FUNDAMENTAL OF SURVEYING 2
FUNSURV2
MODULE # 2
Fundamental of Surveying 2
Holy Angel University
School of Engineering and Architecture
Department of Civil Engineering
Transportation Engineering
Station 2
Horizontal Curve Part 2
Introduction
This is the continuation of horizontal curves. This section will
cover the remaining types of curves, Reverse Curve and
Spiral Curve.
Content
This chapter focuses on
➢ Reverse Curve
➢ Spiral Curve
After this chapter, the student should be able to
➢ Solve complex problem
➢ Differentiate horizontal curve design
Objectives
References
Included here are the references for the chapter. This may include various books
and e-books dated 5 years from the latest update.
Instructor: Instructor’
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Station 2 – Horizontal Curve Part 2
Continuation of Horizontal Curve
“You may place some quote here as an opening remark”
Some quotes’ author
Reverse Curve
A reverse curve is formed by two circular simple curves having a common tangent but lies on
opposite sides. The method of laying out a reversed curve is just similar as the deflection angle method
of laying out simple curves. At the point where the curve reversed in its direction is called the Point of
Reversed Curvature (P.R.C).
Instructor: Instructor’
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Station 2 – Horizontal Curve Part 2
Types of Reversed Curves
a) Reversed Curves with non-parallel tangents
b) Reversed Curves with parallel tangents
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Station 2 – Horizontal Curve Part 2
Converging Tangent
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Station 2 – Horizontal Curve Part 2
Sample Problem:
1. Two parallel tangents 12 m apart are connected by a reverse curve of equal radii. If the length
of the chord from PC to PT is 140 m, determine the total length of the reversed curve.
Solution:
First, illustrate the problem.
To solve for LTotal :
Ltotal = L1 + L2
We need to solve for the value of Radius and Central angle for
both curves.
Using the triangle inside the figure:
ϴ is also equal to I2/2
sin ϴ = 12/140
ϴ = 4.92o = I2 = 9.84o
I1 is also equal to I2, because their tangent is parallel to each
other.
I2 = I1
Therefore, both curves also have the same Radius:
R1 = R2
Using the Chord of the two curves
LCtotal =LC1 + LC2
140 = 2Rsin(I/2) + 2Rsin(I/2)
140 = 4Rsin(9.84/2)
R = 408.09 m
Solve the length of the curve for each curve. Both curves have
the same central angle and radius. therefore: L1 = L2
𝐿
𝐼
Instructor: Instructor’
=
2𝜋𝑅
360
;
𝐿 =
2𝜋𝑅
360
𝐼
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Station 2 – Horizontal Curve Part 2
Ltotal = L1 + L2
𝐿𝑡𝑜𝑡𝑎𝑙 =
𝐿𝑡𝑜𝑡𝑎𝑙 =
2𝜋𝑅
𝐼+
360
2𝜋𝑅
360
2𝜋 (815.43)
360
𝐼
(9.84) +
2𝜋 (815.43)
360
(9.84)
𝐿𝑡𝑜𝑡𝑎𝑙 = 280.08 𝑚
2. Two tangents converged at an angle of 30°. The two direction of the second tangent is due
east. The perpendicular distance of the PC from the second tangent is 116.50 m. The bearing of
the common tangent is S 40° E.
(a) Compute the central angle of the first curve
(b) If a reversed curve is to connect these two tangents, determine the common radius of the
curve.
(c) Compute the stationing of the P.T if PC is at station 10+620
Solution:
First, illustrate the problem.
a) Central angle
Relating all the given to the theory of converging curve
I1 = 20o
I2 = 50o
Instructor: Instructor’
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Station 2 – Horizontal Curve Part 2
a+b = 116.50 m
To solve for the value of a and b, we need to use the triangle in the figure. This problem said
that both curves have common radius. Therefore R1 = R2
cos 30 =
𝑅−𝑎
𝑅
𝑎 = 𝑅𝑐𝑜𝑠30 − 𝑅𝑐𝑜𝑠50
cos 50 =
Eq-1
𝑅−𝑏
𝑅
𝑏 = 𝑅 − 𝑅𝑐𝑜𝑠50
Eq - 2
Relationship of a and b
𝑎 + 𝑏 = 116.50
Eq - 3
Substitute Eq 1 and Eq 2 to Eq 3
(𝑅𝑐𝑜𝑠30 − 𝑅𝑐𝑜𝑠50) + (𝑅 − 𝑅𝑐𝑜𝑠50) = 116.50
𝑅 = 200.86 𝑚
Instructor: Instructor’
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Station 2 – Horizontal Curve Part 2
c) Stationing of PT, given stationing of PC = 10 + 620
We need to find the total length of the curve
𝐿𝑡𝑜𝑡𝑎𝑙 =
𝐿𝑡𝑜𝑡𝑎𝑙 =
2𝜋𝑅
𝐼+
360
2𝜋𝑅
360
2𝜋 (200.86)
360
𝐼
(20) +
2𝜋 (200.86)
360
(50)
𝐿𝑡𝑜𝑡𝑎𝑙 = 245.39 𝑚
𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑃𝑇 = 𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑃𝐶 + 𝐿𝑡𝑜𝑡𝑎𝑙
𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑃𝑇 = 10620 + 245.39
𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑃𝑇 = 10 + 865.39
3. In a rail road layout, the centerline of two parallel tracks are connected with a reversed curve
of unequal radii. The central angle of the first curve is 16° and the distance between parallel
tracks is 27.60 m. Stationing of the PC is 15+420 and the radius of the second curve is 290 m.
