COMPILED PROBLEMS IN ROUTE SURVEYING
ENGR. FRANCIS ARJAY PASTORES LUZ
First Place, CE Board May 2015
BSCE, LPU Cavite – Resident Full Academic Scholar
Review Head and Reviewer in all CE subjects,
FAL Conducive Engineering Review Center
Author of Various Civil Engineering Books
TO THE READER:
I thank God for the wisdom He gave me and to all of us. My goal is to share
this borrowed wisdom from the Lord through book writing and teaching. May He
grant the reader of this compilation of problems the divine wisdom to be able to
understand the principles and concepts written in this book.
How I wish that this compilation if free from errors. However, part of being
a human, we make mistakes that we do not notice. If you find any error in this
compilation, please feel free to tell the author.
Engr. Francis Arjay Luz
Manila, April 2020
Page | 1
ROUTE SURVEYING - CURVES
PART A – HORIZONTAL CURVES
These are the curves used in highway engineering to let the driver change its direction.
These are the curves that we can see at the top view of a highway. The common horizontal
curves are: Simple curve, Compound curve, Reverse curve and Spiral Curve.
I. SIMPLE CURVE
Figure 1. Elements of a Simple Curve
Where:R = Radius
I = Angle of Intersection / Central Angle
C = Long Chord
Lc = Length of Curve
T = Tangent Distance
M = Middle Ordinate
E = External Distance
P. C. = Point of Curve
P. I. = Point of Intersection
P. T. = Point of Tangency
Degree of Curve, D:
The angle formed at the center by 1 chain length is called the degree of curve. The
lengths of 1 chain are often 20 m, 30 m and 100 ft. However, in the Philippines, the standard
length of 1 chain is taken as 20 m.
There are two basis of the Degree of Curve:
a. Arc Basis (considering 20 m length of arc)
S = Rθ
π
20 = R(D°) ( )
180
1145.916
D=
R
b. Chord Basis (considering 20 m length of chord)
D
10
sin ( ) =
2
R
10
D = 2 sin−1 ( )
R
SETTING OUT A SIMPLE CURVE ON THE GROUND
A. Deflection angle method:
A theodolite or a transit is set-up at P.C. and the line of sight is set parallel or along the
tangent through P.C. The degree of curve is then computed and the deflection angle between two
successive stations (1 chain apart) is half of the degree of curve as shown in the figure.
B. Perpendicular offset method:
A line y is measured along the tangent distance and the corresponding perpendicular
offset is computed using the formula:
y2 + (R − x)2 = R2
x = R − √R2 − y2
II. COMPOUND CURVE
A compound curve is a combination of two or more simple curves having the same
direction of concavity. The two curves are joined at a point called the PCC (point of common
curvature). The tangent at PCC is tangent to the first curve and also to the second curve and is
called, the common tangent.
Figure 2. Elements of a Compound Curve
III. REVERSE CURVE
Reverse curves are combined simple curves having opposite direction of concavity. The
two curves are joined at a point called the PRC (point of reverse curvature). The tangent at PRC
is tangent to the first curve and also to the second curve and is called, the common tangent.
Figure 3. Elements of a Reverse Curve
IV. SPIRAL CURVES
Spiral curves are used in horizontal alignment to overcome the abrupt change in
curvature. It is also called, a transition curve. From the figure, the curve T.S. to S.C. is a spiral
curve. It is connected to a simple curve to allow transition for a safe drive towards the simple
curve. The radius of curvature of a spiral curve decreases as it approaches the simple curve.
(A)
(B)
Figure 4. Elements of a Spiral Curve
FORMULA FOR A SPIRAL CURVE
1. Offset distance from tangent to any point of the spiral:
𝐋𝟑
𝐱=
𝐋𝐜
𝟔𝐑 𝐜
At L = Lc, x = xc
𝐱𝐜
𝐋𝟐𝐜
= 𝟔𝐑𝐜
2. Distance along tangent to any point of the spiral:
𝐲 = 𝐋𝐜 −
𝟒𝟎(𝐑 𝐋𝟓
)𝟐(𝐋 )𝟐
𝐜
𝐜
3. Spiral angle from tangent to any point on the spiral (radians)
𝐋𝟐
𝐒=
𝐋𝐜
𝟐𝐑 𝐜
4. Deflection angle:
𝐒
𝐢 =
𝟑
5. Length of throw or Shift:
𝛒=
𝐱𝐜
𝟒
6. Tangent Distance for the spiral:
𝐈𝐬
𝐓𝐬 = 𝐋𝐜 + (𝐑𝐜 + 𝛒) 𝐭𝐚𝐧 ( )
𝟐
𝟐
7. External distance for the
spiral:
𝐈
𝐄𝐬 = (𝐑𝐜 + 𝛒)𝐬𝐞𝐜 ( )𝐬− 𝐑𝐜
𝟐
8. The degree of curve of a spiral.
The radius of curvature of a spiral decreases as it approaches the central curve. Thus, its
degree increases. The degree of curve increases linearly and is given by the relationship:
𝐃 𝐃𝐜
𝐋 = 𝐋𝐜
9. Desirable length for a spiral curve:
One of the geometrical properties of a spiral curve is that, the rate of change in the
centripetal acceleration is constant when a car is travelling at a constant velocity. That is,
da
=C
dt
v2/R
∫
t
da = C ∫ dt
0
0
v2 = Ct
R
The time to travel from T.S. to S.C. with constant velocity is Lc/v. Thus,
v2
Lc
=C( )
R
v
𝐯𝟑
𝐋𝐜 =
𝐂𝐑
Where
m C is a constant mrate of change in centripetal acceleration usually ranging from
0.30 per second to 0.60 per second depending on the design of the spiral. It is often taken
s2
m
s2
that C = 0.60 per second, thus
s2
Lc =
v3
0.60R
Converting v(m/s) to velocity in kph, we get:
𝟎. 𝟎𝟑𝟔𝐊𝟑
𝐋𝐜 =
𝐑
Where K is the design velocity in kph.
PROBLEM 1: The angle of intersection of a simple curve is 60° and its radius is 320 m. Find the
tangent distance, external distance and middle ordinate.
Solution:
a. Find the tangent distance.
60
T
tan ( ) =
2
320
𝐓 = 𝟏𝟖𝟒. 𝟕𝟓 𝐦
b. Find the external distance ordinate.
60
cos ( ) = 320
2
320 + E
𝐄 = 𝟒𝟗. 𝟓𝟎 𝐦
c. Find the middle ordinate.
60
cos ( ) = 320 − M
2
320
𝐌 = 𝟒𝟐. 𝟖𝟕 𝐦
PROBLEM 2: The middle ordinate of a simple curve is 2 m and its long chord is 30 m,
determine its radius.
Solution:
First Solution:
R2 = 152 + (R − 2)2
𝐑 = 𝟓𝟕. 𝟐𝟓 𝐦
Second Solution: by Chord Theorem
(15)(15) = (2)(2R − 2)
𝐑 = 𝟓𝟕. 𝟐𝟓 𝐦
PROBLEM 3: The tangents of a simple curve having a radius of 190.986 m intersect at an angle
of 46°12′. Determine the degree of curve, tangent distance and external distance.
a. D =
b.
1145.916
R
1145.916
= 190.986
I
𝐃 = 𝟔°
46°12′
T = R tan ( ) = 190.986 tan (
2
2
)
𝐓 = 𝟖𝟏. 𝟒𝟔 𝐦
c. cos (I )
=
R
R+E
2
46°12′
cos
(
2
)=
190.986
190.986 + E
𝐄 = 𝟏𝟔. 𝟔𝟓 𝐦
PROBLEM 4: The angle of intersection of a circular curve is 45°30′ and its radius is 198.17 m.
