128 
13. Station Flierenberg. 
■ 
N ormalgleiehungen. 
2 4 A — + 3,755 
24 B = — 3,630 
24 C = 4- 1,760 
24 D= + 11,840 
24 e = — h,345 
24 F = — 2,380 
A = + 0,156 
B = — 0,152 
c = 4- 0,073 
D = 4- 0,493 
E = —0,473 
F = — 0,099 
Ergebnifs mit Einsehlufs sämmtlicher Reduktionen. 
O / II 
53. Hettenheuvel =00 0,000 4- (53) 
54. Kevelaer = 70 25 13,919-4(54) 
55. Venray =1x5 13 58,719 4-(55) 
Gewichtsgleichungen. 
(53) = 0,0417 [53] 
(54) = 0,04x7 [54] 
(55) = 0,0417 [55] 
Fehlerrechnung. 
(VV) 
- 4P k 
580,60 
203,29 
4P 2 ( vv )= 
P (w) = 
377.31 
33,5 8 ; Q v = 
15— 5 = 
10 
2 
m = 
V 
33,58 
IO 
= 2,36; 
m 
V 
=■ i,54 
II II 
p" «- 
P, | 
306,05 
— 50,82 
4 (uu) = 
(uu) = 
*55,*3 
63,81; Q = 
60— 5 = 
55 
2 
m = 
u 
63,81 
55 
= 1,16; 
m 
u 
= 1,08 
(UU): 4 — 
(VV) : 4P = 
(«) = 
76,51 
— 36,29 
40,22; Q = 
60—15 = 
45 
2 
m = 
e 
40,22 
45 
= °, 8 9; 
m 
£ 
=-0,94 
m 
2 
^9^3! P s = 
60' 
2 
1.1 
19^3 
60 
= 0,32; 
ß 
= o,57 
T 2 = 
2 
2 [t 
m — — 
E 2 
= 0,73; 
X 
= 0,85