Heron Alexandrinus, Mechanica, 1999

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 101 > >|
    <archimedes>
      <text>
        <body>
          <chap n="2">
            <pb n="37">
              <s id="A18-2.37.01">37 We want to prove the same for the pentagon <abgde>.</s>
              <s id="A18-2.37.02">Let us draw <be> and determine the center of gravity of the triangle <abe>; let it fall on point <z>; let the center of gravity of the quadrangle <bgde> be at point <h>.</s>
              <s id="A18-2.37.03">Let us connect the two points <z> and <h> and divide the line <zh> in two parts so that <hq> relates to <qz> like the weight of the triangle <abe> to the weight of the quadrangle <bgde>, then the point <q> is the center of gravity of the figure <abgde>.</s>
              <s id="A18-2.37.04">We have to imagine it the same way for all polygons.</s>
            </pb>
          </chap>
        </body>
      </text>
    </archimedes>
Impressum