Problemario de Topografía
Poligonal Cerrada
1
CALCULO DE POLIGONAL CERRADA En los siguientes problemas, se analizarán las cuatro posibilidades del calculo de azimutes, según se tengan ángulos internos o externos, y según se resuelvan los azimutes en sentido horario o antihorario, dependiendo del azimut inicial.
1.- Calcular la poligonal cerrada: 1a.- Caso de tener Azimut de A a B (sentido antihorario, ángulos internos)
E F
αE αD
αF A
αA
αC
AZAB
αB B Datos:
AZAB = 121º 12’ 13” αA = 92º 40’ 44” αB = 132º 27’ 53” αC = 129º 38’ 23” αD = 87º 48’ 34” αE = 133º 12’ 35” αF = 144º 12’ 40”
D
DAB = 52,97 m. DBC = 60,37 m. DCD = 43,01 m. DDE = 63,42 m. DEF = 48,25 m. DFA = 35,32 m.
Calculo de los ángulos internos.
Por definición, Σα = (n – 2) 180º = (6 – 2) 180º = 720º
C
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2
Problemario de Topografía
Poligonal Cerrada
Sumando los ángulos internos medidos tenemos:
Σα = 720º 00’ 49” La diferencia de valor entre los ángulos medidos y el valor teórico es de 49”. La compensación total será en consecuencia de –49”. La compensación para cada ángulo medido es: -49”/6 = -8”,17
Cálculo de las correcciones de los ángulos. αA = 92º 40’ 44” – 8”,17 = 92º 40’ 35”,83 αB = 132º 27’ 53” – 8”,17 = 132º 27’ 44”,83 αC = 129º 38’ 23” – 8”,17 = 129º 38’ 14”,83 αD = 87º 48’ 34” – 8”,17 = 87º 48’ 25”,83 αE = 133º 12’ 35” – 8”,17 = 133º 12’ 26”,83 αF = 144º 12’ 40” – 8”,17 = 144º 12’ 31”,83 Cálculo de Azimutes.
AZAB = 121º 12’ 13” AZBC = AZAB + αB ± 180º = 121º 12’ 13” + 132º 27’ 44”,83 – 180º AZBC = 73º 39’ 57”,83 AZCD = AZBC + αC ± 180º = 73º 39’ 57”,83 + 129º 38’ 14”,83 – 180º AZCD = 23º 18’ 12”,66 AZDE = AZCD + αD ± 180º = 23º 18’ 12”,76 + 87º 48’ 25”,83 + 180º AZDE = 291º 6’ 38”,49 AZEF = AZDE + αE ± 180º = 291º 6’ 38”,59 + 133º 12’ 26”,83 + 180º AZEF = 244º 19’ 5”,32 AZFA = AZEF + αF ± 180º = 244º 19’ 5”,42 + 144º 12’ 31”,83 + 180º AZFA = 208º 31’ 37”,1
Calculo de las proyecciones.
∆ NAB = DAB × cos AZAB = 52,97 × cos 121º 12’ 13” ∆ NAB = -27,44 m. ∆EAB = DAB × sen AZAB = 52,97 × sen 121º 12’ 13” ∆E = 45,31 m.
