Explanation Given: Design hour volume is 5600 veh / h . Percent hour factor is 0.92 . Percentage of trucks is 6 % . Free flow speed is 70 mi / h . Ramp density is 0.75 ramps per mile. Width of lane is 12 ft . Formula used: Write the expression to calculate the f HV . f HV = 1 1 + P T ( E T − 1 ) + P R ( E R − 1 ) ...... (I) Here, P T is the decimal portion of trucks in the traffic stream, P R is the decimal portion of RV's in the traffic stream, E T is passenger car equivalent for trucks, and E R is passenger car equivalent for RV's. Find the number of lanes in each direction using the following relation. v p = V ( P H F ) ( N ) ( f p ) ( f HV ) ...... (II) Here, V is demand volume for the entire peak hour, P H F is peak hour factor, f p is the driver population factor, N is the number of lanes and f HV is the adjustment factor to account for heavy vehicles in the traffic stream. Calculate the free flow speed by using the formula. F F S = B F F S − f LW − f LC − f N − f HD ...... (III) Here, B F F S is the basic free flow speed, f LW is the adjustment for lane width, f LC is the adjustment for right shoulder lateral clearance, f N is the adjustment for number of lanes and f HD is the adjustment for interchange density. Write the expression to calculate the speed. S = F F S − [ 1 9 ( 7 F F S − 340 ) ( v p + 30 F F S − 3400 40 F F S − 1700 ) 2.6 ] ...... (IV) Here, S is the speed. Write the expression to calculate the density using the formula. D = v p S ...... (V) Here, D is the density, v p is the passenger car equivalent and S is the speed. Calculation: Refer table 9.3, "PCE's for heavy vehicles in general terrain segments". In case of rolling terrain, The value of passenger car equivalent for trucks ( E T ) is 2.5 . The value of passenger car equivalent for recreational vehicles is ( E R ) is 2.0 . Substitute 6 % for P T , 2.5 for E T , 0 for P R and 2.0 for E R in equation (I). f HV = 1 1 + 6 % ( 2.5 − 1 ) + 0 ( 2 − 1 ) = 1 1 + 0.06 ( 2.5 − 1 ) = 0.917 Consider the value for the driver population factor, f p = 1.0 . Consider 2 lanes: Substitute 5600 veh / h for V , 0.92 for P H F , 2 for N , 1 for f p and 0.917 for f HV in equation (II). v p = 5600 veh / h ( 0.92 ) ( 2 ) ( 1 ) ( 0.917 ) = 3318.95 pc / h / ln Substitute 70 mi / h for B F F S , 0 for f LW , 0 for f LC , 4.5 mi / h for f N and 0 for f HD in equation (III)