OF DIALING. 371 



CASE I. 



1. Let us suppose that an upright plane at London 

 declines 36 degrees westward from facing the south ; 

 and that it is required to find a place on the globe, to 

 whose horizon the said plane is parallel ; and also the 

 difference of longitude between London and that place. 



Rectify the globe to the latitude of London, and bring 

 London to the zenith under the brass meridian, then 

 that point of the globe which lies in the horizon at the 

 given degree of declination (counted westward from the 

 south point of the horizon) is the place at which the 

 above-mentioned plane would be horizontal. Now, to 

 find the latitude and longitude of that place, keep your 

 eye upon the place, and turn the globe eastward, until 

 it comes under the graduated edge of the brass meridian ; 

 then, the degree of the brass meridian that stands di- 

 rectly over the place, is its latitude ; and the number of 

 degrees in the equator, which are intercepted between 

 the meridian of London and the brass meridian, is the 

 place's difference of longitude. 



Thus, as the latitude of London is 51| degrees north, 

 and the declination of the place is 36 degrees west ; I 

 elevate the north pole 51* degrees above the horizon, 

 and turn the globe until London comes to the zenith, or 

 under the graduated edge of the meridian : then, I count 

 36 degrees on the horizon westward from the south 

 point, and make a mark on that place of the globe over 

 which the reckoning ends, and bringing the mark under 

 the graduated edge of the brass meridian, I find it to be 

 under 30* degrees in south latitude : keeping it there, I 

 count in the equator the number of degrees between the 

 meridian of London and the brazen meridian (which 

 now becomes the meridian of the required place) and 

 find it to be 42*. Therefore an upright plane at Lon- 

 don, declining 36 degrees westward from the south, 

 2 B 2 



