APPENDIX. 



281 



As an illustration, take the data of the example in article 13. 

 Assume = 326, and we find 



log sin 9.74756 n log cose 9.91857 



loge 



log/ 4.70415 

 log e&quot;sin 4.45171 n 



e&quot; sin = 28295&quot; = 7 51 35&quot; 

 = e e&quot; sin e = 333 51 35&quot; 

 M M 4960&quot; 



9.38973 



loge cos 



9.30830 

 cos e = .79662 

 log(l ecos) 9.90125 

 logJf M 3.69548n 



1 e cos s 



e cos 



= _143 46&quot;. 

 And for a second approximation, 



= 326 1 43 46&quot; = 324 16 14&quot; 



log sine 9.7663820 



loge&quot; 4.7041513 

 loge&quot; sine 4.4705333 n 



e&quot; sin = 29548&quot;.36 = 8 12 28&quot;.36 



M = 332 28 42&quot;.36 log (l e COSE) 9.90356 



M M = + 12&quot;.41 log (M M ) 1.09377 



=- = + 15&quot;.50 bg^^- 1.19021 



1 e cos E 1 e cos e 



which gives 



H= 324 16 14&quot; 15&quot;.50 = 324 16 29&quot;.50. 



Putting 



we have 



18. 



q lp =. perihelion distance, 



log x = 8.0850664436, 

 r = 



tan i v -\- i tan 3 i y = x 1 



T = (3 tan i v -f- tan 8 



36 



