Islamic University of Gaza

Civil Engineering Department

Surveying II

ECIV 2332

By

Belal Almassri

Chapter 9

Route Surveying – Part 3

- Linear Methods for setting out simple circular curve.

- Method of offsets on the long chord.

- Method of offsets on tangents.

- Method of radial offsets.

- Method of deflection angles.

- Notes and Examples.

Setting out simple circular curve

using linear methods These methods use the chain surveying

tools only.

These methods are used for the short

curves which doesn’t require high degree

of accuracy.

These methods are used for the clear

situations on the road intersections.

Types of linear method:

There are three types of the linear methods

to set out a simple circular curve.

1. By offsets from the long chord.

2. By offsets from the tangents.

3. By radial offsets.

By offsets from the long chord

Example

By offsets from the tangent

Example

Notes:

In the first method, the value of x = Lc/2

= 7.654 < 8m so we had used x = 7.654m

and y at this point equals zero.

In the second method, the value of x =

8m < T(8.28), we had used x = 8m but we

still know that the circular curve close at

PT when T = 8.28m.

By radial offsets

Example

Notes:

In the second and the third methods the

x value will not exceed the value of T

BUT in the first method the value of x

will not exceed the value of Lc/2.

The third method is used when the

centre of the circular curve is accessible

while the first two methods can be used

when there is an obstacle.

Underground . . .

Laying out simple circular curves by

using the deflection angles method:

Working method:

1. Fix the theodolite device to be at point PC and directed at point PI.

2. Measure the deflection angles d and the chords C.

3. Connect the ends of the chords to draw the curve.

Deflection Angles: the angles between the tangent and the ends of the chords from point PC.

Calculations Steps:

This method is a geometric based method :

1. Calculate the values of T and L. T = R tan (Δ/2)

L = R (Δ in radians)

2. Calculate the chainage of point PC and

point PI. Ch of PC = Ch of PI – T

Ch of PT = Ch of PC + L

3. Calculate the partial chords C, C1,C2. Choose C ≤ R/20 (then round it)

Chainage of the first station: Chainage of PC (rounded) + C

C1 = Chainage of first station – Chainage of PC (original)

C2 = L – ( C1 + n C ); n = number of intermediate chords

4. Calculate the deflection angles d, d1, d2. d = (28.648 C)/R

d1 = (28.648 C1)/R

d2 = (28.648 C2)/R

5. Calculate the cumulative deflection angles.

Notes:

The total d’s will equal Δ/2.

The chainage of the all stations on the

curve should be even number divisible of

5 or 10.

When locating the last point before PT

we measure the distance to the PT if it

was equal C2 then it is correct or the

difference is 5- 10 cm BUT if it is more

than that we should repeat !

Example 9.2

Two tangents intersect at PI with chainage of

2140.00 m and deflection angle of 10 ͦ 35 ` 2``

Da or D˳ = 4 ͦ , Using the deflection angles

method Arrange all the information needed.

Solution !