School of Civil, Environmental, and Geological Engineering

FUNDAMENTALS OF SURVEYING

CE120-02 SECTION E02

PROJECT FOR MODULE 2:

PROBLEM SET WITH SOLUTION

Submitted by:

DIEGO, KRISTINE R.

2019103990
 1.A. TRAVERSING PRINCIPLES

1.A.i. From the given closed traverse shown,

LINES BEARING DISTANCES

A-B N 69O E 261 m
B-C S 10.3O E 217 m
C-D - -
D-E W 239 m
E-A N 28.3O W 225 m
1. What is the length of the line CD?
2. What is the bearing of line CD?
3. Compute for the area of the traverse.
SOLUTION

LINES BEARING DISTANCES LATITUDE DEPARTURE

[Latitude=Distance⋅Cos(β)] [Departure=Distance⋅Sin(β)]
Sign Convention: N (+) S (-) Sign Convention: E (+) W (-)
A-B N 69O E 261 m 93.534 243.664
B-C S 10.3O E 217 m - 213.503 38.800
C-D - - y x
D-E W 239 m 0 - 239
E-A N 28.3O W 225 m 198.107 - 106.670
∑𝐿𝑎𝑡 = 0
93.534 − 213.503 + 𝑦 + 0 + 198.107 = 0; 𝑦 = −78.138 𝑚
∑𝐷𝑒𝑝 = 0
243.664 − 38.800 + 𝑥 − 239 − 106.670 = 0; 𝑥 = 63.206 𝑚

a. What is the length of the line CD?

𝐶𝐷 = √𝐿𝑎𝑡 2 + 𝐷𝑒𝑝2 = √78.1382 + 63.2062 = 100.51𝑚

b. What is the bearing of line CD?
𝐷𝑒𝑝𝑎𝑟𝑡𝑢𝑟𝑒 63.206
𝑡𝑎𝑛𝛽 = | | =| | ; 𝛽 = 𝑆38.969°𝐸
𝐿𝑎𝑡𝑖𝑡𝑢𝑑𝑒 −78.138
c. Compute for the area of the traverse.

LINES LATITUDE DEPARTURE DMD (DMD)(LAT)

A-B 93.534 243.664 243.664 22790.869
B-C - 213.503 38.800 526.128 -112329.906
C-D -78.138 63.206 628.134 -49081.134
D-E 0 - 239 452.34 0
E-A 198.107 - 106.670 106.670 21132.074
2A=∑(DMD)(LAT) -117488.097
A = 58744.049 m2

1.A.ii. From the field notes of a closed traverse shown below, adjust the traverse.

LINES BEARING DISTANCES

A-B N 400 m
B-C S 45O E 800 m
C-D S 60O E 700 m
D-E S 20O W 600 m
E-A S 83O59’ W 966.34 m
a. Compute the correction of latitude on line CD using transit rule.
b. Compute the linear error of closure.
 c. Compute the relative error or precision.
SOLUTION
LINE BEARIN DISTANC LATITUDE DEPARTURE
[Latitude=Distance⋅Cos( [Departure=Distance⋅Sin(
S G E
β)] β)]
Sign Convention: Sign Convention:
N(+) S(-) E(+) W(-)
A-B N 400 m + 400 0
B-C N45oE 800 m + 565.69 + 565.69
C-D S60oE 700 m - 350.00 + 606.22
D-E S20oW 600 m - 563.82 - 205.21
E-A S86o59′ 966.34 m - 50.86 - 965.00
W
P=3466.3 error of closure, e = error of closure, e = +1.7
4 +1.01
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 400 + 800 + 700 + 600 + 966.34 = 3466.34𝑚
𝐸𝑟𝑟𝑜𝑟 𝑜𝑓 𝐶𝑙𝑜𝑠𝑢𝑟𝑒 𝑜𝑓 𝐿𝑎𝑡𝑖𝑡𝑢𝑑𝑒 = 400 + 565.69 − 350 − 563.82 − 50.86 = 1.01
𝐸𝑟𝑟𝑜𝑟 𝑜𝑓 𝐶𝑙𝑜𝑠𝑢𝑟𝑒 𝑜𝑓 𝐷𝑒𝑝𝑎𝑟𝑡𝑢𝑟𝑒 = 0 + 565.69 + 606.22 − 205.21 − 965.00 = 1.7
∑𝑁𝐿 − ∑𝑆𝐿 = (400 + 565.69) − (−350 − 563.82 − 50.86) = 1930.37
a. Compute the correction of latitude on line CD using transit rule.
(𝑒)(𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝐶𝐷)
𝐶𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐿𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝐶𝐷, 𝐶𝐶𝐷 =
∑𝑁𝐿 − ∑𝑆𝐿
(1.01)(350)
𝐶𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐿𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝐶𝐷, 𝐶𝐶𝐷 = = 0.18
1930.37
b. Compute the linear error of closure, LEC.
𝐿𝐸𝐶 = √(𝑒𝐿 )2 + (𝑒𝐷 )2 = √(1.01)2 + (1.7)2 = 1.997
c. Compute the relative error or precision.
𝐸𝑟𝑟𝑜𝑟 𝑜𝑓 𝐶𝑙𝑜𝑠𝑢𝑟𝑒 1.997
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝐸𝑟𝑟𝑜𝑟 = = = 5.762 𝑥 10−4
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑎𝑙𝑙 𝑐𝑜𝑢𝑟𝑠𝑒𝑠 3466.34
1.B. STADIA TACHEOMETRIC SURVEY

