DelMonte, Guidubaldo, Le mechaniche

Table of figures

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            <pb xlink:href="037/01/058.jpg"/>
            <p id="id.2.1.395.0.0" type="main">
              <s id="id.2.1.395.1.0">
                <emph type="italics"/>
              Sia la bilancia AB egualmente diſtante dall'orizonte, il cui centro C ſia ſopra la
                <lb/>
              bilancia, & ſia il perpendicolo CD: & ſiano i centri della grauezza di peſi eguali
                <lb/>
              poſti in AB: & la bilancia ſia moſſa in EF. </s>
              <s id="id.2.1.395.2.0">Dico, che il peſo posto in E ha
                <lb/>
              grauezza maggiore, che il
                <lb/>
              peſo posto in F. </s>
              <s id="N12078">& per­
                <lb/>
              ciò la bilancia EF eſſe­
                <lb/>
              re per ritornare in A B. </s>
              <s id="id.2.1.395.3.0">ſia allungata prima la linea
                <lb/>
              CD fin'al centro del mon
                <lb/>
              do, che ſia S. </s>
              <s id="id.2.1.395.4.0">Dapoi ſia­
                <lb/>
              no congiunte le linee AC,
                <lb/>
              CB, EC, CF, HS;
                <lb/>
              & dai punti EF ſiano ti­
                <lb/>
              rate le linee EKGFL egual
                <lb/>
                <expan abbr="mẽte">mente</expan>
              diſtanti da HS. </s>
              <s id="id.2.1.395.5.0">Per­
                <lb/>
              cioche dunque la diſceſa na
                <lb/>
              turale diritta di tutta la
                <lb/>
              grandezza, cioè della bilan
                <lb/>
              cia EF coſi diſpoſta inſie
                <lb/>
              me co'peſi è ſecondo la gra­
                <lb/>
              uezza del centro H per la
                <lb/>
              diritta linea HS; ſarà
                <expan abbr="parimẽte">pa
                  <lb/>
                rimente</expan>
              la diſceſa de'peſi meſ
                <lb/>
              ſi in EF coſi diſpoſti ſecon
                <lb/>
              do le linee diritte E
                <emph.end type="italics"/>
              K
                <lb/>
                <emph type="italics"/>
              FL egualmente distanti
                <lb/>
              da HS, ſi come di ſopra
                <lb/>
              habbiamo dimoſtrato. </s>
              <s id="id.2.1.395.6.0">La
                <lb/>
              diſceſa dunque, & la ſali­
                <lb/>
              ta de i peſi poſti in EF ſi
                <lb/>
              dirà più, & meno obliqua
                <lb/>
              ſecondo la vicinanza, ò lon
                <lb/>
              tananza diputata ſecondo
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note119"/>
                <emph type="italics"/>
              le linee EK FL. </s>
              <s id="id.2.1.395.7.0">& per­
                <lb/>
              cioche li due lati AD DC
                <lb/>
              ſono eguali a i due lati BD
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.037.01.058.1.jpg" xlink:href="037/01/058/1.jpg" number="41"/>
                <lb/>
                <emph type="italics"/>
              DC; & gli angoli al D ſono retti, ſarà il lato AC eguale al lato CB. </s>
              <s id="N120E4">& eſ­
                <lb/>
              ſendo il punto C immobile; mentre, che i punti AB ſi moueranno, de ſcriueran­
                <lb/>
              no la circonferenza di vno cerchio, il cui mezo diametro ſarà AC. </s>
              <s id="id.2.1.395.8.0">Per laqual co
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="note120"/>
                <emph type="italics"/>
              ſa co'l centro C ſia deſcritto il cerchio AE BF, i punti AB E
                <emph.end type="italics"/>
              F
                <emph type="italics"/>
              ſaranno nel
                <lb/>
              la circonferenza del cerchio. </s>
              <s id="id.2.1.395.9.0">ma eſſendo EF eguale ad AB, ſarà la circonfe­
                <lb/>
              renza EAF eguale alla circonferenza AFB. </s>
              <s id="id.2.1.395.10.0">Onde tolta via la comune AF
                <emph.end type="italics"/>
              </s>
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