Q. the speed of overtaking and overtaken vehicles are 70 kmph and 40kmph respectively on a two-way road. If the acceleration of overtaking vehicle is 0.99 m/sec2
i. Calculate safe overtaking sight distance.
ii. Draw a neat sketch of overtaking zone showing the position of a sign post.
Soln. :
Given: V =Speed of overtaken vehicle =70 kmph
$V_b$ = Speed of overtaken vehicle =40 kmph
A= Average acceleration during overtaking
=0.99 m/sec2

i. To calculate safe overtaking sight distance (OSD)

OSD = Overtaking sight distance for two way traffic
= $(d_1 +d_2+d_3) = (V_2t + V_bt +2s + VT ) $m ….. (i)
Consider reaction time for overtaking, t=2 sec
Therefolre $d_1 = V_2t = 11.1*2 = 22.2$m
$d_2 = (V_bT + 2s)$
$S = ( 0.7 V_b + 6 )= (0.7 * 11.1 +6 ) = 13.8$ m
$T =\sqrt(\frac{4s}{a})=\sqrt(\frac{4*13.8}{0.99})=7.47$ sec
Thereforee $d_2 = ( 11.1 * 7.47 +2*13.85 ) = 110.5$ m
$d_3 =VT = 19.4 *7.47 = 144.9$ m
Substituting the values of $d_1$, $d_2$ and $d_3$ in equation (i), we get
$OSD = (d_1+d_2+d_3) = (22.2+110.5+144.9)$
= 277.6m = 278 m
OSD = 278 m

ii. To draw a neat sketch of overtaking zone showing portion of sign post :
Let SP1 = Sign post “Overtaking sign ahead”
SP2 = Sign post “end of overtaking zone”

a) Minimum length of overtaking zone = 3 * OSD = 3 * 278 = 834m
b) Desirable length of overtaking zone = 5 * OSD = 5 * 278 = 1390m

Ex. Find the stopping sight distances for level, ascending and descending grades on a road with a design speed of 100km/h. Assume total reaction time as 2.5 second and coefficient of friction as 0.36.Take gradient of a road as 1 in 40.
Soln. :
Given :
V = Design speed = 100 kmph
t = Reaction time = 2.5 sec
f = Coefficient of friction = 0.36
n=Ascending grades=(1/40)x100=+2.5%
n+Descinding Grade=(1/40)x100=-2.5%
To find stopping sight distance(S.S.D) :
S.S.D=0.278 Vt +$\frac{V^2}{254(f±0.001n)}$

(i) For ascending grade on a road :

S.S.D = 0.278 Vt +$\frac{V^2}{254(f+0.001n)}$
= 0.278 x 100 x 2.5 +$\frac{100^2}{254(0.36+0.01 * 2.5)}$
=69.5+$\frac{100^2}{97.79}$
= 69.5+102.2
S.S.S = 171.7 m

(ii) For descending grade on a road :

S.S.D = 0.278 Vt +$\frac{V^2}{254(f-0.001n)}$
= 0.278 x 100 x 2.5 +$\frac{100^2}{254(0.36-0.01 * 2.5)}$
=69.5+$\frac{100^2}{85.09}$
= 69.5+117.5
S.S.S = 187 m

Ex. Calculate the S.S.D required to avoid a head on collision of two cars approaching from the opposite direction at 65 and 55 kmph. Assume a reaction time of 2.5 sec,co-efficient of friction as 0.6 and brake efficiency of 50% in either case.
Ans :
Given:$V_1$=65 kmph =$\frac{65 X 1000}{3600}$= 18.05 m/sec
$V_2$=55 kmph =$\frac{55 X 1000}{3600}$= 15.27 m/sec
t=reaction time=2.5 sec
f=coefficent of friction =0.6
brake efficency =50 %

(i) To calculate stopping site distance (SSD)

stopping distance for one of the car (SD in metres)
=Vt+$\frac{V^2}{2gf}$

As the brake efficiency is 50 %,the wheels will skid through 50% of the braking nad rotate through the remaining distance.Hence ,the value of coefficent of friction developed can be taken as 50 % of the coefficient of friction.
i.e. f=50% of 0.6
therefore f=0.5 x 0.6 = 0.3

the stopping distance of the first car ($SD_1$)=$V_1t +\frac{(V^2)_1}{2gf}$
=$18.05 X 2.5+\frac{18.05^2}{2 X 9.8 X 0.3}$
$SD_1$=45.125+55.408 =100.533 m

for second car $SD_2$=$V_2t +\frac{(V^2)_2}{2gf}$
=$15.27 X 2.5+\frac{15.27^2}{2 X 9.8 X 0.3}$
=38.175+39.655=77.83 m

therefore sight distance to avoid head on collision of the two approaching cars=($SD_1+SD_2$)
=(100.533+77.83)=178.363 m

Ex: Calculate the braking distance for a moving vehicle design speed f 100 kmph
sol :

(i)given: V=design speed = 100 kmph=$\frac{100*1000}{3600}$=27.77 m/sec.

(ii) to calculate the braking distance ($l_b$)

IRC recommended the value of the coefficient of friction (f) as 0.35 for a design speed of 100 kmph.
therefore f=0.35 for 100 kmph
we know $l_b =\frac{V^2}{2gf}$
where $l_b$ = breaking distance in 'm' ; g= 9.8 $m/s^2$;
f=0.35;
v= 27.77 m/sec;
therefore $l_b =\frac{27.77^2}{2*9.8*0.35} =112.4$m
therefore $l_b$=112.4 m