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Problemario de Topografía
Poligonal Cerrada
∆ NBC = DBC × cos AZBC = 60,37 × cos 73º 39’ 57”,83 ∆ NBC = 16,98 m. ∆EBC = DBC × sen AZBC = 60,37 × sen 73º 39’ 57”,83 ∆EBC = 57,93 m. ∆ NCD = DCD × cos AZCD = 43,01 × cos 23º 18’ 12”,66 ∆ NCD = 39,50 m. ∆ECD = DCD × sen AZCD = 43,01 × sen 23º 18’ 12”,66 ∆ECD = 17,01 m. ∆ NDE = DDE × cos AZDE = 63,42 × cos 291º 6’ 38”,49 ∆ NDE = 22,84 m. ∆EDE = DDE × sen AZDE = 63,42 × sen 291º 6’ 38”,49 ∆EDE = -59,16 m. ∆ NEF = DEF × cos AZEF = 48,25 × cos 244º 19’ 5”,32 ∆ NEF = -20,91 m. ∆EEF = DEF × sen AZEF = 48,25 × sen 244º 19’ 5”,32 ∆EEF = -43,48 m. ∆ NFA = DEF × cos AZFA = 35,32 × cos 208º 31’ 37”,1 ∆ NFA = -31,03 m. ∆EFA = DEF × sen AZFA = 35,32 × sen 208º 31’ 37”,1 ∆EFA = -16,87 m.
Σ∆ N: -27,44 16,98 39,50 22,84 -20,91 -31,03 Σ = -0,06
Σ∆E: 45,31 57,93 17,01 -59,16 -43,48 -16,87 Σ = 0,74
El error en ∆ N es de –0,06 m, por lo tanto su corrección total será + 0,06 m. El error en ∆E es de 0,76 m, por lo que su corrección total será de – 0,74 m.
Calculo de las correcciones de las proyecciones. La corrección lineal se realiza siguiendo la forma: ∆ Ni corregido = ∆ Ni + (longitud del lado)/(longitud total) × corrección total ∆ N
3
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4
Problemario de Topografía
Poligonal Cerrada
Longitud total = Σ lados = 303,34 m. ∆ NAB corregido = -27,44 + (52,97)/(303,34) × 0,06 = -27,43 m. ∆ NBC corregido = 16,98 + (60,37)/(303,34) × 0,06 = 16,99 m. ∆ NCD corregido = 39,50 + (43,01)/(303,34) × 0,06 = 39,51 m. ∆ NDE corregido = 22,84 + (63,42)/(303,34) × 0,06 = 22,85 m. ∆ NEF corregido = -20,91 + (48,25)/(303,34) × 0,06 = -20,90 m. ∆ NFA corregido = -31,03 + (35,32)/(303,34) × 0,06 = -31,02 m.
∆EAB corregido = ∆EBC corregido = ∆ECD corregido = ∆EDE corregido = ∆EEF corregido = ∆EFA corregido =
45,31 + (52,97)/(303,34) × (– 0,74) = 45,18 m. 57,93 + (60,37)/(303,34) × (– 0,74) = 57,78 m. 17,01 + (43,01)/(303,34) × (– 0,74) = 16,91 m. -59,16 + (63,42)/(303,34) × (– 0,74) = -59,31 m. -43,48 + (48,25)/(303,34) × (– 0,74) = -43,60 m. -16,87 + (35,32)/(303,34) × (– 0,74) = -16,96 m.
Calculo de las coordenadas de los puntos.
NB = NA + ∆ NAB = 1000 + (-27,43) = 972,57 m. EB = EA + ∆EAB = 1000 + 45,18 = 1045,18 m. NC = NB + ∆ NBC = 972,57 + 16,99 = 989,56 m. EC = EB + ∆EBC = 1045,18 + 57,78 = 1102,96 m. ND = NC + ∆ NCD = 989,56 + 39,51 = 1029,07 m. ED = EC + ∆ECD = 1102,96 + 16,91 = 1119,87 m. NE = ND + ∆ NDE = 1029,07 + 22,85 = 1051,92 m. EE = ED + ∆EDE = 1119,87 + (-59,31) = 1060,56 m. NF = NE + ∆ NEF = 1051,92 + (-20,90) = 1031,02 m. EF = EE + ∆EEF = 1060,56 + (-43,60) = 1016,96 m. NA = NF + ∆ NFA = 1031,02 + (-31,02) = 1000,00 m. EA = EF + ∆EFA = 1016,94 + (-16,96) = 1000,00 m.
