Floating car method on 3.2 km road C-D

For each direction, the floating car (moving observer) method gives the flow $q = \dfrac{n_a + y}{t_a + t_w}$ where $n_a$ = vehicles counted from the opposite direction while running against the stream (these are the vehicles of the direction being studied), $y$ = overtaking minus overtaken during the run with the stream, $t_w$ = journey time with the stream and $t_a$ = journey time against it. Odd trips are C-D and even trips are D-C; $L = 3.2$ km.

Direction C-D (stream moving C to D)

Averages of 5 trips: $t_w = 6.538$ min (this direction), $t_a = 7.562$ min (opposite direction), opposing vehicles $n_a = 255.4$, overtaking 4.0, overtaken 3.8 so $y = 0.2$, mean stopped delay 1.586 min.

$$\begin{aligned} q &= \frac{n_a + y}{t_a + t_w} \times 60 = \frac{255.4 + (0.2)}{7.562 + 6.538} \times 60 = 1088\ \text{veh/h}\\ \bar t &= t_w - \frac{y}{q} = 6.538 - \frac{0.2}{18.13} = 6.527\ \text{min}\\ V_j &= \frac{L}{\bar t} = \frac{3.2 \times 60}{6.527} = 29.4\ \text{km/h}\\ V_r &= \frac{L}{\bar t - \text{delay}} = \frac{3.2 \times 60}{6.527 - 1.586} = 38.9\ \text{km/h} \end{aligned}$$

Direction D-C (stream moving D to C)

Averages of 5 trips: $t_w = 7.562$ min (this direction), $t_a = 6.538$ min (opposite direction), opposing vehicles $n_a = 288.0$, overtaking 3.0, overtaken 2.4 so $y = 0.6$, mean stopped delay 1.854 min.

$$\begin{aligned} q &= \frac{n_a + y}{t_a + t_w} \times 60 = \frac{288.0 + (0.6)}{6.538 + 7.562} \times 60 = 1228\ \text{veh/h}\\ \bar t &= t_w - \frac{y}{q} = 7.562 - \frac{0.6}{20.47} = 7.533\ \text{min}\\ V_j &= \frac{L}{\bar t} = \frac{3.2 \times 60}{7.533} = 25.5\ \text{km/h}\\ V_r &= \frac{L}{\bar t - \text{delay}} = \frac{3.2 \times 60}{7.533 - 1.854} = 33.8\ \text{km/h} \end{aligned}$$

Average of the two directions

Answer: C-D: volume 1088 veh/h, journey speed 29.4 km/h, running speed 38.9 km/h; D-C: volume 1228 veh/h, journey speed 25.5 km/h, running speed 33.8 km/h. Average of both: 1158 veh/h, 27.5 km/h and 36.3 km/h.