upon the Antique Dials. 119 



ascension is to be calculated. Draw CH parallel to AB, cutting the line 

 BD in K, and draw the ordinates CE, KF. 



Then FC = n i cos AF by the equation of the developed hectemoria 

 EK = (AB AF + FE) x tan B from the line BD, 



but FC = EK .. n i cos AF = (AB AF + FE) x tan B, 



_, , __,. nicos AF + AF tan B AB tan B 

 or, El (= KC) = tanB 



Putting its differential equal to o, we have 



tan B tan B 



sin AF = r- =. r . 



n i n. tan 1 



1 90 n L' 

 = cos nL, cosec 



n 



where n L' = semidiurnal arc, at the declination D', derived from the equa- 



tion 



cos wL = cot I tan D'. 



For the obliquity of the ecliptic as determined by HIPPARCHUS, and the 

 ancient estimate of the latitude of Athens, viz. 23 51' and 37 30', we find 



ML' = 70 10' 10" 



sin L = sin AF = cos 70 10' 10" cosec 9 49/ 50 " 

 n n 



But the radius of the cylinder from which this curve was developed was , 

 and the arc AF, therefore, when transposed to a cylinder whose radius is a, 

 will measure but th of the same angular magnitude. We shall hence ob- 



Tt 



tain the value of L where the deviation is a maximum, viz. 



1 "' f ! w ,/v Tfv 19 49' 50" 7 



L = sin } cos 70 10' 10 cosec - >- 



n [ n n ) 



From the perfect equality of the two branches of the curve, we readily 

 infer that the same amount of error will be found at the corresponding de- 

 clination in the other hemisphere. We may also infer it from the muta- 



Q2 



