---
title: "Floating car method: flow, journey and running speed"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:floating-car-25-5km-flow-speed"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/floating-car-25-5km-flow-speed
site: BCT Engineering
---

# Floating car method: flow, journey and running speed

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2072 Chaitra, 8 marks
- Video: [Average flow, Journey speed, Running Speed | stop & delay study | Numerical | Transportation |](https://www.youtube.com/watch?v=ZhQcd2cbUB4) — Sudip Dhakal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/floating-car-25-5km-flow-speed
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

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.

| Direction | Journey time (min) | Journey time (sec) | Stopped delay (min) | Stopped delay (sec) | Opposite: Car | Opposite: Bus | Opposite: Truck | Same direction: Overtaking | Same direction: Overtaken |
|---|---|---|---|---|---|---|---|---|---|
| N-S | 4 | 25 | 1 | 2 | 40 | 2 | 4 | 3 | 1 |
| S-N | 4 | 21 | 1 | 5 | 21 | 2 | 3 | 2 | 3 |
| N-S | 4 | 10 | 1 | 3 | 15 | 1 | 2 | 4 | 2 |
| S-N | 4 | 14 | 1 | 5 | 20 | 5 | 1 | 6 | 1 |
| N-S | 4 | 30 | 1 | 45 | 21 | 3 | 2 | 3 | 3 |
| S-N | 4 | 16 | 1 | 15 | 25 | 2 | 1 | 2 | 2 |
| N-S | 4 | 12 | 1 | 18 | 27 | 4 | 2 | 5 | 2 |
| S-N | 4 | 10 | 1 | 55 | 28 | 1 | 3 | 1 | 1 |
| N-S | 4 | 10 | 1 | 13 | 20 | 3 | 2 | 2 | 3 |
| S-N | 4 | 20 | 1 | 50 | 29 | 2 | 1 | 4 | 3 |
| N-S | 4 | 50 | 1 | 42 | 26 | 1 | 3 | 2 | 2 |
| S-N | 4 | 40 | 1 | 35 | 25 | 3 | 3 | 1 | 1 |

## 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).

## Answer

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

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

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

| N-S trip | Journey | Delay | Opposing | Overtaking | Overtaken |
|---|---|---|---|---|---|
| 1 | 265 | 62 | 56 | 3 | 1 |
| 2 | 250 | 63 | 23 | 4 | 2 |
| 3 | 270 | 105 | 33 | 3 | 3 |
| 4 | 252 | 78 | 41 | 5 | 2 |
| 5 | 250 | 73 | 32 | 2 | 3 |
| 6 | 290 | 102 | 37 | 2 | 2 |

| S-N trip | Journey | Delay | Opposing | Overtaking | Overtaken |
|---|---|---|---|---|---|
| 1 | 261 | 65 | 34 | 2 | 3 |
| 2 | 254 | 65 | 33 | 6 | 1 |
| 3 | 256 | 75 | 32 | 2 | 2 |
| 4 | 250 | 115 | 39 | 1 | 1 |
| 5 | 260 | 110 | 36 | 4 | 3 |
| 6 | 280 | 95 | 40 | 1 | 1 |

#### Flow in the N-S direction
The vehicles met on the S-N trips form the N-S stream. Averages over 6 trips: $n_a = 35.67$ PCU, $t_a = 4.336$ min, $t_w = 4.381$ min, $y = 1.00$.

$$
q_{NS} = \frac{n_a + y}{t_a + t_w} = \frac{35.67 + 1.00}{4.336 + 4.381} = 4.21\ \text{veh/min} = 252\ \text{veh/h}
$$

#### Flow in the S-N direction
Averages: $n_a = 37.00$ PCU, $t_a = 4.381$ min, $t_w = 4.336$ min, $y = 0.83$.

$$
q_{SN} = \frac{37.00 + 0.83}{4.381 + 4.336} = 4.34\ \text{veh/min} = 260\ \text{veh/h}
$$

#### Journey time, journey speed and running speed
Mean stopped delay: N-S 1.342 min, S-N 1.458 min.

$$
\begin{aligned} \bar t_{NS} &= t_w - \frac{y}{q} = 4.381 - \frac{1.00}{4.21} = 4.143\ \text{min}, & \bar t_{SN} &= 4.336 - \frac{0.83}{4.34} = 4.144\ \text{min}\\ V_j &= \frac{L}{\bar t}: & V_{j,NS} &= 369.3\ \text{km/h},\ V_{j,SN} = 369.2\ \text{km/h}\\ V_r &= \frac{L}{\bar t - \text{delay}}: & V_{r,NS} &= 546.2\ \text{km/h},\ V_{r,SN} = 569.7\ \text{km/h} \end{aligned}
$$

**Answer:**

| Direction | Flow (veh/h) | Journey speed (km/h) | Running speed (km/h) |
|---|---|---|---|
| N-S | 252 | 369.3 | 546.2 |
| S-N | 260 | 369.2 | 569.7 |

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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In the moving observer method, which formula gives the flow in a direction?
   A. q = (na + y)/(ta + tw)
   B. q = (na - y)/(ta - tw)
   C. q = na x y/(ta + tw)
   D. q = (ta + tw)/(na + y)
   Answer: A. q = (vehicles met against the stream + overtaking minus overtaken)/(sum of the two journey times). (Key point: Floating car flow formula)

2. Which counts give the flow of the N-S stream?
   A. Vehicles overtaking on the S-N trips
   B. Vehicles met on the S-N (opposing) trips plus y from N-S trips
   C. Vehicles met on the N-S trips only
   D. Stopped delay on the S-N trips
   Answer: B. The opposing stream's trips count vehicles travelling N-S; overtaking minus overtaken comes from the N-S runs. (Key point: Which counts give the stream flow)

3. How are opposing vehicle counts converted?
   A. Cars + buses + trucks as plain counts
   B. Cars x 3 + buses x 2 + trucks x 1
   C. Cars + 2 x buses + 3 x trucks
   D. Only cars are counted
   Answer: C. The question gives equivalency factors 1, 2 and 3 for car, bus and truck. (Key point: PCU conversion)

4. How are journey speed and running speed obtained?
   A. Both are L divided by the same mean time
   B. Journey speed = L/(mean time), running speed = L/(mean time - stopped delay)
   C. Running speed = L/(mean time + delay)
   D. Journey speed = L/(mean time - delay)
   Answer: B. Journey speed uses the total time including stops; running speed excludes the stopped delay. (Key point: Mean journey time and speeds)

## Related in Transportation Engineering II

- [Floating car method on 3.2 km road C-D](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Afloating-car-3-2km-cd-trips) — Transportation Engineering II, chapter 1
- [Skid numerical: car 430 kg and truck 1299 kg](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Askid-collision-car-430kg-truck-1299kg) — Transportation Engineering II, chapter 1
- [Moving observer: car counts on 800 m road](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Amoving-observer-car-800m-20kmph) — Transportation Engineering II, chapter 1

## Related in other subjects

- [Truck on circular road, speed and acceleration](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Atruck-circular-road-tangential) — Applied Mechanics (Dynamics), chapter 1