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guia 8 geodesia
Citation preview
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
GUA N 8 GEODESIA I
Estacin Punto visado
Angulo horizontal Angulo vertical Altura de la seal
Distancia inclinada
A hi =
1,40m
C B
00 00 00 37 34 53,10
90 41 30,20 ------------
1,25m ------------
17893,074m ------------
B hi =
1,45m
A C
00 00 00 47 48 24,00
------------ 89 53 37,00
---------- 1,34m
----------- -----------
C hi =
1,46m
B A
00 00 00 94 36 42,00
90 13 51,00 90 41 30,20
1,34m 0,00m
14728,70m ------------
mHmH BA
A
A
89,3720 10,3958
''55,25 '32 70 ''37,12 '37 70
''32,8 '18 31 ''65,29 '30 31
B
B
==-=-=-=-=
lljj
Calcule las coordenadas geogrficas de C y su cota.
Elipsoide de referencia : Internacional 1924, 2971 ; 388.378.6 == fma
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Solucin: Dibujo de la poligonal B
mH B 89,3720
''55,25 '32 70
'32,8 '18 31
B
B
=-=-=
lj
C
A
10,3958
''37,12 '37 70
''65,29 '30 31
mH A
A
A
=-=-=
lj
Cierre angular de la poligonal
''3,0 3
''9,0 ''9,0 '0 0 +=+=+= unitarioErrorError
.estaciones de numero n""con ; ''46,33''2n''2
estacin ''1orden I
=
El error de cierre angular se encuentra dentro la tolerancia por lo tanto se puede compensar.
180 ''10,59 '59 179 '42,30' 36' 94:C ''3,0 '42,00' 36' 94:C
'24,30' 48' 47: '0,3' '24,00' 48' 47:
''40,53 '34 37: ''3,0 ''10,53 '34 37:
=S=S+++
BB
AA
''10,59 '59 179 '42,00' 36' 94:C
'24,00' 48' 47:
''10,53 '34 37:
=S
B
A
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de la superficie de la poligonal Para el calculo de la superficie de la poligonal en este caso un triangulo, utilizaremos la
siguiente expresin: CsenBCAC
Area = 2
; siendo BCy AC , distancias horizontales,
las cuales sern calculadas con la siguiente expresin: 22 hDidhz D-= Calculo de hD A C
mDi
mH
mhbmhb
mhimhi
Z
A
074,17893
10,3958
00,0 25,1
46,1 40,1
''00,16 '27 89 Z ''20,30 '41 90
12
21
21
==
======
Correccin de los ngulos cenitales:
( ) ( )
( ) ( )''83,16
1014,8481368117893,074''00,16 '27 89 46,100,0
''1
''73,11014,8481368117893,074
''20,30 '41 90 40,125,1''1
6-221
2
6-112
1
-=
-=
-=
-=
-=
-
=
senarcDi
senZhihb
senarcDi
senZhihb
q
q
''17,59 '26 89''83,16''00,16 '27 89
''28,47 '41 90''73,1''20,30 '41 90
2222
1111
=-==
=-==
cZZcZ
cZZcZ
q
q
( ) ( ) msencZcZsenDih 85,193''47,28 '41 90''17,59 '26 8921
074,1789321
12 -=-=-=D
Por lo tanto mHhHHmh CAC 25,3764 85,193 =\D=-=D C B
mDi
mH
mhbmhb
mhimhi
Z
C
20,14728
25,3764
34,1 34,1
45,1 46,1
''00,37 '53 89 Z ''00,51 '13 90
12
21
21
==
======
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Correccin de los ngulos cenitales:
( ) ( )
( ) ( )''54,1
1014,8481368114728,20''00,37 '53 98 45,134,1
''1
''68,11014,8481368114728,20
''00,51 '13 90 46,134,1''1
6-221
2
6-112
1
-=
-=
-=
-=
-=
-
=
senarcDi
senZhihb
senarcDi
senZhihb
q
q
''35,46 '53 89''54,1''00,37 '53 89
''32,49 '13 90''68,1''00,51 '13 90
2222
1111
=-==
=-==
cZZcZ
cZZcZ
q
q
( ) ( ) msencZcZsenDih 34,43''32,49 '13 90''46,35 '53 8921
70,1472821
12 -=-=-=D
Por lo tanto mHhHHmh BCB 91,3720 34,43 =\D=-=D HB fijo = 3720,89m HB calculado = 3720,91m Error de cierre = -0,02m Tolerancia para el error de cierre de una nivelacin trigonomtrica.
.621774,32774,32621L INCLINADADISTANCIA Kmm ==S=
=
=
m
m
86,08567321139,0621774,3215,0L15,0
57,05711547426,0621774,321,0L1,0orden I
El error de cierre de altura se encuentra dentro la tolerancia por lo tanto se puede compensar.
