440 
In Opticam 
HC = BC — BH = tg. BAC 
AI — cosec. AIL = ————— 
sin. BCD 
tg. BA1I = 1,29385 — 0,89515 = 0,3987 (BA = r) 
1 = 1,46719 (AL = r) 
37° 44') = cotg. 5° 14' = 10,91778 
1,07362, ergo 
sin. 42° 58 
LC = cotg. ACL = cotg. (42° 58' 
IC = LC — LI. 
= LD — LI. 
LI = cotg. AIL = cotg. 42° 58 
IC = 10,91778 — 1,07362 = 9,84416. 
DI = LC —j— LI = 10,91778 + 1,07362 = 11,9914. 
Cum sit HC axis hyperboles (latus transversum) et dentur HC, AI, IC, DI, parametrum MC 
(figurae latus rectum) computatur proportione AI 3 : DI X IC — HC : CM, ergo 
CM = 21,86 (Kepl. 21,92" „fere“). 
Porro HC bisecta in N, erit NC = 0,19935. CO = */* V(HC X CM) = V* V(0,3987 X 21,86) 
= 1,476, et inde tang. ONC = ^ , /_ ONC = 82° 18' 30" (82° 19' K.). 
99) p. 
NC 2 
Ergo 
366. Cum linea SP tangens sit hyperboles, et N centrum, erit NS = ; 
NC = 0,19935. NR = NC + CR = NC + BR — BC = NC + tg. BAR — tg. BAC, 
linea BA, ut supra = r assumta. Per priora autem Z BAR = 52° 18' 40", = 52° 20' 48", 
= 52° 24', = 52° 29', ergo tg. BAR = 1,29437, = 1,29603, = 1,29853, = 1,30244. 
(Ceterum angulum 52° 20' 48" Keplerus falsum habet, cum sit 84° 43' 20" — 47° 2' = 
37° 41' 20", cujus complementum = 52° 18' 40", et parallaxi auctum = 52° 20' 40"; 
tang. 52° 20' 40" — 1,29592). Tg. BAC = 1,29385. 
I 1,29437 j 
1,29603 
1,29853 
1,30244 ] 
i 0,19987 
nnw ,, >0,20153 
0,039741 . 0 20403 
f 0,20794 
l 0,19987 — 0,19883 j 
0,20153 — 0,19719 f 
0,20403 — 0,19478 ( 
0,20794 — 0,19111 ) 
NR s= 0,19935 + 
NS 
— 1,29385 = 
SR = NR 
NS = 
RP = V 
c 
NR X RS X CM 
HC 
) 
Cot. BPR = — = 
RP 
BR 
z. BPR = 
(5") 
(32") 
Tang. SPR 
SR 
RP ’ 
ergo Z SPR = 
z BPS = BPR — SPR = 
0,10676 i 
0,219 f 
0,32167 i 
0,43804) 
85° 17 
80° 24 
76 n 5' (12") 
71° 24' 40" 
i 0° 33' 
) 1 ° 8 ' 
) 1° 38' 
( 2 ° 12 ' 
84° 43' 
79° 16' 
74° 26' 
69° 12' 
i 0,19987 
1 0,20153 
j 0,20403 ’ 
f 0,20794 
0,19883 
0,19719 
0,19478 
0,19111 
1 0,00104 
_ 10,00434 
0,00925 
0,01683 
0,10676 
0,219 
0,32167 
0,43804 
1,29437 l 
1,29603 f 
1,29853 ( 
1,30244 ) 
et 
30" 
(7") 
50" 
(35") 
(25") 
( 22 ") 
(36") 
Denique 
Keplerus computat ultimum angulum = 69° 16' pro 69° 12', quia falsum numerum 0,01672 
pro 0,01683 errore calculi accipit, nam 20794 — 19111 = 1683, sive assumto radio = 1, 
0,01683, quo factum est, ut ipsi PR = 0,43664 prodierit. Si pro falso angulo 52° 20' 48", 
quem supra indicavimus, verum 52° 20' 40" accepissemus, prodiisset / BPS secundus 
= 79° 32' 48", PR = 0,2134.