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="18" file="0040" n="40" rhead="LA SCIENCE DES INGENIEURS,"/>
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          <head xml:id="echoid-head42" style="it" xml:space="preserve">Remarque premiere.</head>
          <p>
            <s xml:id="echoid-s637" xml:space="preserve">20. </s>
            <s xml:id="echoid-s638" xml:space="preserve">L’on doit remarquer ici que de toutes les figures que l’on
              <lb/>
            peut donner à un profil de muraille qui a quelque pouſſée à ſoûte-
              <lb/>
            nir, il n’y en a point où il faille moins de maçonnerie que dans
              <lb/>
            celle qui eſt triangulaire, parce que le lévier CE, gagne par ſa
              <lb/>
            longueur ce que le poids G, a de moins provenant d’un triangle,
              <lb/>
              <note position="left" xlink:label="note-0040-01" xlink:href="note-0040-01a" xml:space="preserve">
                <emph style="sc">Fig</emph>
              . 15.</note>
            que s’il provenoit d’un paralellograme, ce que je vais démontrer.</s>
            <s xml:id="echoid-s639" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s640" xml:space="preserve">Ayant le paralellograme rectangle AD, dont la hauteur ſoit
              <lb/>
            égale à celle du triangle précédent, & </s>
            <s xml:id="echoid-s641" xml:space="preserve">que la puiſſance qui
              <lb/>
              <note position="left" xlink:label="note-0040-02" xlink:href="note-0040-02a" xml:space="preserve">
                <emph style="sc">Fig</emph>
              . 10.</note>
            pouſſe de K, en C, ou tire de C, en G, ſelon une direction para-
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            lelle à l’horiſon, agiſſe avec la même force que celle du
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            triangle ABC, l’on ſait que pour avoir l’épaiſſeur BD, il faut
              <lb/>
            doubler la puiſſance K, & </s>
            <s xml:id="echoid-s642" xml:space="preserve">en extraire la racine quarrée, puiſqu’a-
              <note symbol="*" position="left" xlink:label="note-0040-03" xlink:href="note-0040-03a" xml:space="preserve">Art. 15.</note>
            près avoir fait les opérations ordinaires, il vient pour derniere
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            équation √2bf\x{0020} = y, & </s>
            <s xml:id="echoid-s643" xml:space="preserve">comme nous venons d’avoir √3bf\x{0020} = y pour
              <lb/>
            la baſe du triangle, l’on peut donc dire que la ſuperficie du profil
              <lb/>
            rectangle AD, ſera à celle du profil triangulaire, comme √2bf\x{0020} eſt
              <lb/>
            à la moitié de √3bf\x{0020}, puiſque ne prenant que la moitié de la baſe
              <lb/>
            du triangle, l’on peut regarder cette moitié comme la baſe du
              <lb/>
            rectangle égal au triangle, mais la moitié de √3bf\x{0020} eſt beaucoup
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            moindre que √2bf\x{0020}, & </s>
            <s xml:id="echoid-s644" xml:space="preserve">pour en être convaincu, il n’y a qu’à faire
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            un triangle rectangle & </s>
            <s xml:id="echoid-s645" xml:space="preserve">iſocelle ABC, & </s>
            <s xml:id="echoid-s646" xml:space="preserve">ſupoſer que chaque quarré
              <lb/>
              <note position="left" xlink:label="note-0040-04" xlink:href="note-0040-04a" xml:space="preserve">
                <emph style="sc">Fig</emph>
              . 14.</note>
            des côtés BA, & </s>
            <s xml:id="echoid-s647" xml:space="preserve">BC, eſt égal à bf, cela étant, l’hypotenuſe AC,
              <lb/>
            ou ce qui eſt la même choſe, √2bf\x{0020}, pourra être regardée comme
              <lb/>
            exprimant la baſe BD, du profil rectangle, & </s>
            <s xml:id="echoid-s648" xml:space="preserve">ſi l’on fait un autre
              <lb/>
            triangle rectangle ACD, dont le côté CD, ſoit égal à CB, l’hy-
              <lb/>
            potenuſe AD, exprimera la baſe AC, du profil triangulaire, & </s>
            <s xml:id="echoid-s649" xml:space="preserve">di-
              <lb/>
            viſant cette hypotenuſe en deux également au point E, ſa moitié
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            AE, ſera la baſe du paralellograme égal au triangle, ainſi la ſuper-
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            ficie du profil rectangle ſurpaſſera autant celle du profil triangulaire,
              <lb/>
            que la ligne AC, ſurpaſſe la moitié de la ligne AD, ce que l’on ne
              <lb/>
            peut pas exprimer en nombre bien exactement à cauſe des incom-
              <lb/>
            menſurables, cependant on peut dire que la maçonnerie du profil
              <lb/>
            triangulaire eſt à celle du profil rectangle, à peu-près comme 11.
              <lb/>
            </s>
            <s xml:id="echoid-s650" xml:space="preserve">à 18. </s>
            <s xml:id="echoid-s651" xml:space="preserve">ce qui fait voir qu’il y a plus d’un tiers moins dans le pre-
              <lb/>
            mier que dans le ſecond.</s>
            <s xml:id="echoid-s652" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s653" xml:space="preserve">Il ne faut pas trouver étrange qu’on ſupoſe ici un profil triangu-
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            laire, nous ſavons bien qu’on ne fait pas de Mur qui ſoit terminé </s>
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