P.C. is at station 0 + 700.
a. Compute the right angle offset from Sta. 0 + 736.58 on the curve to tangent through P.C.
(Similar to CE Board Exam May 2015)
b. Compute the deflection angle of Sta. 0 + 736.58 from the tangent through P.C.
c. Determine the degree of the curve. Use chord basis and assume 20 m length of chain.
Solution:
a. The length of arc from Sta. 0+700 to Sta. 0+736.58 is 36.58 m. Thus,
S = Rθ
36.58 = 198.17(θ)
θ=
180°
36.58
)
rad.
198.17
π
(
θ = 10.576°
To solve for the perpendicular offset, x, let us draw a right triangle by drawing a line parallel to
tangent through P.C.
cos 10.576° =
198.17 − x
198.17
𝐱 = 𝟑. 𝟑𝟕 𝐦
9. The deflection angle of Sta. 0 + 736.58 from the tangent through P.C. is denoted by i in the
figure. That is,
𝐢=
𝟏𝟎. 𝟓𝟕𝟔°
= 𝟓. 𝟐𝟖𝟖°
𝟐
10. Using chord basis, a 20 m length of chain is considered. Thus,
D = 2 sin−1
(
10
) = 5.785°
198.17
𝐃 = 𝟓. 𝟕𝟖𝟓°
PROBLEM 5: The tangents of a simple curve intersect at an angle of 32°. It has a radius of
402.21 m.
a. Compute the degree of the curve using arc basis.
b. Determine the area of the fillet of the curve.
Solution:
a. The degree of the curve by arc basis is derived using a 20 m length of arc. It is given by:
1145.916 1145.916
D=
= 402.21
R
𝐃 = 𝟐. 𝟖𝟒𝟗°
b. The area of the fillet of the curve is shown in the figure
(shaded area).
T = R tan ( I
A
fillet
) = 402.21 tan 16° = 115.33 m
2
1
1
π
= 2 [ (402.21)(115.33)] − (402.21)2(32°) ( )
2
180°
2
𝐀𝐟𝐢𝐥𝐥𝐞𝐭 = 𝟏𝟐𝟏𝟏. 𝟑𝟕 𝐦𝟐
PROBLEM 6: An engineer is setting out a simple curve, having a radius of 191.073 m, along
the FALCON Highway. The instrument was set up at P.C. (Sta. 5 + 060) and deflection angles
were taken to lay out the curve. Due to obstruction in the line of sight through Sta. 5 + 120, the
engineer decided to set up at Sta. 5 + 100 and take a backsight reading through P.C. and rotated
the horizontal angle of the telescope by 180°. Determine the deflection angle of Sta. 5 + 120
from this new line of sight. Use chord basis in the computing the degree of curve.
Solution:
D = 2 sin−1 (
10
191.073
The deflection angle in every station is D = 3°
2
) = 6°
The deflection angle from the new line of sight is,
i = 6° + 3° = 𝟗°
PROBLEM 7: The bearing of two tangents has an angle of intersection of 88°30’. If the degree
of curve is 6° for a chord distance of 20m, find the tangent distance.
Solution:
Let us compute the radius of the curve by chord basis using the degree of curve equal to 6°.
6°
10
sin ( ) =
2
R
Therefore,
R = 191.073 m
88°30′
T
)=
tan (
2
191.073
𝐓 = 𝟏𝟖𝟔. 𝟏𝟑𝟓 𝐦
PROBLEM 8: The tangents of a simple curve intersect at an angle of 60°. This curve shall be
replaced with a new simple curve by rotating the tangents through the P.C. and P.T. by 10° such
that the deviation of the point of intersection of the new curve from the point of intersection of
the old curve is 10 m towards the center of the simple curve.
a. What is the radius of the old curve?
b. Determine the radius of the new simple curve.
c. Find the deviation of the midpoint of the old curve to the midpoint of the new curve.
Solution:
a. By sine law:
10
Told = (
sin 10°
) (sin 110°) = 54.115
54.115
tan 30° = R
old
𝐑𝐨𝐥𝐝 = 𝟗𝟑. 𝟕𝟑 𝐦
b. By sine law, we can get Tnew
10
Tnew = (
sin 10°
) (sin 60°) = 49.872 m
49.872
tan 20° = R
new
𝐑𝐧𝐞𝐰 = 𝟏𝟑𝟕. 𝟎𝟐 𝐦
c. The deviation of the midpoint of the old curve to the midpoint of the new curve.
cos 30 =
93.73
93.73 + E
old
Eold = 14.5 m
cos 20 =
137.02
137.02 + E
new
Enew = 8.793 m
Therefore, the deviation is:
∆= (8.793 + 10) − (14.5) = 𝟒. 𝟐𝟗𝟑 𝐦
PROBLEM 9: The perpendicular offset from a point 75 m from P.C. (measured along the
tangent through P.C.) to a simple curve is 6 m. Determine the radius of the curve.
Solution:
R2 = (R − 6)2 + 752
𝐑 = 𝟒𝟕𝟏. 𝟕𝟓 𝐦
PROBLEM 10: The tangents of a simple curve along the FALCon Highway have bearings of N
20° E and N 80° E, respectively. The radius of this curve is 45m. Compute the stationing of P.T.
if the P.C. is at station 0 + 835.16. Use chord basis. The length of 1 station is 20m.
Solution:
By cosine law:
−1
∆ = cos [
452 + 452 − 4.842
2(45)2
−1
D = 2 sin
(
] = 6.165°
10
) = 25.679°
45
ϕ = 60° − 6.165° − 2(25.679°) = 2.477°
c = √452 + 452 − 2(45)(45) cos(2.477) = 1.945
Therefore, the stationing of P.T. by chord basis is
Sta. P. T. = (0 + 880) + 1.945 = 𝟎 + 𝟖𝟖𝟏. 𝟗𝟒𝟓
PROBLEM 11: A simple curve has a radius of 120 m and a central angle of 42°24′. The curve
is to be revised such that the PT should be moved out a perpendicular distance of 12 m, without
changing the directions of the tangents and the stationing of PC. Find the length of the new
tangent distance.
Solution:
Told = 120 tan
(
42°24′
2
) = 46.545 m
12
c=
sin 42°24′
= 17.796 m
Tnew = Told + c = 𝟔𝟒. 𝟑𝟒𝟏 𝐦
PROBLEM 12: Given a compound curve with a long chord equal to 120 m. forming an angle of
10° and 16° respectively with the tangents. The common tangent is parallel to the long chord.
a. Determine the angle of intersection from of the tangents through P.C. and P.T.
b. Determine the longest radius of the compound curve.
c. Determine the shortest radius of the compound curve.
Solution:
a. Angle of intersection:
(180 − I) + I1 + I2 = 180
I = I1 + I2 = 10° + 16°
𝐈 = 𝟐𝟔°
b. From the triangle drawn, we can solve for C1 and C2 by sine law:
120
C1 = (
) (sin 8°) = 74.212 m
sin 167 °
120
C2 = (
) (sin 5°) = 46.493 m
sin 167 °
From the first curve, we see the relationship between RL and C1. By cosine law,
(74.212)2 = (RL)2 + (RL)2 − 2(RL)(RL) cos 10°
𝐑𝐋 = 𝟒𝟐𝟓. 𝟕𝟒 𝐦
c. From the first curve, we see the relationship between RS and C2. By cosine law,
(74.212)2 = (RL)2 + (RL)2 − 2(RL)(RL) cos 10°
𝐑𝐋 = 𝟒𝟐𝟓. 𝟕𝟒 𝐦
PROBLEM 13: A simple curve having a radius of 150 m and central angle of 84° is to be
replaced by a symmetrical three centered compound curve instead of a spiral curve. The radius of
the end curves of this symmetrical three centered compound curve shall be 200 m and the radius
of its center curve shall be 115 m. If P.C. (Sta. 2 + 236.14) and P.T. is to be retained, find the
new stationing of P.T. Use arc basis.
