610 Prof. W, M. Hicks on the 



To solve this put a 2 = 4a / ( | J f . Then 



4|3 +3?= __ -^=Asay. 



If then sinhK = A, 



f = sinh^-?f, 



a 2 = ^/( 16^/3 ) sinh ^- w, 



and the solution is obtained at once by the use of the 

 Smithsonian tables *. 



In order to get some idea of the general shape o£ the 

 boundaries the case with the node at fc— sin 65° is taken. 

 The curve as calculated is the loop portion of curve VI in 

 fig. 2. Here k = sin 65°; &' = cos65°; X='2173; logF = 

 •36338; log E = -06589, whence aU = 1-8005/a/w; 6/a= '00464; 

 jr^lfji— —-00985. Also the equatorial axial radius expressed 

 in terms or the standard sphere of same volume is a = 21'56c, 



U = '133U. 

 The equation in z is 



**-l-1108X. 3- -01094 = 0; 



with log^ =-04564 + log X, 



log A= 2-54388 + 2 log X, 



log 4V (?/3) = 1:38315. 



It will be sufficient to illustrate the method of calculation 

 by a single example — say where the boundary cuts the 



* If these are not at hand, the following- will enable ordinary tables to 

 be used. The eqn. sinhw = A regarded as an equation in e u gives 



e-<W(A 2 +l)-A. 



If then cot a = A, e u = cot - , e~ u — tan - , 



. , 1 1 i / rt\3 / a 



smh-o u = p < ( cot - j — [ tan - y 



When A is so large as to require values outside the Smithsonian tables, 

 the trigonometrical tables are also inapplicable since a is only a few 

 minutes of arc. In this case the approximation 



riph|«=i |(2A)*-(2A)-*j 



is sufficient. 



