160 Mr Galbraith on Trigonometrical 



Let r be the radius of curvature of the meridian, or normal 

 to the major axis of the elliptical spheroid, then, if a denote the 

 radius of the equator, b the polar semiaxis, and I the latitude. 

 «2 52 



■(a2 cos H-\-b'^ sin H) 



0) 



Again, if g be the radius of curvature of the arc perpendicular 

 to the meridian, or the normal to the minor axis. 



/2 



^"{a? cos H+b^ sin 2/)^ .... (2) 



Lastly, let / be the radius of curvature of an oblique arc, 

 making an angle a with the meridian. 



^^2 • 



^~(«2cos2/+62sin2/)icos2^+62(«2cos2/+62sin2/)|sin2* • (^) 



Now, as the radius of curvature is to an arc R'', equal to the 

 radius in seconds, so is the distance in the same measure with 

 the radius of curvature, to the corresponding arc in seconds. 



From this analogy, and the preceding three equations, will 

 be obtained the factor to convert feet into seconds on the sur- 

 face of the terrestrial spheroid in any given direction. 



Wherefore, if M be the factor to convert a curvilineal dis- 

 tance on the meridian into seconds of arc ; P, that on the per- 

 pendicular to it ; and O, that on any oblique arc, making an 

 angle a with the meridian, there will be obtained 



M=^.(a2 cos 2/4.52 sin ^/) ^ .... (4) '^ 



P=~(a2cos2/+62sin 2/) ^ . . . .(5) 



0=M cos 2^+P sin 2^ (6) 



■p // T> // 



Log-2-^-2=6.0348593, Log -^=0.6731921 



From these formulae, tables I, IV, and V,* have been con- 

 structed. The first table is certainly less extended than might 

 be desirable in some cases, but it is sufficient for most purposes 

 of surveying, especially for nautical men, who are frequently 



* For table V, log -^^=0.0348421, log —=0.6732241, in which a is 

 20921872 feet, and h is 20854078 feet. 