El factor de compensacin ser:L
cierre deerror =FC
710130874428,6774,32621
02,0 --=-=FC
Compensacin de las cotas: recorrida DIn FCC S=
( ) mFCCmFCC
B
C
02,070,14728074,17893
01,080109700189,0074,17893
-=+=--==
Por lo tanto la cota corregida seria:
nncorregidan CHH =
mHH
mHH
BCBC
CCCC
89,372002,091,3720
24,376401,025,3764
=-==-=
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Por lo tanto los hD son:
mhHHh
mhHHh
BCCCBCBC
CAACCCA
35,43
86,193
=D-=D=D-=D
Calculo de distancia horizontal:
mhz
mhz
CB
AC
636,1472835,4370,14728
024,1789286,193074,1789322
22
=-=D
=-=D
Calculo de la superficie:
250,131335952''30,42 '36 94 2
mAsenhzhz
A CBAC =DD
=
calculo del exceso esfrico:
''985,18 '24 312
-=+
= BAmjj
j
( )( )
( )( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
m
62,6384217''985,18 '24 31 220067226700.01
6378388
1
373,6352895''985,18 '24 31 220067226700.01
220067226700,016378388
1
1
2122
122
23
223
22
2
=--
=-
=
=--
-=
-
-=
j
jr
''322,03
''67,0''
''67,010848136811,4373,635289562,6384217
50,131335952''1
'' 6
+==
=
=
= -
U
mm
E
arcNArea
Er
Cierre angular de la poligonal quedando los ngulos compensados y esfricos
'0,67' 0' 180 '0,66' 0' 180 180
''52,42 '36 94: ''52,42 '36 94: 322,0 ''30,42 '36 94:
''52,24 '48 47: ''52,24 '48 47: 322,0 ''30,24 '48 47:
''63,53 '34 37: '0,01' ''62,53 '34 37: 322,0 ''40,53 '34 37:
=S=S=S++++
CCC
BBB
AAA
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de azimut inverso ABa
'46,82' 4' 0 ''33,21 '12 0
'37,12 '37 70 : '65,29 '30 31 :
''55,25 '32 70 : ''32,8 '18 31 :
A
B
=D=D------
ljljlj
A
B
'286,82''' ''33,741'' =D=D lj
''41,1432
'' ''665,370
2'' =D=D lj
''985,18 '24 312
''65,29 '30 31''32,8 '18 312
-=--=+
= mBA
m jjj
j
610848136811,4''1 -=arc
220067226700,02971
2971
22 22
22 =
-=-= effe
( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
m
62,6384217''985,81 '42 31 220067226700.01
6378388
1
373,6352895''985,81 '42 31 220067226700.01
220067226700,016378388
1
1
21
221
22
23
223
22
2
=--
=-
=
=--
-=-
-=
j
jr
( )
( ) ( )
576,993527 21
24''1''82,286
1''1''985,18 '24 31cos62,6384217''82,286
24''1''
1''1cos''21
1
2
2
1
=
D+
--=
D-D=
D+
aa
ljlaa
senS
arcarc
arcarcNsenS mm
( )
( ) ( )
7283,2283221
cos
62,638421724''1''33,741373,6352895
1''1''3,412 '2 0cos373,6352895''33,741
24
''1''1''1
21
cos''21
cos
2
2
22
2
22
2
=
D+
-=
D
-DD=
D+
aa
jrlrjaa
S
arcarc
N
arcarcS
m
mm
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
( )
( )
D
-DD
D-D
=
D+=
D+
D+
=
2
22
2
2
1
21
24
''1''1
21
cos''
24''1''
1cos''
21
21
cos
21
: tienese ; Si
m
mm
mm
N
arc
arcN
tgS
senS
SS
jrlrj
ljl
aaaa
aa
''94,29 '21 183318479255,021
3318479255,07283,22832
99352,757621
1 ==
D+
==
D+
-tg
tg
aa
aa
( )
( )
''73,14 '1 02
''7295,742
''''459,149''
''4594806,149''''67,10 '6 0cos
''985,18 '24 31''82,286
21
cos
''''
=D=D=D
-=D--=D
D=D-
aaa
aj
jla sen
sen m
''67,44 '22 18''73,14 '1 0''94,29 '21 18 =+= aa
Por lo tanto ''67,44 '22 198180 =+= ABAB aaa
( ) mSsensen
S 11,24057''94,29 '21 18
99352,7576
21
99352,757611 ==
D+
=aa
( ) mSS 11,24057''94,29 '21 18cos7283,22832
21
cos
7283,2283222 ==
D+
=aa
21 SS = ; Por lo tanto la distancia geodsica es: mS AB 11,24057=
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de las distancias geodsicas preliminares Calculo de distancia geodsica ACS
''65,29 '30 31-== Am jj ( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
m
017,6384252''65,29 '30 31 220067226700.01
6378388
1
058,6352998''65,29 '30 31 220067226700.01
220067226700,016378388
1
1
2122
122
23
223
22
2
=--
=-
=
=--
-=
-
-=
j
jr
''04,51 '47 160''63,53 '34 37''67,44 '22 198 =-=-= ACABAC A aaa
( ) ( )
mR
sen
Nsen
NR
ACmACm
mm
35,6356364
''04,51 '47 160cos017,6384252''04,51 '47 160058,6352998
058,6352998017,6384252
cos
22
22
=
+=
+
=
a
aarr
a
mR
hmDhCnmm
mHH
hm
mhDiDh
CA
869,1035,6356364
17,3861024,17892
17,38612
24,376410,39582
024,1789286,193074,17893 2222
-=
-=-
=
=+=+
=
=-=D-=
a
mDnmm 155,17881869,10024,17892 =-=
mSS
mR
DnmmCc
161,17881006,0155,17881
006,035,635636424
155,11788124 2
3
2
3
=+=
=
=
=a
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de distancia geodsica CBS
''65,29 '30 31-== Am jj ( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
