CONSTRUCTION

SURVEYING AND LAYOUT

SPECIAL TOPICS I
I. HIGHWAY CURVES
CONSTRUCTION SURVEYING AND LAYOUT
I. SIMPLE CURVE
HIGHWAY CURVES
SIMPLE CURVE
TERMINOLOGIES:
PC = Point of Curvature. It is the beginning of the curve.
PT = Point of tangency. It is the end of curve.
PI = Point of intersection of the tangents. Also called vertex
T = Length of tangent from PC to PI and from PI to PT. It is known as subtangent
R = Radius of simple curve, or simply radius.
L = Length of chord from PC to PT. Point Q as shown below is the midpoint of L.
Lc = Length of curve from PC to PT. Point M in the is the midpoint of Lc.
E = External distance, the nearest distance from PI to the curve.
m = Middle ordinate, the distance from midpoint of curve to midpoint of chord.
I = Deflection angle (also called angle of intersection and central angle). It is the
angle of intersection of the tangents. The angle subtended by PC and PT at O is
also equal to I, where O is the center of the circular curve from the above figure.
x = offset distance from tangent to the curve. Note: x is perpendicular to T.
θ = offset angle subtended at PC between PI and any point in the curve
D = Degree of curve. It is the central angle subtended by a length of curve equal to
one station. In English system, one station is equal to 100 ft and in SI, one station
is equal to 20 m.
Sub chord = chord distance between two adjacent full stations.
 SIMPLE CURVE
FORMULA

Length of Tangent, T Length of Chord, L

𝐼 𝐼
𝑇 =𝑅 𝑡𝑎𝑛 𝐿=2 𝑅 𝑠𝑖𝑛
2 2

External Distance, E Length of the Curve, Lc

𝐼 𝜋 𝑅𝐼
𝐸=𝑅 𝑠𝑒𝑐 −𝑅 𝐿𝑐 =
2 180 ᵒ

If given the stationing of PC and PT

Middle Ordinate, M 𝐿𝑐 =𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑖𝑛𝑔 𝑜𝑓 𝑃𝑇 − 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑖𝑛𝑔 𝑜𝑓 𝑃𝐶
𝐼
𝑚= 𝑅 − 𝑅 𝑐𝑜𝑠
2
 SIMPLE CURVE
FORMULA
DEGREE OF CURVE, D
The degree of curve is the central angle subtended by an arc (arc basis) or chord (chord basis) of one station.
It will define the sharpness of the curve. In English system, 1 station is equal to 100 ft. In SI, 1 station is equal to 20
m. It is important to note that 100 ft is equal to 30.48 m not 20 m.

ARC BASIS
The degree of curve is the central angle subtended by one station of
circular arc. This definition is used in highways. Using ratio and proportion.

SI UNITS ENGLISH UNITS

(1 station = 20m) (1 station = 100ft)
1 𝑠𝑡𝑎𝑡𝑖𝑜𝑛 2 𝜋 𝑅 20 2 𝜋 𝑅 100 2 𝜋 𝑅
= = =
𝐷 360 ᵒ 𝐷 360 ᵒ 𝐷 360 ᵒ
 SIMPLE CURVE
CHORD BASIS
is used in railway design. The degree of curve is the central angle subtended by one
station length of chord. From the dotted right triangle below.
𝐷 h𝑎𝑙𝑓 𝑠𝑡𝑎𝑡𝑖𝑜𝑛
sin =
2 𝑅

SI UNITS (half station = 10m)

𝐷 10
sin =
2 𝑅

ENGLISH UNITS (half station = 50m)

𝐷 50
sin =
2 𝑅
Two tangents adjacent to each other having bearings N 65˚ 30’ E and S 85˚ 10’ E meet at station 11+157.98. If the radius of the
simple curve connecting these two tangents is 249.17m, determine the following:
a. tangent distance d. external ordinate g. stationing of PT
b. long chord e. degree of curve
c. middle ordinate f. length of curve

meet Sta. 11 + 157.98 = Sta. V

R = 249.17 m y = 257.56
y – R = E = 8.39 m

r tan = T D=
T = 249.17 tan D=
T = 65.21 m D = 4.6

L/2 = 249.17 sin ( L=r

L = 126.18 m = 249.17()()
L = 127.57
x = 249.17 cos (
x = 241.05 Sta. V – T + L = 11 + 220.34

