Lista 03- Coordenadas

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LISTA 03 – TOPOGRAFIAEGEOPROCESSAMENTO
01. Calcule o rumo, o azimute e a distância entre os pontos:
M3 (160.800,0000; 9.602,700,000) e M4 (160.960,5455; 9.602,774,4287).
EM3 NM3 EM4 NM4
DH 3-4 = √(ΔE)^2 + (ΔN)^2
DH 3-4 = √(160,5455)^2 + (74,4287)^2
DH 3-4 = 176,959 m
R = arctg (ΔE/ΔN)
R = arctg (160,54555/74,4287)
R = arctg 2,157037541
R = arctg 65,1276179 (º''')
ΔE= FINAL - INICIO
ΔN= FINAL - INICIO
ΔE= (160.960,5455 - 160.800,0000) = + 160,5455 (E)
ΔN= (9.602,774,4287 - 9.602,700,000000) = 74,4287 (N)
02. Calcular azimutes, rumos e distâncias a partir das coordenadas da �igura abaixo.
DH 2-1 = √(ΔE)^2 + (ΔN)^2
DH 2-1 = √((220325,10-220810,00))^2 + (9110424,5-9110755,00)^2
DH 2-1 = √(484,90)^2 + (330,5)^2
DH 2-1 = 586,82 m ΔE = -484,90 m W
R = arctg (ΔE/ΔN) ΔN = -330,5 m S
R 2-1 = arctg(484,90/330,5)
R 2-1 = 55,7222901696
R 2-1 = 55º43'20,24" SW
Az 2-1 = 235º43'20,24"
DH 3-2 = √(ΔE)^2 + (ΔN)^2
DH 3-2 = √((220224,0-220325,10))^2 + (9110657,2-9110424,5)^2
DH 3-2 = √((101,1)^2 + (232,7)^2)
DH 3-2 = 253,71 m ΔE = -101,1 m W
R = arctg (ΔE/ΔN) ΔN = 232,7 m N
R 3-2 = arctg(101,1/232,7)
R 3-2 = 23,5039674547
R 3-2 = 23º30'14,28" NW
Az 3-2 = 293º43'20,24"
DH 4-3 = √(ΔE)^2 + (ΔN)^2
DH 4-3 = √((220342,60-220224,0))^2 + (9110750-9110657,2)^2
DH 4-3 = √((118,6)^2 + (92,8)^2)
DH 4-3 = 150,59 m ΔE = 118,60 m L
R = arctg (ΔE/ΔN) ΔN = 92,80m N
R 4-3 = arctg(118,60/92,80)
R 4-3 = 51,9581683075
R 4-3 = 51º57'29,41" NE
Az 4-3 = 51º57'29,41"
DH 5-4 = √(ΔE)^2 + (ΔN)^2
DH 5-4 = √((220315,10-220342,60))^2 + (9110928,90-9110657,2)^2
DH 5-4 = √((27,5)^2 + (27,17)^2)
DH 5-4 = 38,65 m ΔE = -27,5 m W
R = arctg (ΔE/ΔN) ΔN = 27,17m N
R 5-4 = arctg(27,5/27,17)
R 5-4 = 45,3458455757
R 5-4 = 45º20'45,04" NW
Az 5-4 = 45º20'45,04"
DH 6-5 = √(ΔE)^2 + (ΔN)^2
DH 6-5 = √((220538,40-220315,10))^2 + (9111171,20-9110928,90)^2
DH 6-5 = √((223,3)^2 + (242,3)^2)
DH 6-5 = 329,50 m ΔE = 223,3 m L
R = arctg (ΔE/ΔN) ΔN = 242,3 m N
R 6-5 = arctg(223,3/242,3)
R 6-5 = 42,6631955763
R 6-5 = 42º39'47,5" NE
Az 6-5 = 42º39'47,5"
DH 1-6 = √(ΔE)^2 + (ΔN)^2
DH 1-6 = √((220810-220538,40))^2 + (9110755-9111171,20)^2
DH 1-6 = √((271,6)^2 + (416,2)^2)
DH 1-6 = 496,97 m ΔE = 271,6 m L
R = arctg (ΔE/ΔN) ΔN = -416,2 m S
R 1-6 = arctg(271,6/416,2)
R 1-6 = 33,1272963919
R 1-6 = 33º7'38,27" SE
Az 1-6 = 123º39'47,5"
03. Um levantamento topográ�ico apresenta um alinhamento com ângulo à direita na estaca A-2 de
194° 10' e rumo da estaca A-1 para a estaca A-2 de 16° 40' NW; o rumo da estaca A-2 para A-3 é?