1.B.i. The upper and lower stadia hair readings on a stadia rod held at station B were
observe as 3.50 and 1.00 m, respectively, with the use of a transit with an interval
focusing telescope and having a stadia interval factor of 99.5. The height of the
instrument above the station A is 1.45 m. and the rod reading is taken at 2.25 m. If the
vertical angle observed is −23o34′′, determine the following:

a. Inclined stadia distance.

b. Difference in elevation between the two stations.
c. The elevation of station B, if the elevation of station A is 155.54 m above mean
sea level.

SOLUTION

a. Inclined stadia distance.

𝑓
𝐷= 𝑆 𝑐𝑜𝑠∅ + (𝑓 + 𝑐)
𝑖
𝑓
Stadia Interval Factor, = 99.5
𝑖

Stadia Constant, (𝑓 + 𝑐) = 0 (internal focusing)

Stadia intercept = 250

 𝐷 = (99.5)(2.50) cos (23𝑜 34′ ) + 0 = 228.291 𝑚

b. Difference in elevation between the two stations.

𝑓 𝑠𝑖𝑛2∅ sin(2(23𝑜 34′ ))
𝑉= 𝑆 + (𝑓 + 𝑐)𝑠𝑖𝑛∅ = (99.5)(2.5) + 0 = 91.16 𝑚
𝑖 2 2
𝐷𝐸 = 2.25 + 91.16 = 1.45 = 91.96 𝑚
c. The elevation of station B, if the elevation of station A is 155.54 m above mean
sea level.
𝐸𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 𝑎𝑡 𝐵 = 155.54 − 91.96 = 63.58 𝑚

1.B.ii. A transit with a stadia constant equal to 0.30 is used to determine the horizontal
distance between points B and C, with a stadia intercept reading of 1.85 m, the distance
BC is equal to 182.87 m.

a. Compute the stadia interval factor of the instrument.

b. Using the same instrument, it was used to determine the difference in elevation
between B and D having a stadia intercept reading of 2.42 m at D at a vertical
angle of +6′30′′. Compute for the difference in elevation of B and D.
c. Compute also the horizontal distance between B and D.

SOLUTION

a. Compute the stadia interval factor of the instrument.

𝑓
𝐷= 𝑆 + (𝑓 + 𝑐)
𝑖

Stadia Constant, (𝑓 + 𝑐) = 0.3

Stadia intercept = 1.85

Distance BC = 182.87 m
𝑓
182.87 = (1.85) + 0.30
𝑖
𝑓
= 98.69
𝑖
b. Using the same instrument, it was used to determine the difference in elevation
between B and D having a stadia intercept reading of 2.42 m at D at a vertical
angle of +6′30′′. Compute for the difference in elevation of B and D.

𝑓 𝑠𝑖𝑛2∅ sin(2(6𝑜 30′ ))

𝑉= 𝑆 + (𝑓 + 𝑐)𝑠𝑖𝑛∅ = (98.69)(2.42) + 0.3𝑠𝑖𝑛6𝑜 30′
𝑖 2 2
= 26.90 𝑚

c. Compute also the horizontal distance between B and D.

𝑓
𝐻= 𝑆(𝑐𝑜𝑠 2 ∅ + (𝑓 + 𝑐)𝑐𝑜𝑠∅ = (98.69)(2.42)𝑐𝑜𝑠 2 (6𝑜 30′ ) + 0.3𝑐𝑜𝑠6𝑜 30′
𝑖
= 236.07 𝑚
1.C. HORIZONTAL CURVES

1.C.i. A simple curve has a central angle of 36o and a degree of curve of 6o.

a. Find the nearest distance from the midpoint of the curve to the point of
intersection of the tangents.
b. Compute the distance from the midpoint of the curve to the midpoint of the long
chord joining the point of curvature and point of tangency.
c. If the stationing of the point of curvature is at 10+020, compute the stationing of a
point on the curve which intersects with the line making a deflection angle of 8o
with the tangent through the PC.

