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A history of Greek mathematics - Wilbourhall.org

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'<br />

430 PAPPUS OF ALEXANDRIA<br />

in better logical sequence than his predecessors had attained.<br />

The sort <strong>of</strong> questions to be dealt with are (1) a comparison<br />

between the force required to move a given weight along<br />

a horizontal plane and that required to move the same weight<br />

upwards on an inclined plane, (2) the finding <strong>of</strong> two mean<br />

proportionals between two unequal straight lines, (3) given<br />

wheel with a certain number <strong>of</strong> teeth, to find the<br />

a toothed<br />

diameter <strong>of</strong>, and to construct, another wheel with a given number<br />

<strong>of</strong> teeth to work on the former. Each <strong>of</strong> these things, he says,<br />

will be clearly understood in its proper place if<br />

the principles<br />

'<br />

on which the centrobaric doctrine ' is built up are first set out.<br />

'<br />

It is not necessary, he adds, to define what is meant by heavy<br />

'<br />

'<br />

and light ' or upward, and downward motion, since these<br />

matters are discussed by Ptolemy in his Mathematical but<br />

the notion <strong>of</strong> the centre <strong>of</strong> gravity is so fundamental in the<br />

whole theory <strong>of</strong> mechanics that it is essential in the first<br />

place to explain what is meant by the ' centre <strong>of</strong> gravity<br />

<strong>of</strong> any body.<br />

On the centre <strong>of</strong> gravity.<br />

Pappus then defines the centre <strong>of</strong> gravity as ' the point<br />

within a body which is such that, if the weight be conceived<br />

to be suspended from the point, it will remain at rest in any<br />

position in which it is put '} The method <strong>of</strong> determining the<br />

point by means <strong>of</strong> the intersection, first <strong>of</strong> planes, and then <strong>of</strong><br />

straight lines, is next explained (chaps.<br />

1,2), and Pappus then<br />

proves (Prop. 2) a proposition <strong>of</strong> some difficulty, namely that,<br />

if D, E, F be points on the sides BG, GA, AB <strong>of</strong> a triangle ABG<br />

such that<br />

BD:DC= GE:EA = AF:FB,<br />

then the centre <strong>of</strong> gravity <strong>of</strong> the triangle ABG is also the<br />

centre <strong>of</strong> gravity <strong>of</strong> the triangle DEF.<br />

Let H, K be the middle points <strong>of</strong> BG, GA respectively;<br />

join AH, BK. Join EK meeting DE in L.<br />

Then AH, BK meet in G, the centre <strong>of</strong> gravity <strong>of</strong> the<br />

triangle ABG, and AG = 2 GH, BG = 2 BK, so that<br />

.<br />

GA :AK = AB:HK = BG: GK = AG :<br />

GH.<br />

1<br />

Pappus, viii, p. 1030. 11-13.

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