH: a cuius centro A ad BC basim semicirculorum perpendicularis ducatur AM. Dico ut BM as eam, quæ ex centro circuli EFGH, ita esse in prima figura vtramque simul CB BD ad earum excessum CD; in secunda vero, & tertia figura, ita esse excessum CB BD ad vtramque ipsarum CB </s>
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              <quote>
                <s xml:space="preserve"> Let there be two semicircles BGC and BED, and their touching circle EFGH, from whose centre A to BC, the base of the semicircle, there is drawn the perpendicular AM. I say that as BM is to that line from the centre of the circle EFGH, inthe first figure will be CB and BD togher to their excess, CD; but in the second and third figure, it will be as the excess of CB over BD to both of CB and BD </s>
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              <s xml:space="preserve">]</s>
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          <head xml:space="preserve" xml:lang="lat"> pappus. pag.
            <lb/>
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          [
            <emph style="bf">Translation: </emph>
          Pappus, page ]</head>
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