DelMonte, Guidubaldo, Le mechaniche

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    <archimedes>
      <text id="id.0.0.0.0.3">
        <body id="id.2.0.0.0.0">
          <chap id="N106DF">
            <p id="id.2.1.179.0.0" type="main">
              <s id="id.2.1.179.9.0">
                <pb pagenum="13" xlink:href="037/01/041.jpg"/>
                <emph type="italics"/>
              piu graue, che in D. </s>
              <s id="id.2.1.179.10.0">Similmente mostreraſſi, che quanto il peſo ſarà piu da preſſo
                <lb/>
              ad F, come in L manco grauerà; ma quanto piu da preſſo ſi trouerà al G, co­
                <lb/>
              me in H, eſſere piu graue.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.181.0.0" type="main">
              <s id="id.2.1.181.1.0">
                <emph type="italics"/>
              Che ſe il centro del mondo foſſe in S fra i punti CG; Primieramente ſi moſtrerà nel
                <lb/>
              modo iſteſſo, che il peſo in qualunque luogo poſto starà ſopra il centro C, come in
                <lb/>
              H: peroche tirate le li­
                <lb/>
              nee HG HS, l'angolo
                <lb/>
              che è alla baſe GHC del
                <lb/>
                <expan abbr="triãgolo">triangolo</expan>
              di due lati eguali
                <lb/>
              CHG è ſempre acuto:
                <lb/>
              Per laqual coſa anco SHC
                <lb/>
              minor di lui ſarà parimen
                <lb/>
              te ſempre acuto. </s>
              <s id="id.2.1.181.2.0">ma ſia ti
                <lb/>
              rata dal punto S la linea
                <lb/>
              SK à piombo di CS.
                <lb/>
              </s>
              <s id="id.2.1.181.3.0">Dico che il peſo è piu gra­
                <lb/>
              ue in
                <emph.end type="italics"/>
              K,
                <emph type="italics"/>
              che in alcun'al
                <lb/>
              tro ſito della circonferen
                <lb/>
              za FKG; & quanto
                <lb/>
              piu da preſſo ſarà allo F,
                <lb/>
              ouero al G meno graue­
                <lb/>
              rà. </s>
              <s id="id.2.1.181.4.0">Prendanſi verſo lo
                <lb/>
              F i punti DL, & con
                <lb/>
                <expan abbr="giungãſi">giunganſi</expan>
              le linee LC LS
                <lb/>
              DC DS, & ſiano al­
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.037.01.041.1.jpg" xlink:href="037/01/041/1.jpg" number="24"/>
                <lb/>
                <emph type="italics"/>
              lungate le linee LS DS KS HS fin'alla
                <expan abbr="circõferenza">circonferenza</expan>
              del cerchio in EM NO;
                <lb/>
              & ſiano
                <expan abbr="cõgiunte">congiunte</expan>
              CE, CM, CN, CO. </s>
              <s id="id.2.1.181.5.0">Hor percioche LE DM ſi taglia­
                <lb/>
              no inſieme in S, ſarà il rettangolo LSE eguale al rettangolo DSM. </s>
              <s id="id.2.1.181.6.0">Onde ſi co
                <emph.end type="italics"/>
                <arrow.to.target n="note39"/>
                <lb/>
                <emph type="italics"/>
              me è la LS verſo la DS, coſi ſarà la SM verſola SE; ma è maggior la LS
                <lb/>
              della DS; & la SM di eſſa SE. </s>
              <s id="id.2.1.181.7.0">Dunque LS SE preſe inſieme ſaranno mag­
                <emph.end type="italics"/>
                <arrow.to.target n="note40"/>
                <lb/>
                <emph type="italics"/>
              giori delle DS SM. </s>
              <s id="id.2.1.181.8.0">& per la ragion iſteſſa ſi moſtrerà la KN eſſer minore di DM.
                <lb/>
              </s>
              <s id="id.2.1.181.9.0">Di piu percioche il rettangolo OSH è eguale al rett'angolo KSN; per la medeſi­
                <emph.end type="italics"/>
                <arrow.to.target n="note41"/>
                <lb/>
                <emph type="italics"/>
              ma ragione la HO ſarà maggiore della KN. </s>
              <s id="N11604">& nell'iſteſſo modo in tutto la
                <emph.end type="italics"/>
                <arrow.to.target n="note42"/>
                <lb/>
                <emph type="italics"/>
              KN ſi dimostrerà minore di tutte le altre linee, che paſſino per lo punto S. </s>
              <s id="id.2.1.181.10.0">Et
                <lb/>
              percioche de i triangoli di due lati eguali CLE DCM i lati LC CE ſono e­
                <lb/>
              guali a i lati DC CM; & la baſe LE è maggiore di DM: ſarà l'angolo
                <lb/>
              LCE maggiore dell'angolo DCM. </s>
              <s id="id.2.1.181.11.0">Per laqual coſa gli angoli CLE CEL po
                <emph.end type="italics"/>
                <arrow.to.target n="note43"/>
                <lb/>
                <emph type="italics"/>
              sti alla baſe tolti inſieme ſaranno minori de gli angoli CDM CMD; & le me­
                <lb/>
              tà di queſti, cioè l'angolo CLS ſarà minore dell'angolo CDS. </s>
              <s id="id.2.1.181.12.0">Dunque il peſo po
                <lb/>
              ſto in L ſopra la linea LC grauerà piu, che poſto in D ſopra la DC; & piu
                <lb/>
              ſtarà ſopra il centro in L, che in D. </s>
              <s id="id.2.1.181.13.0">Similmente ſi moſtrerà, che il peſo in D
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
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