21/02/2021
Part 1
HORIZONTAL CURVE
Prepared by:
Engr. Rommel G. Arapo
CE - Faculty
1
H O R I Z O N TA L C U RV E
SIMPLE CURVE
A simple curve is a circular arc extending
from one tangent to the next. The point
where the curve leaves the first tangent is
called the point of curvature and the
point where the curve joins the second
tangent is called the point of tangency.
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S I M P L E C U RV E
Elements of a Simple Curve:
Formulas for a Simple Curve:
PC -
Point of Curvature
PT -
Point of Tangency
PI -
Point of Intersection
R -
Radius of the Curve
D -
Degree of the Curve
3. T = R tan
T -
Tangent Distance
4. E = R sec
I -
Intersection Angle
E -
External Distance
M-
Middle Ordinate
LC -
Length of Curve
C -
Long Chord
1. R =
.
2. R =
(arc basis)
(chord basis)
−1
5. M = R 1 − cos
6. C = 2R sin
7. L =
3
P ro b l e m 1
V
Determine the elements of a simple curve and
I
the stations of the important points of the
E
simple curve. The said curve has a degree of
curvature of 6°, the angle of intersection of 21°
LC
and the stationing of the point of curvature is
80+200. Use arc basis.
PC
C
M
PT
Sta 80+200
R
O
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P ro b l e m 1
Solution:
Using arc basis:
Given:
Degree of Curve (D) = 6°
Angle of Intersection (I) = 21
Required:
Element of Simple Curve
 Radius of the curve (R)

T = R tan


Middle ordinate (M)
I
2
21
M = 190.986 1 − cos
2
M = 3.198 meters
Long chord (C)
C = 2R sin
Station of PC, PT and V:
Sta. @ PC = 80+200
Sta. @ V = Sta. of PC + T
Sta. @ V = (80+200) + 35.397
Sta. @ V = 80+235.397
Sta. @ PT = Sta. of PC + LC
I
2
21
C = 2 190.986 sin
2
C = 69.609 meters
21
T = 190.986 tan
2
T = 35.397 meters
20I
D
20 21
L =
6
LC = 70 meters
L =
21
−1
2
E = 3.253 meters
M = R 1 − cos
I
2
Length of curve (LC)
E = 190.986 sec
1145.916
D
1145.916
R=
6°
R = 190.986 meters
Tangent of the curve (T)

I
E = R sec − 1
2
R=

External distance (E)
Sta. @ PT = (80+200) + 70
Sta. @ PT = 80+270
5
P ro b l e m 2
A simple curve with tangents AV and VE has azimuth from south of 260°48’ and 285°40’
respectively. A certain point “B” is taken along the line AV and point “C” along the line VE. The
azimuth and the distance of BC are 272°16’ and 61.22m respectively. If degree of curvature is 5°
and stationing of point “B” is 8+125, determine the stationing at point of tangency.
V
285°40’
B
C
272°16’
8+125
A
260°48’
E
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P ro b l e m 2
Solution
Given:
Azim AV = 260°48’
Azim VE = 285°40’
Azim BC = 272°16’
Dist. BC = 61.22m
D = 5°
Sta @ B = 8+125
 Angle of intersection (I)
I = Azim VE − Azim AV
I = 285°40 − 260°48
I = 24°52’
V
285°40’
B
C
272°16’
8+125
A
E
260°48’
Required:
Sta @ PT = ?
Sta @ PT = Sta @ PC + LC
Sta @ PC = Sta @ B – ( T – BV )
I
T = R tan
2
 Tangent distance (T)
 Radius of the curve (R)
1145.916
D
1145.916
R=
5°
R = 229.183 meters
T = R tan
R=
I
2
24°52′
2
T = 50.529 meters
T = 229.183 tan
7
P ro b l e m 2

V
Consider ΔBVC
Using sine law:
285°40’
sin V sin C
=
BC
BV
sin 155°08′ sin 13°24′
=
61.22
BV
BV = 33.739 meters
B
C
272°16’
8+125
A
Sta @ PC = Sta @ B – ( T – BV )
E
260°48’
Sta @ PC = (8+125) – ( 50.529 – 33.739)
Sta @ PC = 8+108.21
Sta @ PT = Sta @ PC + LC
 Length of curve (LC)
20I
L =
D
20 (24°52 )
L =
5°
LC = 99.467 meters
Sta @ PT = Sta @ PC + LC
Sta @ PT = (8+108.21) + 99.467
Sta @ PT = 8+207.677
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P ro b l e m 3
The tangent distance of a 3° simple curve is only half
of its radius. Determine the following:
a) Angle of intersection of the simple curve.
