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Horizontal Curves

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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. Horizontal curves
are usually used to change the alignment direction

Horizontal curves
Types of horizontal curves

Simple circular, compound, reversed , and spiral


Simple circular curves
The simple circular curve is described by the radius of the curve R

Length of the curve L

∆rad=

∆ is the deflection angle between the two tangents, R is the radius of the circular curve.

Length of the horizontal curve L= R ∆rad

T= length of the tangent


Length of the long chord C= 2R sin
Example: station calculations
„ Δ = 55

„ R = 1000m station P.I (station of the point of intersection of the two tangents) =20+00

„ find station P.C (first tangent point), P.T (last tangent point)

„ T= R tan ∆/2= 1000x tan 25= 466.31m

„ Station P.C = station P.I –T= 2000- 466.31= 15+33.69

„ L = 1000x 55x = 959.93 m

„ P.T= station P.C + L = 1533.69+ 959.93= 24+93.62

Setting out horizontal curves using tape only

a-offsets from the tangent

(R-y)2 +X2=R2 R-y= √ x

Offset Y=R-√

example: R=100m, ∆=70° y

R R-y

X(m) 0 10 20 30 40 50 60 70.02
Offset 0 0.50 2.02
y(m)
b-offset from the long chord a

ab=y+k y

y=ab-k c/2 x

offset length y=√ -k R k R y+k

k= (constant for the whole curve) x b x

example :
Find the length of offsets from the long chord at distances of 10m

R=90m, ∆=50°

Solution: C= 67.62m, K= 72.51m y=√ -k

x 0 10 20 30 33.81
Offset y 7.50 6.87 4.96 1.66 0.00

Setting out of simple circular curve using theodolite and tape


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.

L= R ∆rad = 349.07m T= R tan ∆/2= 171.99m

Station P.C= 1200- T= 10+18.01

Station P.T= Station P.C+L=13+67.08


.
δ= δ1= =3° 39´ 57˝ δ2, 3,4..= ° ´ ˝

station arc deflection angle δ/2 tangential chord


length δ angle length
(m)
10+18.01
10+50 31.99 3° 39´ 57˝ 1° 49´ 59˝ 1° 49´ 59˝ 31.99
11+00 50 5° 43´ 46˝ 2° 51´ 53˝ 4° 41´ 52˝ 49.97
11+50 50 5° 43´ 46˝ 2° 51´ 53˝ 7° 33´ 45˝ 49.97
12+00 50 5° 43´ 46˝ 2° 51´ 53˝ 10° 25´ 38˝ 49.97
12+50 50 5° 43´ 46˝ 2° 51´ 53˝ 13° 17´ 31˝ 49.97
13+00 50 5° 43´ 46˝ 2° 51´ 53˝ 16° 09´ 24˝ 49.97
13+50 50 5° 43´ 46˝ 2° 51´ 53˝ 19° 01´ 17˝ 49.97
13+67.08 17.08 1° 57´ 26˝ 0° 58´ 43˝ 20° 00´ 00˝ 17.08
∑=349.07 ∑=39° 59´ 59˝ ∑=20° 00´ 00˝
Compound curves : consist of two or more curves joined directly. They are different in
radius. Their centers lies on the same side of the curve. They have common tangent.
Compound curves are used mainly in obtaining desirable shape of the horizontal
alignment…e.g at intersections, ramps, difficult topographic conditions.

R1

R2

Reverse curves
Reverse curves usually consist of two simple curves ( with equal or unequal radii )turning
in opposite direction with a common tangent. Their centers lie on opposite sides of the
curve. They are generally used to change the alignment of the highway .

I1
50°

I2 50°

Equal radius reverse curve : Common tangent I1-I2 =700m find R

Equal radius reverse curve , Common tangent I1-I2 =700m ,∆1=40° ∆2=40°

Straight section on the tangents= 150m … find R


I1

150m

I2

curves or be radii.

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