R – x = m = 8.12 m
COMPOUND CURVE
HIGHWAY CURVES
COMPOUND CURVE
TERMINOLOGIES:
PC = point of curvature
PT = point of tangency
PI = point of intersection
PCC = point of compound curve
T1 = length of tangent of the first curve
T2 = length of tangent of the second curve
V1 = vertex of the first curve
V2 = vertex of the second curve
I1 = central angle of the first curve
I2 = central angle of the second curve
I = angle of intersection = I1 + I2
Lc1 = length of first curve
Lc2 = length of second curve
L1 = length of first chord
L2 = length of second chord
L = length of long chord from PC to PT
T1 + T2 = length of common tangent measured from V1 to V2
θ = 180° - I
x and y can be found from triangle V1-V2-PI.
L can be found from triangle PC-PCC-PT
 SIMPLE CURVE
FORMULA

Finding the stationing of PT

Given the stationing of PC

𝑆𝑡𝑎 . 𝑃𝑇 =𝑆𝑡𝑎 . 𝑃𝐶 + 𝐿 𝑐1 + 𝐿 𝑐 2

Given the stationing of PI

𝑆𝑡𝑎 . 𝑃𝑇 = 𝑆𝑡𝑎 . 𝑃𝐼 − 𝑥 −𝑇 1+ 𝐿 𝑐 1+ 𝐿𝑐 2
A compound curve has the following elements:
I1 = 30˚ I2 = 24˚
D1 = 4˚ D2 = 5˚
If the stationing of the vertex is 4 + 620
a. Determine the stationing of PC b. Find the stationing of PCC c. Determine the stationing of PT

PI L1 = R1 = 286.479 (30)()
I = I 1 + I2
L1 = 150 m
𝜃
x y T1 = 286.479 tan (15) = 76.76 m
Sta. PCC = Sta. PC + L1
T2 = 229.183 tan (12) = 48.71 m
I1 I2 Sta. PCC = (4 + 480.16) +150
V1 PCC V2 Sta. PCC = 4 + 630.16
T1 T2
L1 L2 x 126
PC
L
30
L2 = R2229.1832 (24)()
I2
125.47
L2 = 96 m
R2
x = 63.08 m
Sta. PC = Sta. V - TT Sta. PT = Sta. PCC + L2
R1 I1
= (4 + 620) – (76.76 + 63.08) Sta. PT = (4 + 630) + 96
Sta. PC = 4 + 480.16 m Sta. PT = 4 + 726.16
REVERSE CURVE
HIGHWAY CURVES
 REVERSE CURVE

PRC

I1 I2
R2 R2
T1 T1
LC1 L2
PC
L1 LC2 PT
T2 T2
R1 R1 I2
I1
 REVERSE CURVE

PRC
PC T1
I1 R2 I2
LC1
T1
L1 R2

T2 L2
R1 LC2
I2
R1 T2 PT
I1
The perpendicular distance between two parallel tangents of the reverse curve is 30 m. The azimuth of the back tangent is 270
degrees and the common tangent is 300 degrees. If the radius of the first curve is 175 m, determine the radius of the second
curve.

T1 R2
I1 I2 T
1 +T
R2
30 m 2
30 m
30ᵒ
175 m

I2
T2
I1 175 m 30
sin 30 ᵒ =
𝑇 1 +𝑇 2
30
sin 30 ᵒ =
46.891+𝑇 2

I1 = 300ᵒ - 270ᵒ = 30ᵒ 𝑇 2=13.109 𝑚

𝑇 1 =𝑅1 tan ¿ 𝑅2 =
𝑇2
𝑇 1 =46.891 𝑚
tan ¿ ¿
𝑅2 =48.923 𝑚
The common tangent BC of the reverse curve is 285.75 m and has a bearing of S 48 ᵒ E. AB is tangent of the first curve whose
bearing is N 75ᵒ E. CD is a tangent of the second curve whose bearing is N 40 ᵒ E. Radius of the first curve is 185.66 m and Sta. of
PI of the first curve is at 530.47. Find the stationing of PT.