m
017,6384252''65,29 '30 31 220067226700.01
6378388
1
058,6352998''65,29 '30 31 220067226700.01
220067226700,016378388
1
1
2122
122
23
223
22
2
=--
=-
=
=--
-=
-
-=
j
jr
''52,8 '11 246''52,42 '36 94180''04,51 '47 160 180 =-+=-+= CBACCB C aaa
( ) ( )
mR
sen
Nsen
NR
CBmCBm
mm
622,6379135
''52,8 '11 246cos017,6384252''52,8 '11 246058,6352998
058,6352998017,6384252
cos
22
22
=
+=
+
=
a
aarr
a
mR
hmDhCnmm
mHH
hm
mhDiDh
CB
641,8622,6379135
565,3742636,14728
565,37422
24,376489,37202
636,1472835,4370,14728 2222
-=
-=-
=
=+=+
=
=-=D-=
a
mDnmm 995,14719641,8636,14728 =-=
mSS
mR
DnmmCc
998,14719003,0995,14719
003,0622,637913524
995,1471924 2
3
2
3
=+=
=
=
=a
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de posicin por el problema directo (preliminar) ( A C)
Calculo de latitud
( )444444 3444444 21444 3444 2143421
III
m
mACAC
II
m
mAC
I
m
AC
arcN
tgsenSarcN
tgsenSarc
S
''16
31cos''12''1
cos'' 2
1
223
1
22
+
-
+
=D-r
jaar
jar
aj
Iteracin 1
mSAC
161,17881
''05,51 '47 160
==a
''65,29 '30 311 -== Am jj ( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
017,6384252''65,29 '30 31 220067226700.01
6378388
1
058,6352998''65,29 '30 31 220067226700.01
220067226700,016378388
1
1
2122
1
122
1
23
223
122
2
1
=--
=-
=
=--
-=
-
-=
j
jr
( ) ''252557,5481014,8481368186352998,05''05,51 '47 160cos161,17881
''1cos
6-1
1 -==
=arc
SI
m
AC
ra
( )( )
( ) ''60539144595,0
1014,8481368186352998,0576384252,012''65,29 '30 31 ''05,51 '47 160 161,17881
''12
1
6-
22
11
122
1
-=
-
=
=
II
tgsenarcNtgsenS
IIm
mAC
rja
( )( )
( ) ( )( )
( ) ''0330001649597,0
1014,8481368186352998,0576384252,016''65,29 '30 31 g31''05,51 '47 160cos''05,51 '47 160 161,17881
''16
31cos
1
6-2
223
12
1
1223
1
-=
-+=
+=
III
tsen
arcN
tgsenSIII
m
mACAC
rjaa
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
( ) ( ) ( )
''31,8 '9 0
''306,548000,0''054,0''252,548'' 111
=D\
-=+--=-+=D-
j
j IIIII
''34,21 '21 31 ''31,8 '9 0''65,29 '30 31 -=\+-=D+= CCAC jjjjj
''50,55 '25 312
''34,21 '21 31''65,29 '30 3122
-=--=+
= CAmjj
j
Iteracin 2
( )( )
( )( )( )
mN
msensene
ea
m
m
017,6384252
092,6352922''50,55 '25 31 220067226700.01
220067226700,016378388
1
1
1
23
223
222
2
2
=
=--
-=-
-=j
r
( ) ''25911128,5481014,8481368126352922,09''05,51 '47 160cos161,17881
''1cos
6-2
2 -==
=arc
SI
m
AC
ra
( )( )
( ) ''30537544080,0
1014,8481368126352922,0976384252,012''50,55 '25 31 ''05,51 '47 160 161,17881
''12
2
6-
22
21
222
2
-=
-
=
=
II
tgsenarcNtgsenS
IIm
mAC
rja
( )( )
( ) ( )( )
( ) ''5450001644413,0
1014,8481368126352922,0976384252,016''50,55 '25 31 g31''05,51 '47 160cos''05,51 '47 160 161,17881
''16
31cos
2
6-2
223
22
1
2223
2
-=
-+=
+=
III
tsen
arcN
tgsenSIII
m
mACAC
rjaa
( ) ( ) ( )
''31,8 '9 0
''313,548000,0''054,0''259,548'' 222
=D\
-=+--=-+=D-
j
j IIIIII
''34,21 '21 31 ''31,8 '9 0 ''65,29 '30 31 -=\+-=D+= CCAC jjjjj
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
''50,55 '25 312
''34,21 '21 31''65,29 '30 3123
-=--=+
= CAmjj
j
32 mm jj = ; Por lo tanto se cumple la convergencia Calculo de longitud
( ) ( )( )m
sensene
aN
C
158,6384201''34,21 '12 31 220067226700.01
6378388
1 21
221
222 =
--=
-=
j
''51,42 '3 0''51,222'''''5128601,222
10848136811,4''34,21 '21 31cos158,6384201''05,51 '47 160 161,17881
''1cos'' 6
2
-=D\-=D=
-=
=D- -
ll
ja
l senarcN
senS
C
AC
Por lo tanto la longitud es:
''88,54 '40 70''51,42 '3 0 ''37,21 '37 70 -=\--=D+= CAC llll Calculo de azimut inverso CAa
aaa D+= 180ACCA
''51,222
'155,34 '4 02
''31,8 '9 0
=-=D
-=D-=-=D
CA
CA
lll
jjjj
( )
( )
''04,56 '1 0''04,116''
''0362871,116''''155,34 '4 0cos
''50,55 '25 31''51,222
21
cos
''''
=D=D
-=D--
-=D
D=D-
aa
aj
jla sen
sen m
''09,47 '49 340''04,56 '1 0180''05,51 '47 160180 =++=D+= CAACCA aaaa
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de posicin por el problema directo ( C B)
Calculo de latitud
( )444444 3444444 21444 3444 2143421
III
m
mCBCB
II
m
mCB
I
m
CB
arcN
tgsenSarcNtgsenS
arc
S
''16
31cos''12''1
cos'' 2
1
223
1
22
+
-
+
=D-r
jaar
jar
aj
Iteracin 1
mS
CBCACB998,14719
''57,4 '13 246'42,52' 36' 94'47,09' 49' 340C
==-=-=
''34,21 '21 311 -== Cm jj ( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
