Calculo prolixiori prodit:
8. 150= 3190 49,77 et t.= 215 19,32 r t. 150= 620 37,56. t.= 42 10˙„50 A41 t. 150= 2300 28 ,41.. t.= 4575 21 ,89 3 8. 150= 1560 51,80. 5.= 104 27,455 9, quo valores? ne singula quidem temporis minuta prima mutantur. i
§ 35. His valoribus in aechuatione 42) mu.. 0 6 ositis, sequitne: B9= 745,963+ 0,18160 sin(t. 150+ 1770 17,6)+† 025318 sin(t. 150+†. 1550 6,2) P„ 8: . mm minimum, t= 15*b 21„89= 3* 21 s9 matutina et B00= 745,852 maximum, ,= 21 1932= 9 19,32 antemerid. et B)= 746,331 minimum,,= 4 10,50= 4 10,50 pomerid. et B0)=—= 745,557 maximum,,= 105 27,45= 10* 27 45 vespertin. et B)= 746,123, Est igitur 3 ..*. 2... 2* mm 2 5„ 1 1) differentia inter maximum majus et minimum minus= 0,774 aestus totalis, fluctuatio man variatio diurna;
.„** ⸗** 2. Imm 4 4 1 5. 1
2) differentia inter maximum minus et minimum majus= 0,271, fluctuatio minor, variatio noctum,
ita ut variatio diurna nocturnam ter fere superet;. 3 . mm 3) differentia inter maximum majus(diurnum) et maximum minus(nocturnum)= 0208;
. um..—
4) differentia inter minimum majus(nocturnum) et minimum minus(diurnum)= 0,295, igituy u-
nima inter se magis differunt quam maxima.
§. 36.
Postremo si quaeritur tempus, quo atmosphaera statum medium obtineat sive medium harome-
tricum intret, aequatio u' sin(t. 150+„)+ u“ sin(t. 300+)= 0, et nostris observationibus applican:
0,18160 sin(t. 150+ 1770 17,6)+ 0, 25318 sin(t. 300+ 1550 6,2)= 0 responsum nobis offert.
§. 37.
Posito t. 150=, ergot 300= 2z et adhibita formula ein(2+()= sin a cos b+† sin 5 cosa, ista Mutatdi i u(sin x cos+ cos æ sin v)+ u¹(sin 2 x cos v“+ cos 2 sin)= 0. Est vero cos 22= 2 cos ½2— 1 et sin 2 z= 2 sin x cos x, inde oritur: u¹(sin æ cos"+, cos æ sin 2-)+† u(2 sin æ cos x cos+ 2 cos xe sin v— sin w)= 0.
Est autem sin== /(1— cos*.), inde sequitur:
u[cos v M(1— cos**)+ cos z sin v+ u“[2 cos æ cos v' Nd— cos* † 2 cos e sin— Ssin 6= 0;3 Posito cos= g et uncinis solutis prodit:
u, cos v' N(1— 9 † u’ sin 2*.+ 2 u“ cos vé, g N(L 9)+ 2 u“ sin v“„e— u sin u*= 03 et transpositione radicalium: (u“ cos v+† 2 u“ cos 5, p) V(1i Ge)= un sin w“— u. sin v, i— 2 u“ sin ²“, ge; inde quadratura sive reductione: (u' cos» † 2 u“ cos 9.)?(1— G?)=(u-, sin p“— u“ sin 90. g— 2 u“ sin»“. J²)?; give-
— 4 u's cos w'2. gt— 4 u, u, cos" cos w'. ps †+(A u's cos""*— u*² cos"²). e † A u, u' cos v Gos u",—†. a2 Cog v3 5 J. 4 u,2 sin v“2,* † 4 u' u“ sin v' sin„ ,3—(4 uns sin oe*— urr sin„). Ge— 2 u“ u“ gin e' sin u“.9+† u sin 2*.


