Fulst: Methoden zur Berechnung der Höhe eines Gestirns, 
t = 1st 59m 40s 
= 5°18 N 
d= 10° 27 N 
„_d= 59° 9 
log sem = 8,8236 
log cos = 9,9981 
log cos = 9,9927 
s = 8,8144 
8a = 9,4072 
log cosec == 11,3461 
log tg == 10,7533 
CP 
a (@-— 0) = 2° 35' 
log sin = 8,6539 
log sec = 10,7603 
log sin = 9,4142 
a = 15° 3 
Zz = 30° 6 
h == 59° 54' 
1,74 
; — V/i0s 2 cos Tom! 
8X = sem (@ — d) 
3em z == sem (@ — d) sec? x 
ı = Ist 59m 408 log sem = 8,8236 
= 5° 18‘ N log cos = 9,9981 
d= 10°27 N log cos = 9,9927 
g—d= 5° co log sem == 2,6950 log sem = 7,3050 
8 == 1,5094 = 10,7610 
A 2 log se { r 
log tg == "0,7547 65001 = 10,7610 
z2= 30° Y' log sem == 8,8270 
h = 59° 58 
ig x = coseo £ — 0 Veos g@ cos d sem t 
sin 2/3 == cosec x Vcos @ cos d sem t 
it = 1st 59m 408 log sem = 8,8236 
p= 5°18 N log cos == 9,9981 
d= 10° 27 N log cos = 9,9927 
e—d= PS 8 = "8,8144 
8/9 = 9,4072 
a (g— 0) = 2° 35 log cosec == 11,3461 
log tg = 10,7533 
m 15° 9 
2= 30° 4 
h — 59° 56 
9,4072 
log cosec == 0,0066 
log sin == 9,4138 
sin x = cos 2 sec e Vcos @ cos d 
sin 2/2 == cos ed cos X 
== 5918 N log cos = 9,99814 
d = 10° 27' N log cos = 9,99274 
+0 = 15° 45 s == "9,99088 
5/2 = 9,99544 
[a (9 + 0) =" 7° 538’ log sec = 10,00412 
1a = Ost 59m 508 log cos = 9,98501 
log sin = 9,98457 
= 15° 
z= 30° 4 
h == 59° 56’ 
bO8 X = sec ex: Vcos @ cos 0 cos? t/2 
sin z/2 = tg x cos # cos d’ cos? t/2 
'/2 == Ost 59m 508 2 log cos = 9,97002 
=> 5° 18‘ N log cos = 9,99814 
d= 10° 27 N log cos = 9,99274 
+0 = 15° 45' 8 = 9,96090 
8/2 = ""79,98045 
log sec = 10,00412 
og cos = 09,98457 
Ua (+0) = 7° 58 
U = 15° 
zz = 30° 4 
h = 599° 56’ 
log cos == 9,99588 
log cos == 9,41815 
log sin == 9,41403 
3,98045 
log tg = 9,43358 
log sin == 9,41403