1b.- Caso de tener Azimut de A a F (sentido horario, ángulos internos)
AZAF = 28º 31’ 37”,17
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Problemario de Topografía
Poligonal Cerrada
AZFE = AZAF + αF – 180º = 28º 31’ 37”,17 – 144º 12’ 31”,83 + 180º AZFE = 64º 19’ 5”,34 AZED = AZFE + αE – 180º = 64º 19’ 5”,34 – 133º 12’ 26”,83 + 180º AZED = 111º 6’ 38”,51 AZDC = AZED + αD – 180º = 111º 6’ 38”,51 – 87º 48’ 25”,83 + 180º AZDC = 203º 18’ 12”,6 AZCB = AZDC + αC – 180º = 203º 18’ 12”,6 – 129º 38’ 14”,83 + 180º AZCB = 253º 39’ 57”,7 AZBA = AZCB + αB – 180º = 253º 39’ 57”,7 – 132º 27’ 44”,83 + 180º AZBA = 301º 12’ 12”,8
Calculo de las proyecciones.
∆ NAF = DAF × cos AZAF = 35,32 × cos 28º 31’ 37”,17 ∆ NAF = 31,03 m. ∆EAF = DAF × sen AZAF = 35,32 × sen 28º 31’ 37”,17 ∆EAF = 16,87 m.
∆ NFE = DFE × cos AZFE = 48,25 × cos 64º 19’ 5”,34 ∆ NFE = 20,91 m. ∆EFE = DFE × sen AZFE = 48,25 × sen 64º 19’ 5”,34 ∆EFE = 43,48 m.
∆ NED = DED × cos AZED = 63,42 × cos 111º 6’ 38”,51 ∆ NED = -22,84 m. ∆EED = DED × sen AZED = 63,42 × sen 111º 6’ 38”,51 ∆EED = 59,16 m. ∆ NDC = DDC × cos AZDC = 43,01 × cos 203º 18’ 12”,6 ∆ NDC = -39,50 m. ∆EDC = DDC × sen AZDC = 43,01 × sen 203º 18’ 12”,6 ∆EDC = -17,01 m. ∆ NCB = DCB × cos AZCB = 60,37 × cos 253º 39’ 57”,7 ∆ NCB = -16,98 m.
5
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6
Problemario de Topografía
Poligonal Cerrada
∆ NBA = DBA × cos AZBA = 52,97 × cos 301º 12’ 12”,8 ∆ NBA = 27,44 m. ∆EBA = DBA × sen AZBA = 52,97 × sen 301º 12’ 12”,8 ∆EBA = -45,31 m. Σ∆ N: -27,44 16,98 39,50 22,84 -20,91 -31,03 Σ = -0,06
Σ∆E: 45,31 57,93 17,01 -59,16 -43,48 -16,87 Σ = 0,74
El error en ∆ N es de –0,06 m, por lo tanto su corrección total será + 0,06 m. El error en ∆E es de 0,76 m, por lo que su corrección total será de – 0,74 m.
Calculo de las correcciones de las proyecciones. La corrección lineal se realiza siguiendo la forma: ∆ Ni corregido = ∆ Ni + (longitud del lado)/(longitud total) × corrección total ∆ N ∆Ei corregido = ∆Ei + (longitud del lado)/(longitud total) × corrección total ∆E
Longitud total = Σ lados = 303,34 m.
∆ NAB corregido = ∆ NBC corregido = ∆ NCD corregido = ∆ NDE corregido = ∆ NEF corregido = ∆ NFA corregido =
-27,44 + (52,97)/(303,34) × 0,06 = -27,43 m. 16,98 + (60,37)/(303,34) × 0,06 = 16,99 m. 39,50 + (43,01)/(303,34) × 0,06 = 39,51 m. 22,84 + (63,42)/(303,34) × 0,06 = 22,85 m. -20,91 + (48,25)/(303,34) × 0,06 = -20,90 m. -31,03 + (35,32)/(303,34) × 0,06 = -31,02 m.