158,6384201''34,21 '21 31 220067226700.01
6378388
1
228,6352846''34,21 '21 31 220067226700.01
220067226700,016378388
1
1
2122
1
122
1
23
223
122
2
1
=--
=-
=
=--
-=
-
-=
j
jr
( ) ''7296143,1921014,8481368186352846,22''57,4 '13 246cos998,14719
''1cos
6-1
1 -==
=arc
SI
m
CB
ra
( )( )
( ) ''2811411379,0
1014,8481368186352846,2286384201,152''34,21 '21 31 ''57,4 '13 246 998,14719
''12
1
6-
22
11
122
1
-=
-
=
=
II
tgsenarcNtgsenS
IIm
mCB
rja
( )( )
( ) ( )( )
( ) ''4310003022806,0
1014,8481368186352846,2286384201,156''34,21 '21 3131''57,4 '13 246cos''57,4 '13 246 998,14719
''16
31cos
1
6-2
223
12
1
1223
1
-=
-+=
+=
III
tgsen
arcN
tgsenSIII
m
mCBCB
rjaa
( ) ( ) ( )
''01,13 '3 0
''011,193000,0''281,0''730,192'' 111
=D\
-=+--=-+=D-
j
j IIIII
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
''33,8 '18 31 ''01,13 '3 0''34,21 '21 31 -=\+-=D+= BBCB jjjjj Iteracin 2
''835,44 '19 312
''34,21 '21 31''33,8 '18 3122
-=--=+
= CBmjj
j
( )
( )( )
( )( )
mN
msensene
ea
m
m
158,6384201
549,6352819''835,44 '19 31 220067226700.01
220067226700,016378388
1
1
1
23
223
222
2
2
=
=--
-=-
-=j
r
( ) ''7304237,1921014,8481368196352819,54''57,4 '13 246cos998,14719
''1cos
6-2
2 -==
=arc
SI
m
CB
ra
( )( )
( ) ''2808463838,0
1014,8481368196352819,5486384201,152''835,44 '19 31''57,4 '13 246 998,14719
''12
2
6-
22
21
222
2
-=
-
=
=
II
tgsenarcNtgsenS
IIm
mCB
rja
( )( )
( ) ( )( )
( ) ''5630003019467,0
1014,8481368196352819,5486384201,156''835,44 '19 31 g31''57,4 '13 246cos''57,4 '13 246 998,14719
''16
31cos
2
6-2
223
22
1
2223
2
-=
-+=
+=
III
tsen
arcN
tgsenSIII
m
mCBCB
rjaa
( ) ( ) ( )
''01,13 '3 0
''011,193000,0''281,0''730,192'' 222
=D\
-=+--=-+=D-
j
j IIIIII
''33,8 '18 31 ''01,13 '3 0''34,21 '21 31 -=\+-=D+= BBCB jjjjj
''835,44 '19 312
''34,21 '21 31''33,8 '18 3123
-=--=+
= CBmjj
j
32 mm jj = ; Por lo tanto se cumple la convergencia.
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de longitud
( ) ( )( )m
sensene
aN
B
288,6384183''33,8 '18 31 220067226700.01
6378388
1 21
221
222 =
--=
-=
j
''34,29 '8 0''34,509'''''3413639,509
10848136811,4''33,8 '18 31cos288,6384183''57,4 '13 246 998,14719
''1cos'' 6
2
=D\=D-=
-=
=D- -
ll
ja
l senarcN
senS
B
CB
Por lo tanto la longitud es:
''54,52 '32 70''34,29 '8 0 ''88,54 '40 70 -=\+-=D+= BCB llll
'0,01'- '' ''01,0''
''54,25 '32 70 ''33,8 '18 31
''55,25 '32 70 ''32,8 '18 31
=D=D-=-=
-=-=
ljljlj
CALCULADOBCALCULADOB
FIJOBFIJOB
Error en posicin = sS+ 22 elej
Arco de paralelo = ''1 cos arcN mm j Arco de meridiano = ''1 arcm r
77,19 '23 31
''34,21 '21 31
''33,8 '18 31
''65,29 '30 31
-=-=-=-=
m
C
B
A
jjjj
( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
m
131,6384212''77,19 '23 31 220067226700.01
6378388
1
987,6352878''77,19 '23 31 220067226700.01
220067226700,016378388
1
1
2122
122
2322
322
2
=--
=-
=
=--
-=-
-=
j
jr
Arco de paralelo = marcarcN mm 422,26''1 ''77,19 '23 31cos131,6384212''1 cos =-= j Arco de meridiano = marcarcm 800,30''1 987,6352878''1 ==r
0,26426,422'-0,01'paralelo de Arco''
308,0800,30''01,0meridiano de Arco''
-==D===D=
leljej
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Error lineal = ( ) ( ) m406,0264,0308,0 2222 =-+=+elej
mSSs CBAC 159,32601998,14719161,17881 =+=+=S
Error de posicin = 42118,80298
1
406,0159,32601
406,0406,0
159,32601406,0linealError ===
Ss
Error en posicin = 000.40122
S+s
elej; para que el error de posicin se encuentre
dentro de la tolerancia de I orden.
Error en posicin = 000025,044980000124535,0000.401
42118,802981
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
CALCULO DEFINITIVO Calculo del exceso esfrico
77,19 '23 31
''33,21 '21 31
''32,8 '18 31
''65,29 '30 31
-=-=-=-=
m
C
B
A
jjjj
( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
m
131,6384212''77,19 '23 31 220067226700.01
6378388
1
987,6352878''77,19 '23 31 220067226700.01
220067226700,016378388
1
1
2122
122
2322
322
2
=--
=-
=
=--
-=-
-=
j
jr
''322,03
''67,0''
''67,010848136811,4987,6352878131,6384212
50,131335952''1
'' 6
+==
=
=
= -
U
mm
E
arcNArea
Er
Reduccin de los ngulos horizontales al elipsoide Correccin por efecto de la altura de la estacin observada.
