DelMonte, Guidubaldo, Le mechaniche

List of thumbnails

< >
71
71 		72
72
73
73 		74
74
75
75 		76
76
77
77 		78
78
79
79 		80
80
< >
page |< < of 270 > >|
    <archimedes>
      <text id="id.0.0.0.0.3">
        <body id="id.2.0.0.0.0">
          <chap id="N106DF">
            <p id="id.2.1.244.0.0" type="main">
              <s id="id.2.1.244.14.0">
                <pb pagenum="31" xlink:href="037/01/077.jpg"/>
                <emph type="italics"/>
              peſeranno egualmente. </s>
              <s id="id.2.1.244.15.0">& percioche tutti peſano egualmente, tolti via i peſi HN,
                <lb/>
              iquali peſano egualmente, i reſtanti peſeranno egualmente; cioè i peſi EF, & il pe
                <emph.end type="italics"/>
                <arrow.to.target n="note81"/>
                <lb/>
                <emph type="italics"/>
              ſo LM pendenti dal centro C della bilancia. </s>
              <s id="id.2.1.244.16.0">Ma percioche la parte L è egua­
                <lb/>
              le ad F, & la parte M è eguale alla parte E; ſarà tutto LM eguale a i peſi
                <lb/>
              FE inſieme preſi. </s>
              <s id="id.2.1.244.17.0">& eſſendo CG eguale à CD, ſe i peſi EF ſaranno ſatti
                <lb/>
              pendenti dal punto D, i peſi EF appiccati in D peſeranno
                <expan abbr="egualmẽte">egualmente</expan>
              con LM.
                <lb/>
              </s>
              <s id="id.2.1.244.18.0">Per laqual coſa LM peſerà
                <expan abbr="egualmẽte">egualmente</expan>
                <expan abbr="tãto">tanto</expan>
              ad eßi EF appiccati in AB,
                <expan abbr="quã­to">quan­
                  <lb/>
                to</expan>
              ſe foſſero appiccati nel punto D; peroche la bilancia rimane ſempre nell'iſteſſo
                <emph.end type="italics"/>
                <arrow.to.target n="note82"/>
                <lb/>
                <emph type="italics"/>
              modo. </s>
              <s id="id.2.1.244.19.0">Adunque i peſi EF peſeranno tanto in AB quanto nel punto D; che
                <lb/>
              biſognaua moſtrare.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.245.0.0" type="margin">
              <s id="id.2.1.245.1.0">
                <margin.target id="note73"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              17.
                <emph type="italics"/>
              del quinto.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.246.0.0" type="margin">
              <s id="id.2.1.246.1.0">
                <margin.target id="note74"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              23.
                <emph type="italics"/>
              del quinto.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.247.0.0" type="margin">
              <s id="id.2.1.247.1.0">
                <margin.target id="note75"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              17.
                <emph type="italics"/>
              del quinto.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.248.0.0" type="margin">
              <s id="id.2.1.248.1.0">
                <margin.target id="note76"/>
                <emph type="italics"/>
              Corollario della quarta del quinto.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.249.0.0" type="margin">
              <s id="id.2.1.249.1.0">
                <margin.target id="note77"/>
              11.
                <emph type="italics"/>
              del
                <emph.end type="italics"/>
              5. </s>
            </p>
            <p id="id.2.1.250.0.0" type="margin">
              <s id="id.2.1.250.1.0">
                <margin.target id="note78"/>
              16.
                <emph type="italics"/>
              del
                <emph.end type="italics"/>
              5. </s>
            </p>
            <p id="id.2.1.251.0.0" type="margin">
              <s id="id.2.1.251.1.0">
                <margin.target id="note79"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              6.
                <emph type="italics"/>
              del
                <emph.end type="italics"/>
              1.
                <emph type="italics"/>
              di Archimede delle coſa che
                <expan abbr="egualmẽte">egualmente</expan>
              peſano.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.252.0.0" type="margin">
              <s id="id.2.1.252.1.0">
                <margin.target id="note80"/>
                <emph type="italics"/>
              Per la
                <emph.end type="italics"/>
              2.
                <emph type="italics"/>
              notitia commune di queſto.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.253.0.0" type="margin">
              <s id="id.2.1.253.1.0">
                <margin.target id="note81"/>
                <emph type="italics"/>
              Per la commune notitia di questo.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.254.0.0" type="margin">
              <s id="id.2.1.254.1.0">
                <margin.target id="note82"/>
                <emph type="italics"/>
              Per la commune notitia di questo.
                <emph.end type="italics"/>
              </s>
            </p>
            <p id="id.2.1.255.0.0" type="main">
              <s id="id.2.1.255.1.0">Ma queſte coſe tutte dimoſtreremo in altra maniera, & piu Mechani
                <lb/>
              camente. </s>
            </p>
            <figure id="id.037.01.077.1.jpg" xlink:href="037/01/077/1.jpg" number="65"/>
            <p id="id.2.1.257.0.0" type="main">
              <s id="id.2.1.257.1.0">
                <emph type="italics"/>
              Sia la bilancia AB, & il ſuo centro C, & ſiano, come nel primo caſo, due peſi EF
                <lb/>
              pendenti da i punti BG: & ſia GH ad HB, come il peſo F al peſo E. </s>
              <s id="id.2.1.257.2.0">Di­
                <lb/>
              co che i peſi EF peſeranno tanto in GB, quanto ſe ambidue ſteſſero pendenti
                <lb/>
              dal punto H della diuiſione. </s>
              <s id="id.2.1.257.3.0">Siano diſpoſte le medeſime coſe, cioè facciaſi AC
                <lb/>
              eguale à CH, & dal punto A ſiano appeſi due peſi LM, per modo che il pe
                <lb/>
              ſo E verſo il peſo L ſia come CA verſo CG; & come CB verſo CA, co
                <lb/>
              ſi ſia il peſo M verſo il peſo F. </s>
              <s id="id.2.1.257.4.0">I peſi LM peſeranno egualmente (come è detto
                <lb/>
              di ſopra) con li peſi EF appiccati in GB. </s>
              <s id="id.2.1.257.5.0">Siano dapoi due punti NO li centri
                <lb/>
              della grauezza de' peſi EF; & ſiano congiunte le linee GN BO; & ſia con­
                <lb/>
              giunta NO, laquale ſarà come bilancia; laquale etiandio faccia sì, che le linee
                <lb/>
              GN BO ſiano tra loro egualmente diſtanti; & dal punto H ſia tirata la HP
                <lb/>
              à piombo dell'orizonte, laquale tagli NO nel P, & ſia egualmente distante dal
                <lb/>
              le linee GN BO. </s>
              <s id="id.2.1.257.6.0">In fine congiungaſi GO, laquale tagli HP in R. </s>
              <s id="id.2.1.257.7.0">Percio
                <emph.end type="italics"/>
                <arrow.to.target n="note83"/>
                <lb/>
                <emph type="italics"/>
              che dunque HR è egualmente diſtante dal lato BO del triangolo GBO; ſarà
                <lb/>
              la GH verſola HB, come GR ad RO. </s>
              <s id="id.2.1.257.8.0">Similmente percioche RP è egual
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
Impressum