Bélidor, Bernard Forest de, La science des ingenieurs dans la conduite des travaux de fortification et d' architecture civile

Table of Notes

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          <pb o="27" file="0051" n="52" rhead="LIVRE I. DE LA THEORIE DE LA MAÇONNERIE."/>
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          <head xml:id="echoid-head55" xml:space="preserve">PROPOSITION CINQUIE’ME.</head>
          <head xml:id="echoid-head56" xml:space="preserve">
            <emph style="sc">Proble’me</emph>
          .</head>
          <p style="it">
            <s xml:id="echoid-s777" xml:space="preserve">28. </s>
            <s xml:id="echoid-s778" xml:space="preserve">Ayant comme dans le Probléme précédent un profil
              <lb/>
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                <emph style="sc">Fig</emph>
              . 18.
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              & 20.</note>
            rectangulaire AC, en équilibre par ſon poids avec une puiſ-
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            ſance P, on demande un autre profil GHIK, qui ait la mê-
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            me hauteur, que le précédent, mais dont la ſuperficie n’en
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            ſoit que les trois quarts, avec cette condition que le Mur GHIK,
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            ſoit encore en équilibre par ſa réſiſtance à l’effort de la puiſ-
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            ſance P, qu’on ſupoſe agir toûjours avec la même force.</s>
            <s xml:id="echoid-s779" xml:space="preserve"/>
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          <p>
            <s xml:id="echoid-s780" xml:space="preserve">Nommant les lignes BA, ou HG, c; </s>
            <s xml:id="echoid-s781" xml:space="preserve">AD, a; </s>
            <s xml:id="echoid-s782" xml:space="preserve">HI, ou GL,
              <lb/>
            x; </s>
            <s xml:id="echoid-s783" xml:space="preserve">LK, y; </s>
            <s xml:id="echoid-s784" xml:space="preserve">l’on aura ac, pour le rectangle BD, cx, pour le rectan-
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            gle HL, ou ſi l’on veut pour le poids Q, & </s>
            <s xml:id="echoid-s785" xml:space="preserve">{cy/2} pour le trian-
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            gle ILK, qui eſt la même choſe que le poids P; </s>
            <s xml:id="echoid-s786" xml:space="preserve">or comme le
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            Trapeze GHIK, ne doit être que les trois quarts du rectangle
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            BD, l’on aura donc {3ac/4} = cx + {cy/2}, & </s>
            <s xml:id="echoid-s787" xml:space="preserve">ſi l’on réünit le poids Q,
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            avec le poids P, après les avoir multipliés par leur bras de léviers,
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            l’on aura une quantité égale au produit de la puiſſance P, qui eſt
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            toûjours bf, par le bras de lévier KR, ce qui donne cette ſeconde
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            équation {xxc/2} + xyc + {yyc/3} = bcf, ou en effaçant de tous les termes
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            la lettre c, {xx/2} + xy + {yy/3} = bf, mais ſi dans la premiere équation
              <lb/>
            {3ac/4} = cx + {yc/2} l’on dégage y, l’on aura {ba/4} - 2x = y, & </s>
            <s xml:id="echoid-s788" xml:space="preserve">ſupo-
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            ſant {6a/4} = n, pour plus de facilité, l’on aura n - 2x = y. </s>
            <s xml:id="echoid-s789" xml:space="preserve">Si pre-
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            ſentement l’on ſubſtituë la valeur de y, dans l’équation {xx/2} + yx
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            + {yy/3} = bf, elle ſera changée en celle-cy{xx/2} + nx - 2xx
              <lb/>
            + {nn - 4nx + 4xx/3} = bf, d’où faiſant évanoüir la fraction l’on a
              <lb/>
            3xx + 6nx - 12xx + 2nn - 8nx + 8xx = 6fb, qui étant réduite
              <lb/>
            donne 2nn - xx - 2nx = 6bf, ou bien 2nn - 6bf = xx + 2nx; </s>
            <s xml:id="echoid-s790" xml:space="preserve">or </s>
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