( ) ''1 1coscos
''22
222
arceNseneH
Xm -
= aaj
''49,55 '25 312
''05,51 '47 160
24,3764; ''33,21 '21 31''65,29 '30 31
-=+=
==-=-=
CAm
AC
CCA mH
jjj
ajj
( ) ( )( )m
sensene
aN
m
m 57,6384226''49,55 '25 31 220067226700.01
6378388
1 21
221
22=
--=
-=
j
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
( )
( )( )
''186,0''
10848136811,4220067226700,0157,6384226''05,51 '47 160''05,51 '47 160cos''33,21 '21 31cos220067226700,024,3764
''
''1 1
coscos''
6
2
2
22
-=
--=
-
=
-
C
C
m
ACACCCC
X
senX
arceN
seneHX
aaj
''985,18 '24 312
''67,44 '22 198
89,3720; ''32,8 '18 31''65,29 '30 31
-=+
=
==-=-=
BAm
AB
BBA mH
jjj
ajj
( ) ( )( )m
sensene
aN
m
m 62,6384217''985,18 '24 31 220067226700.01
6378388
1 21
221
22=
--=
-=
j
( )
( )( )
''178,0''
10848136811,4220067226700,0162,6384217''65,29 '22 198''65,29 '22 198cos''32,8 '18 31cos220067226700,089,3720
''
''1 1
coscos''
6
2
2
22
=
--=
-
=
-
B
B
m
ABABBBB
X
senX
arceN
seneHX
aaj
ngulos observados en terreno Estacin Punto visado Angulo ledo X Angulo corregido Angulo reducido
A C 00 00 00 -0,186 359 59 59,81 00 00 00 B 37 34 53,1 1,178 37 34 54,28 37 34 54,47
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
''825,44 '19 312
''57,4 '13 246
89,3720; ''32,8 '18 31''33,21 '21 31
-=+=
==-=-=
BCm
CB
BBC mH
jjj
ajj
( ) ( )( )m
sensene
aN
m
m 22,6384192''825,44 '19 31 220067226700.01
6378388
1 21
221
22=
--=
-=
j
( )
( )( )
''219,0''
10848136811,4220067226700,0122,6384192''57,4 '13 246''57,4 '13 246cos''32,8 '18 31cos220067226700,089,3720
''
''1 1
coscos''
6
2
2
22
=
--=
-
=
-
B
B
m
CBCBBBB
X
senX
arceN
seneHX
aaj
''49,55 '25 312
''09,47 '49 340
10,3958; ''65,29 '30 31''33,21 '21 31
-=+=
==-=-=
ACm
CA
BAC mH
jjj
ajj
( ) ( )( )m
sensene
aN
m
m 17,6384226''49,55 '25 31 220067226700.01
6378388
1 21
221
22=
--=
-=
j
( )
( )( )
''195,0''
10848136811,4220067226700,0117,6384226''09,47 '49 340''09,47 '49 340cos''65,29 '30 31cos220067226700,010,3958
''
''1 1
coscos''
6
2
2
22
-=
--=
-
=
-
A
A
m
CACAAAA
X
senX
arceN
seneHX
aaj
ngulos observados en terreno Estacin Punto visado Angulo ledo X Angulo corregido Angulo reducido
C B 00 00 00 0,219 00 00 0,22 00 00 00 A 94 36 42,00 -0,195 94 36 41,80 94 36 41,58
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
''98,18 '24 312
''21,15 '20 18 ''46,29 '2 0180''67,44 '22 198 180
10,3958; ''65,29 '30 31''32,8 '18 31
-=+
=
=\--=D-==-=-=
ABm
BAABBA
AAB mH
jjj
aaaajj
( ) ( )( )m
sensene
aN
m
m 62,6384219''98,18 '24 31 220067226700.01
6378388
1 21
221
22=
--=
-=
j
( )
( )( )
''188,0''
10848136811,4220067226700,0162,6384219''21,15 '20 18''21,15 '20 18cos''65,29 '30 31cos220067226700,010,3958
''
''1 1
coscos''
6
2
2
22
=
--=
-
=
-
A
A
m
BABAAAA
X
senX
arceN
seneHX
aaj
''825,44 '19 312
''73,39 '8 66 ''52,24 '48 47 ''21,15 '20 18
24,3764; ''33,21 '21 31''32,8 '18 31
-=+=
=\+=+==-=-=
CBm
BCBCBABC
CCB
B
mH
jjj
aaaajj
( ) ( )( )m
sensene
aN
m
m 22,6384192''825,44 '19 31 220067226700.01
6378388
1 21
221
22=
--=
-=
j
( )
( )( )
''222,0''
10848136811,4220067226700,0122,6384192''73,39 '8 66''73,39 '8 66cos''33,21 '21 31cos220067226700,024,3764
''
''1 1
coscos''
6
2
2
22
=
--=
-
=
-
C
C
m
BCBCCCC
X
senX
arceN
seneHX
aaj
ngulos observados en terreno Estacin Punto visado Angulo ledo X Angulo corregido Angulo reducido
B A 00 00 00 0,188 00 00 0,19 00 00 00 C 47 48 24,00 0,222 47 48 24,22 47 48 24,03
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Estacin Punto visado Angulo ledo X Angulo corregido Angulo reducido
A C 00 00 00 -0,186 359 59 59,81 00 00 00 B 37 34 53,1 1,178 37 34 54,28 37 34 54,47
B A 00 00 00 0,188 00 00 0,19 00 00 00 C 47 48 24,00 0,222 47 48 24,22 47 48 24,03
C B 00 00 00 0,219 00 00 0,22 00 00 00 A 94 36 42,00 -0,195 94 36 41,80 94 36 41,58
ngulos interiores reducidos
''08,0 '0 180 ''58,41 '36 94:
''03,24 '48 47:
''47,54 '34 37:
=S
C
B
A
''602,0 3
''08,0 ''08,0 '0 0 -=-=-= unitarioErrorError
.estaciones de numero n""con ; ''46,33''2n''2
estacin ''1orden I
=
El error de cierre angular se encuentra dentro la tolerancia por lo tanto se puede compensar.
180 '59,99' 59' 179 ''08,0 '0 180
'41,55' 36' 94:C '41,55' 36' 94:C ''602,0 '41,58' 36' 94:C
'24,00' 48' 47: '24,00' 48' 47: ''602,0 '24,03' 48' 47:
''45,54 '34 37: '0,01' ''44,54 '34 37: ''602,0 ''47,54 '34 37:
=S=S=S--+-
BBB
AAA
Cierre angular de la poligonal quedando los ngulos compensados y esfricos.