Area of fillet of
curve
b) Length of curve.
c) Area of the fillet of the curve.
9
P ro b l e m 3
Solution
Given:
b. Length of curve, LC:
Degree of curvature, D = 3°
L =
20I
D
L =
20(53°07 )
3°
a. Angle of intersection, I:
note : Tangent distance, T = ½ R
T = R tan
I
2
LC = 354.111 meters
1
I
R = R tan
2
2
1
I
= tan
2
2
I = 53°07’
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P ro b l e m 3
c. Area of fillet of curve:
A =A −A
AT = Area total (Area bounded
by the tangents and radius)
AC = Area of the curve
AT = TR
AT = ½ R(R)
A =
1 1145.916
2
D
A =
πR I
360
1145.916
π
I
D
A =
360
Area of fillet of
curve
1145.916
53°07′
3°
A =
360
AC = 67,630.289 sq.m.
1 1145.916
2
3°
AT = 72,951.304 sq.m.
A =
π
A =A −A
AF = 5, 321.015 sq.m.
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H O R I Z O N TA L C U RV E
COMPOUND CURVE
A compound curve consists of two or
more consecutive simple curves having
different radius, but whose centers lie
on the same side of the curve, likewise
any two consecutive curves must have a
common tangent at their meeting point.
In a compound curve, the point of the
common tangent where the two curves
join is called the point of compound
curvature (PCC).
Where:
dist. AB = common tangent
dist. AB = T1 + T2
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Problem 1
Problem no.1
Two tangents intersect at sta. 25+050. A compound
curve laid on their tangents has the following data:
I1 = 31° , I2 = 36°, D1 = 3°, D2 = 5°.
a) Determine the stationing of PC, PCC and PT.
b) If tangent on the side of the PT is moved 10 meters
out. Compute the stationing of new PT with PCC at
same point.
(25+050)
A
V
31°
36°
B
PCC
PT
PC
I2
I1
13
Problem 1
Solution:
For the value of AV:
a) Sta @ PC = ?
Consider ΔVAB
AV
180.396
=
sin 36° sin 113°
using sine law
𝐀𝐕 = 𝟏𝟏𝟓. 𝟏𝟗𝟏 𝐦
Sta @ PC = Sta @ V – AV – T1
For the value of T1:
I
T = R tan
2
R =
1145.916
D
1145.916
R =
3
R = 381.972 m
31
T = 381.972 tan
2
𝐓𝟏 = 𝟏𝟎𝟓. 𝟗𝟑𝟎 𝐦
AV
AB
=
sin I
sin 113°
AB = common tangent
AB = T + T
T = R tan
T =
I
2
1145.916
I
tan
D
2
1145.916
36°
T =
tan
5°
2
T = 74.466 m
AB = 105.930 + 74.466
AB = 180.396
Sta @ PC = Sta @ V – AV – T1
Sta @ PC = (25+050) – 115.191 – 105.930
Sta @ PC = 24 + 828.879
Sta @ PCC = ?
Sta @ PCC = Sta @ PC + LC1
Sta @ P
= 24 + 828.879 +
Sta @ PCC = 25 + 035.546
20(31°)
3°
Sta @ PT = ?
Sta @ PT = Sta @ PCC + LC2
Sta @ P = 25 + 035.546 +
Sta @ PT = 25 + 179.546
20(36°)
5°
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Problem 1
(25+050)
V
b)Sta @ new PT = ?
Sta @ P ′ = sta @ Pcc + L ′
L ′=
πR ′I
180
A
31°
PT
PC
I1
36
91.479 = R ′ tan
2
R ′ = 281.534 m
π 281.534 (36)
180
L ′ = 176.893 m
B’
PT’
T = 74.466 + 17.013
T ′ = 91.479 m
L ′=
B
PCC
I
T ′ = R ′ tan
2
T = T + BB′
For the value of BB’
10
sin 36° =
BB′
BB′ = 17.013 m
36°
I2
Sta @ P ′ = (25 + 035.546) + 176.893
𝐒𝐭𝐚 @ 𝐏𝐓 ′ = 𝟐𝟓 + 𝟐𝟏𝟐. 𝟒𝟑𝟗
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Problem 2
On a railroad line, two tangents that intersect at
station 10+240 so as to form an angle of 36°28’ are
to be connected by a compound curve consisting of
two simple curve. The simple curve beginning at
the PC which at sta. 10+160 is to be a 4° curve and
is to have a central angle of 17°.
