RESPONSIO AD QUAESTIONEM ASTPvONOMlC AM. s 



Sit (Fig I et II) Z zenitli, P polus, Zffx (Fig. I), Z«i3 (Fig. II) linea vertica- 

 lisj Y£i/3 = //R/3, VPa = ^iv«. Erg» 



«£:(3 = (^R/3 — ^R») = « — tf' = 2oS° r^' 4'" = S 

 Iiliic «P/3 = suppl. (rt — rt'J = 360 — S = 151° 4^' 12". 



Ex magnitudine hvjus anguli tePl3 patet, stellas /3 et « simul sive supra, sive in« 

 fra polum esse non posse, dum sint in eodem verticali , steih « ergo in Transitu su- 

 periori infra polum , in Transitu vero inferiori supra polum eiit. 



Linea PC perpendiculariter verticali Z(2x vel Z»^ insistens ob eamdem caussam ca- 

 det inter /3 et «, V vero inter » et C. 



Sit porro l CPZ = t//, CP« = 0, (SPx = 360° — S, et erit in Transitu superiori 

 (Fig. I.) 



ViiiZ = V£:/3 — /SPZ 

 = a — (3PZ 



Verum /3PZ = «PZ — /3P« 



=: CPZ + «PC — 0P» 



= ^ + <p - (3C0 - 5) 

 Ergo V^Z = rt — (;/, + <j) _ (360 _ S)) 

 sed V£i:Z = /fR. merid. sive puncti culminantis.=:Iior. syd;=/ 

 i = /» - (^ 4- $ -f S — 360°) 



Sic quoque in Transitu inferiore (Fig. 11) 



VZ = VZ/3 — Z£:0 = r 

 = fl - (360°- ZP,J) 



Verum ZP/3 = ZPC + /3P« — *PC 



= ^ + (360°- S) -(p 

 = «^ — S — cp + 3<'o° 

 Ergo VZ = « — ( 360° _ (>^ — s -_ <p -^ 3(5o'^)) 

 = « - ( 360° - ,^ + S + ^ - 360°) 

 = « + ./. — (5 + $) 



A 3 Tang^ 



