a. Calcular el azimut de los lados.Datos: RP-R1 = 591416 ZPR1 = 360 - 591416 = 3004544 RQ-R2 = 135512 ZQR2 = 360 - 135512 = 346448 Entonces: ZPR1 = 3004544 - 180-------------------------------------------------------ZR1P = 1204544 ++ P 1623020------------------------------------------------------ 283164 180-----------------------------------------------------------ZPA = 103164 +A = 1701040------------------------------------------------------- 2732644 - 180-----------------------------------------------------ZAB = 932644 +B = 2481035------------------------------------------------------- 3413719 - 180--------------------------------------------------------ZBC = 1613719 +C = 921950------------------------------------------------------- 253579 - 180-----------------------------------------------------Z CD = 73579 +D = 1293920 +------------------------------------------------------ 2033629 - 180------------------------------------------------------ZDQ = 233629 +Q = 1422834------------------------------------------------------ 16653 + 180------------------------------------------------------ZQR2 = 34653 Por lo tanto el error de azimut es

3460503 3460448 = 0015Ea = 0015b. El error de cierre angular y compensacin de ngulos:

Evaluamos con la tolerancia: 0015 < 0113.48 Trabajo aceptable.

Compensacin de ngulos:VrticeAnguloCorreccinAngulo corregido

P1623020-2.51623017.5

A1701040-2.51701037.5

B2481035-2.52481032.5

C921950-2.5921947.5

D1293920-2.51293917.5

Q1422834-2.51422831.5

-15

Con los ngulos corregidos calculamos los azimuts.

ZR1P = 1204544 ++ P 1623017.5------------------------------------------------------ 283161.5 180-----------------------------------------------------------ZPA = 103161.5 +A = 1701037.5------------------------------------------------------- 2732639 - 180-----------------------------------------------------ZAB = 932639 +B = 2481032.5------------------------------------------------------- 3413711.5 - 180--------------------------------------------------------ZBC = 1613711.5 +C = 921947.5------------------------------------------------------- 2535659 - 180-----------------------------------------------------Z CD = 735659 +D = 1293917.5 +------------------------------------------------------ 2033616.5 - 180------------------------------------------------------ZDQ = 233616.5 +Q = 1422831.5------------------------------------------------------ 166448 + 180------------------------------------------------------ZQR2 = 346448

Ahora si el rumbo final calculado y el rumbo final del dato coinciden.

c) Clculo y compensacin de las coordenadas de los vrtices.Datos: P=(325679.431 , 8626354.293) Q (327967.350 , 8626173.900)

LADOAZIMUT(Z)Distancia(D)Coordenadas parciales

X- XY- Y

PA103161.5350.08340.73780.32

AB932639630.50629.36137.88

BC1613711.5753.25237.515714.82

CD735659946.28909.393261.63

DQ233616.5426.78170.892391.07

X1= 2287. 898 Y1= -180.32 Como no se trata de una poligonal cerrada tenemos que verificar con el dato de la coordenada que nos proporcionaron.P (325679.431, 8626354.293)Q (327967.350, 8626173.900X2= 2287.919 Y2= -180.393 Calculamos el errorEX = X1 - X2 = 2287. 81 - 2287.919 = -0.021EY = Y1 - Y2 = -180.32 (-180.393) = +0.073

= 0.07596

Evaluamos Er < Tr

Trabajo aceptable.

Compensamosx = ( ) L = L = -0.0000067 LCx = 0.0000067 L

y = ( ) L = L = 0.000023 LCy = -0.000023 L

LadoCx = 0.000035 LXCy=-0.000023 LY

PA350.08+340.737 +0.003350.08-80.32-0.008

340.740-80.328

AB630.50629.361+0.004630.50-37.88-0.014

629.365-37.894

BC753.25237.515 +0.005753.25-714.82-0.017

237.520-714.837

CD946.28909.393 +0.006946.28261.63-0.022

909.399261.612

DQ426.78170.892 +0.003426.78391.07-0.016

170.895391.054

Ahora calculamos las coordenadas.VrticeCoordenadas absolutas

X (E)Y (N)

P325679.4318626354.293

A326020.1718626273.965

B326649.5368626236.071

C326887.0568625521.234

D327796.4558625782.846

Q327967.3508626173.900

e) calcularemos el azimut y la distancia de PQ