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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              bre de parties égales, par chaque point de diviſion comme L & </s>
              <s xml:id="echoid-s7402" xml:space="preserve">O,
                <lb/>
              abaiſſer les perpendiculaires LM, OP, &</s>
              <s xml:id="echoid-s7403" xml:space="preserve">c. </s>
              <s xml:id="echoid-s7404" xml:space="preserve">ſur le demi diamêtre CB,
                <lb/>
              tirer les rayons BK, BN, &</s>
              <s xml:id="echoid-s7405" xml:space="preserve">c. </s>
              <s xml:id="echoid-s7406" xml:space="preserve">auſſi bien que les lignes KE, NE, &</s>
              <s xml:id="echoid-s7407" xml:space="preserve">c.
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              </s>
              <s xml:id="echoid-s7408" xml:space="preserve">enſuite chercher aux lignes BC, BI, & </s>
              <s xml:id="echoid-s7409" xml:space="preserve">au ſinus LM, (que nous
                <lb/>
              regarderons comme le plus petit de tous), une quatriéme propor-
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              tionnelle que l’on portera ſur la verticale FH, en commençant
                <lb/>
              du point r, qui répond immediatement au-deſſous du poids G, & </s>
              <s xml:id="echoid-s7410" xml:space="preserve">
                <lb/>
              ſupoſant que rR ſoit égale à la quatriéme proportionnelle qu’on
                <lb/>
              vient de trouver, on élevera au point R la perpendiculaire RS in-
                <lb/>
              definie; </s>
              <s xml:id="echoid-s7411" xml:space="preserve">on cherchera de même aux lignes BC, BI, & </s>
              <s xml:id="echoid-s7412" xml:space="preserve">au ſinus OP,
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              (que nous ſupoſons ſuivre immediatement le plus petit LM,) une
                <lb/>
              quatriéme proportionnelle qu’on portera depuis r juſqu’en T, & </s>
              <s xml:id="echoid-s7413" xml:space="preserve">
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              on élevera encore la perpendiculaire TV.</s>
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              <s xml:id="echoid-s7415" xml:space="preserve">Le triangle CBI étant rectangle & </s>
              <s xml:id="echoid-s7416" xml:space="preserve">izocelle, il ſera bien aiſé de
                <lb/>
              trouver toutes les quatriémes proportionnelles dont nous avons be-
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              ſoin; </s>
              <s xml:id="echoid-s7417" xml:space="preserve">car ſi l’on prend chaque ſinus comme LM ou OP, pour le
                <lb/>
              côté d’un quarré, la diagonalle de ce quarré ſera quatriéme propor-
                <lb/>
              tionnelle aux lignes BC, BI, & </s>
              <s xml:id="echoid-s7418" xml:space="preserve">au ſinus qu’on aura pris pour côté
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              du quarré, ce qui eſt bien évident à cauſe des triangles ſemblables.</s>
              <s xml:id="echoid-s7419" xml:space="preserve"/>
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              <s xml:id="echoid-s7420" xml:space="preserve">Après qu’on aura toutes les perpondiculairoc
                <unsure/>
              , comme R, S, T,
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              V, &</s>
              <s xml:id="echoid-s7421" xml:space="preserve">c. </s>
              <s xml:id="echoid-s7422" xml:space="preserve">on tirera une ligne d e, égale à la longueur de la cor-
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              de AEFG, l’on prendra dans cette ligne (en commençant de l’ex-
                <lb/>
              trémité d,) la partie d f, égale à la diſtance du centre de la poulie
                <lb/>
              F au poids G, c’eſt-à-dire, égale à cette partie de la corde qui eſt
                <lb/>
              paralelle à la verticale FH, quand le levier AB eſt horiſontal; </s>
              <s xml:id="echoid-s7423" xml:space="preserve">on
                <lb/>
              prendra la difference de la ligne KE, quirépond aurayon de la pre-
                <lb/>
              miere diviſion à la ligne AE, & </s>
              <s xml:id="echoid-s7424" xml:space="preserve">on portera cette difference depuis
                <lb/>
              f juſqu’en h, alors on prendra la longueur d h avec un compas,
                <lb/>
              pour décrire un arc qui aura pour centre celui de la poulie F, & </s>
              <s xml:id="echoid-s7425" xml:space="preserve">
                <lb/>
              cet arc venant couper la perpendiculaire RS, donnera le point
                <lb/>
              S qui eſt un de ceux de la courbe, par le moyen duquel on aura
                <lb/>
              l’ordonnée S a & </s>
              <s xml:id="echoid-s7426" xml:space="preserve">ſon abciſſe r a; </s>
              <s xml:id="echoid-s7427" xml:space="preserve">de même prenant la difference
                <lb/>
              des lignes NE & </s>
              <s xml:id="echoid-s7428" xml:space="preserve">AE, pour la porter de f en j, ſi l’on ouvre le com-
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              pas de l’intervale d j, & </s>
              <s xml:id="echoid-s7429" xml:space="preserve">que du centre F de la poulie, on décrive
                <lb/>
              un arc qui vienne couper la perpendiculaire TV, on aura un autre
                <lb/>
              point V de la courbe, qui donnera l’ordonnée V b & </s>
              <s xml:id="echoid-s7430" xml:space="preserve">l’abciſſe BIr;
                <lb/>
              </s>
              <s xml:id="echoid-s7431" xml:space="preserve">enfin le point N étant parvenu en E, toute la ligne AE pourra être
                <lb/>
              priſe pour ſa difference avec zero, & </s>
              <s xml:id="echoid-s7432" xml:space="preserve">le portant depuis F juſqu’en
                <lb/>
              K, ouvrant le compas de l’intervale d K, on décrira du centre or-
                <lb/>
              dinaire, un arc qui venant rencontrer la derniere perpendiculaire
                <lb/>
              h X, donnera le point X qui ſera celui de la courbe, où va ſe ter- </s>
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