STA. PI = 0 + 530.47
75ᵒ
92ᵒ 𝑇 1 =𝑅1 tan ¿ 𝐶 𝑇 =𝑇 1 +𝑇 2
R2 R2
T1
57ᵒ
T1
𝑇 1 =185.66 tan¿ 𝑇 2=184.945 𝑚

75ᵒ 48ᵒ
𝑇 1 =100.805 𝑚 𝑇2
PC 𝑅2 =
PT tan ¿ ¿
T2 48ᵒ 40ᵒ T2 𝑅2 =178.599 𝑚
6
18

5.6

92ᵒ STA. OF PC = STA. OF PI – T1

5.6

18

57ᵒ
6

STA. OF PC = (0 + 530.47) – (0 + 100.805)

STA. OF PC = 0 + 429.665

𝐿 𝐶1 = 𝑅1 𝐼 1 ( 180
𝜋
) 𝐿 𝐶 2 = 𝑅2 𝐼 2 ( 𝜋
180 ) STA. OF PT = STA. OF PC + LC1 + LC2

184.702 m 286.777 m STA. OF PT = 0 + 901.144 m

II. TRAVERSE
CONSTRUCTION SURVEYING AND LAYOUT
DISTANCE BY TACHYMETRY
TRAVERSE
 STADIA METHOD

• Provides a rapid means of determining horizontal distances.

• The precision of the stadia method depends on

• skill of the observer
• length of measurement
• refinement of the instrument
• refraction and parallax (effects)
 STADIA METHOD
VERTICAL CROSS HAIR
LEVEL CROSS HAIR

STADIA HAIR

UPPER STADIA HAIR

STADIA INTERVAL
LOWER STADIA HA
IR

D
 STADIA METHOD
The distance is measured using the formula
D1 = Ks + C

where:
D = distance from the telescope to the rod
C = Stadia constant, distance from the center of the instrument to the
principal focus: usually zero for internal focusing telescopes
K = stadia interval factor of the instrument
s = stadia interval, difference between the upper and lower stadia hair
reading
 STADIA METHOD
The distance is measured using the formula
D1 = Ks + C

where:
D = distance from the telescope to the rod
C = Stadia constant, distance from the center of the instrument to the
principal focus: usually zero for internal focusing telescopes
K = stadia interval factor of the instrument
s = stadia interval, difference between the upper and lower stadia hair
reading
 SUBTENSE BAR METHOD

• The subtense bar is a convenient and practical device used for quick and
accurate measurement of horizontal distance.

• The bar, which is precisely 2 meters long, consists of a rounded steel

tube through which runs a thin invar rod.

• The procedure for determining the distance between two points

consists of setting up the subtense bar at the distant station, and
measuring the horizontal angle subtended by the distance between two
tangents.
AREA OF CLOSED TRAVERSE
TRAVERSE
 AREA BY
COORDINATES AREA BY TRIANGLES
B

( )
1 𝑥1 𝑥2 𝑥3 , , , , 𝑥𝑛 𝑥1 c a
𝐴=
2 𝑦1 𝑦2 𝑦3 , , , , 𝑦𝑛 𝑦1
𝛼
A b C
1
𝐴= 𝑎𝑏 sin 𝜃
2

area of properties whose property intersect each other

Find the area of the inside of the green dimensions as shown
60.06

c Using Heron’s formula:

33.09
a 53.52
𝐴=√ 𝑠(𝑠− 𝑎)(𝑠− 𝑏)(𝑠− 𝑐)
51.55
b
20.43 𝐴=√ 89.108 (89.108 −54.192)(89.108 −55.451)(89.108 −55.451)
8.51
A = 1466.446 m2
𝑎= √8.5 1 +53.52
2 2
a = 54.192 m

