-
FIUSAC. Topografa 2 Tarea 2010-20560 Diego Antonio Seisdedos
Javier 10/10/2012
Transformar un lindero ABCD por una lnea recta (QR) con
Az(Q-R)=20
EST PO Y X
C D 3,9548 -5,3274
D 1 35,0001 -54,99921 2 134,9974 -5,0002
2 A 118,6991 52,0452
A B 101,0000 24,9970
B C 65,9870 35,5740
5 D 3,9548 -5,3274
D C 65,9870 35,5740C B 101,0000 2,9970
B A 118,6991 52,0452A 3 114,9980 64,9997
3 4 64,9996 84,9994
4 5 -15,0000 25,0000
LIBRETA DE CAMPO
DIBUJO POLGONO + LINDERO
-50,00
0,00
50,00
100,00
150,00
-75,00 -25,00 25,00 75,00
Y
X
Polgono 12345
Lindero ABCD1
2
3
4
5
B
A
C
D
-
Az(D-P) = Az (Q-R) = 20
Por el mtodo de punto tangente:
Pendiente de 20 = Pendiente (D-P)
X(D)-X(P)
Y(D)-Y(P) Y(P)= 122,73866
X(P)= 37,9064
Pendiente (A-2) = Pendiente (P-2)X(A)-X(2) X(P)-X(2)
Y(A)-Y(2) Y(P)-Y(2)
A(12PD)= 6072,9411 mA(DPRQ)= A(12ABCD)-A(12PD)= 927,0589 m
-6072,9411
A. Trapecio(DPRQ) = 0,5*(B+b)*h
0,5(B+b)*h = 927,05893 f (h)
B =DH(D-P)= 126,40715 m
b = B QQ' RR'
Az (h)= 20+90 = 110
Az (A-3)= 105,9446 Az(h)>Az(A-3) Dentro: +RR'
Az (D-5)= 122,0056 Az(h)
-
0,5(B+b)*h = 927,05893
0,5*(126,4071+ 126,4071 + 0,141760199*h)*h = 927,05893 h=
7,27456069 m
126.4071*h - 0.1417785*h^2 = 927.05893
QQ'= h*tan(4,0554) = 0,52431
RR'= h*tan(12,0056) = 1,57267
DH(D-Q)= (h+QQ')^0,5= 7,2934311
DH(R-Q)= (h+RR')^0,5= 7,4426147
Y(Q)= Y(D)+yc(DQ)= 0,0893 yc(DQ)= -3,86553642X(Q)= X(D)+xc(DQ)=
0,8574 xc(DQ)= 6,18480117
Y(R)= Y(P)+yc(PR)= 120,69412 yc(PR)= -2,04454865
X(R)= X(P)+xc(PR)= 45,0627 xc(PR)= 7,15627934
YX= 6006,9097 EST PO Y X
XY= -7994,2121 Q 1 35,0001 -54,9992
A(12RQ)= 7000,5609 1 2 134,9974 -5,0002
2 R 120,69412 45,0627
R Q 0,0893 0,8574
EST PO yc xc
Q 1 34,9108 -55,8566
7000,5609 1 2 99,9973 49,999
2 R -14,303285 50,0628702
R Q -120,60485 -44,205269
ARMADO DE POLGONO
COORDENADAS DE Q, R (LINDERO)
EC. CUADRTICA DEL TRAPECIO
SUSTITUYENDO VALORES DE "h"
-
' "
Q 1 N 32 0 20 W 65,8690 34,9108 -55,8566 32,005723
1 2 N 63 26 5 E 111,8005 99,9973 49,999 63,434788
2 R S 15 56 41 E 52,0661 -14,303285 50,06287 15,94498
R Q S 69 52 14 W 128,4509 -120,60485 -44,205269 69,870611
0,3434 0,343426,0873 0,0873
56,6988 0,6988
52,2366 0,2366
y x Rumbo
DIBUJO FINAL
LIBRETA FINAL
RUMBOEST PO DH (m)
-50,00
0,00
50,00
100,00
150,00
-75,00 -25,00 25,00 75,00
Polgono 12345
Linea QR1
2
3
4
5 Q
R
7000,5609 m
-
Az (D-5)= 122,0056
PO Y X Y= -18,9548
1 35,0001 -54,9992 X= 30,3274
2 134,9974 -5,0002 |Y/X|= 0,6250058
3 114,9980 64,9997 Rumbo (D-5)= 32,005621 SE
4 64,9996 84,99945 -15,0000 25,0000 Az (A-3)= 105,9446
1 35,0001 -54,9992 Y= -3,7011
X= 12,9545
|Y/X|= 0,2856999
PO Y X Rumbo (D-5)= 15,944637 SE
A 118,6991 52,0452B 101,0000 24,9970
C 65,9870 35,5740D 3,9548 -5,3274
PO Y X rea Matricial (12PD)
1 35,0001 -54,9992 6072,9411 m 4070,8679
2 134,9974 -5,0002 -8075,0142P 122,738664 37,9064 Distancia
(D-P)
D 3,9548 -5,3274 Y= 118,7839
X= 43,2338
DH(D-P)= 126,40715 m
PO Y X
D 3,9548 -5,3274
P 122,738664 37,9064
PO Y X
R 120,694115 45,0627
Q 0,0893 0,8574
PO Y X
1 35,0001 -54,9992
2 134,9974 -5,0002
R 120,694115 45,0627
Q 0,0893 0,8574
35,0001 -54,9992
Linea QR
Polgono Final
TABLAS AUXILIARES (PARA DIBUJO)
Polgono
Lindero
Linea DP
-
DIBUJO AUXILIAR ( POLGONO + LINDEROS )
-50,00
0,00
50,00
100,00
150,00
-75,00 -25,00 25,00 75,00
Polgono 12345
Lindero ABCD
Linea DP
Linea QR1
2
3
4
5
B
A
C
D
P
Q
R
