812 Transactions.— Miscellaneous. 
The above description is all that is necessary for my purpose. As 
regards the investigation that ensues, I may remark that it is of the same 
degree of exactness as that given in treatises on civil engineering as appli- 
cable to the more ordinary method of structure. In the first place, I have 
been compelled to neglect the weight of the chain, because, though it is 
easy to form the equations of equlibrium when that weight is taken into 
account, yet they are unintegrable, or, at any rate, I believe so. This is 
to be the more regretted in this particular instance, because the weight of 
the chains formed a very appreciable part, rather more than one-seventh, of 
the total weight; in ordinary cases, of course the weight of the chains is 
insignificant, compared with that of the bridge, and no sensible error is 
made, therefore, in leaving their weight out of account. In the next place, 
I have treated the chains as forming a continuous curve, which is a depar- 
ture, though a very slight one, from the case which actually occurs. I 
repeat that these suppositions are those usually made. 
For considering the equilibrium of either of the chains, take its lowest 
point as origin of co-ordinates, the vertical line through that as axis of z: 
the horizontal line through the same point, and parallel to the length of the 
bridge as axis of x; a line at right angles to both of them, that is to say, 
transverse to the bridge, as axis of y. Let w be the weight of a unit of the 
length of the bridge, T the tension of the rope at any point (v y z) in it; s 
thelength of the rope measured from the lowest point up to the point 
(x yz); X Y Z, the resolved parts, parallel to the axes of co-ordinates, of the 
forces acting on the element ds in the neighbourhood of (x y z); then the 
ordinary equations for equilibrium are:— 
X — d+ (7 #) ~0 uc I 
Y ~d3(7T®)=0 MUT 
Z-—ds (7 #) =9 i cog) 
Since X = 0, there being no force in the direction of the axis of x, the : 
first equation gives us at once T "Ww = constant = c suppose (4) 
If T” be the tension along the suspending rod connected with the point 
(x y z) and 6 the inclination of that rod to the vertical ; then 
4=T'cos0; Y = T' sing, 
But, since there are two rods, one on each side of the bridge, which 
between them support a length dz of the bridge, therefore resolving 
vertieally,— è 
2 T' cos 0 = wdx 
' I" =F sec 0 de 
ede e Y= T tan 6 dx 
