Overtaking Sight Distance- Derivation and Numerical Example

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While designing the highway the most important factor to keep in mind is overtaking sight distance. Without the proper use of these factors, there is no point in designing a road as it may cause several accidents.

But engineers have to work always with safety while designing any structure. So, here I have explained the procedure, derivation, and numerical example of overtaking sight distance (OSD).

I highly request you to read each line carefully to get an idea and full understanding of overtaking sight distance. Let us start.  

What is Overtaking Sight Distance?

The minimum distance on a highway required by a vehicle/driver to overtake a slow-moving vehicle ahead i.e. traveling in the same direction safely against approaching traffic in the opposite direction is known as overtaking sight distance (OSD).

Various overtaking maneuvers models have been suggested and compared with actual crossing practice. One of them is discussed below & based on the following simple assumption. 

  • The overtaken (Slow) vehicle travels at a uniform speed. The overtaking (fast) vehicle reduces its speed and follows the overtaken vehicles and prepares for overtaking operation.
  • The driver requires a short period of time (In average 2 sec.) to perceive the situation and start acceleration when the passing operation is called into play.
  • Overtaking is accomplished under a delayed start & early return. Consider the overtaking model as shown on the next page.

Derivation of Overtaking Sight Distance

Overtaking Sight Distance
Overtaking Sight Distance

Let us consider,  

A1, A2, A3 Stands for overtaking vehicle being the various position of the vehicle during overtaking operation. 

B1 Stands for the overtaken vehicle which reached at position B2 during overtaking operation.  

C1, C2 Stands for the vehicle coming from the opposite direction.   

d1: Distance traveled by overtaking vehicle during reaction time ‘t’ taken by the driver to decide he should overtake or not.  

d2: Distance traveled by overtaking vehicle during actual overtaking operation.  

d3: Distance traveled by a vehicle coming from the opposite direction during actual overtaking operation.   

Vb: Speed of slow moving (overtaken) vehicle. In design practice it is taken as (V- 16),

V = Design Speed in kmph or (V-4.5) V in m/s. [If not given in Numerical]  

Vc: Speed of oncoming vehicle, same as design speed.   

Now from figure,        

(Overtaking sight Distance),                               

OSD = d1 + d2 + d3———(i)           

But, d1 = Vb * t  (t=2 sec.)———(ii)  

And, d2 = b + 2S                

= (Vb*T + 2S)——– (iii)              

Where, S= (0.69 Vb + 6.1) is spacing between vehicles.                  

b = Vb*T   Here, T is overtaking time from A2 to A3.  

Then d2 becomes ,
       d2 = {(Vb*T + 2(0.69Vb + 6.1)}  

Also,         

d2 = VbT + 1/2 (aT^2)———–(iv)  

From Solving equation (iii) and (iv)

We get,
          T = √(4S/a) Where (a) is the acceleration of overtaking the vehicle.  

Again,
        d3 = v * T —————-(v)    

Where (v) is designed speed i.e speed of overtaking vehicles while overtaking and (T) is overtaking time between (A2) to (A3).  

Therefore,
        Putting the value of equation (ii), (iii), (iv), and (v) in equation (i)

We get,    

OSD = d1 + d2 + d3          

= Vb.t+Vb.T+1/2.a.√(4S/a)+vT          

= (Vb + VbT+ 2S + vT)  

For single lane road, neglect (d3)

So, OSD = d1 + d2  

Minimum length of overtaking zone

= 3 * OSD

But, desirable length of overtaking zone

= 5 * OSD  

For clear understanding let us take a numerical example on it.  

Overtaking Sight Distance numerical

Q) The speed of overtaking and the overtaken vehicle is 80kmph and 65 kmph respectively on two-way traffic. The acceleration of the overtaking vehicle is 3.6 kmph. Calculate. (i) Safe overtaking sight distance. (ii) Minimum and desirable overtaking zone.  

Solution:-        

Given,      

Design velocity(v) = 80 kmph                                      

=22.22 m/s Slow moving vehicle

(Vb)= 65 kmph                                        

= 18.05 m/s We know, OSD = d1+d2+d3———(i)  

Let us assume total reaction time = 2 seconds.  

Then,         

d1 = Vb * t               

= 18.05 * 2               

= 36.1 meter. ———(ii)                          

d2 = 2S + b                

=2(0.69Vb+6.1) + Vb*T                

= 2*18.55+18.05+8.61                

= 192.51 meter.——–(iii)                             

d3 = v * T                

= 22.22 * 8.61                

= 191.31 meter.—–(iv)

Hence,           

OSD = d1+d2+d3 {from equation (i).}  

Put the value of equation (ii), (iii), and (iv) in equation (i)

We get ,        

(1) Safe overtaking sight distance          

OSD= 36.1+ 292.51 + 192.51                  

= 420 Approx. (2) Minimum overtaking zone              

= 3 * OSD              

= 3 * 420              

= 1260 meter.      

Desirable overtaking zone              

= 5 * OSD              

= 5 * 420              

= 2100 meter.               

Hence, In this way, the overtaking sight distance can be calculated.  

I Hope friends, you liked my written article on “Procedure, Derivation and Numerical example of overtaking sight distance (OSD)” and remains helpful.

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