Solution: Our first goal in this problem is to get the central angle of the end curves and center
curve.
BC = 200 − 150 = 50 m
AB = 200 − 115 = 85 m
By sine law,
Ic
sin ( ) sin(180 − 42)
2
50 =
85
Icenter = 46.358°
We make use of the sum of the interior angles of a triangle to get Iend
138 +
46.358
+ Iend = 180
2
Iend = 18.821°
Therefore,
π
π
Sta. P. T. = (2 + 236.14) + 200(18.821°) (
) (2) + 115(46.358°) ( )
180
180°
°
𝐒𝐭𝐚. 𝐏. 𝐓. = 𝟐 + 𝟒𝟔𝟎. 𝟓𝟖
PROBLEM 14: A compound curve has the following data:
I1 = 28°
D1 = 3°
I2 = 31°
D2 = 4°
STA P.I. 30 + 120.5
a. Find the tangent distance of the first curve.
b. Find the stationing of PCC.
Solution:
a. Tangent distance of first curve.
R1 =
R2 =
1145.916
3
1145.916
= 381.972 m
= 286.479 m
4
𝐓𝟏 = 𝐑𝟏 𝐭𝐚𝐧 𝟏𝟒° = 𝟗𝟓. 𝟐𝟒 𝐦
b. Stationing of PCC
AB = T1 + T2 = R1 tan 14° + R2 tan 15°30′
AB = 174.68 m
By sine law,
AC
sin 31°
=
174.68
sin 121°
AC = 104.96 m
Thus, Sta. P.C. = (30 + 120.5) – 104.96 – 95.24 = 29 + 920.30
π
Sta. PCC = (29 + 920.30) + 381.972(28°) (
)
180
𝐏. 𝐂. 𝐂. = 𝟑𝟎 + 𝟏𝟎𝟔. 𝟗𝟕
PROBLEM 15: The angle of intersection of two tangents is 61°30’. A circular curve joining
these tangents is to pass through the point “A” that is 129 m. from the P.I. and on a line that
makes an angle of 16°20’ with the tangent through the P.C. Determine the radius of the curve.
Solution:
OV = R sec I
2
Solving for angle α.
180° − I
α +
+ 16°20’ + I = 180°
2
180° − 61°30′
α +
2
+ 16°20’ + 61°30’ = 180°
α = 42°55’
By sine law we get angle β,
sin β sin α
=
R
OV
sin β =
sin 42°55′ R sec 30°45′
R
β = 127°35’46.18”
ø = 180° – (42°55’ + 127°35’46.18”) = 9°29’13.82”
R
sin 42°55′
=
129
sin 9°29′13.82"
R = 532.926 m
PROBLEM 16: The P.C. of a 6° curve is inaccessible as well as invisible due to an obstacle.
Point C along the tangent AV is at station 5 + 039.97 and at a distance of 39.44 m. from the point
of intersection of the tangents. Locate the stationing of point B on the curve. Angle of
intersection between tangents is 30°.
Solution: Solving for the radius of the curve,
1145.916
R=
= 190.986 m
6
Solving for the tangent distance,
T = 190.986 tan (
30
) = 51.1745 m
2
y = 51.1745 − 39.44 = 11.7345 m
Solving for θ,
sin θ =
11.7345
190.986
θ = 3.5226°
Therefore, the stationing of B is:
π
Sta. B = 5 + 039.97 − 11.7345 + 190.986(3.5226°) (
°)
180
𝐒𝐭𝐚. 𝐁 = 𝟓 + 𝟎𝟑𝟗. 𝟗𝟖
PROBLEM 17: The azimuths of AB = 240°15’, BC = 284°30’, CD = 340°24’, and DE = 52°58’.
These four tangents shall connect a compound curve consisting of three simple curve. The
stationing of A is 7 + 520, B is 7 + 815, C is 8 + 118, D is 8 + 318, and E is 8 + 564. If the last
curve is four times sharper than the first, compute the shortest radius of the compound curve.
Solution:
Solving for the central
angles,
I1 = 284°30’ – 240°15’ = 44°15’
I2 = 340°24’ – 284°30’ = 55°54’
I3 = 360° – 340°24’ + 52°58’ = 72°34’
Solving for the tangents,
S1 = (8 + 118) – (7 + 815) = 303 m.
S1 = (8 + 318) – (8 + 118) = 200 m.
T1 + T2 = 303
R1 tan I 1 + R2 tan I2 = 303
2
R1 tan
44°15’
2
+ R2 tan
55°54’
2
= 303
2
R1 (0.406566) + R2 (0.53059) = 303
Similarly,
T2 + T3 = 200
R2 tan
55°54’
+ R3 tan
72°34’
2
= 200
2
R2 (0.53059) + R3 (0.73413) = 200
Subtracting the two equations, where R1 = 4R3 we get:
R3(4)(0.406566) + R2 (0.53059) = 303
R2 (0.53059) + R3 (0.73413) = 200
0.892134 R3 = 103
R3 = 115.45 m
PROBLEM 18: Similar to May 2016 CE BOARD EXAM. In a reverse curve, the first radius of
curvature is 200 m. and the central angle is 10°. If the perpendicular distance between the
parallel tangents is 10 m., compute the radius of the second curve.
Solution 1:
From the first curve,
cos 10°
=
200 − x
200
x = 3.03845 m
z = 10 − x = 6.96155 m
From the second curve,
R2 − 6.96155
cos 10° =
R2
𝐑𝟐 = 𝟒𝟓𝟖. 𝟐𝟑 𝐦
Solution 2:
Using the right triangle in the figure,
cos 10° =
200 + R2 − 10
200 + R2
𝐑𝟐 = 𝟒𝟓𝟖. 𝟐𝟑 𝐦
PROBLEM 19: Two diverging tangents AV and CV with an angle of intersection AVC of 38°
are to be connected by a reverse curve. The tangent distance VA to the P.C. is to be 1,630 m.,
and the radii of the branches AB and BC where B is the P.R.C. are to be 818.51 and 954.93 m.,
respectively. Determine the central angle for the branch BC.
Solution:
tan ø =
818.51
1630
ø = 26°39’49.21”
α = 38° – 26°39’49.21” = 11°20’10.79”
818.51
OV = sin 26°39’49.21” = 1823.97 m.
OT = 1823.97 sin 11°20’10.79”
OT = 358.5335 m
PM = R2 – OT
PM = 954.93 – 358.5335 = 596.396 m.
596.396
cos I2 =R1+ R2
I2 =70°21’
PROBLEM 20: Two parallel tangents 18 m. apart are to be connected by a reversed curve. If
the radius of the branch beginning at the P.C. is to be 820 meters and the total length of the chord
from the P.C. to the P.T. is to be about 250 meters, what is the required radius of the branch
ending at the P.T.?
Solution:
By similar triangles ABC and EGF,
R1 + R2 250
BC = 18
820 + R2 250
125 = 18
𝐑𝟐 = 𝟗𝟏𝟔. 𝟏𝟏 𝐦
PROBLEM 21: An 80m long spiral curve is to be connected to a central curve having a radius of
280 m. The angle of intersection between the tangents for the spiral is 60°.
a. Find the offset distance at the first quarter point of the spiral.
b. Find the spiral angle at the third quarter point of the spiral.
c. Find the deflection angle at the third quarter point of the spiral.
d. Find the length of the short tangent.
e. Find the length of the long tangent.
f. Find the tangent distance for the spiral.
g. Find the external distance for the spiral.
