 extrados, and A the crown of the arch ; make also AB such that its square may represent the horizontal thrust there; then, having joined OB and drawn BC perpendicular to it, and meeting the continuation of OA in C, C is the centre of curvature for the crown of the arch. Or, if the radius of curvature and the thickness of the arch at the crown be pre • scribed, we may obtain the horizontal thrust by describing on CO a semicircle, cutting a horizontal line through A in the point B, then the horizontal thrust is equal to the weight of the quantity of the stone-work which would fill up the square on AB. The conditions of the problem require that the curve APQ be so shaped as that the radius of curvature at any point P shall be inversely proportional to the ordinate HP.
Resuming the general equation of conditionSz(H - fy8z)= 8.rfy8x , and observing that in this case y=z, we haveSz(H - .118z)=8*Sx Now the integral .11-8: is 1z2, but as it must be reckoned only from A where 2= A, the equation becomes Sz(11 + 1-A2 - iz2)= 8.rfiSx .
The coefficient of Sz becomes less when z increases, and when 1;z2= 112 + 2A2, this coefficient becomes zero, at which time az also becomes zero in proportion to Sz ; that is to say, the direction of the curve becomes vertical. Wherefore, if we make OD = D such that D2 = A2+ 2112, we shall obtain that depth at which the curve is upright, or at which the horizontal ordinate DQ is.the greatest, and then the equation takes the form-- Sz(D2 - z2) = 28:03.r , difficulty, and therefore it may be convenient to attempt a graphical solution. Since, for any vertical ordinate HP( = z), the horizontal thrust is 1(D2 - 7.2), while the oblique strain is 1(D2 - A2), the obliquity of the curve at P has for - its cosine the value D2 z2 wherefore the angle at which the curve crosses the horizontal line pP is knoWn. Let then a multitude of such lines be drawn in the space between BA and DQ, and let the narrow spaces thus marked be crossed in succession from A downwards by lines at the proper inclination, and we shall obtain a representation of the curve, which will be nearer to the truth as the intervals are more numerous. The beginning of the curve at A may be made a short are of a circle described from the centre O.
Since the minute differentials thus obtained are proportional to the sides of a triangle whose hypotenuse is D2 - A2, and one of whose sides is D2 - z2, we must have - 8x - 4./ ( (D2 _ A2)2_ (D2_ 2,2)2) 8> , and the integration of this would give the value of x. If we put 0 for the inclination of the curve at any point P, D2 - z2 = (D9 - A2) cos 0, ••• z= (D2_ (D2 _ A 2) cos OP, and taking the differential, Sz = RD?. = .A.2) sin st,(D2 - (D2 - A2) cos CI 80, II. cos p. .•. 8x - %/(D2-21-1 cos p) where 211 is put for its equivalent D2 - A2. The integral of this expression may be obtained by developing the radical in terms arranged according to the powers of cos 0, and then integrating each term separately. The result is a series of terms proceeding by the powers of cos 0, the coefficient of each power being itself an interminate series ; and the rate of convergence is so slow as to make the labour of the calculations very great. Such expressions belong to the class of elliptic functions, for which peculiar methods have been devised. Fortunately the actual calculation is not required in the practice of bridge - building, and therefore we shall only refer the reader to the above-named subject.
If the horizontal thrust and the thickness at the crown of the arch be prescribed, the radius of curvature there must be the same whichever of the two hypotheses be adopted ; now, if we sweep an arch from the centre C with the radius CA, the catenarian curve lies outside of it, while the curve which we have just been considering lies inside. Each of these is compatible with sound principles: the one if the inner ends of the arch-stones be dressed with horizontal facets, the other if the ends be dressed to a continuous curve ; wherefore, between these two limits we may have a vast variety of forms, each of which may be 