Bion, Nicolas, Traité de la construction et principaux usages des instruments de mathématique, 1723

Table of figures

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[121] Fig. 6.F E K G A D I H B C
[122] Fig. 7.A D E B
[123] Fig. 8.A B
[124] Fig. 9.A D E F C B
[125] Fig. 10.C A B
[126] Fig. 11.C A B
[127] Fig. 12.C A B
[128] Fig. 13.F D E
[129] Fig. 14.C A B
[130] Fig. 15.G H I
[131] Fig. 16.F D E
[132] Fig. 17.D E C F F A B
[133] Fig. 18.D E A B
[134] Fig. 19.C A E B D
[135] Fig. 1.C A 1 2 3 4 5 B D
[136] Fig. 2.D F B A C H G E
[137] Fig. 3.D E G F A B C
[138] Fig. 4.D A C B
[139] Fig. 5.E K G A C D B F I H
[140] Fig. 6.D A E G F B C
[141] Fig. 7.O P T H C H P I L A F G E B P H D H P
[142] Fig. 8.D E C A B
[143] Fig. 8.d e c a b
[144] Fig. 9.D d E e F c C a b A B
[145] Fig. 10.D E C A B
[146] Fig. 10.d e c a b 20 40 60 80 100 10 30 50 70 90 K L 20 40 60 80 100 10 30 50 70 90 G H
[147] Fig. 11.D E C A B
[148] Fig. 11.I N L G K M H
[149] Fig. 11.d e c a b
[150] Fig. 12.D C A B d c a b
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        <div xml:id="echoid-div412" type="section" level="1" n="189">
          <p>
            <s xml:id="echoid-s4925" xml:space="preserve">
              <pb o="151" file="165" n="165" rhead="POUR LEVER LES PLANS. Liv. IV. Ch. V."/>
            & </s>
            <s xml:id="echoid-s4926" xml:space="preserve">que lâ diſtance meſurée ſoit de 20 toiſes; </s>
            <s xml:id="echoid-s4927" xml:space="preserve">je diſpoſe la regle de
              <lb/>
            proportion en la maniere ſuivante. </s>
            <s xml:id="echoid-s4928" xml:space="preserve">[20. </s>
            <s xml:id="echoid-s4929" xml:space="preserve">100. </s>
            <s xml:id="echoid-s4930" xml:space="preserve">20.</s>
            <s xml:id="echoid-s4931" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s4932" xml:space="preserve">Multipliant 20, par 100, & </s>
            <s xml:id="echoid-s4933" xml:space="preserve">diviſant le produit 2000 par 40,
              <lb/>
            on trouvera pour quatriéme terme de cette regle 50, qui ſignifie
              <lb/>
            que la hauteur de la Tour eſt de 50 toiſes.</s>
            <s xml:id="echoid-s4934" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s4935" xml:space="preserve">Mais ſi le fil du plomp a coupé le côté d'ombre verſe, comme par
              <lb/>
            exemple, au point marqué 60, & </s>
            <s xml:id="echoid-s4936" xml:space="preserve">que la diſtance meſurée ſoit de
              <lb/>
            35 toiſes, diſpoſez les trois premiers termes de la regle de proprion
              <lb/>
            en cette autre maniere. </s>
            <s xml:id="echoid-s4937" xml:space="preserve">[100. </s>
            <s xml:id="echoid-s4938" xml:space="preserve">60. </s>
            <s xml:id="echoid-s4939" xml:space="preserve">35.</s>
            <s xml:id="echoid-s4940" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s4941" xml:space="preserve">Mult
              <unsure/>
            ipliez 35 par 60, & </s>
            <s xml:id="echoid-s4942" xml:space="preserve">le produit 2100, étant diviſé par 100;
              <lb/>
            </s>
            <s xml:id="echoid-s4943" xml:space="preserve">le quotien 21 ſera la hauteur de la Tour.</s>
            <s xml:id="echoid-s4944" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div414" type="section" level="1" n="190">
          <head xml:id="echoid-head289" style="it" xml:space="preserve">Vſage du Treillis ſans calcul.</head>
          <p>
            <s xml:id="echoid-s4945" xml:space="preserve">TOutes ces operations ſe peuvent réſoudre ſans calcul, comme
              <lb/>
            nous allons le faire voir par quelques exemples.</s>
            <s xml:id="echoid-s4946" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div415" type="section" level="1" n="191">
          <head xml:id="echoid-head290" xml:space="preserve">USAGE I.</head>
          <p>
            <s xml:id="echoid-s4947" xml:space="preserve">SUppoſons, comme nous avons déja fait, que le fil du plomb
              <lb/>
              <note position="right" xlink:label="note-165-01" xlink:href="note-165-01a" xml:space="preserve">Fig. G.</note>
            coupe le côté d'ombre droite au point marqué 40, & </s>
            <s xml:id="echoid-s4948" xml:space="preserve">que la diſ-
              <lb/>
            tance meſurée ſoit de 20 toiſes, cherehez dans le treillis celle des
              <lb/>
            perpendiculaires au raïon, qui ſoit de 20 parties depuis le fil; </s>
            <s xml:id="echoid-s4949" xml:space="preserve">cette
              <lb/>
            perpendiculaire coupera le côté du quarré qui aboutit an centre au
              <lb/>
            point marqué 50; </s>
            <s xml:id="echoid-s4950" xml:space="preserve">c'eſt pourquoi en ce cas la hauteur de la Tour
              <lb/>
            ſera de 50 toiſes.</s>
            <s xml:id="echoid-s4951" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s4952" xml:space="preserve">On diviſe quelquefois l'alidade mobile en parties égales à celles
              <lb/>
            du Treillis, & </s>
            <s xml:id="echoid-s4953" xml:space="preserve">par ce moyen on peut connoître la longueur de l'hy-
              <lb/>
            potenuſe ou raïon viſuel, eu rapportant l'alidade diviſée à la place
              <lb/>
            du fil.</s>
            <s xml:id="echoid-s4954" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div417" type="section" level="1" n="192">
          <head xml:id="echoid-head291" xml:space="preserve">USAGE II.</head>
          <p>
            <s xml:id="echoid-s4955" xml:space="preserve">MAis ſi le fil coupoit le côté d'ombre verſe au point marqué
              <lb/>
            60, & </s>
            <s xml:id="echoid-s4956" xml:space="preserve">que la diſtance meſurée fut de 35 toiſes, comptez ſur
              <lb/>
            le raïon pu quart du cercle depuis le centre, 35 parties; </s>
            <s xml:id="echoid-s4957" xml:space="preserve">comptez
              <lb/>
            auſſi les diviſions de la perpendiculaire depuis ce point 35 juſqu'au
              <lb/>
            fil, vous y trouverez 21 parties; </s>
            <s xml:id="echoid-s4958" xml:space="preserve">c'eſt pourquoi en ce cas la hau-
              <lb/>
            teur de la Tour ſeroit de 21 toiſes.</s>
            <s xml:id="echoid-s4959" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s4960" xml:space="preserve">Souvenez-vous qu'en tous les cas il faut ajoûter la hauteur du
              <lb/>
            centre de l'Inſtrument au-deſſus du terrain. </s>
            <s xml:id="echoid-s4961" xml:space="preserve">Si, par exemple, cette
              <lb/>
            hauteur eſt 5 pieds, la hauteur de la Tour dans le dernier exem-
              <lb/>
            ple ſera de 21 toiſes 5 pieds.</s>
            <s xml:id="echoid-s4962" xml:space="preserve"/>
          </p>
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