Solution:
a. Offset Distance at first quarter point, L = 20 m
x=
203
= 𝟎. 𝟎𝟔 𝐦
6(280)(80)
b. Spiral angle at third quarter point of the
spiral, L = 60 m
S=
602
18
2(280)(80) rad. 0 ) = 𝟒. 𝟔𝟎°
(
π
c. Deflection angle at the third quarter point
of the spiral, L = 60 m
i=
S
= 4.60 = 𝟏. 𝟓𝟑°
3
°
3
d. Length of the short tangent.
Sc =
(
802
180
) = 8.185°
2(280)(80)
π
803
xc =
Therefore,
6(280)(80)
= 3.81 m
sin(8.185°) =
3.81
ST
𝐒𝐓 = 𝟐𝟔. 𝟕𝟓 𝐦
e. Length of long tangent.
tan(8.185°) =
3.81
a
a = 26.49 m
Then,
805
yc = 80 −
40(280)2(80)2
= 79.84 m
LT = 79.84 − 26.49 = 𝟓𝟑. 𝟑𝟓 𝐦
f. Tangent Distance for the spiral:
Lc
Is
Ts = + (R + ρ) tan ( )
c
2
2
80
3.81
60
+ (280 +
Ts =
) (tan )
4
2
2
𝐓𝐬 = 𝟐𝟎𝟐. 𝟐𝟎𝟖 𝐦
g. External Distance for the spiral:
3.81
280 +
60
cos (
4
2 ) = 280 + Es
𝐄𝐬 = 𝟒𝟒. 𝟒𝟐 𝐦
PROBLEM 22: A spiral curve having a length of 100 m is connected to a simple curve having a
radius of 350 m. If the angle of intersection for the simple curve is to be 42°, what will be the
angle of intersection for the spiral curve?
COMPILED PROBLEMS IN ROUTE SURVEYING
Solution:
Sc =
(
180°
1002
) = 8.185°
2(350)(100) π
Is = 2Sc + Ic = 2(8.185°) + 42°
𝐈𝐬 = 𝟓𝟖. 𝟑𝟕°
PROBLEM 23: A 6° simple curve is
connected by two tangents having an
azimuth of 240° and 280° respectively. It is
required to replace this curve by
introducing a transition curve 80m long at
each end of a new central curve which is to
be shifted at its midpoint away from the
intersection of the tangents. Determine the
deviation of the new curve from the old
curve at their midpoint if the center of the
old curve is to be retained.
From the old curve:
I = 280° − 240° = 40°
Rold =
1145.916
= 190.986 m
6
cos (
40°
2
)=
190.986
190.986 + E
E = 12.257 m
From the spiral curve,
cos (
But ρ =
xc
4
=
(Lc)3
6RcLc
4
=
(Lc)
2
24Rc
40°
2
)=
190.986
R2 + ρ + 12.257
COMPILED PROBLEMS IN ROUTE SURVEYING
cos (
40°
2
)
=
2
190.986
80
R+ 2+
12.257
24R2
R2 = 189.579 m
Finally, the deviation is
ρ=
802
= 𝟏. 𝟒𝟏 𝐦
24(189.579)
PROBLEM 24: A circular curve of 610 m radius deflects through an angle of 40°30'. This curve
is to be replaced by one of smaller radius so as to admit transitions 110 m long at each end. The
deviation of the new curve from the old at their midpoint is 0.50 m towards the intersection
point. Determine the amended radius assuming the shift can be calculated with sufficient
accuracy on the old radius.
Solution:
From the old curve,
cos 20°15′ =
610
610 + E
E = 40.1872 m
From the spiral curve,
Es = 40.1872 − 0.50 = 39.6872 m
cos 20°15′ =
R2 + ρ
R2 + 39.6872
cos 20°15′ =
2
R2 + 80
24R2
R2 + 39.6872
𝐑𝟐 = 𝟓𝟗𝟓. 𝟏𝟔 𝐦
PROBLEM 24: A reverse curve of common radius is constructed along the FALCON Highway.
The tangents at the P.C. and P.T. intersect at an angle of 90°. The common tangent is 800 m
long. If the central angle of the first curve is 40°, determine the stationing of P.T. if the P.C. is at
Sta. 8
+ 980. Use arc basis.
Solution:
tan 20° = x
R
and
800 − x
tan 65° =
Thus,
R
800
tan 65° =
R
− tan 20°
800
R=
tan 65° + tan 20°
Therefore, the stationing of P.T. is
= 318.9185911 m
π
Sta. P. T. = (8 + 980) + 318.9185911(40°) π
)
) + 318.9185911(130°)
(
180
180°
(
°
𝐒𝐭𝐚. 𝐏. 𝐓. = 𝟗 + 𝟗𝟐𝟔. 𝟐𝟓 𝐦
PROBLEM 25: What is the maximum velocity that a car could pass thru an easement
curve having a length of spiral equal to 120 m. if the radius of the central
curve is 260 m.?
Solution:
Lc =
0.036K3
R
c
140 =
0.036K3
260
𝐊 = 𝟗𝟓. 𝟑𝟒𝐤𝐩𝐡
PROBLEM 26: Given lines AB, BC and CD. A reversed curve is to connect these lines thus
forming the centerline of the new road. Find the length of the common radius of the reverse curve.
LINES
AB
BC
CD
BEARING
Due East
N68°E
S48°E
DISTANCES
57.60 m.
91.50 m.
109.20 m.
Solution:
T1 = R tan 11°
T2 = R tan 32°
T1 + T2 = BC
Thus,
R tan 11° + R tan 32° = 91.50
𝐑 = 𝟏𝟏𝟏. 𝟔𝟗𝐦
PROBLEM 27: A reversed curve with diverging tangents is to pass through three lines to form
the centerline of a proposed road. The first line AB has a bearing of N88°E and a distance of 120
m., BC has a bearing of N62°E, and a distance of 340 m., while that of CD has a bearing of
S40°E and a distance of 230 m. Compute the total length of the reversed curve from P.C. to P.T.
if the first tangent has a distance of only ¼ that of the common tangent measured from the point
of intersection of the first curve.
Solution:
1
T1 = (340) = 85m = R1 tan 13°
4
85 = R1 tan 13°
R1 = 368.18m
T2 = 340 − 85 = 255m = R2 tan 39°
255 = R2 tan 39°
R2 = 314.90m
Lc1 =
R 1I1π
180°
=
2I2π
Lc2 = R180°
=
Total length of curve:
368.18(26°)π
180°
314.90(78°)π
180°
= 167.07m
= 428.69m
Lc = Lc1 + Lc2 = 167.07 + 428.69 = 𝟓𝟗𝟓. 𝟕𝟔𝐦
PROBLEM 28: Given in a compound curve, I1 = 28°, I2 = 34°, D1 = 4°, D2 = 5° and P.C.
is at station 10 + 480.14. What is the stationing of P.T.? Use arc basis.
Solution:
20I1 20(28)
= 140m
D1 =
4
20I2 20(34)
=
= 136m
Lc2 =
5
D2
Lc1 =
Stationing of P.T. = (10 + 480.14) + 140 + 136
Stationing of P.T. = 10+756.14
PROBLEM 29: A 3° curve has a middle ordinate of 8.35 m. What is the central angle?
Use chord basis.
Solution:
sin
D
2
=
10
R
sin 1.5° =
10
R
R = 382.02m
I
M = R − R cos
2
I
8.35 = 382.02 − 382.02 cos ( )
2
I
= 12°
2
𝐈 = 𝟐𝟒°
PROBLEM 30: The azimuth of the back tangent of a simple curve is 200°30’, while that of the
forward tangent is 240°30’. If the external distance of the simple curve that connects these two
tangents is 14.20 m., compute the length of curve. Use arc basis.
