Talks about Simple Curve

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SIMPLE CURVE
SAMPLE PROBLEMS
Engr. Miguel G. Cabanit
SAMPLE PROBLEM 1
Fundamentals of Surveying 2 2
If the central angle of a simple curve is 32° and the degree of curve
is 7°, determine the elements of the simple curve (R, T, E, M, C, Lc,
subchords and subangles and the stations of the curve) if station of
the PC is at 6+038. Illustrate the curve and label each station
STATION SUBTENDED ANGLE DEFLECTION ANGLE REMARKS
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Fundamentals of Surveying 2 3
Radius of the Curve (R)
R =
1145.916
D
R =
1145.916
7
R = 163.70 m
Tangent Distance (T)
T = R tan
I
2
T = 46.94 m
T = 163.70 tan
𝟑𝟐
2
External Distance (E)
E = 6.60 m
𝐄 = 𝐑 𝐬𝐞𝐜
𝐈
𝟐
− 𝟏
𝐄 = 𝟏𝟔𝟑. 𝟕𝟎 𝐬𝐞𝐜
𝟑𝟐
𝟐
− 𝟏
Solution:
Fundamentals of Surveying 2 4
Middle Ordinate (M)
M = 6.34 m
Long Chord (C)
Length of Curve (LC)
Lc = 91.43 m
Solution:
𝐌 = 𝐑 𝟏 − 𝐜𝐨𝐬
𝐈
𝟐
𝐌 = 𝟏𝟔𝟑. 𝟕𝟎 𝟏 − 𝐜𝐨𝐬
𝟑𝟐
𝟐
𝐂 = 𝟐𝐑 𝐬𝐢𝐧
𝐈
𝟐
𝐂 = 𝟐(𝟏𝟔𝟑. 𝟕𝟎) 𝐬𝐢𝐧
𝟑𝟐
𝟐
C = 90.24 m
Lc =
𝟐𝟎𝐈
D
Lc =
𝟐𝟎(𝟑𝟐)
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Fundamentals of Surveying 2 5
Solution:
STATION SUBTENDED ANGLE DEFLECTION ANGLE REMARKS
Sta PC 6+038
Sta 1
Sta 2
Sta 3
Sta 4
Sta 5
Sta PT
𝐝1 =
𝐜1𝐃
𝟐𝟎
Subtended Angle
c1 = (6+040) – (6+038) = 2 m
𝐝1 =
𝟐(𝟕)
𝟐𝟎
𝐝1 = 𝟎. 𝟕°
𝐝₂ =
𝐜₂𝐃
𝟐𝟎
c2 = (6+129.43) – (6+120) = 9.43 m
𝐝₂ =
𝟗. 𝟒𝟑(𝟕)
𝟐𝟎
𝐝₂ = 𝟑. 𝟑°
6+040
6+060
6+080
6+100
6+120
6+129.43
0.7⁰
0.7⁰ + 7⁰ = 7.7⁰
0.7⁰ + 2(7⁰) = 14.7⁰
0.7⁰ + 3(7⁰) = 21.7⁰
0.7⁰ + 4(7⁰) = 28.7⁰
0.7⁰ + 4(7⁰) + 3.3⁰= 32⁰
0.35⁰
3.85⁰
7.35⁰
10.85⁰
14.35⁰
16⁰
Fundamentals of Surveying 2 6
I
PC PT
PI
T
Lc
1
2
3
4
5
0.7⁰
7⁰
7⁰ 7⁰
7⁰
3.3⁰
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SAMPLE PROBLEM 2
Fundamentals of Surveying 2 7
A 5° curve intersects a property line CD at point D. The back tangent
intersects the property line at point C which is 105.72m from the PC which is
at station 2+040. The angle that the property line CD makes with the back
tangent is 110°50’. Compute the following:
(a) The length of curve from the PC to point D
(b) The stationing of point D, and
(c) The distance CD
Fundamentals of Surveying 2 8
Length of Curve (LC)
R =
1145.916
D
R =
1145.916
5
R = 229.183 m
𝐭𝐚𝐧 ∅ =
𝟏𝟎𝟓.𝟕𝟐𝟎
𝟐𝟐𝟗.𝟏𝟖𝟑
∅ = 24. 𝟕𝟔°
∝ = 180° − 90° − 24. 𝟕𝟔°
∝ = 65. 𝟐𝟒°
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Fundamentals of Surveying 2 9
Length of Curve (LC)
𝜷 = 110°𝟓𝟎′ − ∝
𝜷 = 110°𝟓𝟎′ −𝟔𝟓. 𝟐𝟒°
𝜷 = 4𝟓. 𝟓𝟗°
sin ∅ =
𝟏𝟎𝟓. 𝟕𝟐
𝑶𝑪
OC =
𝟏𝟎𝟓.𝟕𝟐
𝟐𝟒.𝟕𝟔°
OC = 𝟐𝟓𝟐. 𝟒𝟐 𝐦
Fundamentals of Surveying 2 10
Length of Curve (LC)
𝑶𝑪
sin 𝜽
=
𝟐𝟐𝟗. 𝟏𝟖𝟑
sin 𝟒𝟓. 𝟓𝟗°
𝜽 = 𝟓𝟏. 𝟖𝟗°
Considering triangle OCD:
𝟐𝟓𝟐. 𝟒𝟐
sin 𝜽
=
𝟐𝟐𝟗. 𝟏𝟖𝟑
sin 𝟒𝟓. 𝟓𝟗°
𝒙 = 𝟏𝟖𝟎° − 𝟓𝟏. 𝟖𝟗° − 𝟒𝟓. 𝟓𝟗°
𝒙 = 𝟖𝟐. 𝟓𝟐°
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Fundamentals of Surveying 2 11
Length of Curve (LC)
𝑪𝑫
sin 𝟖𝟐. 𝟓𝟐°
=
𝟐𝟐𝟗. 𝟏𝟖𝟑
sin 𝟒𝟓. 𝟓𝟗°
𝑪𝑫 = 𝟑𝟏𝟖. 𝟏𝟎 𝐦
Considering triangle OCD:
Lc = 429.12 m
Lc =
𝟐𝟎𝐈
D
Lc =
𝟐𝟎(𝟐𝟒.𝟕𝟔° 𝟖𝟐.𝟓𝟐°)
5°
Fundamentals of Surveying 2 12
Solving for Station D
Station D = Sta PC + Lc
Station D = 2+040 + 429.12
Station D = 2+469.12
Solving for Distance CD
CD = 318.10m
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