𝑏=√ 20.43 +51.55

2 2 b = 55.451 m

𝑐=√ 33.09 +60.06

2 2
c = 68.572 m

1 1
𝑠= (𝑎 +𝑏+𝑐) 𝑠= (54.192+55.451+ 68.572)
2 2

𝑠=89.108 𝑚
DOUBLE MERIDIAN DISTANCE DOUBLE PARALLEL DISTANCE

RULES: RULES:

1. DMD of first course equals departure of 1. DPD of first course equals latitude of
course itself. course itself.
NOTE: East Latitude = + NOTE: North Latitude = +
West Latitude = - South Latitude = -
2. DMD of succeeding course equals DMD 2. DPD of succeeding course equals DPD of
of previous plus departure of previous plus previous plus latitude of previous plus
departure of course itself. latitude of course itself.
3. DMD of last course equals departure of 3. DPD of last course equals latitude of the
the course itself but opposite in sign course itself but opposite in sign
 SAMPLE PROBLEM
A closed traverse has the following data:
LINE BEARING DISTANCE (m)
AB S 70ᵒ 14’ E 32.24
BC S 36ᵒ 30’ W 31.71
CD N 65ᵒ 19’ W 17.54
DE N 80ᵒ 35’ W 36.30
EA N 60ᵒ 80’ E 46.44

a. Find the departure of line CD

b. Compute the DMD of line DE
c. Determine the area of the traverse using DMD method.
d. How much is the total cost of the land if the value of its square meter is 3,200 Php?
 SOLUTION USING DOUBLE MERIDIAN DISTANCE
LINE BEARING DISTANCE (m) LATITUDE DEPARTURE DMD DA
AB S 70ᵒ 14’ E 32.24 -10.90 30.34 30.34 -330.71
BC S 36ᵒ 30’ W 31.71 -25.49 -18.86 41.82 -1065.99
CD N 65ᵒ 19’ W 17.54 7.32 -15.94 7.02 51.39
DE N 80ᵒ 35’ W 36.30 5.94 -35.81 -44.73 -265.70
EA N 60ᵒ 80’ E 46.44 23.13 40.27 -40.72 -931.45
Sum of DA = - 2542.46m2
Latitude = Distance x cos
Departure = Distance x sin
DA = DMD x Lat
AREA = 1,271.23
a. -15.94 m
b. -44.73 m Total cost
c. 1,271.23 m2
d. Php 4,067,936.00 TC = Php 3,200/m2 * 1,271.23
TC = Php 4,067,936.00
 SOLUTION USING DOUBLE PARALLEL DISTANCE
LINE BEARING DISTANCE (m) LATITUDE DEPARTURE DPD DA
AB S 70ᵒ 14’ E 32.24 -10.90 30.34 -10.90 -330.71
BC S 36ᵒ 30’ W 31.71 -25.49 -18.86 -47.29 891.90
CD N 65ᵒ 19’ W 17.54 7.32 -15.94 -65.46 1043.43
DE N 80ᵒ 35’ W 36.30 5.94 -35.81 -52.2 1869.28
EA N 60ᵒ 80’ E 46.44 23.13 40.27 -23.13 -931.45
Sum of DA = 2542.45m2
Latitude = Distance x cos
Departure = Distance x sin
DA = DPD x Dep.
AREA = 1,271.225
a. -15.94 m
b. -52.20 m Total cost
c. 1,271.225 m2
d. Php 4,067,920.00 TC = Php 3,200/m2 * 1,271.225
TC = Php 4,067,920.00
ERRORS OF CLOSED
TRAVERSE
TRAVERSE
 ERRORS OF CLOSED TRAVERSE