SOLUTION

1145.916
𝑅= = 190.99
6
a. Find the nearest distance from the midpoint of the curve to the point of
intersection of the tangents.
 𝐼
𝐸 = 𝑅 (𝑠𝑒𝑐 − 1) = 190.99(𝑠𝑒𝑐18𝑜 − 1) = 9.83 𝑚
2
b. Compute the distance from the midpoint of the curve to the midpoint of the long
chord joining the point of curvature and point of tangency.
𝐼
𝑀 = 𝑅 (1 − 𝑐𝑜𝑠 ) = 190.99(1 − 𝑐𝑜𝑠18𝑜 ) = 9.35 𝑚
2
c. If the stationing of the point of curvature is at 10+020, compute the stationing of a
point on the curve which intersects with the line making a deflection angle of 8o
with the tangent through the PC.
190.00(16)𝜋
𝑆 = 𝑅𝜃 = ( ) = 53.33 𝑚
180
𝑆𝑇 𝐴. 𝑜𝑓 𝐵 = (10 + 020) + (53.33) = 10 + 073.33

1.C.ii. The tangent distance of a 3o simple curve is only 1/2 of its radius.

a. Compute the angle of intersection of the curve.

b. Compute the length of curve.
c. Compute the area of the fillet of the curve.

SOLUTION

a. Compute the angle of intersection of the curve.

𝐼
𝑇 = 𝑅𝑡𝑎𝑛
2
1 𝐼
𝑅 = 𝑅𝑡𝑎𝑛
2 2
𝐼 1
𝑡𝑎𝑛 =
2 2
𝐼 = 53.13𝑜
b. Compute the length of curve.
𝜋 1145.916 𝜋
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝐶𝑢𝑟𝑣𝑒 = 𝑅𝐼 ( )=( ) (53.13) ( ) = 354.20 𝑚
180 3 180
c. Compute the area of the fillet of the curve.
 1 𝜋
𝐴 = 𝑇𝑅 − 𝑅 2 𝐼 ( )
2 180

1145.916
𝑅= = 381.972
3

53.13
𝑇 = 381.972𝑡𝑎𝑛 = 190.986
2

1 𝜋
𝐴 = (190.986)(381.972) − (381.972)2 (53.13) ( ) = 5304.04 𝑠𝑞. 𝑚.
2 180
1.D. EARTHWORKS

1.D.i. Given the cross section notes of an earthwork between station 10+100 to 10+200.
Assume both stations to have the same side slope and width of the base.

STA. 10+100 STA. 10+200

Left Center Right Left Center Right
6.45 0 4.5 6.0 0 6.9
+2.3 +1.5 +1.0 +2.0 x +2.6
a. Compute the width of base.
b. Compute the value of x at Station 10+200 if it has a cross sectional area of 14.64
m2.
c. Compute the volume between stations 10+100 and 10+200 using the end area
method with prismoidal correction.

SOLUTION
a. Compute the width of base.

𝐵
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 = + 2.3𝑆 = 6.45
2
𝐵
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2 = + 1𝑆 = 4.5
2
 Solving the equations simultaneously, we will get S=1.5m and B=6m.

b. Compute the value of x at Station 10+200 if it has a cross sectional area of 14.64
m2.

1 1 1 1
𝐴= 2(3) + 6(𝑥) + 6.9(𝑥) + 3(2.6)
2 2 2 2
1 1 1 1
14.64 = 2(3) + 6(𝑥) + 6.9(𝑥) + 3(2.6); 𝑥 = 1.2 𝑚
2 2 2 2

c. Compute the volume between stations 10+100 and 10+200 using the end area
method with prismoidal correction.
Volume using End Area Method
1 1
𝑉𝐸 = (𝐴1 + 𝐴2 )𝐿 = (13.16 + 14.64)(100) = 1390.125 𝑚3
2 2

1 1 1 1
𝐴1 = 2.3(3) + 1.5(6.45) + 4.5(1.5) + 3(1) = 13.16 𝑚2
2 2 2 2

𝐴2 = 14.64 𝑚2
 Prismoidal Correction

𝐿 100
𝑉𝑝 = (𝐶1 − 𝐶2 )(𝐷1 − 𝐷2 ) = [(10.95 − 12.9)(1.5 − 1.2)] = −4.875 𝑚3
12 12

Volume with Prismoidal Correction

𝑉𝑐𝑝 = (𝑉𝐸 − 𝑉𝑝 ) = 1390.125 − (−4.875) = 1395 𝑚3

1.D.ii. Given the following section of an earthworks for a proposed road construction on
a hilly portion of the route. The width of the road base for cut is 6m for allowance of
drainage canals and 5m for fill. Sides slopes for cut is 1:1 and for fill is 1:5:1.

3.7 0 x
-0.8 0 +1.8
a. Compute the value of x.
b. Compute the area in fill.
c. Compute the area in cut.

SOLUTION

a. Compute the value of x.

𝑥 = 3 + 2.8 = 4.8 𝑚
b. Compute the area in fill.

1
𝐴𝑓𝑖𝑙𝑙 = (2.5)(0.8) = 1.0 𝑚2
2

c. Compute the area in cut.

1
𝐴𝑐𝑢𝑡 = (3)(1.8) = 2.7 𝑚2
2