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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              <s xml:id="echoid-s2076" xml:space="preserve">
                <pb o="7" file="0115" n="118" rhead="LIVRE II. DE LA MECANIQUE DES VOUTES."/>
              le paralellograme ABCD, la puiſſance P, eſt exprimée par le côté
                <lb/>
              AB, la puiſſance Q, par le côté AD, & </s>
              <s xml:id="echoid-s2077" xml:space="preserve">la puiſſance R, par la
                <lb/>
              diagonale CA: </s>
              <s xml:id="echoid-s2078" xml:space="preserve">ou, ce qui revient au même, ſi chaque puiſſance eſt
                <lb/>
              exprimée par un des côtés du triangle ABC, parce qu’à la place de
                <lb/>
              AD, l’on pourra prendre BC, qui lui eſt égal ; </s>
              <s xml:id="echoid-s2079" xml:space="preserve">ſupoſant donc
                <note symbol="*" position="right" xlink:label="note-0115-01" xlink:href="note-0115-01a" xml:space="preserve">V. le C.
                  <lb/>
                Art. 767.</note>
              qu’on ſoit bien prevenu de cette verité, voici une propoſition
                <lb/>
              fondamentale qu’on en peut tirer.</s>
              <s xml:id="echoid-s2080" xml:space="preserve"/>
            </p>
            <p>
              <s xml:id="echoid-s2081" xml:space="preserve">Ayant trois puiſſances P, Q, R, qui tirent ou pouſſent toutes
                <lb/>
              trois enſemble au tour du point A, je dis qu’elles ſeront en équi-
                <lb/>
              libres, ſi la force avec laquelle chacune agit eſt exprimée par un
                <lb/>
              des côtés du triangle EFG, qui couperoit en angles droits la ligne
                <lb/>
              de direction de chaque puiſſance.</s>
              <s xml:id="echoid-s2082" xml:space="preserve"/>
            </p>
            <p>
              <s xml:id="echoid-s2083" xml:space="preserve">Pour le prouver, remarquez que ſi la ligne AO, eſt perpendicu-
                <lb/>
              laire ſur le côté EF; </s>
              <s xml:id="echoid-s2084" xml:space="preserve">& </s>
              <s xml:id="echoid-s2085" xml:space="preserve">la ligne CT, ſur le côté EG (comme nous
                <lb/>
              le ſupoſons) l’on aura les deux triangles AOF & </s>
              <s xml:id="echoid-s2086" xml:space="preserve">FTE, ſemblables,
                <lb/>
              puiſqu’ils ont chacun un angle droit, & </s>
              <s xml:id="echoid-s2087" xml:space="preserve">l’angle OFT, qui leur eſt
                <lb/>
              commun; </s>
              <s xml:id="echoid-s2088" xml:space="preserve">ainſi l’angle E, ſera égal à l’angle O A E. </s>
              <s xml:id="echoid-s2089" xml:space="preserve">Par un ſembla-
                <lb/>
              ble raiſonnement on verra auſſi que le triangle FAS eſt ſemblable
                <lb/>
              au triangle FTG, & </s>
              <s xml:id="echoid-s2090" xml:space="preserve">que de même l’angle G, ſera égal à l’angle
                <lb/>
              FAS; </s>
              <s xml:id="echoid-s2091" xml:space="preserve">mais comme ce dernier l’eſt encore à l’angle alterne BCA, il
                <lb/>
              s’enſuit donc que le triangle ABC eſt ſemblable au triangle EFG:
                <lb/>
              </s>
              <s xml:id="echoid-s2092" xml:space="preserve">ainſi les trois côtés du grand triangle pourront donc être pris à la
                <lb/>
              place de ceux du petit, & </s>
              <s xml:id="echoid-s2093" xml:space="preserve">par conſequent exprimer le raport de
                <lb/>
              chaque puiſſance dont ils coupent la ligne de direction en angles
                <lb/>
              droits; </s>
              <s xml:id="echoid-s2094" xml:space="preserve">mais comme nous avons vû que ces trois puiſſances étoient
                <lb/>
              en équilibre, lorſque leur raport étoit exprimé par les côtés du petit
                <lb/>
              triangle ABC, l’on peut donc dire qu’elles ſeront encore en équilibre
                <lb/>
              quand leur raport ſera exprimé par les côtés du triangle EFG. </s>
              <s xml:id="echoid-s2095" xml:space="preserve">C. </s>
              <s xml:id="echoid-s2096" xml:space="preserve">
                <lb/>
              Q. </s>
              <s xml:id="echoid-s2097" xml:space="preserve">F. </s>
              <s xml:id="echoid-s2098" xml:space="preserve">D.</s>
              <s xml:id="echoid-s2099" xml:space="preserve"/>
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              <emph style="sc">Corollaire</emph>
              <emph style="sc">Premier</emph>
            .</head>
            <p>
              <s xml:id="echoid-s2100" xml:space="preserve">3. </s>
              <s xml:id="echoid-s2101" xml:space="preserve">Il ſuit que quand on aura trois puiſſances P, Q, R, qui tirent
                <lb/>
                <note position="right" xlink:label="note-0115-02" xlink:href="note-0115-02a" xml:space="preserve">
                  <emph style="sc">Fig</emph>
                . 3.</note>
              ou pouſſent au tour du point H, ſi elles ſont en équilibre, on con-
                <lb/>
              noîtra toûjours le raport que ces puiſſances ont entr’elles, puiſqu’on
                <lb/>
              n’aura qu’à couper chaque ligne de direction en angles droits par
                <lb/>
              une ligne tirée à telle diſtance que l’on voudra du point H; </s>
              <s xml:id="echoid-s2102" xml:space="preserve">car ces
                <lb/>
              trois lignes venant à ſe rencontrer, donneront les côtés du trian-
                <lb/>
              gle IKL, qui exprimeront le raport des puiſſances; </s>
              <s xml:id="echoid-s2103" xml:space="preserve">c’eſt-à-dire,
                <lb/>
              que ſil’on ſupoſe que la puiſſance P, ſoit exprimée par IK, la puiſ-
                <lb/>
              ſance Q, le ſera par KL, & </s>
              <s xml:id="echoid-s2104" xml:space="preserve">la puiſſance R, par IL.</s>
              <s xml:id="echoid-s2105" xml:space="preserve"/>
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