B
CONDITION OF CLOSED
TRAVERSE

A L
E LINEAR ERROR OF CLOSURE
L C
A’ 𝐿𝐸𝐶= √ ¿ ¿
ΣD
RELATIVE ERROR
D 𝐸𝑟𝑟𝑜𝑟 𝑜𝑓 𝐶𝑙𝑜𝑠𝑢𝑟𝑒
𝑅𝐸=
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
E
 COMPASS RULE TRANSIT RULE
the correction to be applied to the latitude or
the correction to be applied to the latitude or
departure of any course is to the total correction in
departure of any course is tot the total correction in
latitude or departure as the latitude or departure of that
latitude or departure as the length / distance of the
course is to the arithmetical sum of all latitude or
course is the length / perimeter of the traverse.
departure in the traverse without regards to sign.
LAT.
𝐶𝐿 LAT. 𝐶𝐿
𝐿𝐸𝑁𝐺𝑇𝐻 / 𝐷𝐼𝑆𝑇𝐴𝑁𝐶𝐸 𝐿𝐴𝑇𝐼𝑇𝑈𝐷𝐸
= =
𝐸𝑅𝑅𝑂 𝑅 𝐿 𝑃𝐸𝑅𝐼𝑀𝐸𝑇𝐸𝑅 𝐸𝑅𝑅𝑂 𝑅 𝐿 |Σ 𝐿𝐴𝑇|

DEP.
DEP.
𝐶𝐷 𝐿𝐸𝑁𝐺𝑇𝐻 / 𝐷𝐼𝑆𝑇𝐴𝑁𝐶𝐸 𝐶𝐷 𝐷𝐸𝑃𝐴𝑅𝑇𝑈𝑅𝐸
= =
𝐸𝑅𝑅𝑂 𝑅 𝐷 𝑃𝐸𝑅𝐼𝑀𝐸𝑇𝐸𝑅 𝐸𝑅𝑅𝑂 𝑅 𝐷 |Σ 𝐷𝐸𝑃|
 SAMPLE PROBLEM
LINE BEARING DISTANCE
AB N 45ᵒ 20’ E 410 m
BC S 65ᵒ 10’ E 605 m
CD N 80ᵒ 15’ W 600 m
DA S 55ᵒ 30’ W 280 m

Find the following:

a. Linear Error of Closure.
b. Relative Error.
c. Adjusted departure, adjusted latitude, adjusted bearing and adjusted distance of
line AB and CD using Compass Rule and Transit Rule.
 LINE BEARING DISTANCE LATITUDE DEPARTURE
AB N 45ᵒ 20’ E 410 m 288.22223 291.59551
BC S 65ᵒ 10’ E 605 m -254.08798 549.05765
CD N 80ᵒ 15’ W 600 m 101.60970 - 591.33364
DA S 55ᵒ 30’ W 280 m - 158.59375 - 230.75533

Σ 𝐿𝐴𝑇 =288.22223+ ( −254.08798 ) +101.60970+(− 158.59375)

Σ 𝐿𝐴𝑇 =−22.8498
Σ 𝐷𝐸𝑃=291.59551+549.05765 + ( −591.33364 )+(−230.75533)
Σ 𝐷𝐸𝑃=18.56149
Σ 𝐷𝐼𝑆𝑇 .=410+605+600+280
Σ 𝐷𝐼𝑆𝑇 .=1895𝑚
𝐿𝐸𝐶= √ ¿ ¿ 29.43879 0.01553
𝑅𝐸=
𝐿𝐸𝐶=29.43879 𝑀 1895
 LINE BEARING DISTANCE LATITUDE DEPARTURE
AB N 45ᵒ 20’ E 410 m 288.22223 291.59551
BC S 65ᵒ 10’ E 605 m -254.08798 549.05765
CD N 80ᵒ 15’ W 600 m 101.60970 - 591.33364
DA S 55ᵒ 30’ W 280 m - 158.59375 - 230.75533
Σ 𝐿𝐴𝑇 =−22.8498 COMPASS RULE
Σ 𝐷𝐸𝑃=18.56149 LINE AB LINE CD
Σ 𝐷𝐼𝑆𝑇 .=1895𝑚 𝐶𝐿 410
tan 𝛽=
𝐷𝐸𝑃 . 𝐶𝐿
=
600
tan 𝛽=
𝐷𝐸𝑃 .
= 22.8498 1895 𝐿𝐴𝑇 .
22.8498 1895 𝐿𝐴𝑇 .
β=44 ᵒ 26 ′ 55.92 β=79 ᵒ 40 ′ 15.39
CL = 4.94376 CL = 7.23477