Solution:
cos 20° =
R
R+E
R
cos 20° =
R + 14.20
R = 221.26 m
Lc =
221.26(40°)π
180°
= 𝟏𝟓𝟒. 𝟗𝟎𝐦
PROBLEM 31: Determine the maximum velocity that a car could pass through a spiral easement
curve having a length of 100 m if the design constant rate of change in the centripetal acceleration
is 0.5 m per second. The radius of the central curve equal to 320 m.
s2
Solution:
One of the geometrical properties of a spiral curve is that, the rate of change in the centripetal
acceleration is constant when a car is travelling at a constant velocity. That is,
da
=C
dt
v2/R
∫
t
da = C ∫ dt
0
0
v2
= Ct
R
The time to travel from T.S. to S.C. with constant velocity is Lc/v. Thus,
v2
Lc
=C( )
R
v
3
Lc = v
CR
100 =
v3
0.5(320)
𝐦
𝐯 = 𝟐𝟓. 𝟏𝟗𝟖
𝐨𝐫 𝟗𝟎. 𝟕𝟏 𝐤𝐩𝐡
𝐬
PROBLEM 32: A spiral curve has length of 100 m and the angle of intersection of its tangents
through T.S. and S.T. is 43°30′. It is connected to simple curve having a radius of 300 m. If this
curve should be replaced by a simple curve having a larger radius with T.S. as PC and S.T. as
PT, determine the deviation of the new curve from the old curve at their midpoint.
Solution: From the spiral curve,
But,
I
R+ρ
cos ( ) =
2
R + Es
(Lc)2
xc
Therefore,
ρ= =
4
24R
cos (43°30′
2
)=
= 1.3889 m
300 + 1.3889
300 + E
s
Es = 21.489 m
Lc
I
T = + (R + ρ) tan ( ) = 160.24 m
s
2
2
From the new simple curve of larger radius, T = Ts
tan (
43°30′
2
)
=
160.24
Rnew
Rnew = 401.65 m
The middle ordinate of the new simple curve of larger radius is,
43°30′
cos (
) = 401.65
2
401.65 + E
E = 30.785 m
Therefore, the deviation of the midpoint is = E − Es = 𝟗. 𝟐𝟗𝟔 𝐦
PROBLEM 33: A 3° curve has a middle ordinate of 8.35 m. What is the central angle?
Use chord basis.
Solution:
D
sin
2
10
=
R
10
sin 1.5° =
R
R = 382.02m
I
M = R − R cos
2
I
8.35 = 382.02 − 382.02 cos
2
I
= 12°
2
𝐈 = 𝟐𝟒°
PROBLEM 34: The intermediate tangent of a reverse curve is 500 m. long. The parallel
tangents are 300 m apart. Determine the central angle of the reverse curve if it has a common
radius of 1000 m.
Solution:
x = √2502 + 10002 = 1030.7764 m
θ = tan−1 (
250
) = 14.036°
1000
cos(I + 14.036°) =
1000 − 150
1030.7764
𝐈 = 𝟐𝟎. 𝟒𝟏𝟒°
PROBLEM 35: A reversed curve is to connect two tangents which are parallel to each other and
are 200 m. apart with directions of due east. There is an intermediate tangent of 200 m. in
between the reversed curve and the horizontal distance of the P.C. and P.T. measured parallel to
the tangents is 800 m. long. The P.C. of the reversed curve is on the upper tangent while the P.T.
of the reversed curve is at the lower tangent. Compute the common radius of the reversed curve.
Solution:
x2 = R2 + 1002
Thus,
x2 = 4002 + (R − 100)2
R2 + (100)2 = (R − 100)2 + (400)2
𝐑 = 𝟖𝟎𝟎𝐦
PART B – VERTICAL CURVES
These are the curve that we can see in the longitudinal section of a road. A road may
seem to be straight in aerial view but if we are going to look at its longitudinal view, we can see
a parabolic curve as shown in the following figure.
Parabolic Curve
It is very important in the study of vertical curves used in route surveying to be familiar with the
properties of a parabola. In this topic, we will discuss the geometry of a parabola and derive
some of the useful formula.
y = kx2 (parabola either concave upward or downward)
If we get the first derivative of the equation above, we can get the equation for the slope or grade
at any point. The equation of the grade shows a line.
y′ = 2k ∙ x (linear)
If we again get the first derivative of the grade, we can get the rate of change in grade of a
parabola.
(y′)′ = y′′ = 2k
(constant)
It shows that the rate of change in grade of a parabola is constant. This is the reason why
parabola is the best curve that fits for a vertical curve so that the rate of change in the direction of
the line of sight is constant.
Also, if we have a parabolic curve and we choose two points, x distance apart on the curve, and
draw the tangents at those points, the two tangents will intersect halfway between the two points
as shown in the figure.
A vertical curve implemented in the design of highways for vertical alignment, therefore, is just
a portion of a parabola.
I. Symmetrical Parabolic Curve
It is called symmetrical parabolic curve because the intersection of the two tangents of the
parabola cuts the curve into two equal lengths. It is very important for the student to note that the
length of curve is measured on the horizontal plane. It the case of horizontal curves, the length of
curve is measured along the curve because the curve is already lying on the horizontal plane.
However, in the case of vertical curves, the length of curve is measured horizontally and not on
the curve.
Where:
P.C. = Point of Curve
P.T. = Point of Tangency
V.P.I. = Vertical Point of Intersection
VCL = Vertical Curve Length
L
= Length
If we draw the grade diagram for a parabola, we can see a line as we have proved to be the first
derivative of the parabola. At P.C., the grade is g1, while at P.T., the grade is g2. If we connect
the two points, the curve is just a line.
At x distance from the P.C., the grade is zero. Thus, this point is the highest point on the curve.
In the case of a sag curve, this point is the lowest point on the curve. By similar triangles, we can
get the location of the lowest point or highest point on the curve.
x
=
g1 g
Thus,
L
1 − g2
g1L (location of H. P. or L. P. from P. C. )
x = g1 − g2
We have learned that the slope of the curve is the equation of the grade at any point. Therefore,
dy
= grade
dx
dy = (grade) ∙ dx
y2
∫
x2
dy = ∫ (grade) ∙ dx
y1
x1
Thus, the difference in elevation between two points on the curve is equal to the area under the
grade diagram between those two points.
y2 − y1 = (Area)under the grade diagram from x1 to x2 = (∆Elevation)
The figure below illustrates the area under the curve.
VERTICAL OFFSET
The vertical offset of a point on the curve to the point of intersection is denoted by H.
By Squared Property of a Parabola:
H = z
L 2 L2
( )
2
L
z = 4H = g1 ( ) − g2 (L)
2
2
Thus,
𝐋
𝐇=
𝟖
(𝐠𝟏 − 𝐠𝟐)
II. Unsymmetrical Parabolic Curve
It is a combination of two or more parabolic curves.
PROBLEM 1: A vertical parabolic curve has a forward tangent of 2% and a backward tangent
of -4% intersecting at Sta. 2 + 160. Locate the stationing of the lowest point of the curve if the
length of curve is 240 m.
Solution:
By similar triangles:
x
= 240 − x
0.04
0.02
x = 160 m
Sta. Lowest Point
L
(
)
= Sta. VPI − + x = 2 + 160 – 120 + 160
2
Sta. Lowest Point = Sta. (2 + 200)
PROBLEM 2: Determine the grade at the third quarter point of a vertical parabolic curve having
grades of g1 = −4% and g2 = 6%.
