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8 persons who weigh 960 pounds. To be altogether on the safe side, we willallow 1100 pounds as the maximum load for every foot of floor. There is noother kind of transitory load which would be equal in weight to a crowd ofpeople. Let us suppose, for instance, a herd of full grown oxen, occupying thewhole area of the roadway. A large ox, weighing 1000 pounds, will occupy aspace of 25 superficial feet; there is, therefore, not quite room enough for onehead in one foot length of roadway. A herd of cattle will, therefore, not pressas hard as a crowd, of people. A heavy six horse coal team, loaded with 100bushels of coal, will weigh altogether 7 tons, and occupy a space of 50 feet inlength. Two teams alongside of each other, weighing together 14 tons, or28,000 pounds, will cause a load of 560 pounds for every foot length of floor,therefore little more than one-half of the load caused by a crowd of people.

The following quantities will fully constitute the maximum vertical pres-sure which has to be supported by the cables and stays for every foot oflength of floor:

Maximum transitory load, ------- 1,100 pounds.

Weight of 1 foot of floor, including all materials, - - 1,100 “

Weight of 2 cables and saddles of suspenders, - 600 “

Allowance for wet and snow in winter, - 200 “

Total maximum weight,.- 3,000 pounds.

The length of floor of each span between the abutment and town is 760feet, and the maximum weight for this distance 1140 tons.

It will be observed, that the curve, in which the cables are suspended, doesnot form a complete catenary, but only a portion of it, which may be consideredas being cut off at the abutment. If the curve was continued from the abut-ment until it rises to a point at a level with the summit of the town, which is95 feet above the lowest point of the curve, then it would form a completecatenary with a chord of 1180 feet long and a deflection of 95 feet. Now letus suppose the horizontal length of the span to be equal to the chord of the fullcatenary, or 1180 feet, and the cables equally loaded at the rate of 1 1-2 tons forevery foot, including their own weight, then it is evident that the tension of thecables would be the result of this load of 1770 tons, and which would amountto about 2,850 tons. But it is also apparent that the equilibrium of a catenary,or what is the same, the relative position of all points of a cable, will not bedisturbed if we cut it through at some certain point, and secure this pointimmovably. Or in place of cutting the cable, we may support that point by anabutment, bend the balance of the cable over it, and anchor it in the ground, soas to resist the strain of the suspended portion with the same power it did pre-viously, when in its original position. Thus, it would appear, that the cables ofthe Cincinnati bridge, although they have only to support a floor of 760 feetlong, may be subjected to a tension resulting from a floor of 1180 feet long.Indeed, in place of terminating the floor at the abutment, it might be continuedthe remainder of the distance of 1180 feet, and without requiring any increasein the size of the cables.

The above is a correct exposition of a mathematical law, which I hope willhave been rendered intelligible enough without the aid of analytical deductions.Those who have a desire of testing its truth practically, can easily do so by asimple experiment. We have, therefore, to base our calculations upon a fullydeveloped catenary curve, with a chord of 1180 feet and 95 feet deflection. Butthe figure of this catenary will not be employed in its purity, but changed sothat the ends for some distance will form tangents, or run nearly straight, while6