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="25" file="0047" n="47" rhead="LIVRE I. DE LA THEORIE DE LA MAÇONNERIE."/>
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          <head xml:id="echoid-head51" xml:space="preserve">PROPOSITION QUATRIE’ME.</head>
          <head xml:id="echoid-head52" xml:space="preserve">
            <emph style="sc">Proble’me.</emph>
          </head>
          <p style="it">
            <s xml:id="echoid-s744" xml:space="preserve">26. </s>
            <s xml:id="echoid-s745" xml:space="preserve">Ayant le profil ABCD, d’un Mur élevé à plomb des
              <lb/>
            deux côtés, & </s>
            <s xml:id="echoid-s746" xml:space="preserve">dont l’épaiſſeur BC, eſt tellement proportion-
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              <note position="right" xlink:label="note-0047-01" xlink:href="note-0047-01a" xml:space="preserve">Fig. 18.
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              & 19.</note>
            née à la hateur CD, que ce Mur ſoit en équilibre par ſonpoids
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            avec la puiſſance P, qui tire de C, en E, on demande de chan-
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            ger ce profil-là en un autre IGHL, qui lui ſoit égalen ſuperſi-
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            cie, & </s>
            <s xml:id="echoid-s747" xml:space="preserve">en hauteur, & </s>
            <s xml:id="echoid-s748" xml:space="preserve">dont le côté GI, ſoit perpendiculaire,
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            pour que ce ſecond ſoit en équilibre par ſa réſiſtance à une
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            puiſſance Q, dont la force ſeroit double de la puiſſance P.</s>
            <s xml:id="echoid-s749" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s750" xml:space="preserve">Pour cela nous nommerons BC, a; </s>
            <s xml:id="echoid-s751" xml:space="preserve">CD, de même que GI,
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            c; </s>
            <s xml:id="echoid-s752" xml:space="preserve">GH, ou IK, x; </s>
            <s xml:id="echoid-s753" xml:space="preserve">KL, y; </s>
            <s xml:id="echoid-s754" xml:space="preserve">la puiſſance P, ſera bf, comme à
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            l’ordinaire, & </s>
            <s xml:id="echoid-s755" xml:space="preserve">la puiſſance Q, 2bf; </s>
            <s xml:id="echoid-s756" xml:space="preserve">cela poſé, la ſuperſicie du rec-
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            tangle IGHK, ou ſi l’on veut le poids N, ſera xc, & </s>
            <s xml:id="echoid-s757" xml:space="preserve">celle du
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            triangle KHL, ou le poids S, ſera {yc/2}, & </s>
            <s xml:id="echoid-s758" xml:space="preserve">ces deux poids étant
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            multipliés par leur bras de lévier, réüniſſant leur produit, on
              <note symbol="*" position="right" xlink:label="note-0047-02" xlink:href="note-0047-02a" xml:space="preserve">Art. 23.</note>
            aura une quantité égale au produit de la puiſſance par ſon bras de
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            lévier, c’eſt-à-dire {xxc + 2yxc/2} + {yyc/3} = 2bfc, ou diviſant tous les
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            termes par c, l’on aura {xx+2yx/2} + {yy/3} = 2bf; </s>
            <s xml:id="echoid-s759" xml:space="preserve">mais comme le
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            rectangle BD, (ac) eſt ſupoſé égal au Trapezoïde IGHL, il vien-
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            dra encore cette équation ac = cx + {cy/2}, d’où dégageant l’in-
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            connuë y, l’on aura y = 2a - 2x, & </s>
            <s xml:id="echoid-s760" xml:space="preserve">ſubſtituant la valeur de y,
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            dans l’équation {xx + 2xy/2} + {yy/3} = 2bf, cela donne {xx/2} + 2ax
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            - 2xx + {4aa - 8ax + 4xx/3} = 2bf, qui, étant réduite, donne
              <lb/>
            4aa - 2ax - {xx/2} = 6bf, ou bien {xx/2} + 2ax = 4aa - 6bf, & </s>
            <s xml:id="echoid-s761" xml:space="preserve">faiſant
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            évanoüir la fraction l’on a xx + 4ax = 8aa - 12bf, à quoiajoûtant
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            4aa de part & </s>
            <s xml:id="echoid-s762" xml:space="preserve">d’autre pour rendre le premier membre un quarré
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            parfait, il viendra xx + 4ax + 4aa = 12aa - 12bf, d’où l’on tire
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            x = √12aa - 12bf\x{0020} - 2a, après avoir extrait la racine quarrée.</s>
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