Solution:
From the grade diagram, we may use similar triangles
g + 0.04 0.06 + 0.04
=
0.75L
L
𝐠 = 𝟎. 𝟎𝟑𝟓 𝐨𝐫 𝟑. 𝟓%
PROBLEM 3: A parabolic sag curve has a length of 180 m. The tangents are −2% and 4%
respectively. Determine the vertical offset at the first quarter point.
Solution: Let us first determine the vertical offset at the midpoint of the curve.
L
180
(−0.02 − 0.04)
H = (g1 − g2) = 8
8
H = −1.35 m
By squared property of parabola,
y = −1.35
2
L 2
L
( )
( )
4
2
𝐲 = −𝟎. 𝟑𝟑𝟕𝟓 𝐦
PROBLEM 4: A 6% downgrade meets a 4% upgrade at elevation 102 m. at station 0 + 100. If the
length of the parabolic curve is 200 m. long,
a. Find the stationing of the lowest point of the curve.
b. Find the elevation of the lowest point of the curve.
c. Find the rate of change in grade per 20m length.
Solution:
a. Sta. P.C. = (0 + 100) – 100 = 0 + 000
By similar triangle:
x
= 200 − x
0.06
0.04
x = 120 m
Therefore,
Sta. Lowest Point = (0 + 000) + 120 = 𝟎 + 𝟏𝟐𝟎
b. Elev. Of P.C. = 102 + 0.06(100) = 108 m
1
Elev. of L. P. = 108 + ) (120)(−0.06) = 𝟏𝟎𝟒. 𝟒 𝐦
2
(
c. The rate of change in grade is constant for a parabolic curve,
4% −
(−6%)
R=
x 20 = 𝟏% 𝐩𝐞𝐫 𝟐𝟎 𝐦
200m
PROBLEM 5: Below are the data from a level note:
Station
0+000
0+050
0+100
0+150
0+200
Elevation
100.00 m
100.58 m
98.00 m
101.50 m
102.00 m
A vertical parabolic sag curve is to be constructed so that the point of tangencies are at station
0+000 and 0+200 respectively. If the intersection of the two tangents lies on the natural ground,
determine the elevation of the lowest point of the curve.
Solution:
g1 =
98 − 100
g2 =
100
102 − 98
100
x 100% = −2%
x 100% = 4%
By similar triangles,
x
= 200 − x
0.02
0.04
x = 66.667 m
1
Elev. Lowest Point = 100 + ( ) (66.667)(−0.02)
2
𝐄𝐥𝐞𝐯. 𝐋𝐨𝐰𝐞𝐬𝐭 𝐏𝐨𝐢𝐧𝐭 = 𝟗𝟗. 𝟑𝟑 𝐦
PROBLEM 6: A symmetrical parabolic curve has a length of 120 m. The highest point of curve
is located at a distance of 70.59 m. from the P.C. If the back tangent of the curve is +2%,
compute the grade of the forward tangent.
Solution:
g 1L
x = g1 − g2
0.02(120)
70.59 = 0.02 − g
2
g2 = −0.014 = −𝟏. 𝟒%
PROBLEM 7: Two tangents are connected with a vertical parabolic curve passing thru P of
elevation 20.175 m. at station 1 + 020. If the forward tangent is +2% grade and the backward
tangent is +7%, and the vertex of elevation 20.00 m. is at station 1 + 000, find the total length of
the curve.
Solution:
Elevation of P′ = 20 + 20(0.07) = 21.40m
y = 21.40 − 20.175 = 1.225m
y
0.05x
=
2
(x + 20)
(2x)2
1.225
(shift solve)
(x +
20)2
0.05x
(2x)2
x = 50m
=
𝐋 = 𝟐𝐱 = 𝟏𝟎𝟎𝐦 (𝐋𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝐭𝐡𝐞 𝐜𝐮𝐫𝐯𝐞)
PROBLEM 8: A parabolic curve has an upgrade of +4% and a downgrade of -3% intersecting at
station 10+020 at an elevation of 120.16 m. If the length of the parabolic curve is 600 m. long,
compute the distance from the P.C. to the highest point of curve.
Solution:
g1L = 0.04(600)
x = g1 + g 2
0.04 + 0.03
𝐱 = 𝟑𝟒𝟐. 𝟖𝟔 𝐦
PROBLEM 9: An unsymmetrical parabolic curve has a forward tangent of -10% and a back
tangent of +6%. The length of the curve on the left side (P.C.) is 60 m., If the vertex is at
elevation of 100 m, and the rate of change in grade of the first curve is -2% per 20m length,
a. Determine the elevation of P.C.
b. Determine the common grade.
c. Find the elevation of P.T.
d. Find the length of the second curve.
Solution:
a. The elevation of P.C.
Elev. P. C. = 100 − 0.06(60) = 𝟗𝟔. 𝟒 𝐦
b. The common grade:
Rate of change in grade, R1 = g3 − g1 x 20
L1
−2% g3 − 6%
R1 =20 m = 60 m
𝐠𝟑 = 𝟎%
c. The elevation of P.T.
Note that the common grade of either a symmetrical and unsymmetrical parabolic curve is
parallel to the chord from P.C. to P.T. Thus, if g3 = 0, P.C. would have the same elevation as the
P.T.
𝐄𝐥𝐞𝐯. 𝐏. 𝐓. = 𝟗𝟔. 𝟒 𝐦
d. Length of the second curve:
We make use of the elevation of PT to get the length of the curve 2.
Elev. PT. = 100 − 0.1(L2) = 96.4
𝐋𝟐 = 𝟑𝟔 𝐦
PROBLEM 10: An unsymmetrical parabolic curve is to be constructed with a backward tangent
of +6% and a forward tangent of +4%. The common grade is -2%. The first curve has a length of
160m. The desired rate of change in grade for the second curve is 0.6% per 20m.
a. Find the rate of change in grade of the first curve per 20m length.
b. Find the required length of the second curve.
c. If the point of inflection is at elevation 102.3 m, find the elevation of the highest point on the
curve.
d. If the point of inflection is at elevation 102.3 m, find the elevation of the lowest point on the
curve.
Solution:
a. Rate of change in grade of curve 1.
R1 =
−2% − 6%
x 20 = −𝟏% 𝐩𝐞𝐫 𝟐𝟎 𝐦
160 m
Note that the rate of change in grade of a parabolic curve is the slope of the grade diagram.
b. Length of the second curve.
0.6% 4% − (−2%)
=
20m
L2
𝐋𝟐 = 𝟐𝟎𝟎 𝐦
c. The highest point on the curve is at x distance to the left of the point of inflection.
x
= 160 − x
0.02
0.06
Thus,
x = 40 m
1
Elev. Highest Pt. = 102.3 + ( ) (40)(0.02) = 𝟏𝟎𝟐. 𝟕 𝐦
2
d. The lowest point on the curve is at z distance to the right of the point of inflection.
z
= 200 − z
0.02
0.04
Thus,
z = 66.667 m
1
Elev. Lowest Pt. = 102.3 − ( ) (66.667)(0.02) = 𝟏𝟎𝟏. 𝟔𝟑 𝐦
2
PROBLEM 11: An unsymmetrical parabolic curve has a forward tangent of -8% and a back
tangent of +5%. The length of curve on the left side of the curve is 40 m. long while that of the
right side is 60 m. long. The P.C. is at station 5 + 280.
a. Compute the difference in elevation between P.C. and P.T.
b. Find the common grade.
c. Compute the stationing of the highest point of the curve.
