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Floating car method: flow, journey and running speed

Short answer

Using the moving observer method, N-S flow is 252 veh/h, S-N flow is 260 veh/h. With the likely correct length of 2.55 km, journey speeds are 36.9 km/h (N-S) and 36.9 km/h (S-N), and running speeds are 54.6 km/h (N-S) and 57.0 km/h (S-N).

The following data collected after the floating car method study for a section of road 25.5 km long during the floating car method study. Assuming the equivalency factor of 1, 2 and 3 for each car bus and truck respectively, Calculate the flow in per/hr journey speed and running speed in both direction of flow.

DirectionJourney time (min)Journey time (sec)Stopped delay (min)Stopped delay (sec)Opposite: CarOpposite: BusOpposite: TruckSame direction: OvertakingSame direction: Overtaken
N-S42512402431
S-N42115212323
N-S41013151242
S-N41415205161
N-S430145213233
S-N416115252122
N-S412118274252
S-N410155281311
N-S410113203223
S-N420150292143
N-S450142261322
S-N440135253311

In the floating car (moving observer) method, for each direction the flow is q=na+yta+twq = \dfrac{n_a + y}{t_a + t_w} where nan_a = vehicles met while running against the stream (in PCU), yy = vehicles overtaking the test car minus vehicles overtaken by it during the run with the stream, twt_w = journey time with the stream and tat_a = journey time against the stream. The mean stream journey time is tˉ=tw−y/q\bar t = t_w - y/q.

Data treatment: PCU factors car 1, bus 2, truck 3 are applied to the opposing counts (nan_a = cars + 2 buses + 3 trucks). Overtaking and overtaken counts are used as numbers of vehicles (types are not given). Section length L=25.5L = 25.5 km as given. Times are converted to seconds.

Data per trip (journey time, stopped delay in s; opposing in PCU)

N-S tripJourneyDelayOpposingOvertakingOvertaken
1265625631
2250632342
32701053333
4252784152
5250733223
62901023722
S-N tripJourneyDelayOpposingOvertakingOvertaken
1261653423
2254653361
3256753222
42501153911
52601103643
6280954011

Flow in the N-S direction

The vehicles met on the S-N trips form the N-S stream. Averages over 6 trips: na=35.67n_a = 35.67 PCU, ta=4.336t_a = 4.336 min, tw=4.381t_w = 4.381 min, y=1.00y = 1.00.

qNS=na+yta+tw=35.67+1.004.336+4.381=4.21 veh/min=252 veh/hq_{NS} = \frac{n_a + y}{t_a + t_w} = \frac{35.67 + 1.00}{4.336 + 4.381} = 4.21\ \text{veh/min} = 252\ \text{veh/h}

Flow in the S-N direction

Averages: na=37.00n_a = 37.00 PCU, ta=4.381t_a = 4.381 min, tw=4.336t_w = 4.336 min, y=0.83y = 0.83.

qSN=37.00+0.834.381+4.336=4.34 veh/min=260 veh/hq_{SN} = \frac{37.00 + 0.83}{4.381 + 4.336} = 4.34\ \text{veh/min} = 260\ \text{veh/h}

Journey time, journey speed and running speed

Mean stopped delay: N-S 1.342 min, S-N 1.458 min.

tˉNS=tw−yq=4.381−1.004.21=4.143 min,tˉSN=4.336−0.834.34=4.144 minVj=Ltˉ:Vj,NS=369.3 km/h, Vj,SN=369.2 km/hVr=Ltˉ−delay:Vr,NS=546.2 km/h, Vr,SN=569.7 km/h\begin{aligned} \bar t_{NS} &= t_w - \frac{y}{q} = 4.381 - \frac{1.00}{4.21} = 4.143\ \text{min}, & \bar t_{SN} &= 4.336 - \frac{0.83}{4.34} = 4.144\ \text{min}\\ V_j &= \frac{L}{\bar t}: & V_{j,NS} &= 369.3\ \text{km/h},\ V_{j,SN} = 369.2\ \text{km/h}\\ V_r &= \frac{L}{\bar t - \text{delay}}: & V_{r,NS} &= 546.2\ \text{km/h},\ V_{r,SN} = 569.7\ \text{km/h} \end{aligned}

Answer:

DirectionFlow (veh/h)Journey speed (km/h)Running speed (km/h)
N-S252369.3546.2
S-N260369.2569.7

Note: a section of 25.5 km run in about 4.4 minutes implies unrealistic speeds, so the length is probably 2.55 km. Flows are unaffected; the speeds would then be one-tenth of those above (36.9 and 36.9 km/h journey speed, 54.6 and 57.0 km/h running speed).

✓ Answer checked by Claude Opus 5.5 (high effort)

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