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CBLAMSIS UNIVERSITY OF THE CORDILLERAS REVERSED CURVES
REVERSED CURVE
Definition:
Reversed curve is a curve formed by two simple curves with their center of curvature
located on opposite side of their common tangent.
Types of Reversed Curve
- Reversed curve with parallel tangents
- Reversed curve with converging tangents
Reversed curve with parallel tangents
The azimuth or bearing of the back and forward tangents are the same.
Definition of elements:
PRC (Point of Reversed Curve) – the point where the two simple curves meet at the
common tangent.
P = perpendicular (shortest) distance between the parallel tangents
All other elements with subscript 1 and 2 refer to the elements of the first simple curve
and the second simple curve, respectively.
If the common tangent will intersect the back and forward tangents, which are parallel to
each other, the common tangent will make the same angle of intersection with the two
parallel tangents. Thus I 1 = I 2 = I.
Also, the line 𝑃𝐶, 𝑃𝑅𝐶
and line 𝑃𝑅𝐶, 𝑃𝑇
makes the same angle with the common tangent,
thus the line 𝑃𝐶, 𝑃𝑅𝐶, 𝑃𝑇
(long chord of the reversed curve) is a straight line.
The length of the common tangent is equal to the sum of the tangent distances of the two
simple curves. That is V 1 V 2 = (T 1 + T 2 ).
Also, the long chord of the reversed curve (LC) is equal to the sum of the long chords of
the two simple curves. That is LC = (LC 1
+ LC
2
PC
PT
PRC
O 1
O 2
R 1
R 2
V 1
V 2
R
2
R
1
P
I
2
I
2
I
1
I
1
I 1
/
I 2
/
T 1
T 2
X
CBLAMSIS UNIVERSITY OF THE CORDILLERAS REVERSED CURVES
Construct a line passing through V 2
perpendicular to the two parallel tangents, right
triangle V 2 XV 1 is formed. From the right triangle formed, P = (T 1 + T 2 )(tan I).
Reversed Curve with Converging Tangents
The back and forward tangents of the reversed curve will intersect at the vertex, V.
Elements:
T
L
= tangent distance from point PC to point V (vertex)
TS = tangent distance from point PT to point V
I = angle of intersection of the back and forward tangents
All other elements with subscript 1 and 2 refer to the elements of the first simple curve
and the second simple curve, respectively.
Isolate triangle V 1 V 2 V
0
1
2
2
1
Sine Law:
𝐿
1
2
𝑆
2
1
1
2
𝐿
1
2
𝐿
1
2
2
1
PC
PRC
PT
V
V 1
V 2
O 1
O 2
R 1
R 2
R 1
I 1
I 2
I 1
I 2
I
T
1
V
1
V
V
2
I
I
1
I 2
(180 – I 2
)
CBLAMSIS UNIVERSITY OF THE CORDILLERAS REVERSED CURVES
1
1
2
2
1
1
1
2
OR
Construct a line passing through point PRC parallel to the parallel tangents dividing the
parallel distance between the parallel tangents into two line segments (i.e. X and Y) and
forming two right triangles (i.e. right triangles DAO 1
and DBO 2
). Then
From right triangle DAO 1
1
1
1
1
1
1 − cos 𝐼
From right triangle DBO 2
2
2
2
2
2
( 1 − cos 𝐼)
Substitute
1
1 − cos 𝐼
2
1 − cos 𝐼
1
2
1 − cos 𝐼
2
2
( 1 − cos 44 )
And
2
2
0
The other requirements will be same above.
PC
PT
PRC
O 1
O 2
R 1
R
2
V 1
V 2
R
2
R
1
I
I
I
I
T 1
T 2
N
260
0
304
0
A
B
D
CBLAMSIS UNIVERSITY OF THE CORDILLERAS REVERSED CURVES
OR:
Construct a line passing through one of the centers of curvature (i.e. line O 1
AB) of
the simple curves parallel to the long chord of the reversed curve and a line
extension (i.e. line PTB) forming an isosceles triangle (i.e. triangle O 1 O 2 B). Draw
the perpendicular bisector O 2 V 2 A forming two right triangles with one acute equal
to I/2. Construct a line passing through PT perpendicular to the parallel tangents
forming a right triangle with one acute angle equal to I/2 and side measurements
equal to P and LC respectively.
Triangles DCPC and O 2 AB are similar right tringles. Using ratio and proportion with
angle I/2 as reference
2
1
2
1
2
2
From right triangle DCPC
1
2
sin
1
2
sin
2
PC
PT
PRC
O 1
O 2
R 1
R 2
V 1
V 2
R
2
R
1
I
I
I
I
T 1
T 2
N
260
0
304
0
A
B
I/
R 1
C
D
I/
CBLAMSIS UNIVERSITY OF THE CORDILLERAS REVERSED CURVES
2
2
2
Isolate triangle V 1 V 2 V
0
1
2
2
1
Sine Law:
𝐿
1
2
𝑆
2
1
1
2
𝐿
1
2
𝐿
1
2
2
1
1
1
1
- 338 sin 34 − 110. 266 sin 60
sin 34 + sin 60
𝑆
1
2
1
2
𝑺
𝟐
𝟐
1
1
1
1
1
1
1
1
1
1
1
0
V
1
V
V
2
I
I
1
I 2
(180 – I 2
)
CBLAMSIS UNIVERSITY OF THE CORDILLERAS REVERSED CURVES
2
𝑆
2
2
𝑆
- The azimuths of the two converging tangents of a reversed curve are 275
0
and
0
. The tangential distance from the point of curvature of the reversed curve to
the point of intersection of the two converging tangents is 339.338 meters. The
degree of the simple curve tangent to the back tangent is 4
0
while the degree of
curve of the curve tangent to the forward tangent is 6
0
. Determine the central
angles of the two simple curve and the tangent distance from PT to V. Use arc
basis. (variation of sample problem no. 2)
Given:
D 1 = 4
0
D
2
0
I = 275 – 241 = 34
0
Required: I 1 , I 2 , TS
Solution:
Convert the reversed with converging tangents into a reversed curve with parallel
tangents by extending or shortening a simple curve. For example, the back tangent
direction will be retained, the simple curve tangent to the forward tangent will be
shortened until such that its tangent will be parallel to the back tangent. And if the
forward tangent will be one to be retained, the simple curve tangent to the back
tangent will be extended beyond PC until its tangent will be parallel to the forward
tangent (this sample problem). Use the perpendicular distance between the
parallel tangents to establish an equation using identity (whole = sum of its parts).
P = X + Y
PC
PRC
PT
V
V 1
V 2
O 1
O 2
R 1
R 2
R 1
I 1
I 2
I 1
I 2
I
T
1
N
275
0
241
0
X
Y
P
A
B
C
D
E
F
G
I
I 2
R 1
CBLAMSIS UNIVERSITY OF THE CORDILLERAS REVERSED CURVES
From right triangle DEV
cos 𝐼 =
𝐿
𝐿
cos 𝐼
From right triangle DBO 1
sin 𝐼 =
1
1
sin 𝐼
From right triangle FAO 1
2
1
1
2
From right triangle FGO 2
2
2
2
2
Substitute
𝑆
𝑆
𝐿
cos 𝐼 + 𝑅
1
sin 𝐼 − 𝑅
1
2
2
2
𝑆
= 339. 338 cos 34 + 286. 479 sin 34 − 286. 479 𝑠𝑖𝑛 60 − 190. 986 𝑠𝑖𝑛 60
𝑺
