Samtgemeinde Duingen Landkreis Hildesheim Gemarkung Weenzen

Samtgemeinde Duingen Landkreis Hildesheim Gemarkung Weenzen

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 .

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    1 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us