PC
(10+160)
a) What should be the radius of other simple curve
that ends at PT?
b) Determine the stationing of the PCC.
c) What is the length of the tangent from the point
of intersection to the PT of the compound
curve?
(10+240) V
36° 28’
A
B
17°
PCC
PT
D1 = 4°
I1
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Problem 2
a) R2 = ?
I
T = R tan
2
I2 = 36° 28’ - 17°
I2 = 19° 28’
For the value of T2:
For the value of AB:
AB = T1 + T2
use ΔVAB
66. 304 = 42.823 + T2
T2 = 23.829 m
AV
AB
=
sin I
sin 143°32′
AV = P V − T
23.481 = R tan
AB = T1 + T2
P V = 10 + 240 − 10 + 160
I
T = R tan
2
P V = 80 m
10
17°
T =
tan
D
2
sin
2
10
17°
T =
tan
4
2
sin
2
T = 42.823m
19°28′
2
𝐑 𝟐 = 𝟏𝟑𝟔. 𝟖𝟗𝟎 𝐦
AV = 80 − 42.823
AV = 37.177 m
37.177
AB
=
sin 19°28′ sin 143°32′
AB = 66.304 m
17
Problem 2
b.) sta @ PCC = ?
c.) dist VPT = ?
sta @ PCC = sta @ PC + Lc1
sta @ P
= (10 + 160) +
𝐬𝐭𝐚 @ 𝐏𝐂𝐂 = 𝟏𝟎 + 𝟐𝟒𝟓.017
VP = VB + T
π 286.537 17°
180
For the value of VB:
use ΔVAB
VB
37.177
=
sin 17° sin 19°28′
VB = 32.616 m
VP = 32.616 + 23.829
𝐕𝐏𝐓 = 𝟓𝟔. 𝟒𝟒𝟓 𝐦
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Problem 3
V
Given the following compound curve with the
vertex V, inaccessible. Angles VAD and VDA are
equal to 16°20’ and 13°30’ respectively, Stationing
of A is 1+120, degree of curve are 3°30’ for the
first curve and 4° for the second curve.
a) It is desired to substitute the compound curve
with a simple curve that shall end with the
same PT, determine the total length of the
simple curve.
b) It is desired to substitute the compound curve
with a simple curve that shall be tangent to the
two tangent lines as well as the common
tangent. What is the radius of the simple
curve?
(1+120)
A
16°20’
13°30’
D
PCC
PC
D1 = 3°30’
D2 = 4°
PT
c) What is the stationing of the new PC?
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Problem 3
a.) Required:
Lc’ of the new simple curve
L ′=
πR′I′
180
T′ = R′ tan
I′
2
I’ = I1 + I2
Note:
VD + T = T′
Where T’ – Tangent of new
curve
For the value of VD:
Using ΔVAD
VD
AD
=
sin 16°20′ sin 150°10′
AD = T + T
I
I
AD = R tan + R tan
2
2
VD
80.892
=
sin 16°20′ sin 150°10′
VD = 45.728 m
T′ = 45.728 +
1145.916
13°30′
tan
4°
2
T’ = 79.635 m
79.635 = R′ tan
29°50′
2
R’ = 298.904 m
π 298.904 29°50
180
𝐋𝐂 = 𝟏𝟓𝟓. 𝟔𝟑𝟔 𝐦
L =
1145.916
16°20′
1145.916
13°30′
tan
+
tan
3°30′
2
4°
2
AD = 80.892 m
AD =
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Problem 3
b.) Required:
R” = radius of the new curve
Note:
AD = T + T
AD = 80.892 m (from previous)
AD = R tan
I
I
+ R tan
2
2
80.892 = R′ tan
16°20′
13°30′
+ R′ tan
2
2
𝐑" = 𝟑𝟒𝟑. 𝟐𝟕𝟓 𝐦
c.) Required:
Sta @ Pc’ = ?
sta @ P = Sta @ A − T"
I
T" = R" tan
2
T" = (343.275) tan
16°20′
2
T” = 49.263 m
sta @ P = (1 + 120) − 49.263
sta @ PC = 1+070.737
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H O R I Z O N TA L C U RV E
REVERSED CURVE
A reversed curve is formed by two circular simple curves having
a common tangent but their centers lies on opposite sides. At the
point where the curve reversed in its direction is called the point of
reversed curvature (PRC).
