Module 2: Geometric design of highways

Lecture 13 Sight distance

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Overtaking sight distance

Figure 1: Time-space diagram: Illustration of overtaking sight distance

The overtaking sight distance is the minimum distance open to the vision of the driver of a vehicle intending to overtake the slow vehicle ahead safely against the traffic in the opposite direction. The overtaking sight distance or passing sight distance is measured along the center line of the road over which a driver with his eye level 1.2 m above the road surface can see the top of an object 1.2 m above the road surface.

The factors that affect the OSD are:

• Velocities of the overtaking vehicle, overtaken vehicle and of the vehicle coming in the opposite direction.
• Spacing between vehicles, which in-turn depends on the speed
• Skill and reaction time of the driver
• Rate of acceleration of overtaking vehicle
• Gradient of the road

The dynamics of the overtaking operation is given in the figure which is a time-space diagram. The x-axis denotes the time and y-axis shows the distance traveled by the vehicles. The trajectory of the slow moving vehicle (B) is shown as a straight line which indicates that it is traveling at a constant speed. A fast moving vehicle (A) is traveling behind the vehicle B. The trajectory of the vehicle is shown initially with a steeper slope. The dotted line indicates the path of the vehicle A if B was absent. The vehicle A slows down to follow the vehicle B as shown in the figure with same slope from $t_0$ to $t_1$. Then it overtakes the vehicle B and occupies the left lane at time $t_3$. The time duration $T = t_3- t_1$ is the actual duration of the overtaking operation. The snapshots of the road at time $t_0, t_1$, and $t_3$ are shown on the left side of the figure. From the Figure 1, the overtaking sight distance consists of three parts.

• $d_1$ the distance traveled by overtaking vehicle A during the reaction time $t = t_1 - t_0$
• $d_2$ the distance traveled by the vehicle during the actual overtaking operation $T = t3 - t1$
• $d_3$ is the distance traveled by on-coming vehicle C during the overtaking operation ($T$).

Therefore:

$$OSD=d_1+d_2+d_3$$ (1)

It is assumed that the vehicle A is forced to reduce its speed to $v_b$, the speed of the slow moving vehicle B and travels behind it during the reaction time $t$ of the driver. So $d_1$ is given by:

$$d_1=v_bt$$ (2)

Then the vehicle A starts to accelerate, shifts the lane, overtake and shift back to the original lane. The vehicle A maintains the spacing $s$ before and after overtaking. The spacing $s$ in $m$ is given by:

$$s=0.7v_b+6$$ (3)

Let $T$ be the duration of actual overtaking. The distance traveled by B during the overtaking operation is $2s+v_bT$. Also, during this time, vehicle A accelerated from initial velocity $v_b$ and overtaking is completed while reaching final velocity $v$. Hence the distance traveled is given by:

$$d_2 = v_bT+\frac{1}{2}aT^2$$

$$2s+v_bT = v_bT+\frac{1}{2}aT^2$$

$$2s = \frac{1}{2}aT^2$$

$$T = \sqrt{\frac{4s}{a}}$$

$$d_2 = 2s+v_b\sqrt{\frac{4s}{a}}$$ (4)

The distance traveled by the vehicle C moving at design speed $v$ $m/sec$ during overtaking operation is given by:

$$d_3=vT$$ (5)

The the overtaking sight distance is (Figure 1)

$$OSD=v_bt+2s+v_b\sqrt{\frac{4s}{a}}+vT$$ (6)

where $v_b$ is the velocity of the slow moving vehicle in $m/sec^2$, $t$ the reaction time of the driver in $sec$, $s$ is the spacing between the two vehicle in $m$ given by equation 3 and $a$ is the overtaking vehicles acceleration in $m/sec^2$. In case the speed of the overtaken vehicle is not given, it can be assumed that it moves 16 kmph slower the the design speed.