CL = 293.16599 CL = 108.84447
DIST2 = LAT.2 + DEP.2 𝐶𝐷 DIST2 = LAT.2 + DEP.2
𝐶𝐷 410 600
= = DIST = 607.04831 m
18.56149 1895 DIST = 410.66812 m 18.56149 1895

CD = 4.01594 CD = 5.87699
CD = 287.57957 CD = -597.21063
 LINE BEARING DISTANCE LATITUDE DEPARTURE
AB N 45ᵒ 20’ E 410 m 288.22223 291.59551
BC S 65ᵒ 10’ E 605 m -254.08798 549.05765
CD N 80ᵒ 15’ W 600 m 101.60970 - 591.33364
DA S 55ᵒ 30’ W 280 m - 158.59375 - 230.75533
Σ 𝐿𝐴𝑇 =−22.8498 TRANSIT RULE
Σ 𝐷𝐸𝑃=18.56149 LINE AB LINE CD
|Σ 𝐿𝐴𝑇|=802.51366 𝐶 𝐿 288.22223 tan 𝛽=
𝐷𝐸𝑃 . 𝐶𝐿 101.60971 tan 𝛽=
𝐷𝐸𝑃 .
= =
|Σ 𝐷𝐸𝑃|=1662.74213 22.8498 802.51366 𝐿𝐴𝑇 . 22.8498 802.51366 𝐿𝐴𝑇 .

Σ 𝐷𝐼𝑆𝑇 .=1895𝑚 C = 8.20649

L
β=44 ᵒ 12′ 27.19
CL = 2.89311
β=80 ᵒ 4 ′ 39.64

CL = 296.42872 CL = 104.50282
𝐶𝐷 291.59551
DIST2 = LAT.2 + DEP.2 𝐶𝐷 591.33364
DIST2 = LAT.2 + DEP.2
= =
18.56149 1662.74213 DIST = 413.53375 m 18.56149 1662.74213 DIST = 606.99824 m
CD = 3.25513 CD = 6.60116
CD = 288.34038 CD = - 597.3948
III. EARTHWORKS
CONSTRUCTION SURVEYING AND LAYOUT
 FORMULA IN EARTHWORKS
END AREA METHOD PRISMOIDAL FORMULA PRISMOIDAL CORRECTION
𝐿 𝐿 𝐿
𝑉 𝑒= ( 𝐴 1+ 𝐴 2 ) 𝑉 𝑝 = ( 𝐴1 + 4 𝐴𝑚 + 𝐴 2) 𝑉 𝑝 = (𝑐 1 − 𝑐2 )(𝑑 1 − 𝑑 2)
2 6 12
it is assumed that the volume
between successive cross sections
𝑉 𝑝 =𝑉 𝑒 − 𝐶 𝑝
is he average of their areas
multiplied by the distance between Am is determined by averaging the
them corresponding linear dimensions of
the end sections and NOT by
averaging the end areas A1 and A2
Given the following data of he cross section of an earthworks.
A1 = ½ (4)(2.6) + ½ (1.5)(9.3) + ½ (1.5)(5) + ½ (4)(0.7)
STA 2 + 080 A1 = 17.325 m2
2.6 1.5 0.7
9.3 0 5.0
STA. 2 + 100 CL
STA 2 + 100
2.3 1.2 0.9
9.0 0 5.7 2.3

the width of the base is equal to 8 m. 0.9

Compute the volume between stations using end area method, 1.2
prismoidal method and get the prismoidal correction. 2
STA. 2 + 080 CL 3
1
4
2.6 4 4
9.0 5.7

1.5 A2 = ½ (4)(2.3) + ½ (1.2)(9) + ½ (1.2)(5.7) + ½ (4)(0.9)

2 0.7 A2 = 15.22 m2
3
L = 2100m – 2080m = 20m
1
4
Ve = L/2 (A1 + A2) Ve = 325.45 m3
4 4
9.3 5
Given the following data of he cross section of an earthworks.
STA 2 + 090
STA 2 + 080 2.45 1.35 0.8
2.6 1.5 0.7 9.15 0 5.35
9.3 0 5.0
STA 2 + 100 CL
2.3 1.2 0.9
9.0 0 5.7 2.45

the width of the base is equal to 8 m.

