Samtgemeinde Duingen Landkreis Hildesheim Gemarkung Weenzen
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6 5/ ,2 0 Ah 3 5 5 3 5 5 5 5 5 5 5 5 4 4 6 5 6 4 5 0 7 7 7 7 7 7 0 7 7 0 0 AcGhsraed.== 6 0G1_ W W 7 n 6 6 6 6 6 6 6 6 Q e 6 4 4 4 4 4 4 5 z 5 5 6 .035 n 4 5 6 7 Km= 60+325 8 9 1 2 0 B e 0 E 135/2 72 N 0 0 0 = 60_ WW 0 0 0 ( 0 0 Achse = G1 G rad. e + 00 78 k 5740 H = 35 0 A m 0 0 0 2 0 0 0 0 0 V W C022 _ a Achse = 01 n 4 G rad. = G6 0 o 1 67 t = 12.438 1 48/20 2 g i 2. h 2 24 n 1 e 0+ 6 u c m= Acker 43 d r , , K e 73 9 5 = 0.022 a hw 0 f d 0 s n 6 20 - 35 H = 26 w e 044.0 Q 2 = 2+ S 190.556 m T k 1 K s 56 c . 9 f Q ei = 2 = t 2 Acker / Q m o d 5780 Q + Q 0 8 00 h 1 10 de Q = 0.2 H = 2 f h 1 M + 2 e 6 4 2 76 6 . l 1 9 d 8 + 1 2 W 0_W + 0 se= 6 1 S 8 Ach d. = G 8 B G ra T B 4.5 0 201/71 2 u + = 1 t 3 0 r + e 2 667% 3 s =1. , i a 7 9 a a t 0 7 0 l 56% 2 1 s =0.9 7 1 5790 r / a 6 5 ß u 0 u 0 4 , + 0 h 1 6 0 1 7 B + .7 + 0 5 0 f = r 8 p 6 0 F - , m 0 , = 7.101 a 0 l =6 , s 0 m 0 0 k K , 8m 0 76 2 i 0 l =82. Acker A % 3 i , HFP 4 PB n 0 0 56 0 0 9 0 . B 5 1 5 0 m L = e 0 2 5 a O e - s .45 u a 0 1 0 9 = 1 e f 3 TS n 0 k 0 H d 0 , e a 0% 7 0 0 0 i 191,771m ü. NN 0 0 n . 5 e Acker 0 .0 =-2 m 2 n B u 5 s 68 6 2 7 a n = 82. u e - U = k t = l m . 0 = 5 2 6 g B 01 0+ A 9 . 7m % h 0 6 3 p s - 2 . 9 0 + Q = =97 2 ,1 r Q l , % f 1 0 5 , / 3 0 3 3 i 149/1 , 71 6 E e n . 76/1 0 i 0 0 n 1 , t 9 z 0 4 e t 1 W 7 a lb 3 1 S s a 2 T D 0 3 um 0 r 3 N s SW314 SW315 ü + c 5 3 h 5 4 u 0 1 g 6 % t + Acker 7 z h 4 6 8 0 6 3 e . 0 1. 0 , 0 s = 3 OKG. = 189,91 T N A e h 6 0 0 D. = 187,27 R % 5% 0 0 8 .12 e p 5800 s =-1 6 0 0 c + 44 0 K 2.0 D A SW313.2 m=65+03 s Querdurchlaß - p = 6,4 % 13 = 4 0 m 0 D. = 189,06 e , = , s T 1 1 S. = 185,24 2 8 8 .3 5 0 0 8 /5 A 0 0 49 n , 6 + k = E 3 DN 300 / 400 B K m 5 l i 1 H P n OKG. = 190,14 =9 0 7 z 97,5 122 A 6 71 ch se e G = 2 r a 60 m n d. _ . 3 S. = 185,56 = W 6m l G1 W 9.02 b 3 7 52 2 l = a 5 0 9 5 0 % = u = 0 m Einlauf = 187,46 4 Q 0 l ) e H 6 6 3 .8 s 2 , c / 6 1 8 h t=11 % 6 - D. = 188,23 ,413 0 = u 0 7 = s tz 0 0 0 = 8 z + 3 Auslauf = 187,01 + 0 1 t - 0 , m , = 1 0 25 . f= 0 u 1 0 4 , Q 6 S. = 185,63 06 + 0 4 . 5 m 6 e 1 i 9 0 3 10 Bau-km 60+355,750 6 K h n f = 2 i 0 7 . 0 . l 0, 5 0 9 1 n A = 3 0 b 6 7 7 /3 TS 5 = 0 6 + 136 4 , T -0 1 5 Zusatzfahr 926 n e 0 Durchlaß DN 400 B streifen 2 Längsdurchlaß 0 . 