∆EAB corregido = ∆EBC corregido = ∆ECD corregido = ∆EDE corregido = ∆EEF corregido = ∆EFA corregido =
45,31 + (52,97)/(303,34) × (– 0,74) = 45,18 m. 57,93 + (60,37)/(303,34) × (– 0,74) = 57,78 m. 17,01 + (43,01)/(303,34) × (– 0,74) = 16,91 m. -59,16 + (63,42)/(303,34) × (– 0,74) = -59,31 m. -43,48 + (48,25)/(303,34) × (– 0,74) = -43,60 m. -16,87 + (35,32)/(303,34) × (– 0,74) = -16,96 m.
Calculo de las coordenadas de los puntos.
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Problemario de Topografía
Poligonal Cerrada
EF = EA + ∆EAF = 1000 + 45,18 = 1045,18 m. NC = NB + ∆ NBC = 972,57 + 16,99 = 989,56 m. EC = EB + ∆EBC = 1045,18 + 57,78 = 1102,96 m. ND = NC + ∆ NCD = 989,56 + 39,51 = 1029,07 m. ED = EC + ∆ECD = 1102,96 + 16,91 = 1119,87 m. NE = ND + ∆ NDE = 1029,07 + 22,85 = 1051,92 m. EE = ED + ∆EDE = 1119,87 + (-59,31) = 1060,56 m. NF = NE + ∆ NEF = 1051,92 + (-20,90) = 1031,02 m. EF = EE + ∆EEF = 1060,56 + (-43,60) = 1016,96 m. NA = NF + ∆ NFA = 1031,02 + (-31,02) = 1000,00 m. EA = EF + ∆EFA = 1016,94 + (-16,96) = 1000,00 m.
1c.- Caso de tener Azimut de A a B (sentido antihorario, ángulos externos)
αE αF
E F
A
D
AZAB C
αA B
αD
7
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8
Problemario de Topografía
αF = 215º 47’ 20”
Poligonal Cerrada
DFA = 35,32 m.
Calculo de los ángulos externos.
Por definición, Σα = (n – 2) 360º = (6 – 2) 360º = 1440º Sumando los ángulos internos medidos tenemos:
Σα = 1439º 59’ 11” La diferencia de valor entre los ángulos medidos y el valor teórico es de –49”. La compensación total será en consecuencia de +49”. La compensación para cada ángulo medido es: +49”/6 = +8”,17
Cálculo de las correcciones de los ángulos.
αA = 267º 19’ 16” + 8”,17 = 267º 19’ 24”,17 αB = 227º 32’ 07” + 8”,17 = 227º 32’ 15”,17 αC = 230º 21’ 37” + 8”,17 = 230º 21’ 45”, 17 αD = 272º 11’ 26” + 8”,17 = 272º 11’ 34”,17 αE = 226º 47’ 25” + 8”,17 = 226º 47’ 33”, 17 αF = 215º 47’ 20” + 8”,17 = 215º 47’ 28”, 17
Cálculo de Azimutes.
AZAB = 121º 12’ 13” AZBC = AZAB – αB ± 180º = 121º 12’ 13” – 227º 32’ 15”,17 + 180º AZBC = 73º 39’ 57”,83 AZCD = AZBC – αC ± 180º = 73º 39’ 57”,83 – 129º 38’ 14”,83 – 180º AZCD = 23º 18’ 12”,66
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Problemario de Topografía
Poligonal Cerrada
1d.- Caso de tener Azimut de A a F (sentido horario, ángulos externos)
AZAF = 28º 31’ 37”,17 Como ya tenemos los angulos compensados, se procede a calcular los azimutes.