Compute the volume between stations using end area method, 1.35
prismoidal method and get the prismoidal correction. 2 0.8
A1 = 17.325 m2 3
A2 = 15.22 m2 1
4
Ve = 325.45 m3 4 4
9.15 5.35
Prismoidal Correction
Am = ½ (4)(2.45) + ½ (1.35)(9.15) + ½ (1.35)(5.35) + ½ (4)(0.8)
Vp = Ve - Cp Cp = L/12 (c1 - c2)(d1 - d2) Am = 16.2875 m2
Cp = Vp – Ve Vp = L/6 (A1 + 4Am + A2)
Cp = -0.2
Cp = 325.65 – 325.45
Vp = 20/6 (17.325 + 4(16.2875) + 15.22)
Cp = 0.2 Vp = 325.65 m3
 The areas bounded by the waterline
of a lake and contours 1, 2, 3, 4 are: A1 = 9225 m2
A1 = 9225 m2
2m A2 = 7515 m2
A2 = 7515 m2
A3 = 6975 m2 A3 = 6975 m2
A4 = 6210 m2 8m
A5 = 4725 m2 A4 = 6210 m2
The contour interval is 2m. A5 = 4725 m2
Determine the volume of the water
in the lake in m3 using the prismoidal
formula.

PRISMOIDAL FORMULA SIMPSON’S 1/3 RULE

Vp = L/6 (A1 + 4Am + A2) A = d/3 (h1 + hn + 2hodd + 4heven)

Vp = 4/6 (9225 + 4(7515) + 6975) + 4/6 (6975 + 4(6210) + 4725) V = d/3 (A1 + An + 2odd + 4even)
Vp = 55,200 m3 V = 2 /3 (9225 + 4725 + 2(6975) + 4(7515 + 6210))

V = 55,200 m3
V. TWO-PEG TEST
CONSTRUCTION SURVEYING AND LAYOUT
 PEG METHOD OF ADJUSTMENT OF DUMPY OR
WYE LEVEL
TWO PEG TEST
 Method for checking and recalibrating a level or transit

 The two-peg test is very simple, but provides a way to test the accuracy of a level, and if you know
which screw to turn (for analog instruments) or menu to follow (for the digital level), you can adjust it
to remove the error.

 All instruments are subject to errors. The checking of the instrument (level) is therefore important.

 The main error is where the line of sight is not parallel to the horizontal line of collimation. In this
case you levels will not be correct.

 A test for checking the level is known as the two peg test. This test determines the amount of error
and if an error occurs notify the technician (the level must be serviced).
 SAMPLE PROBLEM
In the two peg test of a dumpy level, the following observations were taken.

ROD READING INSTRUMENT @ A INSTRUMENT @ B

AT A 1.203 0.324
AT B 2.523 1.445

a. Find the difference in elevation between A and B

b. What is the line error in the line of sight from A to B?
c. With instrument at A, what should be the rod reading at B to make line of sight level?
TWO PEG TEST SAMPLE PROBLEM
In the two peg test of a dumpy level, the following observations were taken.
ROD READING INSTRUMENT @ A INSTRUMENT @ B
AT A 1.203 0.324
AT B 2.523 1.445
a. Find the difference in elevation between A and B?
HORIZONTAL LINE
ACTUAL LIN
E OF SIGHT e

1.203

2.523
A
h

B
h + 1.203 = e + 2.523
h – e = 2.523 – 1.203
h – e = 1.320 EQN 1
TWO PEG TEST SAMPLE PROBLEM
In the two peg test of a dumpy level, the following observations were taken.
ROD READING INSTRUMENT @ A INSTRUMENT @ B
AT A 1.203 0.324
AT B 2.523 1.445
a. Find the difference in elevation between A and B?
b. What is the line error in the line of sight from A to B?
HORIZONTAL LINE
e E OF SIGHT
ACTUAL LIN

