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

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        <div xml:id="echoid-div137" type="section" level="1" n="85">
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            <pb o="8" file="0116" n="119" rhead="LA SCIENCE DES INGENIEURS,"/>
          </div>
          <div xml:id="echoid-div143" type="section" level="2" n="3">
            <head xml:id="echoid-head110" xml:space="preserve">
              <emph style="sc">Corollaire</emph>
              <emph style="sc">Second</emph>
            .</head>
            <p>
              <s xml:id="echoid-s2106" xml:space="preserve">4. </s>
              <s xml:id="echoid-s2107" xml:space="preserve">Il ſuit encore que connoiſſant les trois côtés du triangle IKL,
                <lb/>
                <note position="left" xlink:label="note-0116-01" xlink:href="note-0116-01a" xml:space="preserve">
                  <emph style="sc">Fig</emph>
                . 4.</note>
              avec une des trois puiſſances, on pourra connoître les deux autres
                <lb/>
              puiſſances; </s>
              <s xml:id="echoid-s2108" xml:space="preserve">car ſi (par exemple) l’on a la puiſſance P, & </s>
              <s xml:id="echoid-s2109" xml:space="preserve">qu’on
                <lb/>
              veüille connoître la ſeconde Q, on n’aura qu’à dire comme le côté
                <lb/>
              KI, eſt au côté KL, ainſi la puiſſance P, eſt à la puiſſance Q, que
                <lb/>
              l’on trouvera par la régle de proportion auſſi-bien que la troiſiéme
                <lb/>
              puiſſance R.</s>
              <s xml:id="echoid-s2110" xml:space="preserve"/>
            </p>
          </div>
          <div xml:id="echoid-div145" type="section" level="2" n="4">
            <head xml:id="echoid-head111" xml:space="preserve">
              <emph style="sc">Corollaire</emph>
              <emph style="sc">Troisie’me</emph>
            .</head>
            <p>
              <s xml:id="echoid-s2111" xml:space="preserve">5. </s>
              <s xml:id="echoid-s2112" xml:space="preserve">Dans les triangles les ſinus des angles étant dans la même rai-
                <lb/>
              ſon que leurs côtés opoſés, on peut ajoûter encore que ſi l’on avoit
                <lb/>
              un triangle IKL, dont les trois côtés fuſſent en mêmé raiſon que
                <lb/>
              les puiſſances PQR, ſi on ne connoiſſoit pas ces côtés, il ſuffiroit
                <lb/>
              de connoître la valeur des angles qui leur ſont opoſés, parce que
                <lb/>
              les ſinus de ces angles pouvant être pris pour les côtés mêmes, ils
                <lb/>
              exprimeront plus exactement le raport en nombre, & </s>
              <s xml:id="echoid-s2113" xml:space="preserve">par conſé-
                <lb/>
              quent les puiſſances, deſorte que ſi on connoiſſoit la valeur de la
                <lb/>
              puiſſance Q, & </s>
              <s xml:id="echoid-s2114" xml:space="preserve">les trois angles I, K, L, on trouvera les deux au-
                <lb/>
              tres puiſſances P & </s>
              <s xml:id="echoid-s2115" xml:space="preserve">R, en ſe ſervant des Tables de Sinus.</s>
              <s xml:id="echoid-s2116" xml:space="preserve"/>
            </p>
          </div>
          <div xml:id="echoid-div146" type="section" level="2" n="5">
            <head xml:id="echoid-head112" xml:space="preserve">
              <emph style="sc">Corollaire</emph>
              <emph style="sc">Quatrie’me</emph>
            .</head>
            <p>
              <s xml:id="echoid-s2117" xml:space="preserve">6. </s>
              <s xml:id="echoid-s2118" xml:space="preserve">Il ſuit enfin que ſi on a trois puiſſances, dont deux priſes
                <lb/>
              enſemble ſoient plus grandes que la troiſiéme, connoiſſant le
                <lb/>
              raport de ces trois puiſſances, on pourra déterminer ſelon quelle
                <lb/>
              direction chaque puiſſance doit tirer ou pouſſer, pour qu’agiſſant
                <lb/>
              toutes enſemble autour d’un point, elles ſoient en équilibre, puiſ-
                <lb/>
              que pour cela il ne faut que ſe donner trois lignes qui ayent entre
                <lb/>
              elles le même raport que les trois puiſſances en queſtion, enſuite
                <lb/>
              faire un triangle de ces trois lignes; </s>
              <s xml:id="echoid-s2119" xml:space="preserve">après quoi ſi d’un point quel-
                <lb/>
              conque pris dans la ſuperficie du triangle, l’on abbaiſſe des perpen-
                <lb/>
              diculaires ſur les côtés, elles détermineront les directions, ou, ce qui
                <lb/>
              eſt la même choſe, les angles que les puiſſances doivent former
                <lb/>
              entr’elles.</s>
              <s xml:id="echoid-s2120" xml:space="preserve"/>
            </p>
          </div>
          <div xml:id="echoid-div147" type="section" level="2" n="6">
            <head xml:id="echoid-head113" style="it" xml:space="preserve">Remarque premiere.</head>
            <p>
              <s xml:id="echoid-s2121" xml:space="preserve">7. </s>
              <s xml:id="echoid-s2122" xml:space="preserve">Il n’eſt pas neceſſaire que les trois puiſſances P, Q, R, ti-
                <lb/>
              rent ou pouſſent toutes trois enſemble le point H, pour être en </s>
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