Bošković, Ruđer Josip, Abhandlung von den verbesserten dioptrischen Fernröhren aus den Sammlungen des Instituts zu Bologna sammt einem Anhange des Uebersetzers

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          <p>
            <s xml:id="echoid-s984" xml:space="preserve">
              <pb o="85" file="0089" n="89" rhead="Von verbeß. Fernröhren."/>
            E X = A H - E H - A X = {1/2}a - {e
              <emph style="super">2</emph>
            /4a}
              <lb/>
            - {2a
              <emph style="super">2</emph>
            t + e
              <emph style="super">2</emph>
            t/4a t + 8e} - {e
              <emph style="super">2</emph>
            /2a} (das letzte Glied {e
              <emph style="super">2</emph>
            /2a}
              <lb/>
            iſt der Werth von A X vermöge (21)) =
              <lb/>
            {2a
              <emph style="super">2</emph>
            e - 2a e
              <emph style="super">2</emph>
            t - 3e
              <emph style="super">3</emph>
            /2a
              <emph style="super">2</emph>
            t + 4a e}. </s>
            <s xml:id="echoid-s985" xml:space="preserve">Man ſetze dieſen
              <lb/>
            = z, und die Länge A B = c, ſo ſtehet wie-
              <lb/>
            derum E X (z): </s>
            <s xml:id="echoid-s986" xml:space="preserve">E B (A B + E X + A X,
              <lb/>
            oder c + z + {e
              <emph style="super">2</emph>
            /2 a}) = M X (e): </s>
            <s xml:id="echoid-s987" xml:space="preserve">B D =
              <lb/>
            {c e/z} + e + {e
              <emph style="super">3</emph>
            /2 a z}. </s>
            <s xml:id="echoid-s988" xml:space="preserve">Nennen wir den itzt ge-
              <lb/>
            fundenen Werth r, wird z = {c e + e
              <emph style="super">3</emph>
            /2 a}/r - e} =
              <lb/>
            {4 a
              <emph style="super">2</emph>
            e - 4 a e
              <emph style="super">2</emph>
            t - 6 e
              <emph style="super">3</emph>
            /4a
              <emph style="super">2</emph>
            t + 8 a e}, oder {c + {e
              <emph style="super">2</emph>
            /2a}/r - e} =
              <lb/>
            {a - e t - {3e
              <emph style="super">2</emph>
            /2a}/a t + 2 e}, welche Gleichung den geſuch-
              <lb/>
            ten halben Durchmeſſer a giebt, ſo fern man
              <lb/>
            aus dem Verſuche c, e, r, das iſt A B, B D
              <lb/>
            und MX, oder die halbe Oeffnungsbreite weiß,
              <lb/>
            wie auch t, den Sinus des halben ſcheinbaren
              <lb/>
            Durchmeſſers der Sonne, nach dem halben
              <lb/>
            Durchmeſſer = 1 gerechnet: </s>
            <s xml:id="echoid-s989" xml:space="preserve">und weil dieſer
              <lb/>
            bey nahe 15{1/2} Minuten faſt allezeit </s>
          </p>
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      </text>
    </echo>
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