Libr. I. Sect. II.

6 6

65 .

Exempli caussa calculum in art. 5 i vsque ad locum heliocentrlcuxn produ-ctum vlterius continuamus. Respondeat illi loco longitudo heliocentrica terrae2-i* lg 7 4905 = 2 /, atque log /2 == 9,9980979; latitudinem 5 statuimus = o. Habe-mus itaque X — L =— 17°24 , 2 o"o7 , log/t'— logH, adeoque secundum methodum II,

il

log—7-.

.9,67298x3

Jog (1 — Q).g,6026258

log sin (X — L).

1 —Q — 0, 44 g 5 g 25

log cos (X — L).

. 959 7 96445

Q = 0,5506075

log-P .

log Q .

Hinc 1 ‘— X =■

— i 4 ° 2 i' 6" 7 5

vnde l = 352 * 54 , 22 *s 3

T A'

lo g r ' .

ynde log A’...0,0797286

log tang/9.

log cos b .9,9973144

logtangb .

log A.0,0824159

b = —

r>0 t r r u

•D 21 05 07

Secundum methodum III ex Iogtang£ = 9,6729813 habetur £= 25 ° 1 5 7 6” 5 1, adeoque

logtang(45’+£).o,444iogi

log tang j( 2 ‘ — L) ..9,1 848958 n

logtang(Z —i X—i L)... 8,6290029 n

l—iX — %L = — 25 ° 5 , i 6"79 1

xji.t _ , r r. ' f- t vnde / = 55 2 0 34'22’'22 5

— 100709,010 J

64 .

Circa problema art. 62 sequentes adhuc obserualiones adiicimus.

I. Statuendo in aequatione secunda illic tradita N=X, N=Z/, N=zzl, proditXi' sin{X~L)~A' sin (/— X); r sin (X. — L) — A' sin (l — L) j r' sin (l-X) — i?’ sin (l - L) ;aequatio prima aut secunda commode ad calculi confirmationem applicatur, si me-thodus I aitt II. art. 62 adhibita est. Ita habetur in exemplo nostro

logsin(A— L ).9,4758650 n l—Iu = — 5 i° 45 / 26"82

. A'

l °g-y~ .9,7546x17

9,7212556 nlogsin(t—~Is) .9,7212556 n