1835.] 



Distance of objects at sea. 



341 



3000. t= 3.47712 == 11.45692 



4,93404 4.91313 

 2 2 



9.86808 9.82626 



, i-2r 2.28195 2.28195 



^ns log • 2 K ' „ 



141. feet = 2.15003 2.10821= 128ft 



3000. 3000. 

 128. 129. 



2872 = 3.45818 2871 ^ 3.45803 



11.45692 11.45692 



4.91510 4.91495 



9 9 



9.83020 9.82990 

 2.28195 2.28195 



129. = 2.11215 2.11185 ==129, 



the result then is log. 4.91495 = 82210 feet to turn 

 which into nautical miles, 



4.91495 



7.99453 



2.90948 = 812" = 13' 32" which is the same 

 result as in the last example. In computing by this for- 

 mula logarithmxS to 5 places are sufficient, and the result 

 will always come out correct, even if errors may have 

 been committed. From this last formula a very conve- 

 nient one may be derived. < 



Let the distance sought be == x 



the height of the mountain — « 