''322,03
''67,0'' +==UE
'0,67' 0' 180 '0,66' 0' 180 180
''77,41 '36 94: ''77,41 '36 94: 322,0 ''55,41 '36 94:
''22,24 '48 47: ''22,24 '48 47: 322,0 ''00,24 '48 47:
''68,54 '34 37: '0,01' ''67,54 '34 37: 322,0 ''45,54 '34 37:
=S=S=S++++
CCC
BBB
AAA
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de alturas definitivas A C
mS
mH
mhbmhb
mhimhi
Z
A
161,17881
10,3958
00,0 25,1
46,1 40,1
''00,16 '27 89 Z ''20,30 '41 90
12
21
21
==
======
Correccin de los ngulos cenitales:
( ) ( )
( ) ( )''84,16
1014,8481368117881,16146,100,0
''1
''73,11014,8481368117881,161
40,125,1''1
6-21
2
6-12
1
-=
-=
-=
-=
-=
-
=
arcShihb
arcShihb
q
q
''16,59 '26 89''84,16''00,16 '27 89
''28,47 '41 90''73,1''20,30 '41 90
2222
1111
=-==
=-==
cZZcZ
cZZcZ
q
q
( ) ( ) mtgcZcZtgSh 73,193''47,28 '41 90''16,59 '26 8921
161,1788121
12 -=-=-=D
''49,55 '25 312
-=+
= CAmjj
j
( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
m
57,6384226''49,55 '25 31 220067226700.01
6378388
1
089,6352922''49,55 '25 31 220067226700.01
220067226700,016378388
1
1
2122
122
23
223
22
2
=--
=-
=
=--
-=
-
-=
j
jr
''99,49 '47 160''68,54 '34 37''67,44 '22 198 =-=-= ACABAC A aaa
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
( ) ( )
mR
sen
Nsen
NR
ACmACm
mm
897,6356293
''99,49 '47 160cos57,6384226''99,49 '47 160089,6352922
089,635292257,6384226
cos
22
22
=
+=
+
=
a
aarr
a
000000659,1987,635629312
161,178811
121
000015239,1897,63562932
73,1931
21
000622706,1897,6356293
10,395811
2
2
2
2
1
=
+=
+=
=
+=
D
+=
=
+=
+=
a
a
a
RS
C
R
hB
Rh
A
mCBAhh 85,193000000659,1000015239,1000622706,173,193' -=-=D=D
mHhHH CAC 25,376485,19310,3958' =-=D=
C B
mS
mH
mhbmhb
mhimhi
Z
C
998,14719
25,3764
34,1 34,1
45,1 46,1
''00,37 '53 89 Z ''00,51 '13 90
12
21
21
==
======
Correccin de los ngulos cenitales:
( ) ( )
( ) ( )''54,1
1014,8481368114719,99845,134,1
''1
''68,11014,8481368114719,998
46,134,1''1
6-21
2
6-12
1
-=
-=
-=
-=
-=
-
=
arcShihb
arcShihb
q
q
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
''35,46 '53 89''54,1''00,37 '53 89
''32,49 '13 90''68,1''00,51 '13 90
2222
1111
=-==
=-==
cZZcZ
cZZcZ
q
q
( ) ( ) mtgcZcZtgSh 31,43''32,49 '13 90''46,35 '53 8921
998,1471921
12 -=-=-=D
''825,44 '19 312
-=+
= BCmjj
j
( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
m
22,6384192''825,44 '19 31 220067226700.01
6378388
1
546,6352819''825,44 '19 31 220067226700.01
220067226700,016378388
1
1
2122
122
23
223
22
2
=--
=-
=
=--
-=
-
-=
j
jr
''32,5 '13 246''77,41 '36 94''09,47 '49 340 =-=-= CBCACB C aaa
( ) ( )
mR
sen
Nsen
NR
ACmACm
mm
472,6379069
''32,5 '13 246cos22,6384192''32,5 '13 246 546,6352819
546,635281922,6384192
cos
22
22
=
+=
+
=
a
aarr
a
000000444,1472,637906912
998,147191
121
000003395,1472,63790692
31,431
21
000590094,1472,6379069
25,376411
2
2
2
2
1
=
+=
+=
=
+=
D
+=
=
+=
+=
a
a
a
RS
C
R
hB
Rh
A
mCBAhh 34,43000000444,1000003395,1000590094,131,43' -=-=D=D
mHhHH BCB 91,372034,4325,3764' =-=D=
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
HB fijo = 3720,89m HB calculado = 3720,91m Error de cierre = -0,02m Tolerancia para el error de cierre de una nivelacin trigonomtrica.
.601159,32159,32601L GEODSICADISTANCIA Kmm ==S=
=
=
m
m
86,08564613696,0601159,3215,0L15,0
57,05709742464,0601159,321,0L1,0orden I
El error de cierre de altura se encuentra dentro la tolerancia por lo tanto se puede compensar.
