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Surveying Computations

This document provides calculations to determine the area of a plot of land using coordinate geometry. It includes the latitude and departure corrections, corrected latitude and departure values, bearing calculations, single and double area calculations by coordinates and distance-motion-distance (DMD), and the final area value of 154.87 square meters.

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Louisse Abuan
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36 views2 pages

Surveying Computations

This document provides calculations to determine the area of a plot of land using coordinate geometry. It includes the latitude and departure corrections, corrected latitude and departure values, bearing calculations, single and double area calculations by coordinates and distance-motion-distance (DMD), and the final area value of 154.87 square meters.

Uploaded by

Louisse Abuan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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CL = ∑N+ LAT + (-S LAT) DEPARTURE CORRECTION CORRECTED DISTANCE

∑N+ LAT = 12.01 + 1.37 = 13.38 𝒅


CD = | CD | ( ) DISTANCE= √LAT2 + DEP2
𝑫
∑S- LAT = -1.58-5.49-6.32 = - 13.39 12.07 1 – 2 = √ (12.02)2 +(1.20)2 = 12.08m
1–2=0( )=0
49.81
= 13.38 + (-13.39) 2 – 3 = √ (-1.58)2 +(12.85)2 = 12.95m
12.95
CL = -0.01 2–3=0( )=0 3 – 4 = √ (-5.49)2 +(-0.56)2 = 5.52m
49.81
5.52
3–4=0( )=0 4 – 5 = √ (-6.32)2 +( -0.64)2 = 6.35m
49.81
CD = ∑E+ DEP + (-W DEP) 6.35 5 – 1 = √ (1.37)2 +(-12.85)2 = 12.92m
4–5=0( )=0
49.81
∑E+ DEP = 1.20+12.85 = 14.05
12.92
5–1=0( )=0
∑W- DEP = -0.56-0.64-12.85 = -14.05 49.81 CORRECTED BEARING
= 14.05 + (-14.05) CORRECTED LAT (L’) BEARING = TAN-1 (
𝐷𝐸𝑃
)
𝐿𝐴𝑇
CD = 0 1 – 2 = 12.01 + 0.01 = 12.02 1.20
1 – 2 = TAN-1 ( ) º = N 05º 42′ E
2 – 3 = -1.58 + 0.00 = -1.58 12.02
12.85
LATITUDE CORRECTION 3 – 4 = -5.49 + 0.00 = -5.49 2 – 3 = TAN-1 ( ) º = S 82º 59′ E
−1.58
𝑑
CL = | CL | ( ) 4 – 5 = -6.32 + 0.00 = -6.32 3 – 4 = TAN-1 (
−0.56
) º = S 05º 49′ W
𝐷 −5.49
12.07
5 – 1 = 1.37 + 0.00 = 1.37 −0.64
1 – 2 = 0.01 ( ) = 0.002 OR 0.00 + 0.01 = 4 – 5 = TAN-1 ( ) º = S 05º 47′ W
−6.32
49.81
∑= 0
0.01 − 12.85
CORRECTED DEP (D’) 5 – 1 = TAN-1 ( ) º = N 83º 55′ W
12.95 1.37
2 – 3 = 0.01 ( ) = 0.002 OR 0.00
49.81
1 – 2 = 1.20 + 0 = 1.20 AREA BY COORDINATES
5.52
3 – 4 = 0.01 ( ) = 0.001 OR 0.00 2 – 3 = 12.85+ 0 = 12.85
1
AREA= (∑ PRODUCT OF DOWN TO RIGHT)-
49.81
2
4 – 5 = 0.01 (
6.35
) = 0.001 OR 0.00 3 – 4 = -0.56 + 0 = -0.56 (∑ PRODUCT OF UP TO RIGHT)
49.81
00 1.20 14.05 13.49
12.92 4 – 5 = -0.64+ 0 = -0.64 = ∑1 = ( ) ↘ x ( )↘ x ( ) ↘x ( )↘ x
5 – 1 = 0.01 ( ) = 0.002 OR 0.00 12.85
00
00
12.02 10.44 4.95
49.81
5 – 1 = - 12.85 + 0 = - 12.85 ( )↘x ( )
−1.37 00

∑= 0
= ∑1 = (00) (12.02) + (1.20) (10.44) + (14.05) DOUBLE AREA
(4.95) + (13.49) (-1.37) + (12.85) (00) = 63.59m2
DA= DMD (LAT)
1 – 2 = 1.20 (12.02) = 14.42
00 1.20 14.05 13.49
= ∑1 = ( ) ↗ x ( )↗ x ( )↗x( )↗ x 2 – 3 = 15.25 ( -1.58) = -24.10
00 12.02 10.44 4.95
12.85 00
( )↗x( ) 3 – 4 = 27.54 ( -5.49) = -151.19
−1.37 00

4 – 5 = 26.34 ( -6.32) = -166.47


= ∑2 = (00) (1.20) + (12.02) (14.05) + (10.44) 5 – 1 = 12.85 (1.37) = 17.60
(13.49) + (4.95) (12.85) + (-1.37) (00) =372.78m2
∑= -309.74

1
A= [ 63.59 – 372.78] 1
2
A= (∑EDA + WDA)
2
A= 154.595 m2
1
A= ( -309.74)
2

= 154.87 m2

AREA BY DMD
DMD
1 – 2 = 1.20
2 – 3 = 1.20 + 1.20 + 12.85= 15.25
3 – 4 = 15.25 + 12.85 -0.56 = 27.54
4 – 5 = 27.54 -0.56 – 0.64 = 26.34
5 – 1 = 26.34 -0.64 – 12.85 = 12.85

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