Solution:
a. Difference in elevation
z = 0.08(60) − 0.05(40)
𝐳 = 𝟐. 𝟖𝟎 𝐦
b. Common grade
−2.80
g3 =
(100%)
40 + 60
𝐠𝟑 = −𝟐. 𝟖%
c. Stationing of the highest point:
The highest point of the curve is at curve 1 since the common grade is negative. From a positive
grade to a negative grade means the highest point is somewhere along that curve. From the grade
diagram,
x
= 40 − x
0.05 0.028
x = 25.64 m
Sta. H. P. = (5 + 280) + 25.64
𝐒𝐭𝐚. 𝐇. 𝐏. = 𝟓 + 𝟑𝟎𝟓. 𝟔𝟒
PROBLEM 12: A forward tangent having a slope of -4% intersects the back tangent having a
slope +7% at station 6 + 300. The length of curve on the side of the back tangent is 60 m. The
vertical distance from the curve at the point of intersection of the tangent is 1.32 m. Compute the
total length of curve.
Solution:
1.32 = 4H1
60
1.32 = 4 [ (0.07 − g3)]
8
g3 = 0.026 or + 2.60%
On the second curve,
1.32 = 4H2
1.32 = 4 [
L2
8
(0.026 − (−0.04))]
L2 = 40 m
Therefore, the total length is:
LT = 60 + 40 = 𝟏𝟎𝟎 𝐦
Another Solution:
L1L2(g1 − g2)
H = 2(L1 + L2)
1.32 =
60(L2)(0.07 + 0.04)
2(60 + L2)
L2 = 40m
Total length of curve = 60 +
40
Total length of curve = 100m
PROBLEM 13: An unsymmetrical parabolic curve has a back tangent of +7% and a forward
tangent of -4% intersecting at station 5 + 060. The length of curve on the side on the back
tangent is 60 m., while that on the right side on the forward tangent it is 40 m. long. Compute the
stationing of the highest point of curve.
Solution:
H=
H=
L1L2(g1 − g2)
2(L1 + L2)
60(40)(0.07 + 0.04)
2(60 + 40)
H = 1.32 m
However, from the first curve, H = 4H1. Thus, H1 = 0.33
60
(0.07 − g3)
8
g3 = 0.026 or 2.6%
0.33 =
Since the common grade is still positive, the highest point is on the right side of the curve.
x
= 40 − x
0.026
0.04
x = 15.76 m
Therefore, the stationing of the highest point is 5 + 060 +15.76
𝐒𝐭𝐚. 𝐇. 𝐏. = 𝟓 + 𝟎𝟕𝟓. 𝟕𝟔
PROBLEM 14: An unsymmetrical parabolic curve has a forward tangent of +6% and a back
tangent of -3% intersecting at station 10 + 020. The length of curve on the left side of the curve
is 200., while that on the right side is 100 m. If the elevation of the P.C. is 100 m., compute the
elevation of the curve at station 9 + 940.
Solution:
L1L2(g1 − g2)
H=
H=
2(L1 + L2)
200(100)(−0.03 − 0.06)
2(200 + 100)
H = −3m
y
x2
=
y
(120)2
H
(L1 )2
=
3
(200)2
y = 1.08m
Elevation of A = 100 – 0.03(120)
Elevation of A = 96.4m
Elevation of B = 96.4 + 1.08
Elevation of B = 97.48m
PROBLEM 16: A descending grade of -6% intersects another descending grade of -2% at
station 9 + 100. The two grade lines are to be connected by a 160 m. vertical parabolic curve
which starts at station 9 + 020. From the given tabulated data, compute the amount of cut or fill
at station 9 + 080.
Elev. of P.C. = 105.37 m.
Station Ground Elevation Station
9 + 020
105.37
9 + 120
9 + 040
102.48
9 + 140
9 + 060
102.50
9 + 160
9 + 080
103.18
9 + 180
9 + 100
103.80
Ground Elevation
100.57
98.36
99.00
98.25
Solution:
Elevation of A = 105.37 – 60 (0.06)
Elevation of A = 101.77m
L
H = (g1 − g2)
8
160
(−0.06 + 0.02)
H=
8
H =−0.80m
By squared property of parabola
y
0.80
=
(60)2 (80)2
y = 0.45m
Elevation of B = 101.77 + 0.45
Elevation of B = 102.22m
Depth of cut = 103.18 – 102.22
Depth of cut = 0.96m
TRAIN YOURSELF!
1. The intermediate tangent of a reversed curve is 600m long. The tangent of the reverse curve
has a distance of 300m which are parallel to each other. Determine the central angle of the
reverse curve if it has a common radius of 1000m.
a. 𝟏𝟖°𝟒𝟖′
b. 19°48′
c. 21°38′
d. 20°48′
2. Two tangents intersecting at V with bearings N75⁰12’E and S78⁰36’E are connected with a 4⁰
simple curve. Without changing the direction of the two tangents and with the same angle of
intersection, it is required to shorten the curve to 100m starting from P.C. By how much shall the
P.T. be moved?
a. 28.56 m
b. 9.86 m
c. 15.76 m
d. 30.73 m
3. The deflection angles of two intermediate points A and B of a simple curve are 6°15’ and
11°15’ respectively from the tangents thru the P.C. If the cord distance between A and B is 20
m., determine the radius of the simple curve.
a. 114.74 m.
b. 122.64 m.
c. 136.72 m.
d. 142.44 m.
4. The bearing of the back tangent of a simple curve is N70⁰E, while the forward tangent has a
bearing of S82⁰30’E. The degree of curve is 4.5⁰. Stationing P.C. is at 10+345.43. It is proposed
to decrease the central angle by changing the direction of the forward tangent by an angle of 7⁰,
in such a way that the position of the P.T. of the forward tangent and the direction of the back
tangent shall remain unchanged. Determine the new radius of the curve.
a. 454.35 m
b. 652.13 m
c. 756.23 m
d. 563.12 m
5. Compute the desirable length of a spiral curve for a velocity of 80 kph if the degree of central
curve is 6°.
a. 96.51 m
b. 110.25 m
c. 87.80 m
d. 121.60 m
6. The tangents of a compound curve intersect at an angle of 104°. The first curve has a radius of
420 m. and a central angle of 38°. If the common tangent is 200 m. long, what is the radius of the
second curve?
a. 85.28 m
b. 92.4 m
c. 79.8 m
d. 69.64 m
7. The perpendicular distance between two parallel tangents of a reversed curve is 7.5m and the
length of the long chord is equal to 65m. Compute the common radius of the reversed curve.
a. 156.20
b. 160.40
c. 140.87
d. 149.20
8. A reverse curve connects two parallel tangents 8 m apart. If the central angle is 10° determine
the common radius.
a. 234.84
c. 321.543
b. 266.67
d. 211.453
9. A reversed curve is to connect two tangents which are parallel to each other and are 200 m.
apart with directions of due east. There is an intermediate tangent of 200 m. in between the
reversed curve and the horizontal distance of the P.C. and P.T. measured parallel to the tangents
is 800 m. long. The P.C. of the reversed curve is on the upper tangent while the P.T. of the
reversed curve is at the lower tangent. Compute the common radius of the reversed curve.
a. 800m
b. 780m
c. 820m
d. 810m
10. Two parallel tangents have directions of due east and are 200 m. apart, are connected by a
reversed curve having equal radius of 800 m. The P.C. of the curve is on the upper tangent while
the P.T. is at the lower tangent. If the horizontal distance parallel to the tangent from the P.C. to
the P.T. of the reversed curve is 800 m., compute the distance of the intermediate tangent
between the curves.
a. 210m
b. 190m
c. 200m
d. 220m
11. Two tangents having directions of due east and azimuth of 291°30’ respectively meet at point
A is to be connected by a reversed curve. The P.C. of the first curve is located at a distance of
200
m. from A along the tangent line which has an azimuth of 291°30’ and has a radius of 330 m.
