Name: Decy Raye M. Cimeni
Course & Year: BSEM - 3
Subject: GEE140
ACTIVITY 5
a) Computing Latitudes (Lat = length x cos α)
LAT1-2
LAT2-3
LAT3-4
LAT4-1
b) Computing Departures (Dep = length x sin α)
= 134.53m x cos (5°59') = -133.80m
= 205.54m x cos (72°31') = 61.75m
= 129.44m x cos (5°59') = 128.73m
= 208.10m x cos (73°47') = -58.12m
DEP1-2
DEP2-3
DEP3-4
DEP4-1
= 134.53m x sin (5°59') = 14.02m
= 205.54m x sin (72°31') = -196.04m
= 129.44m x sin (5°59') = -13.49m
= 208.10m x sin (73°47') = 199.82m
c) Tabulated Solution
Line
Distance (m)
Bearing
1-2
2-3
3-4
4-1
SUM
134.53
205.54
129.44
208.10
677.61
S 5°59' E
N 72°31' W
N 5°59' W
S 73°47' E
Computed Latitude
+N
-S
133.80
61.75
128.73
58.12
+190.48
-191.92
d) Determining Total Closure in Latitude and Departure
∑NL = +61.75 + 128.73
= +190.48m
∑ED = +14.02 + 199.82
= +213.84m
∑SL = -133.80 – 58.12
= -191.92m
∑WD = -196.04 – 13.49
= -209.53m
Computed Departure
+E
-W
14.02
196.04
13.49
199.82
+213.84
-209.53
CL = ∑NL + ∑SL
= +190.48 + (-191.92)
= -1.44m (total closure in latitude)
CD = ∑ED + ∑WD
= +213.84 + (-209.53)
= 4.31m (total closure in departure)
D = d1 + d2 + d3 + d4
= 134.53 + 205.54 + 129.44 + 208.10
= 677.61m (length of the traverse)
e) Determining Corrections for Latitude => [ K1 =
𝐶𝐿
𝐷
; c𝑙 = 𝑑(K1 ) ]
−1.44
K1 = 677.61 = −0.00213
c1-2 = 134.53 x -0.00213 = -0.29m
c2-3 = 205.54 x -0.00213 = -0.44m
c3-4 = 129.44 x -0.00213 = -0.28m
c4-1 = 208.10 x -0.00213 = -0.43m
Solution Check:
Sum of Latitude = CL
-0.29 + (-0.44) + (-0.28) + (-0.43) = -1.44
-1.44 = -1.44 ✓
f) Determining Corrections for Departure => [ K 2 =
K2 =
4.31
677.61
𝐶𝐷
𝐷
; c𝑑 = 𝑑(K 2 ) ]
= 0.00636
c1-2 = 134.53 x 0.00636 = 0.86m
c2-3 = 205.54 x 0.00636 = 1.31m
c3-4 = 129.44 x 0.00636 = 0.82m
c4-1 = 208.10 x 0.00636 = 1.32m
Solution Check:
Sum of Latitude = CL
0.86 + 1.31 + 0.82 + 1.32 = 4.31
4.31 = 4.31 ✓
g) Adjusting the Latitude => Adj Lat = Computed Lat ± c𝑙
Adj Lat1-2 = -133.80 – (-0.29) = -133.51
Adj Lat2-3 = +61.75 – (-0.44) = 62.19
Adj Lat3-4 = +128.73 – (-0.28) = 129.01
Adj Lat4-1 = -58.12 – (-0.43) = -57.69
Solution Check:
Algebraic Sum of Adjusted Latitudes = 0
-133.51 + 62.19 +129.01 – 57.69 = 0
0=0✓
h) Adjusting the Departure => Adj Dep = Computed Dep ± c𝑑
Adj Dep1-2 = +14.02 – 0.86 = 13.16
Adj Dep2-3 = -196.04 – 1.31 = 62.19
Adj Dep3-4 = -13.49 – 0.82 = 129.01
Adj Dep4-1 = +199.82 – 1.32 = -57.69
Solution Check:
Algebraic Sum of Adjusted Latitudes = 0
-133.51 + 62.19 +129.01 – 57.69 = 0
0=0✓
i) Tabulated Solution
Line
1-2
2-3
3-4
4-1
Correction
LAT
-0.29
-0.44
-0.28
-0.43
DEP
0.86
1.31
0.82
1.32
Adjusted Latitude
+N
-S
-133.51
62.19
129.01
-57.69
(∑NL + ∑SL = 0)
Adjusted Departure
+E
-W
13.16
62.19
129.01
-57.69
(∑ED + ∑WD = 0)
j) Determining Linear Error of Closure, Bearing of the Side of Error, and Relative Error of Closure
𝐿 = √𝐶𝐿 2 + 𝐶𝐷 2
= √(−1.44)2 + (4.31)2
𝐿 = 4.54𝑚 (linear error of closure)
tan α =
=
−𝐶𝐷
−𝐶𝐿
−4.31
−(−1.44)
α = 71°31′
𝑅𝑃 =
(Bearing of Side Error = S 71°31′ E
𝐿𝐸𝐶
𝐷
=
4.54
677.61
=
1
149.25
𝑅𝑃 = 𝑠𝑎𝑦
1
100
(Relative Precision)