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

List of thumbnails

< >
31
31 (9)
32
32 (10)
33
33 (11)
34
34 (12)
35
35 (13)
36
36 (14)
37
37 (15)
38
38 (16)
39
39 (17)
40
40 (18)
< >
page |< < (17) of 695 > >|
    <echo version="1.0RC">
      <text xml:lang="fr" type="free">
        <div xml:id="echoid-div45" type="section" level="1" n="31">
          <pb o="17" file="0039" n="39" rhead="LIVRE I. DE LA THEORIE DE LA MAÇONNERIE."/>
        </div>
        <div xml:id="echoid-div46" type="section" level="1" n="32">
          <head xml:id="echoid-head40" xml:space="preserve">PROPOSITION PREMIERE.</head>
          <head xml:id="echoid-head41" xml:space="preserve">
            <emph style="sc">Proble’me</emph>
          .</head>
          <p style="it">
            <s xml:id="echoid-s620" xml:space="preserve">19. </s>
            <s xml:id="echoid-s621" xml:space="preserve">Ayant un profil de Muraille ABC, triangulaire dont
              <lb/>
            le point d’apui eſt en C, & </s>
            <s xml:id="echoid-s622" xml:space="preserve">qu’une puiſſance pouſſe de K, en
              <lb/>
            B, pour la renverſer du côté opoſé, on demande quelle épaiſ-
              <lb/>
            ſeur il faudra donner à la baſe AC, pour que le poids G,
              <lb/>
            qu’on ſupoſe équivalent à la ſuperficie du triangle, ſoit en
              <lb/>
            équilibre avec la puiſſance K.</s>
            <s xml:id="echoid-s623" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s624" xml:space="preserve">Pour bien entendre ce Probléme, il faut conſiderer les côtés
              <lb/>
              <note position="right" xlink:label="note-0039-01" xlink:href="note-0039-01a" xml:space="preserve">
                <emph style="sc">Fig</emph>
              . 15.</note>
            CB, & </s>
            <s xml:id="echoid-s625" xml:space="preserve">CE, de l’angle BCE, comme formant un lévier recourbé
              <lb/>
            dont le point d’apui eſt en C, que la puiſſance K, étant apliquée
              <lb/>
            à l’extrêmité B, bu bras CB, pouſſe ſelon une direction paralelle à
              <lb/>
            l’horiſon, & </s>
            <s xml:id="echoid-s626" xml:space="preserve">par conſequent oblique au bras de lévier, & </s>
            <s xml:id="echoid-s627" xml:space="preserve">que le
              <lb/>
            poids G, eſt apliqué à l’extrêmité E, de l’autre bras CE, qui eſt
              <lb/>
            terminé par la ligne de direction IL, tirée du centre de gravité I,
              <lb/>
            du triangle. </s>
            <s xml:id="echoid-s628" xml:space="preserve">Or comme c’eſt la même choſe que la puiſſance K,
              <lb/>
            pouſſe de K, en B, ou qu’elle tire de B, en H, ſelon une direction toû-
              <lb/>
            jours paralelle à l’horiſon, nous ſupoſerons pour plus de facilité que
              <lb/>
            le poids F, eſt équivalent à cette puiſſance, & </s>
            <s xml:id="echoid-s629" xml:space="preserve">abaiſſant la perpendi-
              <lb/>
            culaire CD, ſur la ligne BH, la longueur du bras de lévier oblique
              <lb/>
            CB, par raport à la puiſſance, ſera réduite à la ligne CD, par l’arti-
              <lb/>
            cle 18
              <emph style="sub">e</emph>
            , & </s>
            <s xml:id="echoid-s630" xml:space="preserve">par-làla puiſſance K, ou F, pourra être admiſe dans ſon
              <lb/>
            entier, en ſupoſant qu’elle eſt apliquée à l’extrêmité D, de la perpen-
              <lb/>
            diculaire CD, que nous regarderons préſentement comme un des
              <lb/>
            bras de lévier. </s>
            <s xml:id="echoid-s631" xml:space="preserve">Sil’on nomme ce bras de lévier, c; </s>
            <s xml:id="echoid-s632" xml:space="preserve">auſſi-bien que la
              <lb/>
            hauteur BA, qui lui eſt égale, y, la baſe CA; </s>
            <s xml:id="echoid-s633" xml:space="preserve">l’on aura {2y/3} pour l’au-
              <lb/>
            tre bras CE, (puiſque par l’article 7
              <emph style="sub">e</emph>
            la partie AE, eſt le tiers
              <lb/>
            de toute la baſe AC,) cela étant, le poids G, ſera {yc/2}, ainſi l’on
              <lb/>
            aura bf, {yc/2} :</s>
            <s xml:id="echoid-s634" xml:space="preserve">: {2y/3}, c, qui donne cette équation {2yyc/6} = bcf, qu’on
              <lb/>
            rendra plus ſimple en faiſant la réduction, puiſqu’on n’aura plus
              <lb/>
            que{yy/3} = bf, ou bien y = √3bf,\x{0020} qui fait voir qu’on trouvera la
              <lb/>
            baſe AC, en triplant la puiſſance K, ou F, & </s>
            <s xml:id="echoid-s635" xml:space="preserve">en extrayant la ra-
              <lb/>
            cine quarrée de ce produit.</s>
            <s xml:id="echoid-s636" xml:space="preserve"/>
          </p>
        </div>
      </text>
    </echo>