SECT. 2.

OF THE ARCHES.

37

the three hk, oh, ok; and so on; in all these cases theweights of the stones being proportional to the bases bd, df,Jh, hk, of the triangles obd, odf, ofh, ohk. But as these tri-angles have all the same common altitude ob, they have thesame ratios as their bases bd, df, &c, which bases, it hasbeen shown, are proportional to the weights of the stones,which have also been found proportional to the quadrilateralareas ad, cf, &c; therefore the quadrilaterals ad, cf, eh, gk,are respectively proportional to the triangles. o6c/,o£//,0/7;,oA/r.

But, as these small triangles have their angles respectivelyequal to the angles of the corresponding sectors, becausetheir sides are parallel by the construction; that is, the an-gle bod = the angle bod, &c; their areas are therefore pro-portional to the squares of their corresponding sides;viz. the sectors obd, oac, obd, ■proportional to ob 2 , oa 2 , Ob' 1 ;and the sectors odf, oce, odf,

proportional to od 2 , oc 2 , od 2 ; and so on.

Therefore, by taking the differences,

ad : obd : : ob’ — oa 2 : ob 2 ,

and cf : odf : : od’ - oc 2 : od 1 ,

and eh : ofh : : of 2 — oe 2 : of 1 ,

and gk : ohk :: oh 2 — og 2 : oh 1 , &c.

Hence, if ob 2 be taken = ob 2 — oa 2 ,then od 2 is = od 2 — oc 2 ,and of 2 is = of 2 - oe 2 ,and old is = oh 2 — og 2 , &c.

Or, by transposing, ob 2 oa 2 + ob 1 ,and od 2 = oc 2 + od 1 ,and or ! = oe 2 + of 1 ,and oh 2 = og 2 + old, &c.

Which gives us the following geometrical construction,viz, Produce the joints till oa,oc, oe, og, &c, be equal tothe several radii of curvature at the corresponding points,A, c, E, &c; to wh ich a lso draw the parallels ob, od, of, &c.Then take ob = ./ob 2 - oa 2 , and draw bdfhk perpendicular