Linear speed-density: free speed 80, jam 130

(i) Speed-density and flow-density equations

For a linear (Greenshields) relation, speed falls from the free-flow speed $v_f$ at zero density to zero at jam density $k_j$:

$$v = v_f\left(1-\frac{k}{k_j}\right) = 80\left(1-\frac{k}{130}\right) = 80 - 0.6154\,k$$

Since $q = v\,k$:

$$q = v_f\,k\left(1-\frac{k}{k_j}\right) = 80\,k - 0.6154\,k^2$$

Critical values and diagrams

Setting $dq/dk = 0$ gives $k_m = k_j/2$, $v_m = v_f/2$ and $q_{max} = v_f k_j/4$:

$$\begin{aligned} k_m &= 130/2 = 65\ \text{veh/km}\\ v_m &= 80/2 = 40\ \text{km/h}\\ q_{max} &= \frac{80 \times 130}{4} = 2600\ \text{veh/h} \end{aligned}$$

v-k diagram (straight line)
 v (km/h)
   80 |*
      |   *
   40 |- - - *   <- at q max
      |         *
      +-----------*--- k (veh/km)
      0    65     130

q-k diagram (parabola)
 q (veh/h)
 2600 |        _ * _   <- q max
      |     *         *
      |   *             *
      +--*-------|-------*-- k
      0        65       130

v-q diagram (half parabola)
 v (km/h)
   80 |*
      |  *
   40 |- - - *  <- q max (nose)
      |     *
      |   *   (congested branch)
      +--------|--------- q (veh/h)
      0       2600

(ii) Speed and density for a flow of 1000 veh/h

Substitute in $q = v_f k - (v_f/k_j)k^2$:

$$0.6154k^2 - 80k + 1000 = 0 \quad\Rightarrow\quad k = \frac{80 \pm \sqrt{80^2 - 4(0.6154)(1000)}}{2(0.6154)}$$ $$k_1 = 14.01\ \text{veh/km},\quad k_2 = 115.99\ \text{veh/km}$$

• Uncongested (stable) state: $k = 14.0$ veh/km and $v = 80(1-14.01/130) = 71.4$ km/h.
• Congested (forced) state: $k = 116.0$ veh/km and $v = 8.6$ km/h.

Answer: for 1000 veh/h, speed = 71.4 km/h at density = 14.0 veh/km (stable flow); the congested alternative is 8.6 km/h at 116.0 veh/km.

(iii) Headway, spacing, clearance and gap at maximum flow

At $q_{max} = 2600$ veh/h, $k_m = 65$ veh/km, $v_m = 40$ km/h:

$$\begin{aligned} \text{Average headway}\ h &= \frac{3600}{q_{max}} = \frac{3600}{2600} = 1.38\ \text{s}\\ \text{Average spacing}\ s &= \frac{1000}{k_m} = \frac{1000}{65} = 15.38\ \text{m}\\ \text{Clearance}\ &= s - L = 15.38 - 6 = 9.38\ \text{m}\\ \text{Gap (time)}\ &= h - \frac{L}{v_m} = 1.38 - \frac{6}{11.111} = 0.84\ \text{s} \end{aligned}$$

Answer: headway = 1.38 s, spacing = 15.38 m, clearance = 9.38 m, gap = 0.84 s.