AZFE = AZAF + αF – 180º = 28º 31’ 37”,17 + 215º 47’ 28”,17 – 180º AZFE = 64º 19’ 5”,34 AZED = AZFE + αE – 180º = 64º 19’ 5”,34 + 226º 47’ 33”,17 – 180º AZED = 111º 6’ 38”,51 AZDC = AZED + αD – 180º = 111º 6’ 38”,51 + 272º 11’ 34”,17 – 180º AZDC = 203º 18’ 12”,6 AZCB = AZDC + αC – 180º = 203º 18’ 12”,6 + 230º 21’ 45”,17 – 180º AZCB = 253º 39’ 57”,7 AZBA = AZCB + αB – 180º = 253º 39’ 57”,7 + 227º 32’ 15”,17 – 180º AZBA = 301º 12’ 12”,8
2.- Calcular la poligonal cerrada:
1
AZ12
9
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10
Problemario de Topografía
Poligonal Cerrada
Datos:
AZ12 = 195º 23’ 42” α1 = 76º 34’ 42”,96 α2 = 73º 57’ 51”,12 α3 = 130º 22’ 03” α4 = 162º 03’ 51”,48 α5 = 97º 00’ 54”,72
D12 = 94,792 m. D23 = 27,853 m. D34 = 43,988 m. D45 = 35,487 m. D51 = 50,272 m.
Calculo de los ángulos internos.
Por definición, Σα = (n – 2) 180º = (5 – 2) 180º = 540º Sumando los ángulos internos medidos tenemos: tene mos:
Σα = 539º 59’ 23”,2 La diferencia de valor entre los ángulos medidos y el valor teórico es de -36”,72. La compensación total será en consecuencia de +36”,72. La compensación para cada ángulo medido es: 36”,72/5 = 7”,34
Cálculo de las correcciones de los ángulos. α1 = 76º 34’ 42”,96 + 7”,34 = 76º 34’ 50”,30 α2 = 73º 57’ 51”,12 + 7”,34 = 73º 57’ 58”,46 α3 = 130º 22’ 03” + 7”,34 = 130º 22’ 10”,34 α4 = 162º 03’ 51”,48 + 7”,34 = 162º 03’ 58”,82 α5 = 97º 00’ 54”,72 + 7”,34 = 97º 01’ 02”,06
Cálculo de Azimutes.
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Problemario de Topografía
Poligonal Cerrada
AZ45 = 21º 47’ 49”,62 AZ51 = AZ45 + α5 ± 180º = 21º 47’ 49”,62 + 97º 01’ 02”,06 + 180º AZ51 = 298º 48’ 51”,68
Calculo de las proyecciones.
∆ N12 = D12 × cos AZ12 = 94,792 × cos 195º 23’ 42” ∆ N12 = -91,391 m. ∆E12 = D12 × sen AZ12 = 94,792 × sen 195º 23’ 42” ∆E12 = -25,165 m. ∆ N23 = D23 × cos AZ23 = 27,853 × cos 89º 21’ 40”,46 ∆ N23 = 0,311 m. ∆E23 = D23 × sen AZ23 = 27,853 × sen 89º 21’ 40”,46 ∆E23 = 27,851 m.
∆ N34 = D34 × cos AZ34 = 43,988 × cos 39º 43’ 50”,80 ∆ N34 = 33,829 m. ∆E34 = D34 × sen AZ34 = 43,988 × sen 39º 43’ 50”,80 ∆E34 = 28,116 m. ∆ N45 = D45 × cos AZ45 = 35,487 × cos 21º 47’ 49”,62 ∆ N45 = 32,950 m. ∆E45 = D45 × sen AZ45 = 35,487 × sen 21º 47’ 49”,62 ∆E45 = 13,177 m. ∆ N51 = D51 × cos AZ51 = 50,272 × cos 298º 48’ 51”,68 ∆ N51 = 24,230 m. ∆E51 = D51 × sen AZ51 = 50,272 × sen 298º 48’ 51”,68 ∆E51 = -44,048 m.