Four Types of Reversed Curve
a) Reversed curve with equal radii and parallel tangents
b) Reversed curve with unequal radii and parallel tangents
c) Reversed curve with equal radii and converging tangents
d) Reversed curve with unequal radii and converging tangents
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Problem 1
Two parallel tangents 20 meters apart
O2
are to be connected by a reversed curve
with equal radii at the PC and PT. The
total length of chord from the PC to the PT
10 + 200
10+200.
PC
a) Find the radius of the reversed curve.
A
R2
I1
b) Find the length of the chord from PC
to PRC.
PRC
c) Find the stationing of the PT.
20m
is 150m and the stationing of PC is
I2
R1
PT
B
O1
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Problem 1
Note:
This is Type A reversed curve.

Reversed curve with equal radii
and parallel tangents
Properties of Type A.
I 1 = I2
R1 = R2
*all elements of curve are equal.
C = R sin
I
2
c) Sta @ PT = ?
For the value of I1:
Consider ΔPCCPT
sin
Sta @ P = sta @ P + Lc + Lc
Sta @ P = (10 + 200) + 2
I
20
=
2 150
π 562.519 15.324
180
Sta @ PT = 10 + 500. 896
I1 = 15.324°
.
Solution:
75 = R sin
Given:
R1 = 562. 519 m
y = 20m
C1 + C2 = 150m
b) length of chord from PC to
Sta @ PT = 10 + 200
PRC =?
Required:
C1 = ?
a) R = ?
C1 = 75 m
R = R1 = R2
C1 = C2 = 75 m
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Problem 2
In a railroad layout, the centerline of two
O2
curve of unequal radii. The central angle of
the first curve is 16° and the distance
between parallel tracks is 27.5 meters, sta. @
15 + 420
A
PC
PC is 15+420 and radius of the second curve is
C
16°
290 meters.
27.5 m
PRC
a) Find the length of long chord from PC and
PT.
I2
R1
b) Find the radius of the first curve.
R2 = 290 m
parallel tracks are connected with a reversed
B
PT
c) Find the stationing of PT.
O1
25
Problem 2
Note: I1 = I2
For the value of AB:
a) Long chord, C = ?
use ΔADB
consider Δ PCCPT
P P =C
I
27.5
sin =
2
C
16° 27.5
sin
=
2
C
C = 197.596 m
b) R1 = ?
I
T = R tan
2
AB – common tangent
AB = T + T
I
T = R tan
2
16°
T = 290 tan
2
T = 40.757 m
sin I =
27.5
AB
sin 16° =
c) Sta @ PT = ?
sta @ P = sta @ P + Lc + Lc
sta @ P = sta @ P +
27.5
AB
πR I
πR I
+
180
180
sta @ P = (15 + 420) +
π(419.892)(16) π(290)(16)
+
180
180
𝐬𝐭𝐚 @ 𝐏𝐓 = 𝟏𝟓 + 𝟔𝟏𝟖. 𝟐𝟑𝟗
AB = 99.769 m
AB = T + T
99.769 = T + 40.757
T1 = 59.012 m
T = R tan
I
2
59.012 = R tan
16°
2
R1 = 419.892 m
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Problem 3
Note: common radius,
Two tangents converge at an 30° and direction of the
second tangent is due east. The distance of the PC from the
second tangent is 116.50 meters. The bearing of the
R1 = R 2
a) I1 = ?
common tangent is S 40° E. If a reversed curve is to connect
these two tangents:
I2 = I1 + θ
50 = I1 + 30
I1 = 20°
a) Determine the central angle of first curve.
b) Determine the radius of the curve.
c) Determine the stationing at PT, if the PC is
at 10+600.
27
Problem 3
b) R=?
R = R1 = R2
From the figure:
a + b = 116.50 m
For the value of b:
Consider ΔDPRC02
cos I =
R −b
R
b = R (1 − cos I )
(1)
For the value of a:
Consider ΔCPRCO1
c
cos I =
R
c = R cos I
Consider ΔEPCO1
cos θ =
a + b = 116.50 m
c+a
R
R cos θ − cos I
Since R1 = R2
a = R cos θ − c
R cos 30° − cos 50° + R(1 − cos 50°) = 116.50
a = R cos θ − R cos I
a = R cos θ − cos I
+ R (1 − cos I ) = 116.50
(2)
R = 200.706 m
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Problem 3
c) sta @ Pc = 10 + 600
sta @ PT = ?
sta @ P = sta @ Pc + Lc + Lc
sta @ P = 10 + 600 +
π(200.706)(20) π(200.706)(50)
+
180
180
sta @ PT = 10 + 845.209
29
Problem 4
O2
A reversed curve connects two converging
tangents intersecting at an angle of 30°.
