Heron Alexandrinus - Mechanica

Heron Alexandrinus, Mechanica, 1999

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[8] As for the case that a point that is moving in two motions of constant speed can follow unequal lines, we are going to prove that now.Let us assume a rectangle, namely <abgd>, and let the line <ad> be a diagonal; further let the point <a> run in steady motion on the line <ab> and let the line move in steady motion on the two lines <ag>, <bd>, so that it is always parallel to line <gd>; also let the time in which point <a> runs to <b> be equal to the time in which line <ab> reaches <gd>; thus I say that point <a> in a certain time covers two unequal lines.Proof for this: If line <ab> moves for a certain time and is placed on line <ez>, then the point moving on line <ab> in the same time comes to rest on line <ez> and a constant relation sets in.The relation between line <ag> and line <ab>, i.e., with line <gd>, is, namely, equal to the relation of line <ae> with the line that lies between point <e> and the point moving on it.Line <ag>, however, is to line <gd> as <ae> is to <eh>.Thus the point moving on line <ab> after <h> falls on the line <ad>, which is the diagonal.Similarly we prove that the point running on line <ab> constantly moves forward on line <ad> and is at the same time moving on lines <ad> and <ab>.The two lines <ad> and <ab> are, however, different; thus, the point moving forward in steady motion covers unequal lines in the same time.But, as I said, the motion of the point on line <ab> is simple, its motion on the diagonal <ad> however is derived from the motion of <ab> on the two lines <ag> and <bd> and the motion of <a> on line <ab>.Thus the one point <a> covers in steady motion two unequal lines. q.e.d.

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