Horizontal alignment is the
design of the directional transition of a highway in a horizontal plane. Basically,
a horizontal alignment consists, of a horizontal arc and two transition curves
forming a curve which joins two straights. In certain situations, the
transition curve may have zero length. Before designing the curve itself, the
positions of the two straights which the curve will join must be fixed. These
two lines are related by a basic parameter called the intersection angle or angle
of intersection.

Minimum
permitted horizontal radius is dependent on the design speed and the super-elevation
of the carriageway.  In the UK, the
maximum allowable value is 7%, with designs in most cases using a value of 5%.

EXAMPLE: Consider the
following bearings and distances taken by a surveyor along the centre-line of a
proposed road alignment and design the horizontal alignment of curves 1 &
2.

The road alignment is
a street road with a proposed design speed of 38mph.

STEP 1: ADOPT A MINIMUM RADIUS OF CURVATURE FOR THE ALIGNMENT

Adopted design speed is
38mph(60.8km/hr). From Table 1-301  (II)
of Nigerian Highway Designers Manual, 1973, we have the following data:

Interpolating to get R

                                   [begin{array}{l}frac{{40 – 38}}{{38 – 30}} = frac{{395 – R}}{{R – 214}}\frac{2}{8} = frac{{395 – R}}{{R – 214}}\2R – 428 = 3160 – 8R\10R = 3588\R = 358.8ft\R = 358.8 x 0.3048m\R = 109.4mend{array}]

Hence, adopted R = 110m

The coordinate points of the
various points are obtained by scaling it from the Northings and Eastings
established by the surveyor on the drawing; using a scale rule or special
computer programme.

                                   Using
such  a method, the coordinates of all
the points are established  along the
proposed road profile. Also, when the bearing and distance between two points
are known,  the coordinates of other
points could be established using these formulae:

E = L sinθ

N = L cosθ

E₂=
E₁ + E

N₂=
N₁ + N

Where
E₁ and N₁ are initial or known coordinates,

E₂
and N₂ are the new coordinates, while

E
and N are the changes in coordinates

The distance D, between two points
can be established by the relation:

[;D = sqrt {{{(N2 – N1)}^2} + {{(E2 – E1)}^2}} ]

For example: for the straight CD,

[begin{array}{l}D = sqrt {{{(1004177.8 – 1004101.20)}^2} + {{(326483.5 – 326410.00)}^2}} \D = 106.16mend{array}]

The tangent formular can also be
used to calculate the bearing between two points.

Tanθ = ΔΕ/ΔΝ

[begin{array}{l}tan theta  = frac{{(326483.5 – 326410.00)}}{{(1004177.8 – 1004101.20)}}\tan theta  = 0.9595\theta  = 43.8^circ end{array}]

STEP 2: CALCULATE THE TANGENT LENGTHS

[TL = Rtan {textstyle{alpha  over 2}}]

Where TL = Tangent Length,

And α = Deflection angle

For curve 1, deflection angle = 180⁰ – 138⁰27’ = 41⁰33’

[begin{array}{l}TL = 110xtan frac{{41^circ 33′}}{2}\   = 41.73mend{array}]

For curve 2, deflection angle = 180⁰
– 136⁰35’
= 43⁰25’

Let R for curve 2 =
120m

[begin{array}{l}TL = 120xtan frac{{43^circ 25′}}{2}\   = 47.78mend{array}]

STEP 3: CALCULATE CURVE LENGTHS

Curve Length, [CL = Ralpha frac{pi }{{180}}]

For Curve 1, 

[begin{array}{c}CL = 110×41^circ 33’xfrac{pi }{{180}}\79.77mend{array}]

For Curve 2,

[begin{array}{c}CL = 120×43^circ 25’xfrac{pi }{{180}}\90.94mend{array}]