El factor de compensacin ser:L
cierre deerror =FC
710134751222,6159,3260102,0 --=-=FC
Compensacin de las cotas: recorrida Sn FCC S=
( ) mFCCmFCC
B
C
02,0998,14719161,17881
01,0010964743,0161,17881
-=+=--==
Por lo tanto la cota corregida seria:
nncorregidan CHH =
mHH
mHH
BCBC
CCCC
89,372002,091,3720
24,376401,025,3764
=-==-=
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de distancias geodsicas definitivas Calculo de distancia geodsica ACS
''49,55 '25 312
-=+
= CAmjj
j
( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
m
57,6384226''49,55 '25 31 220067226700.01
6378388
1
089,6352922''49,55 '25 31 220067226700.01
220067226700,016378388
1
1
2122
122
23
223
22
2
=--
=-
=
=--
-=
-
-=
j
jr
''99,49 '47 160''68,54 '34 37''67,44 '22 198 =-=-= ACABAC A aaa
( ) ( )
mR
sen
Nsen
NR
ACmACm
mm
897,6356293
''99,49 '47 160cos57,6384226''99,49 '47 160089,6352922
089,635292257,6384226
cos
22
22
=
+=
+
=
a
aarr
a
mR
hmDhCnmm
mHH
hm
mhDiDh
CA
869,10987,6356293
17,3861024,17892
17,38612
24,376410,39582
024,1789286,193074,17893 2222
-=
-=-
=
=+=+
=
=-=D-=
a
mDnmm 155,17881869,10024,17892 =-=
mSS
mR
DnmmCc
161,17881006,0155,17881
006,035,635636424
155,11788124 2
3
2
3
=+=
=
=
=a
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de distancia geodsica CBS
''825,44 '19 312
-=+
= BCmjj
j
( )( )
( )( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
m
22,6384192''825,44 '19 31 220067226700.01
6378388
1
546,6352819''825,44 '19 31 220067226700.01
220067226700,016378388
1
1
2122
122
2322
322
2
=--
=-
=
=--
-=-
-=
j
jr
''32,5 '13 246''77,41 '36 94''09,47 '49 340 =-=-= CBCACB C aaa
( ) ( )
mR
sen
Nsen
NR
ACmACm
mm
472,6379069
''32,5 '13 246cos22,6384192''32,5 '13 246 546,6352819
546,635281922,6384192
cos
22
22
=
+=
+
=
a
aarr
a
mR
hmDhCnmm
mHH
hm
mhDiDh
CB
641,8472,6379069
565,3742636,14728
565,37422
24,376489,37202
636,1472835,4370,14728 2222
-=
-=-
=
=+=+
=
=-=D-=
a
mDnmm 995,14719641,8636,14728 =-=
mSS
mR
DnmmCc
998,14719003,0995,14719
003,0622,637913524
995,1471924 2
3
2
3
=+=
=
=
=a
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de posicin por el problema directo (definitivo) ( A C)
Calculo de latitud
( )444444 3444444 21444 3444 2143421
III
m
mACAC
II
m
mAC
I
m
AC
arcN
tgsenSarcN
tgsenSarc
S
''16
31cos''12''1
cos'' 2
1
223
1
22
+
-
+
=D-r
jaar
jar
aj
Iteracin 1
mSAC
161,17881
''49,49 '47 160
==a
''65,29 '30 311 -== Am jj ( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
017,6384252''65,29 '30 31 220067226700.01
6378388
1
058,6352998''65,29 '30 31 220067226700.01
220067226700,016378388
1
1
2122
1
122
1
23
223
122
2
1
=--
=-
=
=--
-=
-
-=
j
jr
( ) ''2515849,5481014,8481368186352998,05''99,49 '47 160cos161,17881
''1cos
6-1
1 -==
=arc
SI
m
AC
ra
( )( )
( ) ''0539160356,0
1014,8481368186352998,0576384252,012''65,29 '30 31 ''99,49 '47 160 161,17881
''12
1
6-
22
11
122
1
-=
-
=
=
II
tgsenarcNtgsenS
IIm
mAC
rja
( )( )
( ) ( )( )
( ) ''330001649642,0
1014,8481368186352998,0576384252,016''65,29 '30 31 g31''99,49 '47 160cos''99,49 '47 160 161,17881
''16
31cos
1
6-2
223
12
1
1223
1
-=
-+=
+=
III
tsen
arcN
tgsenSIII
m
mACAC
rjaa
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
( ) ( ) ( )
''31,8 '9 0
''306,548000,0''054,0''252,548'' 111
=D\
-=+--=-+=D-
j
j IIIII
''34,21 '21 31 ''31,8 '9 0''65,29 '30 31 -=\+-=D+= CCAC jjjjj
Iteracin 2
''50,55 '25 31''495,55 '25 312
''34,21 '21 31''65,29 '30 3122
--=--=+
= CAmjj
j
( )
( )( )
( )( )
mN
msensene
ea
m
m
017,6384252
092,6352922''50,55 '25 31 220067226700.01
220067226700,016378388
1
1
1
23
223
222
2
2
=
=--
-=-
-=j
r
( ) ''2581407,5481014,8481368126352922,09''99,49 '47 160cos161,17881
''1cos
6-2
2 -==
=arc
SI
m
AC
ra
( )( )
( ) ''0537559794,0
1014,8481368126352922,0976384252,012''50,55 '25 31 ''99,49 '47 160 161,17881
''12
2
6-
22
21
222
2
-=
-
=
=
II
tgsenarcNtgsenS
IIm
mAC
rja
( )( )
( ) ( )( )
( ) ''70001644458,0
1014,8481368126352922,0976384252,016''50,55 '25 31 g31''99,49 '47 160cos''99,49 '47 160 161,17881
''16
31cos
2
6-2
223
22
1
2223
2
-=
-+=
+=
III
tsen
arcN
tgsenSIII
m
mACAC
rjaa
( ) ( ) ( )
''31,8 '9 0
''312,548000,0''054,0''258,548'' 222
=D\
-=+--=-+=D-
j
j IIIIII
''34,21 '21 31 ''31,8 '9 0 ''65,29 '30 31 -=\+-=D+= CCAC jjjjj
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
''50,55 '25 312
''34,21 '21 31''65,29 '30 3123
-=--=+
= CAmjj
j
32 mm jj = ; Por lo tanto se cumple la convergencia Calculo de longitud
( ) ( )( )m
sensene
aN
C
158,6384201''34,21 '12 31 220067226700.01
6378388
1 21
221
222 =
--=
-=
j