11
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12
Problemario de Topografía
Poligonal Cerrada
El error en ∆E es de 0,069 m, por lo que su corrección total será de + 0,069 m.
Calculo de las correcciones de las proyecciones.
La corrección lineal se realiza siguiendo la forma: ∆ Ni corregido = ∆ Ni + (longitud del lado)/(longitud total) × corrección total ∆ N ∆Ei corregido = ∆Ei + (longitud del lado)/(longitud total) × corrección total ∆E Longitud total = Σ lados = 252,392 m.
∆ N12 corregido = ∆ N23 corregido = ∆ N34 corregido = ∆ N45 corregido = ∆ N51 corregido =
-91,391 + (94,792)/(252,392) × 0,071 = -91,364 m. 0,311 + (27,853)/(252,392) × 0,071 = 0,319 m. 33,829 + (43,988)/(252,392) × 0,071 = 33,841 m. 32,950 + (35,487)/(252,392) × 0,071 = 32,960 m. 24,230 + (50,272)/(252,392) × 0,071 = 24,244 m.
∆E12 corregido = -25,165 + (94,792)/(252,392) × 0,069 = -25,139 m. ∆E23 corregido = 27,851 + (27,853)/(252,392) × 0,069 = 27,859 m. ∆E34 corregido = 28,116 + (43,988)/(252,392) × 0,069 = 28,128 m. ∆E45 corregido = 13,177 + (35,487)/(252,392) × 0,069 = 13,187 m. ∆E51 corregido = -44,048 + (50,272)/(252,392) × 0,069 = -44,034m.
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Problemario de Topografía
Poligonal Cerrada
3.- Calcular la poligonal cerrada:
6 AZ16
1
α1
α2
5
α6
α5
α4
4
α3 3
2 Datos:
AZ12 = 44º 30’ 05”
13
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14
Problemario de Topografía
Poligonal Cerrada
Cálculo de las correcciones de los ángulos.
α1 = 109º 23’ 42” – 0”,5 α2 = 111º 13’ 21” – 0”,5 α3 = 63º 25’ 10” – 0”,5 α4 = 242º 55’ 28” – 0”,5 α5 = 73º 12’ 32” – 0”,5 α6 = 119º 49’ 50” – 0”,5
= = = = = =
109º 111º 63º 242º 73º 119º
23’ 41”,5 13’ 20”,5 25’ 09”,5 55’ 27”,5 12’ 31”,5 49’ 49”,5
Cálculo de Azimutes.
AZ16 = 44º 30’ 05” AZ65 = AZ16 – α6 ± 180º = 44º 30’ 05” – 119º 49’ 49”,5 + 180º AZ65 = 104º 40’ 15”,5 AZ54 = AZ65 – α5 ± 180º = 104º 40’ 15”,5 – 73º 12’ 31”,5 + 180º AZ54 = 211º 27’ 44” AZ43 = AZ54 – α4 ± 180º = 211º 27’ 44” – 242º 55’ 27”,5 + 180º AZ43 = 148º 32’ 16”,5
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Problemario de Topografía
∆ N54 = D45 × cos AZ54 = 85,85 × cos 211º 27’ 44” ∆ N54 = –73,23 m. ∆E54 = D45 × sen AZ54 = 85,85 × sen 211º 27’ 44” ∆E54 = –44,81 m. ∆ N43 = D34 × cos AZ43 = 68,75 × cos 148º 32’ 16”,5 ∆ N43 = –58,64 m. ∆E43 = D34 × sen AZ43 = 68,75 × sen 148º 32’ 49”,5 ∆E43 = 35,88 m. ∆ N32 = D23 × cos AZ32 = 122 × cos 265º 7’ 7” ∆ N32 = –10,38 m. ∆E32 = D23 × sen AZ32 = 122 × sen 265º 7’ 7” ∆E32 = –121,56 m. ∆ N21 = D12 × cos AZ21 = 84,4 × cos 333º 53’ 46”,5 ∆ N21 = 75,79 m. ∆E21 = D12 × sen AZ21 = 84,4 × sen 333º 53’ 46”,5 ∆E21 = –37,14 m.