The distance of this intersection from the
PC
point of intersection of the second curve is
I2
4 + 450
A
D2 = 6°
150 meters. The azimuth from south of the
I1
common tangent is 320° and the degree of
the point of intersection of the first curve is
I2
4+450.
30°
B
a) Determine the radius of the first curve.
40°
PT
V
150 m
b) Determine the stationing of the PRC.
c) Determine the stationing of the PT.
D
PRC
the second curve is 6° and the stationing of
R2
I1
O1
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Problem 4
a) R1 = ?
T = R tan
I
2
For the value of I1:
For the value of AB:
I2 = 50°
Consider ΔABV
Consider ΔABV
I1 = 180 – 30 – 130
AB
150
=
sin 30 sin I
I1 = 20°
AB = 219.285 m
AB = T1 + T2
For the value of T1:
219.285 = T1 + 89.058
AB – common tangent
T1 = 130.227 m
AB = T1 + T2
T = R tan
I
2
T = R tan
1145.916
50
tan
6
2
T2 = 89.058 m
I
2
130.227 = R tan
T =
20
2
R1 = 738.554 m
31
Problem 4
b) Sta @ PRC = ?
Sta @ PRC = sta @ PC + Lc1
sta @ PC = sta @ A – T1
sta @ PC = (4 + 450) – 130.227
sta @ PC = 4 + 319.773
sta @ P
= (4 + 319.773) +
π(738.554)(20)
180
sta @ PRC = 4 + 577.577
c) sta @ PT =?
sta @ PT = sta @ PRC + Lc2
π
sta @ P = (4 + 577.577) +
1145.916
(50)
6
180
sta @ PT = 4 + 744.244
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H O R I Z O N TA L C U RV E
SPIRAL CURVE
Spiral curve is a transitional curve that should be placed
between tangents and each end of a simple curve. A spiral
curve increases in curvature gradually, thus avoiding an
abrupt change in the rate of lateral displacement of cars. It
also provides a mean of gradually elevating the far end of the
road in proper relation to the degree of curvature.
33
SPIRAL CURVE
Elements of a spiral curve:
SC – Spiral to Curve
TS – Tangent distance of the Spiral Curve
CS – Curve to Spiral
YC – Distance along the tangent from TS to XC
ST – Spiral to Tangent
X – Offset from the tangent at any point on the spiral
TS – Tangent to Spiral
Y – Distance along the tangent at any point on the spiral
I – Angle of intersection of spiral easement curve
SC – Spiral Angle at SC (θs)
IC – Angle of intersection of a simple curve.
LS – Length of the spiral curve
RC – Radius of a simple curve.
L – length of the spiral curve from TS to any point along the
spiral
DC – Degree of simple curve.
LT – Distance of Long Tangent
ST – Distance of Short Tangent
i – deflection angle at any point on the spiral.
P – length of throw
ES – External Distance of Spiral Curve
LC – Long Chord of Spiral Curve
XC – Offset from tangent at SC
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SPIRAL CURVE
Formulas for spiral curve:
1)
L
180
S=
x
2R L
π
2)
D L
S =
40
3)
L
180
S =
x
(spiral angle at SC)
2R
π
4)
X =
5) X =
T =
(spiral angle at any
point on the spiral)
(spiral angle at SC, arc
basis)
L
X
I
+ R +
tan
2
4
2
(tangent distance)
X
I
sec
−R
4
2
(external distance)
10. E = R +
(angle of intersection of
simple curve)
11. I = I − 2S
12. P =
X
(length of throw)
4
13. e =
0.0079K
R
L
14. e =
0.004K
R
S
(deflection angle at SC)
3
15. L =
L
6R
(offset distance from
tangent at SC)
X L
6) i =
7) Y = L −
8)
9.
L
40R L
Y =L −
L
40R
(superelevation)
(considering 75% of K to
counteract superelevation)
0.036K
(desirable length of spiral)
R
(distance along the
tangent at any point in
spiral)
16.
i
L
=
i
L
(deflection angle varies as the
square of the length from TS)
(distance along the
tangent at SC from TS)
17.
