FEU – INSTITUTE OF TECHNOLOGY
P.Paredes St. Sampaloc, Manila
College of Engineering
Civil Engineering Department
Fieldwork # 9
Title: Simple Horizontal Curve
Group # Members:
Calderon, Julian
Javier, Erika
Crismundo, Maxine
Malaluan, Luis
Cuevas, Emmanuel
Paraggua,Sean
De Castro, Shylla
Ignacio, Bob
Reyes, Ethan
Ruby, Nikolai
ENGR. JUN JUN MORENO
Professor
CE0015L – Fundamentals of Surveying
Field Work 9 – Simple Horizontal Curve Objectives:
1. To prepare data set for a simple horizontal curve.
2. To determine the elements of a simple horizontal curve.
Activity Guide:
1. Refer to the figure below of a road section in Sablayan, Mindoro Occidental.
2. The stationing of Point PC is 12+100 and the stationing of point PT is 12+250.
3. Estimate the bearing of the tangent lines at Point PC and PT using the visual estimate.
The estimate should be accurate as possible in degrees, minutes, and seconds.
Bearing of the tangent line to PC: N/A
Bearing of the tangent line to PT: N/A
2.
Determine the basic elements of the simple
curve. Radius of the curve: 208.348m in meters.
3.
Internal Angle: 31.983° in degrees
Degree of the Curve: 5°30’ in degrees (use arc definition)
3. Determine the other basic elements of the simple curve.
Chord:
114.796m
Tangent:
59.71m
Middle Ordinate:
8.0625m
External Ordinate:
8.387m
6. Write your calculations and sketches here in the box.
STATION
CHORD
DEF. ANGLES
TOTAL DEF.
REMARKS
2+341.50
0
0
0
STA.PC
2+360
18.5m
2.54375°
2.54375°
2+380
20m
2.75°
5.29375°
2+400
20m
2.75°
8.04375°
2+420
20m
2.75°
10.79375°
2+440
20m
2.75°
13.54375°
2+457.80
17.8m
2.4475°
15.99125°
STA.PT
Given:
STA.PC = 2+341.50
Solve for tangent distance
T = Rtan
( 2 ) = 208.348tan( 31.983°2 ) = 59.71m
D = 5°30’
LC = 116.30m
Solve for Chord Distance
(
) = 2(208. 348) ( 31.983° )
C = 2Rsin
2
2
Solve for R (Radius)
3600
R=
π
3600
=
= 208.348m
π(5°30')
Solve for middle ordinate
M = R⎡ −
Solve for I (Central Angle)
L=
=>I=
π
( )⎤ = 204.348⎡
−
( 31.983° )⎤
⎣2⎦⎣2⎦ M = 8.0625m
(180°)
180°
I=
STA.PT = STA.PC + LC = (2+341.50) + (0+116.30)
π
=
(180°)
116.30(180°)
π
= 31.983°
π(208.348)
STA.PT = 2+457.90
Solve for External Distance
⎡
1
⎤
− 1⎥
E=R⎢
⎣
(
2
⎦
)
⎡
1
⎤
− 1⎥
E = 208.348⎢
31. 383°
⎣
2
(
⎦
)
E = 8.387m
STA.Vertex = STA.PC+Td
4. (2+341.50 + (0+59.71)
STA.Vertex = 2+401.21
STA.PC = 2 + 341.50
First full station : 2 + 360
STA.PT = 2 + 457.80
| C1sub = ( 2 + 360 ) - ( 2 + 341.50) C1sub = 18.5m
Last full station : 2 + 440
No. of full stations =
| C2sub = ( 2 + 457.80 ) - ( 2 + 440) C1sub = 17.8m
=
=4
(2+440)−(2+360)
80
20
D1 =
=
1
18.5(5°30')
20
D2 =
20
= 5.0875
20
=
2
17.8(5°30')
20
= 4.895
20
No. of alpha = no. of full station + 2
=4+2=6
φ1
=
1
5.0875
=
2
2
φ2
=φ1+
2
2
φ3
=φ2+
2
2
φ4
=φ3+
2
2
φ5
= 2.54375°
2
=φ4+
2
= 2.54375° +
= 5.29375° +
= 8.04375° +
5°30'
= 5.29375°
2
5°30'
= 8.04375°
2
5°30'
= 10.79375°
2
= 10.79375° +
5°30'
2
= 13.54375°
φ6 =φ5+
φ6
=
1
2
2
2
= 13.54375° +
5°30'
2
= 15.99125°
(I)
15.99125° = 12 (31.983°)
31.983° ≚ 31.983°
III. Observation
The group observed that in an actual field work involving the simple horizontal curve certain
adjustments must be made since there are some possibilities of discrepancies and errors that
may affect the data gathered but since the data was given by the professor such adjustments
were not made, and the solving of the data became much easier without such possibilities of
data error.
IV. Recommendation
the recommendation for field work 9, Simple Horizontal Curve Objectives, the students suggest
executing the field work on a real horizontal curve on any location with safety guidelines to
ensure and teach the students the hazards on a real surveying field and to measure the actual
given radius and internal angle on a given horizontal curve of a chosen Horizontal Curve