0.324
1.445
A
h

B
h + 0.324 + e = 1.445 EQN 1 h – e = 1.320 h – e = 1.320
h + e = 1.445 – 0.324 EQN 2 h + e = 1.121 1.2205 – e = 1.320
h + e = 1.121 EQN 2 2h = 2.441 e = - 0.0995
h = 1.2205 e = 0.0995
TWO PEG TEST SAMPLE PROBLEM

INTRUMENT @ A E OF SIGHT e
ACTUAL LIN
HORIZONTAL LINE

1.203
2.523

A
h

c. With instrument at A, what should be the rod reading at B to make line of sight level? B

RRB = 2.523 – e
RRB = 2.523 – 0.0995
RRB = 2.4235
DIFFERENTIAL LEVELING
CONSTRUCTION SURVEYING AND LAYOUT
 DIFFERENTIAL LEVELING

 It is a technique used to determine differences in elevation between points that are remote form each
other

 Requires the use of surveyor’s level together with graduated measuring rods

 ELEVATION – vertical distance above or below a reference datum

DIFFERENTIAL LEVELING

FORESIGHT

BACKSIGHT
HEIGHT B
OF THE
INSTRUMENT
(HI)
A
ELEV. B

ELEV. A

BM TP
Elev. A + BS = HI
Elev. A + BS – FS = Elev. B
Elev. A + ∑ BS = HI
Elev. A + ∑ BS - ∑ FS = Elev. B
DIFFERENTIAL LEVELING: SAMPLE
PROBLEM

USING THE FOLLOWING FIELD NOTES:

Elev. A + BS – FS = Elev. B
STATION BACKSIGHT FORESIGHT ELEVATION
BM 01 4.64 209.65
1 5.80 5.06 209.65 + 4.64 – 5.06 = 209.23
2 2.25 5.02 210.01
BM 02 6.02 5.85 206.41
3 8.96 4.34 208.09
4 8.06 3.22 213.83
5 9.45 3.71 218.18
6 12.32 2.02 225.61
BM 03 1.98 235.95
DIFFERENTIAL LEVELING: SAMPLE PROBLEM

STATION BACKSIGH FORESIGHT ELEVATION

T Elev. A + ∑ BS - ∑ FS = Elev. B
BM 01 4.64 209.65
∑ BS = 4.64 + 5.80 + 2.25 + 6.02 + 8.96 + 8.06 + 9.45 + 12.32
1 5.80 5.06 ∑ BS = 57.50
2 2.25 5.02
∑ FS = 5.06 + 5.02 + 5.85 +4.34 + 3.22 + 3.71 +2.02 + 1.98
BM 02 6.02 5.85 ∑ FS = 31.20
3 8.96 4.34
4 8.06 3.22 Elev. BM1 + ∑ BS - ∑ FS = Elev. BM3
209.65 + 57.50 – 31.20 = Elev. BM3
5 9.45 3.71
Elev. BM3 = 235.95 m
6 12.32 2.02
BM 03 1.98
DIFFERENTIAL LEVELING: SAMPLE PROBLEM

STATION BACKSIGH FORESIGHT ELEVATION What if Elev.BM2 is required?

T
BM 01 4.64 209.65
∑ BS = 4.64 + 5.80 + 2.25
1 5.80 5.06 ∑ BS = 12.69
2 2.25 5.02
∑ FS = 5.06 + 5.02 + 5.85
BM 02 6.02 5.85 ∑ FS = 15.93
3 8.96 4.34
4 8.06 3.22 Elev. BM1 + ∑ BS - ∑ FS = Elev. BM2
209.65 + 12.69 – 15.93 = Elev. BM2
5 9.45 3.71
Elev. BM3 = 206.41 m
6 12.32 2.02
BM 03 1.98