A K C H 329 > = /t , wherefore the strain at Q is OX 2 + A 2 ), or exactly the same as that at P. This result might have been obtained from the consideration that the thrust upon the surface PQ is perpendicular to the oblique strain, and can tend neither to augment nor to diminish it. Hence, as a characteristic of this arrangement, we have the law that the tension across the joints of the arch-stones is the same all along, and therefore is equal to H, the horizontal tension at the crown of the arch. From this it at once follows that if r be the radius of curvature at the point P, y being the vertical thickness of the mason-work there, H = ry, so that if R be the radius of curvature at the crown of the arch, and A the thickness there, the horizontal thrust there, or the strain transmitted along the arch-stones, is H = RAH, being measured in square units of surface ; hence also A : y : : r : R, or the thickness at any place, is inversely proportional to the radius of curvature there. When the form of the intrados is given, its curvature at any point is known, and from that the thickness of the stone-work and the shape of the extrados can be found. The most useful case of the converse problem is, again, that in which the extrados is a horizontal straight line. Let OH, figure G, be the horizontal extrados, and A the crown of the arch ; make also AB such G that its square may represent the hori zontal thrust there; then, having joined OB and drawn BG perpendicular to it, and meeting the con tinuation 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 Fig. 6. 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 condition and observing that in this case y = z, we have Now the integral fzSz is f 2 2 , but as it must be reckoned only from A where z = A, the equation becomes The coefficient of Sz becomes less when z increases, and yhen z- = H 2 + ^A 2 , this coefficient becomes zero, at which time &x also becomes zero in proportion to Sz ; that is to say, the direction of the curve becomes vertical. Wherefore, if we make OD = B such that D 2 = A 2 + 2H 2 , 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 by help of which we should be able to find x in terms of z. 3 he computation, however, is attended with considerable difficulty, and therefore it may be convenient to attempt a graphical solution. Since, for any vertical ordinate HP( = 2), the horizontal thrust is (D 2 z~), while the oblique strain is %(D- A 2 ), the obliquity of the curve at P has for D 2 z~ its cosine the value 77 , wherefore the angle at which D- A 2 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 arc of a circle described from the centre (J. Since the minute differentials thus obtained are pro portional to the sides of a triangle whose hypotenuse is D 2 A 2 , and one of whose sides is D 2 z* t we must have and the integration of this would give the value of x. If we put < for the inclination of the curve at any point P, D 2 -s 2 = (D 2 -A 2 )cos<, .-. 2 = {D 2 -(D 2 -A 2 )cos0}, and taking the differential, 8z = i(D 2 = A 2 ) sin 0{D 2 -(D 2 -A 2 ) cos <}* 80, s H. cos <p. B(p . * . ox = /TTvT~ x /{D 2 -2Hcos<p} where 2H is put for its equivalent D 2 - A 2 . The integral of this expression may be obtained by developing the radical in terms arranged according to the powers of cos <, j and then integrating each term separately. The result is a j series of terms proceeding by the powers of cos </>, 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 made consistent with the laws of equilibrium by merely dressing the inner ends of the arch-stones at the appro priate angles. Hence an entirely new field of inquiry, in which we may find the complete solution of the general problem : " The intrados and extrados of an arch being both prescribed, to arrange the parts consistently with the laws of equilibrium." Let PQ represent the inner end of one of the arch-stones, the part Qq being vertical, and P^- being sloped at some angle which is to be found ; put t for the tangent of the inclina tion of the joint P to the vertical, 6 Fig. for that of Pg- to the horizontal line, then the horizontal
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