long. The radius of the second curve whose P.T. lies along the other tangent line is equal to 240
m. Compute the central angle of each curve.
a. 55°17’
b. 35°17’
c. 25°17’
d. 15°17’
12. Compute the degree of the curve of a simple curve having a perpendicular offset of 63.46 m.
from the tangent, with a tangent distance of 264.35 m. from the P.C. to the point where the
perpendicular offset is located. Use Arc Basis.
a. 1°58’
c. 2°58’
b. 3°58’
d. 6°58’
13. The intermediate tangent of a reversed curve is 600m long. The tangent of the reverse curve
has a distance of 300m which are parallel to each other. Determine the central angle of the
reverse curve if it has a common radius of 1000m.
a. 19°48’
b. 15°48’
c. 18°48’
d. 28°48’
14. A long chord from the P.C. to P.T. of a compound curve is 150 m. long and the angle it
makes with the longer and shorter tangents are 6° and 10° respectively. Find the radius of the
first curve so that the common tangents will be parallel to the long chord.
a. 687.42 m.
b. 778.42 m.
c. 586.46 m.
d. 897.47 m.
15. The deflection angles of two intermediate points A and B of a simple curve are 6°15’ and
11°15’ respectively from the tangents thru the P.C. If the cord distance between A and B is 20
m., determine the radius of the simple curve.
a. 114.74 m.
b. 122.64 m.
c. 136.72 m.
d. 142.44 m.
16. A spiral having a length of 80 m., has a central curve having a radius of 280 m. Determine
the offset distance from the tangent to the first quarter point of the spiral.
a. 0.02
b. 0.06
c. 0.04
d. 0.08
17. The length of the throw of a spiral is 1.42 m. What is the corresponding offset distance at the
S.C.?
a. 3.42 m
b. 5.68
c. 4.48
d. 2.92
18. What is the appropriate radius of the central curve if an 80 m. spiral easement curve has a
distance along the tangent of 59.46 m. at its third quarter point.
a. 75m
b. 68
c. 84
d. 59
19. A spiral curve has an external distance of 16.6 m. and a length of throw of 1.6 m. Determine
the radius of the central curve if the angle of intersection of the spiral tangent is 42°.
a. 194 m.
b. 209.92 m.
c. 280.64 m.
d. 265.64 m.
20. An unsymmetrical parabolic curve has a forward tangent of -8% and a back tangent of +5%.
The length of the curve on the left side (P.C.) is 60 m., what is the elevation of P.C. if the vertex
is at elevation of 100 m.
a. 88 m.
b. 96 m.
c. 89 m.
d. 97 m.
21. A vertical parabolic curve has an approach grade of +3.6%. The curve is to be 260 m. long
and is to have a highest point at a distance of 120 m. from the P.C. Compute the grade of the
forward tangent.
a. 4.2%
b. -4.2%
c. -3.8%
d. -5.4%
22. A grade of -5% meets a grade of 3% at vertex V. It is required to connect those grade lines
with a vertical parabolic curve that shall pass 0.80 m. directly above the vertex. Determine the
length of this curve.
a. 100 m.
b. 80 m.
c. 110 m.
d. 90 m.
23. A vertical highway curve has an approach grade of +8% meeting -6% grades at the vertex. It
is required to connect the two tangents with unsymmetrical vertical parabolic curve. The curve
has an offset of 1.50 m. directly below the intersection of the tangents. Locate the position of the
summit from the P.T. The vertical curve has a length of 120 m. on the side of the P.C. and 90 m.
from the P.T.
a. 60 m.
b. 50 m.
c. 30 m.
d. 45 m.
24. A grade descending at the rate of -4% intersects another grade ascending at the rate of +8%
at station 2 + 000, elevation 100 m. A vertical curve is to connect the two such that the curve will
clear a boulder located at station 1 + 980, elevation 101.34 m. Determine the necessary length of
the curve.
a. 80 m
b. 100 m
c. 120 m
d. 160 m
25. A 6% downgrade meets a 3% upgrade at elevation 100 m. at station 10 + 000. If the length of
the parabolic curve is 160 m. long, find the stationing of the lowest point of the curve.
a. 9 + 946.67
c. 10 + 042.34
b. 9 + 834.22
d. 10 + 026.67
26. The centerline tangents of a proposed street makes an intersection of 47°20’R. From the P.I.
the center of a sanitary sewer manhole is located 9°36’R of the forward tangent at a distance of
272.25 m. from the P.I. Determine the radius of the centerline curve that will pass through the
center of the manhole.
a. 2,525.46m
b. 2,515.46m
c. 2,535.46
d. 2,585.46m
27. An unsymmetrical parabolic curve has a forward tangent of -8% and a back tangent of +5%.
The length of curve on the left side of the curve is 40 m. long while that of the right side is 60 m.
long. The P.C. is at station 10 + 040. Compute the stationing of the highest point of the curve.
a. 10+075.64
b. 10+065.64
c. 10+085.64
d. 10+045.64
28. A forward tangent having a slope of -4% intersects the back tangent having a slope +7% at
station 6 + 300. The length of curve on the side of the back tangent is 60 m. The vertical distance
from the curve at the point of intersection of the tangent is 1.32 m. Compute the total length of
curve.
a. 100m
c. 90m
b. 120m
d. 130m
29. An unsymmetrical parabolic curve has a back tangent of +7% and a forward tangent of -4%
intersecting at station 10 + 060. The length of curve on the side on the back tangent is 60 m.,
while that on the right side on the forward tangent it is 40 m. long. Compute the stationing of the
highest point of curve.
a. 10+075.76
b. 10+065.76
c. 10+055.76
d. 10+095.76
30. An unsymmetrical parabolic curve has a forward tangent of +6% and a back tangent of -3%
intersecting at station 10 + 020. The length of curve on the left side of the curve is 200., while
that on the right side is 100 m. If the elevation of the P.C. is 100 m., compute the elevation of the
curve at station 9 + 940.
a. 98.48m
c. 95.48m
b. 94.48m
d. 97.48m
31. A symmetrical parabolic curve has a length of 120 m. The highest point of curve is located at
a distance of 70.59 m. from the P.C. If the back tangent of the curve is +2%, compute the grade
of the forward tangent.
a. -1.6%
b. -1.2%
c. -1,4%
d. -1.8%
32. A symmetrical parabolic curve has a forward tangent of -4% and a back tangent of +6%
intersecting at station 12 + 020. The length of curve is 120 m. If the elevation of station 12 + 020
is 360.20 m., determine the elevation of the curve at station 11 + 980.
a. 357.63m
b. 387.63m
c. 347.63m
d. 397.63m
33. A symmetrical parabolic curve has a forward tangent of -4% and a back tangent of +6%
intersecting at station 8 + 120 and elevation of 460.20 m. The curve passes through a point A at
station 8 + 080 with an elevation of 457.633 m. Compute the length of the parabolic curve.
a. 110m
b. 90m
c. 120m
d.80m
34. A descending grade of 4.2% intersects an ascending grade of 3% at station 10 + 088 at
elevation of 120.80 m. These two gradelines are to be connected by a 260-meter parabolic curve.
A drainage pipe 900 mm diameter is installed at the lowest point of the curve with an invert
elevation of 121.82
m. Compute the depth of cover of the drainage pipe.
a. 0.35m
b. 0.45m
c. 0.70m
d. 0.90m
35. A parabolic curve has a descending grade of -0.80% which meets an ascending grade of
+0.40% at station 10 + 020. Compute the length of curve if the maximum allowable change in
grade per station having a length of 20 m. is 0.15.
a. 150m
b. 130m
c. 160m
d. 120m