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CE 321
Civil Engineering (Bicol University)
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CE 321: SURVEYING II
PROBLEM 1 : SIMPLE CRUVES
The tangents of a simple curve have bearings of N 20° E and N 80° E respectively. The radius of
the curve is 300m.
a. Compute the external distance of the curve
b. Compute the middle ordinate of the curve
c. Compute the stationing of point A on the curve having a deflection angle of 6° from the P.C.
which is at 10+020.
FIGURE:
GIVEN:
Radius = 300m
Sta. A = 10+020
FORMULAS:
I
• E = R(sec − 1)
2
I
• M = R (1 − cos )
2
πRθ
• L=
180°
SOLUTION:
I
a. E = R(sec − 1)
2
60°
E = (300m)(sec − 1)
2
𝐄 = 𝟒𝟔. 𝟒𝟏𝟎𝐦
I
b. M = R (1 − cos )
2
60°
M = (300m)(1 − cos )
2
𝐌 = 𝟒𝟎. 𝟏𝟗𝟐𝐦
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CE 321: SURVEYING II
πRθ
c. L=
180°
π(300m)(12°)
L=
180°
L = 62.831m
Sta. A = (10 + 020) + 62.831m
𝐒𝐭𝐚. 𝐀 = 𝟏𝟎 + 𝟎𝟖𝟐. 𝟖𝟑𝟏
Notes:
✓ To compute for the Intersecting Angle(I), you must grasp the concept of angle theorems
✓ Place a cartesian plane at the vertex of the curve for better visual of how the angle works
✓ I = 80° – 20° = 60°
πRθ
✓ In the formula L = , θ is the central angle of the desired sector
180°
✓ In computing the stationing of a point, treat the + sign of that station as a comma and just use
simple addition
PROBLEM 2 : COMPOUND CURVES
The long chord from the P.C. to the P.T. of a compound curve is 300 meters long and the angle
it makes with the longer and shorter tangents are 12° and 15° respectively. If the common tangent is
parallel to the chord:
a. Find the radius of the first curve
b. Find the radius of the second curve
c. If stationing of P.C. is 10+204.30, find the stationing of P.T.
FIGURE:
ENFORCE: Empowerment through Notes and Formulas Organized to Revitalize Civil Engineers
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CE 321: SURVEYING II
GIVEN:
Long Chord = 300m
Sta. of P.C. = 10+204.30
I1 = 12°
I2 = 15°
FORMULAS:
a b c
• sin A
=
sin B
=
sin C
I
• LC = 2R sin
2
πRθ
• L=
180°
SOLUTION:
a. Consider Triangle ABC:
300 AC
=
sin 166°30′ sin 7°30′
AC = 167.739m
I
LC = 2R sin
2
12°
167.739 = 2R1 sin
2
𝐑 𝟏 = 𝟖𝟎𝟐. 𝟑𝟔𝟎𝐦
b. Consider Triangle ABC:
300 BC
=
sin 166°30′ sin 6°
BC = 134.329m
I
LC = 2R sin
2
15°
134.329 = 2R 2 sin
2
𝐑 𝟐 = 𝟓𝟏𝟒. 𝟓𝟔𝟕𝐦
πRθ
c. L = 180°
π(802.360m)(12°)
L1 =
180°
L1 = 168.046m
π(514,567m)(15°)
L2 =
180°
L1 = 134.713m
Sta. of P. T. = (10 + 204.30) + 168.046 + 134.713
𝐒𝐭𝐚. 𝐨𝐟 𝐏. 𝐓. = 𝟏𝟎 + 𝟓𝟎𝟕. 𝟎𝟓𝟗
ENFORCE: Empowerment through Notes and Formulas Organized to Revitalize Civil Engineers
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CE 321: SURVEYING II
Notes:
✓ In compound curves, it is usual to form triangles within the curve. Find a triangle that you can
work with
✓ Use sine law or cosine law in solving this kind of triangles
✓ If the common tangent is parallel to the long chord. The deflection angle of the long chord of
the first or second curve to the long chord of the whole curve is half of the intersecting angle of
that curve
PROBLEM 3 : REVERSED CURVES
Two parallel tangents 10 m. apart are connected by a reversed curve . The chord length from the
P.C. to the P.T. equals 120m.
