450 Thickness of Wall necessary to support a given Arch. [JUNE, 
Corollary. When FN = RS, then FG= /2KM= J 
Bim _ ., 68 x 18983 _ 3.957 fot, 
c "546d 
Second Solution. 
If F L, as before, represent the area AB HD, LA will denote 
the force acting at A, at right angles to A G, and is thus deter- 
mined. KL: LA:: m: ae 
Then, per mechanics, 
: sth mxLA . 1} QmxLAxAG 
| RS xFGxiFG=AG~x a oe ae 
2 x 13'333 x 3: 
In numbers, F G = fee Wie 464496 feet, the 
required thickness according to this solution. 
Corollary.—When AL = LK, thn FG= / es = 
20 x 13'333 
uw 12:8" 
= horizontal force acting at A, 
= 4°5643 feet. 
The thickness of the wall alluded to in the question was three 
feet and two inches. 
Hence it appears that the thickness 2°878 feet, as determined by 
the first solution, would not have been sufficient to resist the force 
of the arch; but if the walls had been made 4:4426 feet thick, as 
determined by the second solution, I am of opinion the failure 
would not have happened. Notwithstanding the second solution 
gives a thickness for the walls the most likely to ensure stability in 
the present instance, it must not be understood that this method is 
preferable to the former in all cases; for when the extrados is hor7- 
zontal, Dr. Hutton, on referring to the principles contained in the 
Jirst solution, makes the following remark :— 
‘© It may be presumed that this theorem brings out the thickness 
of the piers very near the truth, and very near what would be 
allowed in practice by the best practical engineers, as may be 
gathered from a comparison of the two cases of Westminster and 
Blackfriars Bridges; in the former of which the centre arch is a 
semicircle ef 76 feet span and 17 feet thickness of piers; and in the 
latter it isa semiellipse of 100 feet span, and 40 feet in height, and 
19 feet thickness of piers.’ (Dr. Hutton’s Tracts, vol. i. p. 81.) 
14 2°8> not widely ¢ 2°S78 
+ of is is {2 differing 1-000} as herein stated, 
100 (20:0! from 19°000 
