NAVIGATION CHART, &C, 307 



miles falls on the parallel of 48° 26', extended equally on each side 

 the central tine ; when the course will be found S. 41° \V. and 

 the difference of longitude 2° 50'. Q. E. I. 



CASE V. 



Given one Latitude, Course, and Difference of Longitude, to 

 find the other Longitude, and distance. 



EXAMPLE. 



A ship from latitude 47° 30' N. sails S. 5 1° W. and then 

 finds her difference of longitude by observation to be 9° 40' 

 W. required the distance run, and the latitude come to ? 



N. B. As this difference of longitude exceeds the extent 

 of the chart; it requires, (unless performed by the general 

 Chart,) a double operation. 



1st. Set the index C" on the given parallel of 47° 30' at the 

 right hand of the Central Meridian, (sailing towards the Equa- 

 tor) and the given course 5 1° will give the distance 264 miles, 

 and latitude 44° 44' at 5° difference of longitude; (2° 30' on 

 each side of the central meridian.) 



2nd. Set the Index on the last found parallel of latitude 44° 

 44', and to the same course; when the distance corresponding 

 to 4° 40' (the remainder of the difference of longitude) setting 

 2° 20' on each side the central meridian, will give a farther 

 distance of 265 miles, making the whole distance 52y miles, 

 and the latitude come to 41° 57'. Q. E. I. 



*CASE VI. 



Given one Latitude, Distance and Difference of Longitude, to 

 find the other Latitude and Course. 



EXAMPLE. 



A ship sails from the latitude of 50° 20' N. between the 

 North and East, 300 miles, and finds by a chronometer her 



* This is Dr. Halley's celebrated problem. See Baron Masseres' "Scriptores Loga- 

 rithmic!." vol 4lh. if may be seen on the general chart K that this problem mill sometimes ad- 

 mit of two answers. 



