74 Methods of describing Various Curves for Arches, 



For five centers, divide by 0.794. 



For seven centers, " " 0.771. 



For nine centers, " " 0.758. 



For eleven centers, " " 0.749. 



The method of finding these divisors will be given hereafter. It 



may be observed that the last divisor is nearly =0.75, hence when 



eleven centers are used, multiply the above difference of rise and half 



span by 4, and divide by 3, the result will be the distance CD. 



Having found CD, make CH = 3CD. Take one from the number 



of centers to be used, and half the remainder will be the number of 



parts into which CH and CD are to be divided ; CH into equal 



parts, and CD into unequal parts, increasing from D as 1, 2, 3, Sic. 



Join these points of division, as in the figure, by straight lines, whose 



intersections will give the centers H, G, F, &:c. Thus, when nine 



centers, are used, as In the figure, CH is divided into four equal parts, 



and CD into the same number of unequal parts, increasing as 1, 2, 



3, 4, from the point D. 



To find the above divisors, put CD=?/, AD=a? and the given 

 quantities KO—a, and BC=^. Now when the number of centers 

 is given, the broken line HD is equal to CD multiplied by a constant 

 quantity ; put this constant quantity =c, then HD=c!/, and since 

 the broken line AH must be equal to BH, we have 

 cc +€?/=£? +3 J/, whence 

 ■r=c?-|-3/ (3 — c), and since 

 AC==AD+CD, 

 a=y-\-d-\-y['i — c), hence 

 a — d 

 4^' 



. In order to apply this general equation, c must be calculated for 

 'the required number of centers. For five centers, take CD -.any 

 assumed quantity, say three; then by trigonometry we find the surti of 



HD 



the lines that constitute HD = 9.6 19, hence 0=7^= 3.206. Inthe 



same way we find for seven centers, c = 3.229, and for nine centers 

 c= 3.242, and for eleven centers c= 3.251. Hence we have for 



a — d 

 Five centers, CD=^ „^' 



a-d , 

 Seven centers, CD= ^ „„ • 



y-=^r-= CD. 



