Aplicacion de curvas de nivel

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Aplicacion de curvas de nivel

  1. 1. APLICACIONDE CURVASDE NIVEL Dibuje lascurvasde nivel a cada100 metros,de la siguientecuadricula. Puntos Cotas (m) A1 2400,87 A2 2200,10 A3 2380,52 B1 2500,55 B2 2320,32 B3 2550,10 C1 2204,87 C2 2100,90 C3 2100,90 321 C B A 10.0 10.0 10.0 10.0
  2. 2. Métodode la Cuadricula Recta A1-A2 Longitud = 10,00 m Δ (m) = (2400.87-2200.10) m Δ (m) = 200,77m 10,00…………………………..200, 77 X1…………………………. (2300-2200, 10) 𝑥1 = 99,9 ∗ 10 200,7 = 4,98 𝑚 2400,87 2200,10 2320,322500,50 10.0 10.0 23002400 2500 2400 2300 8.2 4.35.6
  3. 3. 10,00…………………………..200, 77 X2…………………………. (2400-2300) 𝑥2 = 100 ∗ 10 200,7 = 4,98 𝑚 10,00…………………………..200, 77 X2…………………………. (2400, 87-2400) 𝑥2 = 0.87 ∗ 10 200,7 = 0.04 𝑚 Recta A1-B1 Longitud = 10,00 m Δ (m) = (2500.50-2400.87) m Δ (m) = 96,63m 10,00…………………………..99,63 X1………………………………..99,13 𝑥1 = 99,13 ∗ 10 99,63 = 9,95 𝑚 10,00…………………………..99,63 X2……………………………….0.50 𝑥2 = 0,5 ∗ 10 99,63 = 0,05 𝑚 Recta B1-B2 Longitud = 10,00 m Δ (m) = (2500.50-2320.32) m Δ (m) = 180,18m
  4. 4. 10,00…………………………..180,18 X1………………………………..79,68 𝑥1 = 79,68 ∗ 10 180,18 = 4,26 𝑚 10,00…………………………..180,18 X2……………………………….100 𝑥2 = 100 ∗ 10 180,18 = 5.55𝑚 Recta A2-B2 Longitud = 10,00 m Δ (m) = (2320,32-2200,10) m Δ (m) = 120,22 m 10,00…………………………..120,22 X1………………………………..99,1 𝑥1 = 99,1 ∗ 10 120,22 = 8,24 𝑚 Recta A1-B2 Longitud = 14,14 m Δ (m) = (2400,87-2320,32) m Δ (m) = 80,55 m 14,14…………………………..80,55 X1………………………………..79,68 𝑥1 = 79,68 ∗ 14,14 80,55 = 13,99 𝑚
  5. 5. A1=2400,87 A2=2200,10 A1=2380,52 C3=2100,90C2=2100,90C1=2204,87 B3=2550,10 B2=2320.32B1=2500.5

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