Σ∆ N:
Σ∆E:
Poligonal Cerrada
15
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16
∆ N16 corregido = ∆ N65 corregido = ∆ N54 corregido = ∆ N43 corregido = ∆ N32 corregido = ∆ N21 corregido =
Problemario de Topografía
Poligonal Cerrada
87,37 + (122,5)/(569) × 0,74 = 87,53 m. –21,65 + (85,5)/(569) × 0,74 = –21,54 m. m. –73,23 + (85,85)/(569) × 0,74 = –73,12 m. –58,64 + (68,75)/(569) × 0,74 = –58,55 m. –10,38 + (122)/(569) × 0,74 = –10,22 m. 75,79 + (84,4)/(569) × 0,74 = 75,90 m.
∆E16 corregido = 85,86 + (122,5)/(569) × –0,94 = 85,66 m. ∆E65 corregido = 82,71 + (85,5)/(569) × –0,94 = 82,57 m. ∆E54 corregido = –44,81 + (85,85)/(569) × –0,94 = –44,95 m. ∆E43 corregido = 35,88 + (68,75)/(569) × –0,94 = 35,77 m. ∆E32 corregido = –121,56 + (122)/(569) × –0,94 = –121,76 m. ∆E21 corregido = –37,14 + (84,4)/(569) × –0,94 = –37,28m.
Calculo de las coordenadas de los puntos.
N6 = N1 + ∆ N16 = 1.000 + 87,53 = 1.087,53 m. E6 = E1 + ∆E16 = 1.000 + 85,66 = 1.085,66 m. N5 = N6 + ∆ N65 = 1.087,53 + (–21,54) = 1.065,99 m. E3 = E6 + ∆E65 = 1.085,66 + 82,57 = 1.168,23 m.
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Problemario de Topografía
Poligonal Cerrada
4.- Calcular la poligonal cerrada:
C
B
αC AZAB
A
αA
αB
αD
α
D
17
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18
Problemario de Topografía
Poligonal Cerrada
La diferencia de valor entre los ángulos medidos y el valor teórico es de -18”. La compensación total será en consecuencia de +18”. La compensación para cada ángulo medido es: 18”/6 = 3”
Cálculo de las correcciones de los ángulos.
αA = 81º 48’ 36” + 3” αB = 130º 59’ 48” + 3” αC = 103º 32’ 06” + 3” αD = 152º 55’ 29” + 3” αE = 128º 27’ 56” + 3” αF = 122º 15’ 47” + 3”
= 81º 48’ 39” = 130º 59’ 51” = 103º 32’ 09” = 152º 55’ 32” = 128º 27’ 59” = 122º 15’ 50”
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Problemario de Topografía
∆ NBC = 10,216 m. ∆EBC = DBC × sen AZBC = 231,366 × sen 87º 28’ 09” ∆EBC = 231,140 m.
∆ NCD = DCD × cos AZCD = 198,640 × cos 163º 56’ ∆ NCD = -190,881 m. ∆ECD = DCD × sen AZCD = 198,640 × sen 163º 56’ ∆ECD = 54,975 m. ∆ NDE = DDE × cos AZDE = 179,017 × cos 191º 00’ 28” ∆ NDE = -175,723 m. ∆EDE = DDE × sen AZDE = 179,017 × sen 191º 00’ 28” ∆EDE = -34,182 m.
Poligonal Cerrada
19
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20
Problemario de Topografía
Poligonal Cerrada
Longitud total = Σ lados = 1.291,521 m.