a. Compute the length of tangent with common direction
b. Determine the equal radius of the reversed curve
FIGURE:
GIVEN:
Long chord = 120m
Distance between tangents = 10m
FORMULAS:
opp
• sin A = hyp
• T = T1 + T2
I
• T = R tan
2
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CE 321: SURVEYING II
SOLUTION:
I 10
a. sin =
2 120
I
= 4.780°
2
I = 9.560°
10
sin 9.560° =
AB
𝐀𝐁 = 𝟔𝟎. 𝟐𝟏𝟐𝐦
b. T1 = T2
T = T1 + T2
60.212 = 2T1
T1 = 30.085m
I
T = R tan
2
9.560°
30.085 = R tan
2
𝐑 = 𝟑𝟓𝟗. 𝟕𝟕𝟗𝐦
Notes:
✓ There are four types of reversed curve: equal radius and parallel tangents, unequal radius and
parallel tangents, equal radius and converging tangents, unequal radius and converging
tangents
✓ Triangles are also formed inside reversed curves, find one that you can work with
✓ If the tangents are parallel then the intersecting angle of the first and second curves are equal
PROBLEM 4 : SPIRAL CURVES
A simple curve having a radius of 180 m. connects two tangents intersecting at an angle of 50°.
It is to be replaced by another curve having 80 m. spirals at its ends such that the point of tangency shall
be the same
a. Determine the radius of the new circular curve
b. Determine the distance that the curve will nearer the vertex
c. Determine the offset from tangent at the end point of the spiral
d. Determine the distance along the tangent at the mid-point of the spiral
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CE 321: SURVEYING II
FIGURE:
GIVEN:
Radius of simple curve = 180m
Intersecting Angle = 50°
Length of spiral = 80m
FORMULAS:
Lc Xc I
• Ts = + (R c + ) tan
2 4 2
X I
• Es = (R c + 4c ) sec 2 − R c
Lc 2
• Xc = 6Rc
L5
• y =L−
40Rc 2Lc 2
SOLUTION:
Ts
a. tan 25° =
180
Ts = 83.935m
Lc Xc I
Ts = + (R c + ) tan
2 4 2
80 802 50°
83.935 = + (R c + ) tan
2 24R c 2
266.667
94.219 = R c +
Rc
𝐑 𝐜 = 𝟗𝟏. 𝟐𝟗𝟖𝐦
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CE 321: SURVEYING II
180
b. cos 25° =
OV
OV = 198.608m
E = 198.608 − 180
E = 18.608m
Xc I
Es = (R c + ) sec − R c
4 2
802 50°
Es = [91.298 + ] sec − 91.298
24(91.298) 2
Es = 3.223m
Distance = 18.608 − 3.223
𝐃𝐢𝐬𝐭𝐚𝐧𝐜𝐞 = 𝟏𝟓. 𝟑𝟖𝟓𝐦
Lc 2
c. Xc =
6Rc
802
Xc =
6(91.298)
𝐗 𝐜 = 𝟏𝟏. 𝟔𝟖𝟑𝐦
L5
d. y = L −
40Rc2 Lc 2
405
y = 40 −
40(91.2982 )(802 )
𝐲 = 𝟑𝟗. 𝟗𝟓𝟐𝐦
Notes:
✓ Spiral curves is the most complex curve in solving because of its elements, be sure to familiarize
yourself with its elements and formulas
✓ Triangles are also formed inside spiral curves, find one that you can work with
✓ Analysis of the figure is highly recommended
PROBLEM 5 : VERTICAL CURVES
A vertical summit parabolic curve has a vertical offset of 0.375 m. from the curve to the grade
tangent at sta 10+050. The curve has a slope of +4% and -2% grades intersecting at the P.I. The offset
distance of the curve at P.I. is equal to 1.5 m. If the stationing of the P.C. is at 10+000
a. Compute the required length of curve
b. Compute the horizontal distance of the vertical curve turning point from the point of intersection
of the grades
ENFORCE: Empowerment through Notes and Formulas Organized to Revitalize Civil Engineers
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CE 321: SURVEYING II
FIGURE:
GIVEN:
Vertical offset at sta. 10+050 = 0.375m
Grade 1 = +4%
Grade 2 = -2%
Vertical offset at point of intersection = 1.5m
FORMULAS:
𝐲𝟏 𝐲𝟐
• =
(𝐱𝟏 )𝟐 (𝐱𝟐 )𝟐
g1 L
• s1 =
g1 −g2
SOLUTION:
𝐲𝟏 𝐲𝟐
a. =
(𝐱𝟏 )𝟐 (𝐱𝟐 )𝟐
0.375 1.5
=
(50)2 (L⁄ )2
2
𝐋 = 𝟐𝟎𝟎𝐦
g1 L
b. s1 =
g1 −g2
(0.04)(200)
s1 =
0.04 − (−0.02)
s1 = 133.333m
x = 133.333 − 100
𝐱 = 𝟑𝟑. 𝟑𝟑𝟑𝐦
Notes:
𝐲𝟏 𝐲𝟐
✓ = is known as the square property of a parabola
(𝐱𝟏 )𝟐 (𝐱𝟐 )𝟐
✓ The formula for the highest point of the curve depends on the reference point you chose
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