speedy
A (E277, 545.277; N8’670,799.371)
B (E277, 477.380; N8’670,551.380)
C (E277, 198.000; N8’670, 778.000)
Solución:
Calculamos el Angulo “α”:
41°42’23” + 41°42’26” + 41°42’22” + 41°42’19” =166°49’30”
= 41°42’22.5” ± 5”
41°42’22.5 + 5 “= 41°42’27.5”
41°42’22.5 – 5 “= 41°42’17.5”
α=
α=41°42’22.5”
Calculamos el Angulo “β”:
112°5’56” + 112°5’59” + 112°6’1” + 112°5’56.5” = 448°23’52.5”
= 112°5’58.12” ± 5”
112°5’58.12” + 5 “= 112°6’3.12”
112°5’58.12” – 5 “= 112°5’53.12”
w=
w =112°5’58.12
Entonces:
β = w – α
β= 112°5’58.12 - 41°42’22.5”
β=70°23’30.62”
Calculamos el Angulo “B”:
Tg (AB) = = =
ZAB = arctg (AB) = 15°15’28.56”
ZBA = ZAB + 180°
ZBA = 195°15’28.56”
Tg (BC)= = =
BC = -50°58’21.75”
ZBc = 180° -50°58’21.75”
ZBc =129°1’38.25”
Entonces:
B = ZBA - ZBc
B = 195°15’28.56” - 128°1’38.25”
B = 66°13’50.31”
Calculamos los ángulos “g “y “h “:
g + h = 360° - (α+β+B)
g + h = 181°40’16.57”… (l)= 90°50’8.28”
Tg ( = -68.56083251
Sabemos:
Tan (θ) =
Ademas:
Tg (= ctg (θ + 45°)*tg (
Hallamos distancias:
a= =
a = 257.0518483
b = =
b = 359.8910957
Calculamos “θ”:
Tg (θ) = = = 1.011524672
= 45°19’41.74”
Reemplazamos:
Tg (= ctg (θ + 45°)*tg (
Tg (= ctg (45°19’41.74” + 45°) (-68.56083251)
Tg (= (-5.729299882*) (-68.56083251)
Tg (=0.3928055696
= 21°23’42.64”
42°53’25.27”… (ll)
Sumamos (l) y (ll)
g + h = 181°40’16.57”
42°53’25.27”
2g = 224°33’41.84”
g = 112°16’50.92” ᴧ h = 69°23’25.65”
Calculamos los ángulos “i” Y “j”:
i = 180°-(α + g)
i = 180°-(41°42’22.5” + 112°16’50.92”
I = 26°0’46.58”
j = 180° - (β + h)
j = 180° - (70°23’30.62” + 69°23’25.65” )
j =40°13’3.73”
Calculamos distancias:
AP = = = 169.4487989
BP = = = 357.5158643
CP = = = 246.6847251
Calculamos azimut:
ZAP = ZAB + g
ZAP = 15°15’28.56” + 112°16’50.92”
ZAP = 127°32’19.48”
ZBP = ZBA- i
ZBP = 195°15’28.56” - 26°0’46.58”
ZBP = 169°14’41.98”
ZCP = ZBC + 180° - h
ZCP = 129°1’38.25” + 180° - 69°23’25.65”
ZCP = 239°38’12.6”Calculamos coordenadas de “P” :
Coordenadas de P desde A:
Xp= XA +AP*sen (ZAP)
Xp = 277545.227 + 169.4487989*sen (127°32’19.48”)
Xp = 277679.59
Yp = YA + AP*cos (ZAP)
Yp = 8670799.371 + 169.4487989*cos (127°32’19.48”)
Yp = 8670696.126
Coordenadas de P desde B:
Xp =XB + BP*sen (ZBP)
Xp = 277477.580 + 357.5158643*sen (169°14’41.98”)
Xp = 277544.2965
Yp = YB +BP*cos (ZBP)
Yp = 8670551.380 + 357.5158643*cos (169°14’41.98”)
Yp = 8670200.144
Coordenadas de P desde C:
Xp = XC + CP*sen (ZCP)
XP = 277198.000 + 246.6847251*sen (239°38’12.6”)
Xp = 276985.1508
Yp = YC + CP*cos (ZCP)
Yp = 8670778.000 + 246.6847251*cos (239°38’12.6”)
Yp= 8670653.306
Las coordenadas promedio de P serán :
XP = = 277403.0124
YP = = 8670516.525
P(277403.0124; 8670516.525)
4) Libreta de nivelación
P.V.
V.a
V.A
Cota
A
1.35
101.35
-
100.00
B
2.16
101.69
1.82
99.53
C
2.08
100.72
3.05
98.64
D
1.95
98.77
Libreta taquimétrica
Est.
P.V.
hi
hs
Lv
Cota
E
109.521
i = 1.50
D
1.20
1.80
100°30’
98.77
N
1.70
2.30
84°35’
114.66
Solución :
Hallamos la cota“E”:
V = (G)*sen2α
V = (K*S)*sen2 (90°-100°30’)
V = (k) (hs-hi)*sen (-21°)
V = (100) (0.6}*sen (-21°)
V = -10.751
Cota D = cota E + alt.inst. - hm ± V
98.77 = cota E + 1.50 - 1.50 ± V
98.77 = cota E + 1.50 - 1.50 - 10.751
109.521 = cota E...
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