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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.
๐ฟ
๐
๐ฟ
๐
Using sine law:
๐ฟ
1
๐
๐
2
๐ฟ
1
2
1
2
sin ๐ผ
๐ฟ
1
2
)(sin ๐ผ
๐
sin ๐ผ
1
๐
1
2
)(sin ๐ผ
๐ฟ
sin ๐ผ
2
Using cosine law
2
๐ฟ
2
๐
2
๐ฟ
๐
Using sine law
๐ฟ
sin ๐ฝ
๐
sin ๐
โ 1
๐ฟ
sin ๐ผ
โ 1
๐
Check
V
1
I
I L
I S
V
V
2
(180 โ I)
PC
PT
V
I
ฮธ
ฮฒ
CBLAMSIS UNIVERSITY OF THE CORDILLERAS COMPOUND CURVES
Isolate triangle V 1
VV
2
, solve T L
and T S
Using sine law:
๐ฟ
1
๐
๐
2
๐ฟ
1
2
1
2
sin ๐ผ
๐ฟ
1
2
sin ๐ผ
๐
sin ๐ผ
1
0
0
๐
1
2
)(sin ๐ผ
๐ฟ
sin ๐ผ
2
0
0
Isolate triangle PC V PT, solve LC, ฮธ, and ฮฒ
Using cosine law
2
๐ฟ
2
๐
2
๐ฟ
๐
2
2
2
Using sine law
๐ฟ
sin ๐ฝ
๐
sin ๐
โ 1
๐ฟ
sin ๐ผ
โ 1
[
0
0
]
โ 1
๐
โ 1
[
0
0
]
Check
- 319 + 31. 681 = 70 ok!
V 1
I
I L
IS
V
V
2
(180 โ I)
PC
PT
V
I
ฮธ
ฮฒ
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
๐
From figure 1 above:
2
2
๐
๐
1
1
๐ฟ
1
From right triangle O 1 CPC
1
๐ฟ
1
๐ฟ
cos ๐ผ
sin ๐ผ =
๐ฟ
๐ฟ
sin ๐ผ
๐ฟ
๐ฟ
cos ๐ผ โ ๐ถ๐ท
๐ฟ
( 1 โ cos ๐ผ) โ ๐๐ถ๐น
ฮธ =
๐ผ ๐
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 ๐ผ =
๐ฟ
๐ฟ
sin ๐ผ
cos ๐ผ =
๐ฟ
๐ฟ
cos ๐ผ
๐ฟ
๐ฟ
cos ๐ผ โ ๐ถ๐ท
๐ฟ
( 1 โ cos ๐ผ) โ ๐๐ถ๐น
๐ฟ
๐ฟ
cos ๐ผ โ ๐ถ๐ท
๐ฟ
1 โ cos ๐ผ
๐ฟ
sin ๐ผ
And from figure 1:
2
2
2
๐ฟ
sin ๐ผ โ ๐
๐
๐ฟ
cos ๐ผ
Substitute
2
๐
๐ฟ
1 โ cos ๐ผ
๐ฟ
sin ๐ผ
๐ฟ
sin ๐ผ โ ๐
๐
๐ฟ
cos ๐ผ
Where:
TL = 126.334 m
TS = 105.917 m
D
L
0
๐ฟ
3600
๐๐ท
๐ฟ
I = 60
0
tan
๐
- 183 ( 1 โ cos 60
0
0
0
0
๐
โ 1
0
๐
๐ฟ
๐ฟ
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 โ RS). i.e. right triangle O 1 AO 2.
cos ๐ผ
๐ฟ
1
1
2
1
๐ฟ
๐
๐ฟ
2
๐ฟ
๐
๐ด๐_ 2
๐ฟ
๐
๐ฟ
From Figure 1:
1
1
๐ฟ
๐ฟ
2
sin ๐ผ =
๐
๐
sin ๐ผ
A
O 1
O 2
I
L
RL - RS
V
H
PT
I
0
CBLAMSIS UNIVERSITY OF THE CORDILLERAS COMPOUND CURVES
And from
๐ฟ
๐
๐
๐ฟ
0
And from Figure 1
๐ฟ
2
๐ฟ
๐ฟ
๐
๐ฟ
๐
๐
๐ฟ