D
L
=
D
L
(degree of curve varies directly
with the length from TS)
35
Problem 1
A spiral curve was laid out in a certain portion of a highway.
It has a length of 80m and an angle of intersection of 42°. If
the degree of curvature is 6°, determine the following:
a. Length of long tangent
b. Length of throw
c. External distance
d. Allowable speed for the length of the spiral.
36
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Problem 1
a) LT = ?
X =
L
6R
X =
(80)
1145.916
6
6
from the figure:
LT = YC – h
Y =L −
L
40R
X = 5.585 m
80
Y = 80 −
40
1145.916
6
tan S =
Y = 79.649 m
5.585
h
h = 26.275 m
tan 12 =
For the value of h:
Use ΔABSC
tan S =
LS = 80 m
I = 42°
DC = 6°
LT = YC – h
X
h
LT = 79.649 – 26.275
LT = 53.374 m
S =
L
180
x
2R
π
S =
80
180
1145.916 π
2
6
Given:
X
h
b) length of throw, P = ?
X
P=
4
5.585
P=
4
P = 1.396 m
SC = 12°
37
Problem 1
c) external distance, ES = ?
E = R +
E =
X
I
sec
−R
4
2
1145.916 5.585
42
1145.916
+
sec
−
6
4
2
6
𝐄𝐒 = 𝟏𝟓. 𝟎𝟖𝟑 𝐦
d) allowable speed, K = ?
L =
89 =
0.036K
R
0.036K
1145.916
6
K = 77.869 kph
38
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Problem 2
Two tangents having an azimuth of 240° and
282° are connected by an 80 meter spiral
curve with 5° circular curve. The width of the
a) superelevation, e = ?
Solution:
Given:
R=
I = 42°
LS = 80 m
R = 229.183 m
roadway is 10 meters. If the design speed is
DC = 5°
60 kph, determine the following:
K = 60 kph
a) Superelevation at quarter points
1145.916
5
e=
0.0079K
R
e=
0.0079(60)
229.183
W = 10 m
b) Deflection angle at endpoints
e = 0.124 m/m (width of road)
c) Tangent distance of spiral
super-elevation at quarter
points:
e4 = 0.124 (10)
e2 = 1.24 (0.50)
e4 = 1.24 m
e2 = 0.62 m
e1 = 1.24 (0.25)
e3 = 1.24 (0.75)
e1 = 0.31 m
e3 = 0.93 m
39
Problem 2
b) deflection angle at endpoints, iC =?
i =
S
3
S =
D L
40
i =
and
c) Tangent distance, TS = ?
T =
L
X
I
+ R +
tan
2
4
2
T =
Ls
L
I
+ R +
tan
2
24R
2
T =
80
80
42
+ 229.183 +
tan
2
(24)229.183
2
so;
D L
(3)40
(5)(80)
i =
(3)40
iC = 3.333°
TS = 128.422 m
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Problem 3
A simple curve having a radius of 280 meters connects two tangents
intersecting at an angle of 50°. It is to be replaced by another curve
having 80 meters spiral 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 central angle of the new circular curve.
c) Determine the deflection angle of the endpoint of the spiral.
d) Determine the offset from tangent at the endpoint of spiral.
e) Determine the distance along the tangent at the midpoint of spiral.
41
Problem 3
Solution:
Given:
Old simple curve:
R = 280 m
I = 50°
New spiral curve:
LS = 80 m
L
X
I
+ R +
tan
2
4
2
T =
L
L
I
+ R +
tan
2
24R
2
Note: Tangent of new spiral curve
is equal to the tangent distance of
the old simple curve.
80
80
50
+ R +
tan
2
24R
2
I
2
T = 280 tan
50
2
T = 130.566 m
𝐈𝐂 = 𝟐𝟔. 𝟐𝟑𝟐°
c) iC = ?
d) XC = ?
11.884
3
L
6R
X =
80
6(192.84)
𝐗 𝐂 = 𝟓. 𝟓𝟑𝟏 𝐦
b) IC = ?
and;
e) Y40 = ?
L
S =
L
180
x
2R
π
Y
=L−
S =
80
180
x
2(192.84) π
Y
= 40 −
SC = 11.884°
X =
iC = 3.961°
RC = 192.84 m
I = I − 2S
Ts = T
I = 50 − 2(11.884)
i =
130.556 =
a) RC = ?
T = R tan
T =
40R L
40
40(192.84) (80)
𝐘𝟒𝟎 = 𝟑𝟗. 𝟗𝟖𝟗 𝐦
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