1817.) mecessary to support a given Arch. 449 
AO =7 feet, height AG = 10 feet, line RS = 12°8 feet, and 
BH =AD = 2°75 feet; from hence it is required to find the 
thickness of the wall F G necessary to support the said arch and 
work on its top. 
The data in the question are collected from a building that fell 
just as completed. 
First Solution. 
Let R (Fig. 10)* represent the centre of the space ABH D and 
I the centre of the are AB; draw K L perpendicular, and IP 
parallel to AO; join IK, and draw K N perpendicular thereto; 
draw LM, FN, parallel to 1 K, and K Q parallel to AO. ‘Then 
if K Lrepresent the area ABH D, KM will represent the force 
acting at N, at right angles to F N, and by the similar triangles 
IKP, KLM, CKQ, and CFN, we have, after making 
Wa roe cea s sletevatats =p’ Feet 
OT = OBC. = 4:0414 Ditto 
Ee ak Shinde oes as = 7°8414 Ditto = @ 
BE a OS. es esse ans = 34 Ditto = b 
Pe a ds = a+ /*Ditto = 
Me NE os ce a ree == S56 Ditto = d 
Ara ABHD ...... =13°333 Ditto = m 
FQ = GA + AT:. =137 Ditto = f 
a Palin mgr peel =12°8 Ditto = g 
it al a ge ata = 2 Ditto 
KQ=KT+TQ.. = d+ 2 Ditto 
ees. Tie re. Bee oe 
TK coats 
. bes : _ KQ~x« PI _ (d+2z)d 
KP: PI: KQ:QC=—— = 
Pe —0C = re ft 2) aise? 
a a 
f = , _ FExKP _af— d+ 2z)b Pere 
IK: KP: FC:FN= —~—= > 4 Saher 
af—(d+a)b 
ore Ngee 
Then, per mechanics, 
RS x FG xiFG=FN x KM; that is, 
gz?  eaef—(d+x)b bm , 2m ot 2bhm 
eee te a ie rr thet bd) Fai 
In numbers x«* + *3297 x = 9:23074 .°. 2 = 2°878 feet, the re- 
quired thickness according to this solution. 
* The centre of gravity K was found as follows: I constructed the mixed lined 
space A BH D very accurately to a large scale, and subdivided it by lines drawn 
parallel to BH, equidistant from each other, and one foot apart; then found 
geometrically the centre of gravity of each subdivision, by considering them as so 
many trapeziums. From the same scale, and from the consideration of moments, 
4 found O L = 3-4 feet, and L K = 3°S feet, 
Vox. 1X. N° VI. 2F 