''52,42 '3 0''52,222'''''5161433,222
10848136811,4''34,21 '21 31cos158,6384201''99,49 '47 160 161,17881
''1cos'' 6
2
-=D\-=D=
-=
=D- -
ll
ja
l senarcN
senS
C
AC
Por lo tanto la longitud es:
''89,54 '40 70''52,42 '3 0 ''37,21 '37 70 -=\--=D+= CAC llll Calculo de azimut inverso CAa
aaa D+= 180ACCA
''52,222
'155,34 '4 02
''31,8 '9 0
=D-D=D
-=D-=-=D
CA
CA
l
jjjj
( )
( )
''04,56 '1 0''04,116''
''0418095,116''''155,34 '4 0cos
''50,55 '25 31''52,222
21
cos
''''
=D=D
-=D--
-=D
D=D-
aa
aj
jla sen
sen m
''09,47 '49 340''04,56 '1 0180''05,51 '47 160180 =++=D+= CAACCA aaaa
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de posicin por el problema directo ( C B)
Calculo de latitud ( )
444444 3444444 21444 3444 2143421III
m
mCBCB
II
m
mCB
I
m
CB
arcN
tgsenSarcNtgsenS
arc
S
''16
31cos''12''1
cos'' 2
1
223
1
22
+
-
+
=D-r
jaar
jar
aj
Iteracin 1
mS 998,14719
''32,5 '13 246'41,77' 36' 94'47,09' 49' 340C CBCACB=
=-=-=
''34,21 '21 311 -== Cm jj ( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
158,6384201''34,21 '21 31 220067226700.01
6378388
1
228,6352846''34,21 '21 31 220067226700.01
220067226700,016378388
1
1
2122
1
122
1
23
223
122
2
1
=--
=-
=
=--
-=
-
-=
j
jr
( ) ''7280241,1921014,8481368186352846,22''32,5 '13 246cos998,14719
''1cos
6-1
1 -==
=arc
SI
m
CB
ra
( )( )
( ) ''2811420388,0
1014,8481368186352846,2286384201,152''34,21 '21 31 ''32,5 '13 246 998,14719
''12
1
6-
22
11
122
1
-=
-
=
=
II
tgsenarcNtgsenS
IIm
mCB
rja
( )( )
( ) ( )( )
( ) ''1770003022791,0
1014,8481368186352846,2286384201,156''34,21 '21 31 g31''32,5 '13 246cos''32,5 '13 246 998,14719
''16
31cos
1
6-2
223
12
1
1223
1
-=
-+=
+=
III
tsen
arcN
tgsenSIII
m
mCBCB
rjaa
( ) ( ) ( )
''01,13 '3 0
''009,193000,0''281,0''728,192'' 111
=D\
-=+--=-+=D-
j
j IIIII
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
''33,8 '18 31 ''01,13 '3 0''34,21 '21 31 -=\+-=D+= BBCB jjjjj Iteracin 2
''84,44 '19 31''835,44 '19 312
''34,21 '21 31''33,8 '18 3122
--=--=+
= CBmjj
j
( )
( )( )
( )( )
mN
msensene
ea
m
m
158,6384201
551,6352819''84,44 '19 31 220067226700.01
220067226700,016378388
1
1
1
23
223
222
2
2
=
=--
-=-
-=j
r
( ) ''7288334,1921014,8481368116352819,55''32,5 '13 246cos998,14719
''1cos
6-2
2 -==
=arc
SI
m
CB
ra
( )( )
( ) ''280847299,0
1014,8481368116352819,5586384201,152''84,44 '19 31 ''32,5 '13 246 998,14719
''12
2
6-
22
21
222
2
-=
-
=
=
II
tgsenarcNtgsenS
IIm
mCB
rja
( )( )
( ) ( )( )
( ) ''4980003019452,0
1014,8481368116352819,5586384201,156''84,44 '19 31 g31''32,5 '13 246cos''32,5 '13 246 998,14719
''16
31cos
2
6-2
223
22
1
2223
2
-=
-+=
+=
III
tsen
arcN
tgsenSIII
m
mCBCB
rjaa
( ) ( ) ( )
''01,13 '3 0
''01,193000,0''281,0''729,192'' 222
=D\
-=+--=-+=D-
j
j IIIIII
''33,8 '18 31 ''01,13 '3 0''34,21 '21 31 -=\+-=D+= BBCB jjjjj
''835,44 '19 312
''34,21 '21 31''33,8 '18 3123
-=--=+
= CBmjj
j
32 mm jj = ; por lo tanto se cumple la convergencia.
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Calculo de longitud
( ) ( )( )m
sensene
aN
B
288,6384183''33,8 '18 31 220067226700.01
6378388
1 21
221
222 =
--=
-=
j
''34,29 '8 0''34,509'''''341834,509
10848136811,4''33,8 '18 31cos288,6384183''32,5 '13 246 998,14719
''1cos'' 6
2
=D\=D-=
-=
=D- -
ll
ja
l senarcN
senS
B
CB
Por lo tanto la longitud es:
''55,52 '32 70''34,29 '8 0 ''89,54 '40 70 -=\+-=D+= BCB llll
'0,00' '' ''01,0''
''55,25 '32 70 33,8 '18 31
''55,25 '32 70 ''32,8 '18 31
=D=D-=-=-=-=
ljljlj
CALCULADOBCALCULADOB
FIJOBFIJOB
Error en posicin = sS+ 22 elej
Arco de paralelo = ''1 cos arcN mm j Arco de meridiano = ''1 arcm r
77,19 '23 31
''34,21 '21 31
''33,8 '18 31
''65,29 '30 31
-=-=-=-=
m
C
B
A
jjjj
( )
( )( )
( )( )
( ) ( )( )m
sensene
aN
msensene
ea
m
m
m
m
131,6384212''77,19 '23 31 220067226700.01
6378388
1
987,6352878''77,19 '23 31 220067226700.01
220067226700,016378388
1
1
2122
122
2322
322
2
=--
=-
=
=--
-=-
-=
j
jr
Arco de paralelo = marcarcN mm 422,26''1 ''77,19 '23 31cos131,6384212''1 cos =-= j Arco de meridiano = marcarcm 800,30''1 987,6352878''1 ==r
00,026,422'0,00'paralelo de Arco''
308,0800,30''01,0meridiano de Arco''
==D===D=
leljej
Profesor: Matas Saavedra A. Ayudante: Andrs Romn E.
Error lineal = ( ) ( ) m308,00308,0 2222 =+=+elej
mSSs CBAC 159,32601998,14719161,17881 =+=+=S
Error de posicin = 9188,105847
1
308,0159,32601
308,0308,0
159,32601308,0linealError ===
Ss
Error en posicin = 000.40122
S+s
elej; para que el error de posicin se encuentre
dentro de la tolerancia de I orden.
Error en posicin = 000025,0168840000094475,0000.401
9188,1058471