. An encyclopaedia of architecture, historical, theoretical, & practical. New ed., rev., portions rewritten, and with additions by Wyatt Papworth. Fig. Fig. 53S. ; triangle. 127fi. To find the centre of gravity of any irregular rectilinear surface, .such as thepentagon,/^. 538., let it be divided into the three triangles, AED, ABC, ADC fa. 538.),and by the preceding rule determine their centres of gravity F, G, H. Then draw thetwo lines NO, OP, which form a right angle surrounding the polygon. Multiply thearea of each triangle by the distance of its centre of gravity on the line ON, indi
. An encyclopaedia of architecture, historical, theoretical, & practical. New ed., rev., portions rewritten, and with additions by Wyatt Papworth. Fig. Fig. 53S. ; triangle. 127fi. To find the centre of gravity of any irregular rectilinear surface, .such as thepentagon,/^. 538., let it be divided into the three triangles, AED, ABC, ADC fa. 538.),and by the preceding rule determine their centres of gravity F, G, H. Then draw thetwo lines NO, OP, which form a right angle surrounding the polygon. Multiply thearea of each triangle by the distance of its centre of gravity on the line ON, indicated byVf, Gy, Uh, and divide the sum of these products by the entire area of the pentagon, andthis will give a mean distance through which an indefinite line IK parallel to ON is to bedrawn. Conducting a similar operation in respect of the line OP, we obtain a new meandistance for drawing another line LQ parallel to OP, which will intersect the first in thepoint M, the centre of gravity of the pentagon. nie centre of gravity of a sector of a circle AEBC (fir/. 539.) must be upon the radiusCE which divides the arc into two equal parts. To determine f
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