(a) Compute the length of the chord between PC to PT
(b) Compute the radius of the first curve
(c) Compute the stationing of the PT
Solution:
First, illustrate the problem
Instructor: Instructor’
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Station 2 – Horizontal Curve Part 2
a) Length of the chord from PC to PT (L1 +L2)
Using the triangle
sin 8 =
27.60
𝐿1+𝐿2
L1 + L2 =
27.60
sin 8
L1 + L2 = 198.31 𝑚
b) Radius of First Curve
Instructor: Instructor’
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Station 2 – Horizontal Curve Part 2
Relationship a and b
Eq - 1
a + b = 27.20
We need to find the value of a and b in terms of radius. We will use the
triangle and trigonometry.
cos 16 =
𝑅1−𝑎
𝑅1
a = 𝑅1 − 𝑅1 𝑐𝑜𝑠 16
cos 16 =
Eq - 2
𝑅2−𝑏
𝑅2
b = 𝑅2 − 𝑅2 𝑐𝑜𝑠 16
Instructor: Instructor’
Eq-3
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Station 2 – Horizontal Curve Part 2
Substitute Eq 2 and Eq 3 in Eq 1
(R1 − R1 cos 16) + (R2 − R2 cos 16) = 27.20
(R1 − R1 cos 16) + (290 − 290 cos 16) = 27.20
R1 = 422.47 𝑚
c) Stationing of PT
First, we need to find the total length of the curve
𝐿𝑡𝑜𝑡𝑎𝑙 =
𝐿𝑡𝑜𝑡𝑎𝑙 =
2𝜋𝑅1
360
𝐼+
2𝜋(422.4)
360
2𝜋𝑅2
360
𝐼
(16) +
2𝜋 (290)
360
(16)
𝐿𝑡𝑜𝑡𝑎𝑙 = 198.96 𝑚
𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃𝑇 = 𝑆𝑡𝑎𝑡𝑖𝑜𝑛𝑖𝑛𝑔 𝑜𝑓 𝑃𝐶 + 𝐿𝑡𝑜𝑡𝑎𝑙
𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃𝑇 = 15420 + 198.96
𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃𝑇 = 15 + 618.96
Instructor: Instructor’
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Station 2 – Horizontal Curve Part 2
Spiral Curves
Spirals are used to overcome the abrupt change in curvature and superelevation that occurs
between tangent and circular curve. The spiral curve is used to gradually change the curvature and
superelevation of the road, thus called transition curve.
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Station 2 – Horizontal Curve Part 2
Formulas:
Instructor: Instructor’
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Station 2 – Horizontal Curve Part 2
Sample Problem:
1. The radius of the central angle of a spiral easement curve is equal to 230 m. The central curve
has a central angle of 36°.
(a) Compute for the offset distance at the S.C if the external distance is 13.20 m.
(b) Compute the length of the spiral curve.
(c) Compute for the spiral angle from tangent to SC
2. An 80 m spiral easement curve has a 6° curve for its central curve.
(a) Determine the radius of the central curve.
(b) Compute the length if throw of the spiral curve.
(c) If the central angle of the central curve is 42°, compute the external distance of the central curve of
a spiral curve.
3. The radius if the interior curve of a spiral easement curve is 190 m. The central angle of the
interior curve is 42° and its external distance is 15.98 m. (a) Compute the length of throw of the spiral
curve. (b) Compute the length of the spiral. (c) Compute the max velocity in kph that a car could pass
thru the easement curve where: (Lc =0.036V3/Rc )
Instructor: Instructor’
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Station 2 – Horizontal Curve Part 2
References: This part includes online references. This is to avoid plagiarism in the content of the
modules.
Chapter 2:
Charles D. Ghilani and Paul R. Wolf, 2015. Elementray Surveying: An Introduction to Geomatics, 14 th
Edition
Jerry Nathanson, Michael T. Lanzafama, and Philip Kissam, 2011. Surveying Fundamentals and Practices,
6th Edition
, Jethi, 2015
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