∆ NAB corregido = ∆ NBC corregido = ∆ NCD corregido = ∆ NDE corregido = ∆ NEF corregido = ∆ NFA corregido =
169,728 + (216,775)/( 1.291,521) × 0,010 = 169,364 m. 10,216+ (231,366)/( 1.291,521) × 0,010 = 10,319 m. -190,881 + (198,640)/( 1.291,521) × 0,010 = -190,841 m. -175,723 + (179,019)/( 1.291,521) × 0,010 = -175,960 m. -18,430 + (39,969)/( 1.291,521) × 0,010 = -18,244 m. 205,080 + (406,754)/( 1.291,521) × 0,010 = 205,244 m.
∆EAB corregido = ∆EBC corregido = ∆ECD corregido = ∆EDE corregido = ∆EEF corregido =
134,847 + (216,775)/( 1.291,521) × -0,044 = 134,139 m. 231,140 + (231,366)/( 1.291,521) × -0,044 = 231,859 m. 54,975 + (198,640)/( 1.291,521) × -0,044 = 54,128 m. -34,182 + (179,019)/( 1.291,521) × -0,044 = 34,187 m. -35,466 + (39,969)/( 1.291,521) × -0,044 = -35,034m.
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Problemario de Topografía
Poligonal Cerrada
AZFE = AZAF + αF ± 180º = 120º 16’ 39” + 122º 15’ 50” - 180º AZFE = 62º 32’ 29” AZED = AZFE + αE ± 180º = 62º 32’ 29” + 128º 27’ 59” - 180º AZED = 11º 00’ 28” AZDC = AZED + αD ± 180º = 11º 00’ 28” + 152º 55’ 32” + 180º AZDC = 343º 56’ 00” AZCB = AZDC + αC ± 180º = 343º 56’ 00” + 103º 32’ 09” + 180º AZCB = 267º 28’ 09”
AZBA = AZCB + αB ± 180º = 267º 28’ 09” + 130º 59’ 51” + 180º AZBA = 218º 28’ 00”
21
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22
Problemario de Topografía
∆ NCB = DCB × cos AZCB = 231,366 × cos 267º 28’ 09” ∆ NCB = -10,216 m. ∆ECB = DCB × sen AZCB = 231,366 × sen 267º 28’ 09” ∆ECB = -231,140 m. ∆ NBA = DBA × cos AZBA = 216,775 × cos 218º 28’ 00”
Poligonal Cerrada
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Problemario de Topografía
Poligonal Cerrada
∆EED corregido = 54,975 + (198,640)/( 1.291,521) × -0,044 = 54,128 m. ∆EDC corregido = -34,182 + (179,019)/( 1.291,521) × -0,044 = 34,187 m. ∆ECB corregido = -35,466 + (39,969)/( 1.291,521) × -0,044 = -35,034m. ∆EBA corregido = -134,847 + (406,754)/( 1.291,521) × -0,044 = -134,034m.
23
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Problemario de Topografía
Poligonal Abierta
CALCULO DE POLIGONAL ABIERTA
1.- Calcular la poligonal abierta Datos: AZ 75º 33’ 55” (inicial)
AZ = 42º 37’ 50” (final)
24
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Problemario de Topografía
Poligonal Abierta
25
La diferencia de valor entre los ángulos medidos y el valor teórico es de 8”. La compensación total será en consecuencia de – 8”. La compensación para cada ángulo medido es: – 8” /5 = – 1”,6
Cálculo de las correcciones de los ángulos.
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Problemario de Topografía
∆ N12 = -10,9299 m. ∆E12 = D12 × sen AZ12 = 87,45 × sen 97º 10’ 47”,4 ∆E12 = 86,7643 m. ∆ N23 = D23 × cos AZ23 = 55,40 × cos 69º 3’ 4”,8 ∆ N23 = 19,8072 m.
Poligonal Abierta
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Problemario de Topografía
Poligonal Abierta
27
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Problemario de Topografía
Poligonal Abierta
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Problemario de Topografía
Poligonal Abierta
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Problemario de Topografía
Areas
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Areas
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