SECT. 2.

OF THE ARCHES.

43

dividing them in any ratio whatever, so as that it may beevery where dv : r>v : : ci : c i; then if acdm be an arch ofequilibration to the wall akvlm, it will be an arch of equili-bration to the inner wall a ktlu also.

prop. x.

Having given the Intrados or Soffit, of a Balanced Arch; tofind the Extrados. That is, having given the nature or formof an arch ; from thence to find the nature of the line formingthe top of the seperincumbent wall, by the pressure of which thearch is kept in equilibria.

The solution of this problem is to be made out generallyfrom the last proposition and its corollaries, by adopting ge-neral values of the lines there employed, which belong to allcurves whatever: or otherwise by making use of the peculiarvalues proper to any individual curve, for the solution ofparticular cases.

For the general solution, in fig. pa. 41, kvl represents theextrados, the form of which is required, and abcdm the givenintrados or soffit of the arch, the vertex of which is d, and

dv the height or thickness of the wall there, which is com-jnonly a dimension that is known from the particular circum-

stances of the case. Now if we make the arch dc = s, its

element c c — k, the absciss dh = x, its element ca — x, the

ordinate ch = y, its element ca =y, the height or thickness

of wall at the vertex dv = a, and the radius of curvature at

any point c = r, that at the vertex d being = r.

Then, because the height cx, at any point c, is as

sec 3 , be h or of cca , . . . , , .

----, by the last proposition, and because the

C6‘ Z

secant of cca is =— = —, the radius being 1, thereforeca y °

£3 ___,

cx is as— — , or as -j because k = cc = fca z -f car =

rr

f x 7 ' + j* or (i* + T'fi