Method of tracing Oval Arches. 305 
twelve oval arches; four of which are, of forty feet span, one of 
twenty five, one of twenty, &c. In fact the great facilities afforded 
for tracing the curves, and cutting the joints of the voussoirs normal 
to them, must cause their adoption wherever they are known. 
Five Centres. (Fig. 3.) Seven Centres. (Fig. 4.) 
ab=1., AB= 6, 
bA— 2, AD=18 
AB=4.50 ab= 1 
AC=9.00 be=. 2 
ca=2.704 cA 
(h=2.305 De=13.252 
bd=0.399 ec= 8.111 
Ce=9.619 eb= 8.667 
Ae=0.619 db= 1.854 
b=9.220 da= 2.425 
dea=21° 9 40” bf= 0.567 
eCd=12° 31’ 44” eg= 1.124 
AaB=56° 18/ 36” Ah= 1.376 
ADc=9° 27’ 45” 
ACb= noe 37! 12” 
ABa=4 
pet 9/ 37” ve 
bda=23° 22/ 48” : 
Nine Centres. (Fig. 5.) 
Aa=10 t= 271s 
AB= 7.5 bh= 0.683 
At =30 * c= 1.507 
ab= 1 dk= 2.155 
bc= 2 Am= 2.423 
ca Rid = 7° 35/ 41” 
dA= 4 dgc=.9 41.12 
Ed=30.26S ofb=13 40 57 
gd=17.025 bea=22 - 9 58 
ge=17.673 Aab=—36- 52 12°5 
fe= 7.250 “AbC=59 2 10 
fb= 8.074 AcD=72 43 7 
eb= 1.590 | AdE=82 24 19 
