Answer:
•Estimated density = 39.685Pc/mi/en
•Level of service, LOS frequency = LOSC
Explanation:
We are given:
•Freeway current lane width,B = 12ft
• freeway current shoulder width,b = 6ft
• percentage of heavy vehicle, Ptb = 10℅
• peak hour factor, PHF = 0.9
Let's consider,
•Number of lanes N = 4
• flow of traffic V = 7500vph
• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph
• cars equivalent for recreational purpose Er= 2
•cars to be used for trucks and busses Etb= 2.5
Let's first calculate for the heavy adjustment factor.
We have:
![F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}](https://tex.z-dn.net/?f=%20F_H_v%20%3D%20%5Cfrac%7B1%7D%7B1%2BP_t_b%28E_t_b-1%29%2BPr%28Er-1%29%7D%20)
Substituting figures in the equation we have:
![= \frac{1}{1+0.1(2.5-1)+0(2-1)}](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B1%7D%7B1%2B0.1%282.5-1%29%2B0%282-1%29%7D)
= 0.75
Let's now calculate equivalent flow rate of the car using:
![Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}](https://tex.z-dn.net/?f=Vp%20%3D%20%5Cfrac%7BV%7D%7B%28P_H_F%29%2AN%29%2A%28F_H_v%29%2A%28F_p%29%7D%20)
![= \frac{7500}{0.9*4*0.75*1}](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B7500%7D%7B0.9%2A4%2A0.75%2A1%7D%20)
= 2777.7 pc/h/en
Calculating for traffic density, we have:
![D = \frac{Vp}{FFS}](https://tex.z-dn.net/?f=%20D%20%3D%20%5Cfrac%7BVp%7D%7BFFS%7D%20)
![D = \frac{2777.7}{70}](https://tex.z-dn.net/?f=%20D%20%3D%20%5Cfrac%7B2777.7%7D%7B70%7D%20)
D = 39.685 Pc/mi/en
Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC