216 Mr. Bowditeh’s estimate of the height Sc. 
EXAMPLE. 
Suppose the azimuth of the meteor at Wenham PW Me=117° 339’ 
54”, azimuth at Weston PSM=3°, altitude at Wenham 5° 50’ 4 40", 
co-latitude of Wenham PW=47? 19’ 45", co-latitude of Weston PS 
48° 45', difference of meridians WPS=2° 3645". It is required to 
find the latitude and longitude of the meteor, its distance from Wen- 
ham and Weston, and its vertical height above the level of the sea. 
This corresponds to Example 10 in Table 1. 
In the Page PSW. 
3PS = 24°22’ 30” 
IPW=23 39 52 
tal 
Sum 48 2 22Cos.AC 0:1748214| - - - - Sine AC 0°1286575 PSW 52°56/34" SineACh™ 
Diff. 0 42 38 Cos.. 99999666!,- - - - Sine 8°0934643IPW 47 19 45 
LWPS 1 18 22 Cotan. 11°6420673| - - - - Cot. —11°6420673|WPS 2 36 48 Sine 
4Sum 89 7 36 Tang. 11-6166555)|fait36°11'2""Tan.9 864189i|SW 2 24 25 Sine 
iDiff. 36 11 2 PWS 125 18 38 PSW 52 56 34 
— ; ~PWM117 35 54 Azim. PSM 3 
Sum 125 18 383=PWS —-——_ —— 
Diff. 52 56 34—=PSW SWM 7 42 44 WSM 55 56 34 
In the triangle WSM. 
4WSM 27 58 17 
ISWM 351 22 
Sum 31 49 39 Cos. AC 0:0707652 Sine AC i -2773898 
Diff. 24 655 Cos. 9°9603401 Sine 6112706 
-3SW = 1:12.12} Tang. 8-3223773 Tang. . "3093773 
— ee 
iSum 11734 Tang. 8 8°3534826 Adif. 0° 5557” Tang. & 2115377 
qDiff. Sete cn cement res | 
ae Cwm=95° 41’ 50” = Alt. 5° 50/40” 90° 
Sum 21331=WM=mCw - - - = 48S 
Diff. 021 37=SM=sCm Cmw=82 4 39 
Sumis0 oO 0O 
——— 
, in 
* The aii 3;WM, by the above calculation is 9’ 32”. The value made oer 
finding Cw 50", estimated by a rough calculation, and it was thought 
to repeat ‘he aebtaen on account of this small difference. 
