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="60" file="0086" n="88" rhead="LA SCIENCE DES INGENIEURS,"/>
            même choſe que {9ch/25}, l’on voit que ſuprimant ch, qui eſt inutile, & </s>
            <s xml:id="echoid-s1467" xml:space="preserve">
              <lb/>
            retranchant le numerateur du dénominateur, il vient {9/16} qui mar-
              <lb/>
            que le raport de l’épaiſſeur qu’il faut donner aux contreforts avec
              <lb/>
            l’intervalle dont ils doivent être éloignés les uns des autres; </s>
            <s xml:id="echoid-s1468" xml:space="preserve">c’eſt-
              <lb/>
            à-dire, par exemple, que ſi l’on donnoit 4 pieds {1/2} d’épaiſſeur
              <lb/>
            aux contreforts, il faudroit les conſtruire a 8 pieds les uns des
              <lb/>
            autres.</s>
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        <div xml:id="echoid-div117" type="section" level="1" n="77">
          <head xml:id="echoid-head95" xml:space="preserve">PROPOSITION QUATRIE’ME.
            <lb/>
            <emph style="sc">Proble’me</emph>
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              <emph style="bf">48.</emph>
            Ayant déterminé la longueur GA, des contreforts,
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              <note position="left" xlink:label="note-0086-01" xlink:href="note-0086-01a" xml:space="preserve">
                <emph style="sc">Fig</emph>
              . 6.</note>
            leur épaiſſeur & </s>
            <s xml:id="echoid-s1471" xml:space="preserve">leur diſtance, de même que la ligne de
              <lb/>
            ED, & </s>
            <s xml:id="echoid-s1472" xml:space="preserve">la hauteur CE, l’on demande qu’elle épaiſſeur il
              <lb/>
            faudra donner au ſommet BC, du revêtement pour qu’il
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            ſoit en équilibre par ſon poids avec une puiſſance qui tire-
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            roit de C, en Q.</s>
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            <s xml:id="echoid-s1474" xml:space="preserve">Nous nommerons GA, h; </s>
            <s xml:id="echoid-s1475" xml:space="preserve">ED, d; </s>
            <s xml:id="echoid-s1476" xml:space="preserve">la hauteur CE, c; </s>
            <s xml:id="echoid-s1477" xml:space="preserve">l’épaiſſeur
              <lb/>
            BC, ou AE, x; </s>
            <s xml:id="echoid-s1478" xml:space="preserve">& </s>
            <s xml:id="echoid-s1479" xml:space="preserve">la puiſſance bf, comme à l’ordinaire; </s>
            <s xml:id="echoid-s1480" xml:space="preserve">or com-
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            me l’on ſupoſe que l’eſpace occupé par les contreforts eſt à toute
              <lb/>
            l’étenduë LMNO, comme 2 eſt à 5, la reduction des contreforts,
              <lb/>
            ou ſi l’on veut, la valeur du poids L, ſera donc {2hc/5}, le poids M,
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            ſera xc, & </s>
            <s xml:id="echoid-s1481" xml:space="preserve">le poids N, {cd/2}; </s>
            <s xml:id="echoid-s1482" xml:space="preserve">ſi préſentement l’on réünit ces trois
              <lb/>
            poids dans un ſeul O, & </s>
            <s xml:id="echoid-s1483" xml:space="preserve">qu’on multiplie enſuite ce poids par le bras
              <lb/>
            ID, l’on aura comme ci-devant un produit égal à celui de la puiſ-
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            ſance P, par ſon bras de lévier DQ. </s>
            <s xml:id="echoid-s1484" xml:space="preserve">par conſéquent cette équation
              <lb/>
            {xxc/2} + xcd + {2xhc/5} + {hhc/5} + {2hdc/5} + {ddc/3} = bfc, d’où faiſant paſſer du pre-
              <lb/>
            mier membre dans le ſecond les termes où l’inconnuë ne ſe trouve
              <lb/>
            point, & </s>
            <s xml:id="echoid-s1485" xml:space="preserve">diviſant le tout par c, l’on aura {xx/2} + xd + {2xh/5} = bf - {hh/5}
              <lb/>
            - {2hd/5} - {dd/3}, mais ſi l’on ſupoſe n = d + {2h/5} l’on aura nx,
              <lb/>
            = dx + {2hx/5} & </s>
            <s xml:id="echoid-s1486" xml:space="preserve">mettant nx, à la place de ſa valeur dans l’équation </s>
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