Horizontal Alignment: Highway Engineering/Chapter Five Dr. Raad Awad Kattan
Horizontal Alignment: Highway Engineering/Chapter Five Dr. Raad Awad Kattan
Horizontal Alignment: Highway Engineering/Chapter Five Dr. Raad Awad Kattan
Horizontal alignment
The horizontal alignment consists of straight sections of the roads, known as tangents,
connected by horizontal curves. the curves usually segments of circles.
Cricular Curve
Tangent
Horizontal curves
Reverse Curve
1
Highwa
ay Engineeering/Ch
hapter Fiive Dr. Raaad Awad Kattan
Sim
mple circullar curves
PC M
PT
C
R
Δ/2 Δ/2
L=
= Length of the currve
∆rad= L= R ∆rad
T
T= length of
o the tang
gent
T=R tan
M
M= the midd
dle ordinate
M = R - R cos ( ))
E
E=the extern
nal distance
E = - R
∆
Len
ngth of thee long cho
ord C= 2R
R sin )
Exaample
Δ = 50ْ , R = 100
00m , sttation P.I =20+00. Find statiion : P.C,, P.T
an ∆/2= 10
T= R ta 000* tan 25= 466.31m
Station
n P.C= stattion P.I –T=
– 2000-- 466.31= 15+33.699
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Highway Engineering/Chapter Five Dr. Raad Awad Kattan
L = 1000* 55* = 959.93 m
Measure the deflection angle from the tangent at point P.C and chord joining consecutive
whole station
Example: a horizontal curve having R= 500m, ∆=40°, station P.I= 12+00 ,prepare a setting
out table to set out the curve using deflection angle from the tangent and chord length
method, dividing the arc into 50m stations. PI
50m arc
δ/2
PT
PC
δ=
3
Highway Engineering/Chapter Five Dr. Raad Awad Kattan
.
δ1= =3° 39´ 57˝
δ2, 3,4..= ° ´ ˝
Compound curves : consist of two or more curves in succession turning in the same
direction with any two successive curves having common tangent. Compound curves are
used mainly in obtaining desirable shape of the horizontal alignment…e.g : at intersections,
ramps, difficult topographic conditions.
The second curve is set out from point PCC. The theodolite
read 360-Δ1/2 ( 360- 17=343ْ in the next example). The theodolite rotate untie it reads zero
then transit on the vertical plan to be on tangent to the second curve.
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Highway Engineering/Chapter Five Dr. Raad Awad Kattan
60ْ
26ْ
Example : ∆1=34° ∆2=26° st PI= 611+30 T1 d1 d2
34ْ
t1=500 tan 17°=152.87m L1=269.71m t1+t2
t1+t2=233.67m
500 26ْ 350
.
= d1= 118.28m
T1=t1+d1=152.87+118.28=271.15m 34ْ
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Highway Engineering/Chapter Five Dr. Raad Awad Kattan
Reverse curves
Reverse curves usually consist of two simple curves turning in opposite direction with a
common tangent . They are generally used to change the alignment of the highway .
Reverse curves are seldom recommended because of the sudden change in alignment.
When it is necessary to reverse alignment it is better to use two simple curves separated by
sufficient tangent between them.
I1
50°
I2 50°
Example
Example
Equal radius reverse curve , Common tangent I1-I2 =700m , ∆1=40° ∆2=40°
I1
150m
I2
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Highway Engineering/Chapter Five Dr. Raad Awad Kattan
Calculating R depending on the distances between the beginning and end of the reverse
curve d (between the tangents) and D (along the tangents)
tangent
common tangent
tangent
∆ Δ/2
R (R- R cos ∆)
R cos ∆
(R- R cos ∆) R d
R cos ∆ R
Δ/2 Δ
D is the distance between the beginning and end of the reverse curve (along the tangents)
d= 2R(1-cos ∆) R=
∆