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="31" file="0139" n="142" rhead="LIVRE II. DE LA MECANIQUE DES VOUTES."/>
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          <div xml:id="echoid-div187" type="section" level="2" n="30">
            <head xml:id="echoid-head143" style="it" xml:space="preserve">Troiſiéme Principe.</head>
            <p>
              <s xml:id="echoid-s2688" xml:space="preserve">27. </s>
              <s xml:id="echoid-s2689" xml:space="preserve">Si au point H, où une tengente HI, touche l’Ellipſe on éle-
                <lb/>
              ve une pérpendiculaire HK qui aille rencontrer l’axe AB au point
                <lb/>
              K, je dis que FG eſt à GK comme le quarré de AF eſtau quarré de
                <lb/>
              FD, ou, ce qui revient au même, comme le rectangle de AG par
                <lb/>
              GB eſt au quarré GH.</s>
              <s xml:id="echoid-s2690" xml:space="preserve"/>
            </p>
            <p>
              <s xml:id="echoid-s2691" xml:space="preserve">Pour le prouver, conſiderés que les triangles IGH & </s>
              <s xml:id="echoid-s2692" xml:space="preserve">GHK, ſont
                <lb/>
              ſemblables, par conſéquent IG ({aa-xx/x}), GH (yy) : </s>
              <s xml:id="echoid-s2693" xml:space="preserve">: </s>
              <s xml:id="echoid-s2694" xml:space="preserve">GH
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              (y); </s>
              <s xml:id="echoid-s2695" xml:space="preserve">GK{(yy)/aa-xx;</s>
              <s xml:id="echoid-s2696" xml:space="preserve">/x} ou ce qui eſt la même choſe {yyx/aa-xx}; </s>
              <s xml:id="echoid-s2697" xml:space="preserve">comme
                <lb/>
              nous avons l’expreſſion de KG, il n’eſt donc queſtion que de prouver
                <lb/>
              que GF (x) eſt à GK ({yyx/aa-xx}) commele rectangle de AG par GB
                <lb/>
              (aa-xx) eſt au quarré de GH (yy), ce qui eſt bien évident, puiſ-
                <lb/>
              que le produit des extrémes & </s>
              <s xml:id="echoid-s2698" xml:space="preserve">celui des moyens donnent l’un & </s>
              <s xml:id="echoid-s2699" xml:space="preserve">
                <lb/>
              l’autre yyx; </s>
              <s xml:id="echoid-s2700" xml:space="preserve">car on remarquera que c’eſt multiplier le ſecond terme
                <lb/>
              yyx par aa-xx que de ne le pas diviſer par la même quantité.</s>
              <s xml:id="echoid-s2701" xml:space="preserve"/>
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              <s xml:id="echoid-s2702" xml:space="preserve">Comme les propriétés de l’Ellipſe ſont toujours les mêmes, ſoit
                <lb/>
              que la tengente aille rencontrer le grand axe AB prolongé, ou le
                <lb/>
              petit axe DE auſſi prolongé, l’on verra par une démonſtration ſem-
                <lb/>
              blable à la précédente, que ſi la perpendiculaire élevée ſur la tengen-
                <lb/>
              te IO alloit rencontrer le petit axe ED au point L, l’on auroit encore
                <lb/>
              le quarré de EF eſt au quarré de AF comme la coupée MF eſt à la
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              ligne ML.</s>
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              <emph style="sc">Corollaire</emph>
              <emph style="sc">Premier</emph>
            .</head>
            <p>
              <s xml:id="echoid-s2704" xml:space="preserve">28. </s>
              <s xml:id="echoid-s2705" xml:space="preserve">Il ſuit du premier principe, que quand on connoîtra les deux
                <lb/>
              diamêtres AB & </s>
              <s xml:id="echoid-s2706" xml:space="preserve">ED d’une Ellipſe, & </s>
              <s xml:id="echoid-s2707" xml:space="preserve">la diſtance du centre F au point
                <lb/>
              G où on aura mené une ordonnée GH, qu’on connoîtra toujours
                <lb/>
              la valeur de cette ordonnée en nombre, en diſant ſi le quarré du
                <lb/>
              demi diamêtre AF donne tant pour le quarré du diamêtre FD, que
                <lb/>
              donnera la difference du quarré de AF au quarré FG, pour le quarré
                <lb/>
              GH que l’on cherche? </s>
              <s xml:id="echoid-s2708" xml:space="preserve">lequel étant trouvé, on n’aura qu’à en ex-
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              traire la racine quarrée, qui ſera la perpendiculaire GH.</s>
              <s xml:id="echoid-s2709" xml:space="preserve"/>
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          <div xml:id="echoid-div189" type="section" level="2" n="32">
            <head xml:id="echoid-head145" xml:space="preserve">
              <emph style="sc">Corollaire</emph>
              <emph style="sc">Second</emph>
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            <p>
              <s xml:id="echoid-s2710" xml:space="preserve">29. </s>
              <s xml:id="echoid-s2711" xml:space="preserve">Il ſuit auſſi du troiſiéme principe, que ſi on avoit beſoin de </s>
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