0 S 0 Bau-km 2+020,000 K , f / f + T i . R 2 5 R 0 e , ,3 Lz = 86 3 0 ,5 % 0 1 0 0m b L = 20,00 m /78 ; i l 0 DN 400 B s 9 2 39 i = 0,25 m 75 m 3 , P n = 2 3 = R = 0 = Durchlaß DN 800 B 1 S . 8 = u 3 0 - 7 % 5 1 0,462 Einlauf = 187,35 0 Einlauf li : 189,95 S= , m . 5 2 l 1 3 2 Sperrfl 39 9 , 9 0 äch = 9 L = 18,00 m 6 enmarkierung 80 , = = 2 5 R= 0+ L Auslauf = 187,01 3 8 3 % 3 Auslauf re : 189,15 2 -25 372 L=11,413m Bau-km 60+200,000 .1 Lz = 86 5 3 . 7 6 ,50m ; Q 3 75 Einlauf li : 188,90 8 i li = 3,25 m lt 5 7 V 5 6 / 5 . ) Lw x Lh = 1,25 x 1,00 m 5 a 0 0 m k 2 n 7 - Auslauf re : 188,64 Zusatzfahrs e h 3 0 treifen + l , + L 0 0 A=20,00m l p 0 LV= e 1 L = 9,00 m Acker 20,00 L 0 + m z = s 3 K 2 86,50 m w i 4 i 0 1 n , A ife 2 2 Es 0,20/4 s 42 Arbeitsstre 5/6 SW313.1 5 h L 9 Einlauf li : 188,95 c 8 i li = 2, 212/69 = + 50 0 m s Schutzeinrichtung 2 . 6 - 2 z 1 8 2 l , l 24 2 . 5 Arbeitsstreifen D. = 189,10 + + , = 8 Auslauf re : 188,85 R o 0 8 7 6 + 5 0 1 0 0 , gemäß RPS 4 / 2 H 0 0 0 3 2 6 6 35 2 , 35 4 3 1 % 75 % 0% = f 5 0, , S. = 185,76 0 . 6 6 49 u 3 8 4 8 6 + 2 R 0% 0 a , 2 n 6, , Schutzeinrichtung fe 1 R i 1 5 e 7 tr 3 0 ss r 1 e t te 0 54 m t 0 = 49 , i o 0 h 0 , e 0 c 0 i S + b 0 s 0 3 n , = % i r 0 3, 3 i g A 8 8 5 n , % fa gemäß RPS 1 uan 5 e L 0 Ba 5 1 l k 4 7 , 8 0 0 0 , 0 ,00 0 6 m 0 G 0 5 1 +01 . r 0 0 2 m 6 5 0 7 . - m -k ,7 0 0 u 0 0 Ba 0 1 7 e tt t % 0 k 0 1:5 4,00 % 2 t e . 1:5 , 0 m 5 WE 2 Sch b 0 o ott r 2 er 6 0 e / / 2 h tt , 0 4 H 6 0 0 c 0 0 5,0 o 3 36 = 1 0 m Baum 0,20/3 0 h 4, 6 , S c 00 o . 121 + 3 G R 3 0 S 0 R 6 = i r 0 ( 0 0 0 n , % N L R Mast 17 0 R 6 2 8 + 0 m 6 0 0 + , 0 a 2 3 0 . 0 2 3 D R 8 = = 7 = m - 6 2 5 50/5 3 . 0 A a c p 5 D 0 8 54 , , 4 8 0 5 5 ,1 d 0 1 F 5/5 k H 0 , n N 3 . 3 37 % n fe 0 h e P 3 i 4 1 re 0 a = t 2 6 s 0 R 3 ts 6, B r i 65 6 e % R R 0 rb . w R Rohrdurchlaß A A 5 BS 10 0 p 5 0 = ,0 0,2 = 6 R 0 + e + 0 / = 5 ,0 3 m 5 1 / 2 B 3 00 m % 11 = gem. Hydraulischer - 0 0 g 5 39 2 5730 34 8 8 0 ,0 0 8 2 5 r Acker 9 Dimensionierung te 5 0 t Baum 0,20/3 o ,8 . ch BS 9 6 HFP 1 PB S 5 BS 11 5 Anschlußplanung 67 % 0 ,0 R=30 - + Acker 193,366m ü. NN 6 % 250/120 R 1 % 3,0 lappe = 3 m / 200 m m "B 240 OU Weenzen Nord" % roschk 1 u 1 0 4 5 , F 2 3,6 6 Schutzeinrichtung a e 5 . 5 p . = lap ,7 z 4 5640 k 0 4 lb t 7 A % h . 0 2 c 0 / ros > nach E 0 u F gemäß RPS ime e s ,5 05 5 249/120 % 52 h 5 = G z 3 p , R , , 5 n c h 5 , i 0 BS 8 4 7 s a 6 R 0 1 F 0 E 5630 > = h 6 .