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CBLAMSIS UNIVERSITY OF THE CORDILLERAS COMPOUND CURVES
COMPOUND CURVES
A compound curve is a combination of two simple curves of unequal radii where their
centers of curvature are located on the same side of the common tangent. This curve can
be used to connect two tangents either in horizontal or vertical alignments.
Elements of a compound curve
T
L
= tangent distance on the side of the larger curve (simple curve having the
longer radius)
T
S
= tangent distance on the side of the smaller curve (simple curve having the
shorter radius)
RS = radius of the smaller curve
R
L
= radius of the larger curve
I = angle of intersection of the back and forward tangents
LC = long chord of the compound curve
PCC = point of Compound Curve
All elements with subscript 1 and 2 refer to the elements of the first curve (simple
tangent to the back tangent) and second curve (simple curve tangent to the forward
tangent) respectively.
PC
PT
V
RL
R
S
I
I
L
IL
IS
I
S
V
1
V 2
O
1
O
2
CBLAMSIS UNIVERSITY OF THE CORDILLERAS COMPOUND CURVES
Relationship of the central angles of the simple curves and the angle of intersection of
the back and forward tangents and other elements.
𝐿
𝑆
𝐿
𝑆
Using sine law:
𝐿
1
𝑆
𝑆
2
𝐿
1
2
1
2
sin 𝐼
𝐿
1
2
)(sin 𝐼
𝑆
sin 𝐼
1
𝑆
1
2
)(sin 𝐼
𝐿
sin 𝐼
2
Using cosine law
2
𝐿
2
𝑠
2
𝐿
𝑆
Using sine law
𝐿
sin 𝛽
𝑆
sin 𝜃
− 1
𝐿
sin 𝐼
− 1
𝑆
Check
V
1
I
I L
I S
V
V
2
(180 – I)
PC
PT
V
I
θ
β
CBLAMSIS UNIVERSITY OF THE CORDILLERAS COMPOUND CURVES
Isolate triangle V 1
VV
2
, solve T L
and T S
Using sine law:
𝐿
1
𝑆
𝑆
2
𝐿
1
2
1
2
sin 𝐼
𝐿
1
2
sin 𝐼
𝑆
sin 𝐼
1
0
0
𝑆
1
2
)(sin 𝐼
𝐿
sin 𝐼
2
0
0
Isolate triangle PC V PT, solve LC, θ, and β
Using cosine law
2
𝐿
2
𝑠
2
𝐿
𝑆
2
2
2
Using sine law
𝐿
sin 𝛽
𝑆
sin 𝜃
− 1
𝐿
sin 𝐼
− 1
[
0
0
]
− 1
𝑆
− 1
[
0
0
]
Check
- 319 + 31. 681 = 70 ok!
V 1
I
I L
IS
V
V
2
(180 – I)
PC
PT
V
I
θ
β
CBLAMSIS UNIVERSITY OF THE CORDILLERAS COMPOUND CURVES
- The azimuths of the back and forward tangents of a compound curve are 320
0
and
0
respectively. The degree of curve of the simple curve tangent to the back tangent
(larger curve) is 5
0
. The tangential distance on the side of the larger curve is 126. 334
meters while on the side of the smaller curve is 105.917 meters. Determine the central
angles of the two simple curves and the degree of curve of the simple curve tangent
to the forward tangent. Use arc definition.
Given (figure)
Figure 1
TL = 126.334 m
T
S
= 105.917 m
D
L
0
Required: IL, IS, DS
Solution:
The solution requires the construction of parallel lines, perpendicular lines, right
triangles, and extension of arcs. The hypotenuse and one acute angle of the right
triangle that will be constructed must known so that we can make use of it.
CBLAMSIS UNIVERSITY OF THE CORDILLERAS COMPOUND CURVES
2
𝑆
From figure 1 above:
2
2
𝑆
𝑆
1
1
𝐿
1
From right triangle O 1 CPC
1
𝐿
1
𝐿
cos 𝐼
sin 𝐼 =
𝐿
𝐿
sin 𝐼
𝐿
𝐿
cos 𝐼 − 𝐶𝐷
𝐿
( 1 − cos 𝐼) − 𝑃𝐶𝐹
θ =
𝐼 𝑆
2
O 2
A
B
0
O
1
PC
C
I = I
L
+ I
S
RL
CBLAMSIS UNIVERSITY OF THE CORDILLERAS COMPOUND CURVES
From right triangle PCFV
sin 𝐼 =
𝐿
𝐿
sin 𝐼
cos 𝐼 =
𝐿
𝐿
cos 𝐼
𝐿
𝐿
cos 𝐼 − 𝐶𝐷
𝐿
( 1 − cos 𝐼) − 𝑃𝐶𝐹
𝐿
𝐿
cos 𝐼 − 𝐶𝐷
𝐿
1 − cos 𝐼
𝐿
sin 𝐼
And from figure 1:
2
2
2
𝐿
sin 𝐼 − 𝑇
𝑆
𝐿
cos 𝐼
Substitute
2
𝑆
𝐿
1 − cos 𝐼
𝐿
sin 𝐼
𝐿
sin 𝐼 − 𝑇
𝑆
𝐿
cos 𝐼
Where:
TL = 126.334 m
TS = 105.917 m
D
L
0
𝐿
3600
𝜋𝐷
𝐿
I = 60
0
tan
𝑆
- 183 ( 1 − cos 60
0
0
0
0
𝑆
− 1
0
𝑠
𝐿
𝐿
0
PC
F
V
I
I
TL
CBLAMSIS UNIVERSITY OF THE CORDILLERAS COMPOUND CURVES
- The back and forward tangents of a compound curve intersect at an angle equal to
0
. The degree of curve of the simple curve tangent to the back tangent is 5
0
while
the simple curve tangent to the forward tangent is 8
0
. If the tangential distance on the
side of the smaller curve is 105.917 meters determine the central angles of the two
simple curves and the tangential distance on the side of the larger curve. Use arc
definition.
Given: (figure)
Figure 1
I = 60
0
DL = 5
0
D
S
0
T
S
= 105.917 m
Required: IL, IS, TL
CBLAMSIS UNIVERSITY OF THE CORDILLERAS COMPOUND CURVES
Solution:
Extend the smaller curve on the side of the larger curve since TS is given. Draw
line perpendicular to the back tangent passing through point PT (line PTH), line
parallel to the back tangent passing through PT (line CGPT), line parallel to the
back tangent passing through O 2 (line AO 2 ), line perpendicular to the back tangent
passing through O 2 (line EGO 2 ). These drawn lines form right triangles with known
distances as their hypotenuse (i.e. TS, RS, and RL).
Since D L
and D S
(R
L
and R S
) are given, the key right triangle will be a right triangle
with hypotenuse equal to (RL – RS). i.e. right triangle O 1 AO 2.
cos 𝐼
𝐿
1
1
2
1
𝐿
𝑆
𝐿
2
𝐿
𝑆
𝐴𝑂_ 2
𝐿
𝑆
𝐿
From Figure 1:
1
1
𝐿
𝐿
2
sin 𝐼 =
𝑆
𝑆
sin 𝐼
A
O 1
O 2
I
L
RL - RS
V
H
PT
I
0
CBLAMSIS UNIVERSITY OF THE CORDILLERAS COMPOUND CURVES
And from
𝐿
𝑆
𝑆
𝐿
0
And from Figure 1
𝐿
2
𝐿
𝐿
𝑆
𝐿
𝑆
𝑆
𝐿
