of the Problem of Shortest Twilight, 



181 





problem is one of some historical celebrity, and its solution 

 calculated to enforce an important truth, — on all these ac- 

 counts its occupying a place in the Philosophical Magazine 

 may perhaps be justified. 



Problem. — Let EH RQ be the horizon of the place ; 7j the 

 zenith; ? the pole ; ESQ the equator ; MUP', LNT any two 

 tmacantars ; LNPM a circle of declination; then it is re- 

 red to find the position or polar distance o/*LNPM, so that 

 the arc of it intercepted between the given almacantars may be 

 a maximum or a minimum. 



Such is the more general geometrical statement of the pro- 

 blem in the aspect under which 

 the Bernoullis viewed it: the 

 angle NPP' is the measure of 

 the time elapsed between the 

 sun's passing the two almacan- 

 tars. In the common problem 

 for one of these almacantars is 

 substituted the horizon itself. 



Let NP' be the intercepted 

 portion of the circle of declina- 

 tion. Denote the colatitude, or 

 PZ, by A : put g, g t , g u the sphe- 

 rical radii of the circles LRN, 

 LTN, and MUP' respectively: 

 and denote the angles ZPF and ZPN by W and 6,. Then, 



cos 



— i cos g /~ cos § cos A 

 sin g sin A 



A „„„ - 1 cos §u — cos Q cos A 



V.. = COS * 2JJ : — 5 . 



sin g sin A 



From which we have the polar angle or measure of the time, 



A a i_ cosq u — COS Q cos A 

 0. — 6, = COS 1 5J i r- 5 



sin g sin A 



, cos p.— cos o cos A 

 — cos 1 **-, ? (i) 



sin g sin A v ' 



But that the time may be a maximum or a minimum, we must 

 have 



'{ 



cos 



cos 



cos g u — cos g cos A 

 sin g sin A 



cos Qi — cos g cos A 1 _ 

 sin g siri A J " 



(2) 



