Traffic Engineering

Driver characteristics are the personal attributes that decide how a driver senses, decides and acts in traffic. The critical ones are physical, mental and psychological.

Physical characteristics

• Vision: acuity (about 6/6 needed for a licence), the cone of clear vision (3-5 degrees), the field of vision (clear about 10 degrees, fairly clear about 12-14 degrees, peripheral up to about 120-180 degrees), colour perception (colour blindness for red/green is critical for signals), glare sensitivity, glare recovery and depth perception. Speed narrows the field of vision (at 100 km/h only about 40 degrees).
• Hearing: horn and siren detection; deafness is partly compensated by vision, but affects awareness of emergency vehicles.
• Reaction ability: muscle coordination, strength and speed of hand/foot response; fatigue, age and disability reduce it.
• Age and health: reaction slows and vision/hearing fade in old age; illness, epilepsy and heart conditions are hazards.
• Fatigue, alcohol and drugs: lengthen reaction time, impair judgement and cause drowsiness; Nepal's Motor Vehicles and Transport Management Act 2049 prohibits driving under the influence.

Mental characteristics

• Intelligence and judgement: ability to judge speed, distance, gaps and the intentions of other users.
• Knowledge and training: familiarity with traffic rules, signs and the route; unfamiliar drivers hesitate and make errors.
• Attention and concentration: a driver can attend to only a limited amount of information; distractions (mobile phone, passengers) reduce attention.
• Memory and learning: helps in recognising signs and routes and in forming habits.
• Perception-reaction process (PIEV): perception, intellection, emotion and volition, which together determine the total reaction time.

Psychological characteristics

• Emotional state and temperament: anger, stress, worry or elation lead to aggressive driving, overtaking in unsafe places and over-speeding.
• Attitude and behaviour: attitude towards rules, other users and risk; over-confident or risk-taking drivers have more crashes.
• Anxiety and fear: a nervous driver may brake suddenly or hesitate at intersections.
• Social habits: lane discipline, courtesy, and tendency to follow the crowd.
• Driving habits and experience: young drivers take more risks; experienced drivers anticipate hazards better.
• Response to stimuli: drivers often react to expected events but are slow to react to unexpected ones (for example a stationary vehicle on a bend).

Design implication

Because drivers vary, design values are taken for an average or conservative driver: perception-reaction time of 2.5 s for sight distance (IRC:66, IRC:SP:23; Nepal Road Standard 2070), letter sizes of signs matched to acuity, and lighting and markings to help night vision.

A collision on a flat road is solved by (i) finding each vehicle's speed just after impact from its post-collision skid, (ii) applying conservation of linear momentum as a vector equation to get the speeds just before impact, and (iii) adding back the energy lost in the pre-collision skid.

Given: $W_A = 4$ t, $W_B = 6$ t, $f = 0.52$, $g = 9.81\ \text{m/s}^2$. Pre-collision skids: $d_A = 32$ m, $d_B = 18$ m. Post-collision skids: $d_A' = 14$ m, $d_B' = 33$ m.

Vehicle A comes from the North (moving South). Vehicle B comes from the West (moving East). After impact A slides 47 degrees East of North and B slides 56 degrees South of East.

where $\theta$ is the bearing from North, clockwise: $\theta_A = 180^\circ$, $\theta_B = 90^\circ$ (travel directions) and $\theta_A' = 47^\circ$, $\theta_B' = 146^\circ$ (skid directions after impact).

   before impact          after impact
   A: bearing 180 deg       A': bearing 47 deg
   B: bearing 90 deg       B': bearing 146 deg
   (bearing measured clockwise from North)

Step 1: Speeds just after collision

From work-energy, $v' = \sqrt{2 g f d'}$:

$$\begin{aligned} v_A' &= \sqrt{2(9.81)(0.52)(14)} = 11.95\ \text{m/s}\\ v_B' &= \sqrt{2(9.81)(0.52)(33)} = 18.35\ \text{m/s} \end{aligned}$$

Step 2: Momentum balance (vector form)

Take $x$ = East and $y$ = North, and let $u_A, u_B$ be the speeds just before impact. Weights are proportional to masses, so they can be used directly.

$$\begin{aligned} W_A u_A \sin\theta_A + W_B u_B \sin\theta_B &= W_A v_A' \sin\theta_A' + W_B v_B' \sin\theta_B' \\ W_A u_A \cos\theta_A + W_B u_B \cos\theta_B &= W_A v_A' \cos\theta_A' + W_B v_B' \cos\theta_B' \end{aligned}$$

Substituting the numbers:

$$\begin{aligned} 4\,u_A(0.0000) + 6\,u_B(1.0000) &= 4(11.951)(0.7314) + 6(18.349)(0.5592) = 96.526 \\ 4\,u_A(-1.0000) + 6\,u_B(0.0000) &= 4(11.951)(0.6820) + 6(18.349)(-0.8290) = -58.668 \end{aligned}$$

Solving: $u_A = 14.67\ \text{m/s} = 52.8\ \text{km/h}$ and $u_B = 16.09\ \text{m/s} = 57.9\ \text{km/h}$ at the instant of impact.

Step 3: Original speeds (before the pre-collision skid)

$v_0 = \sqrt{u^2 + 2 g f d}$:

$$\begin{aligned} v_A &= \sqrt{(14.67)^2 + 2(9.81)(0.52)(32)} = 23.27\ \text{m/s}\\ v_B &= \sqrt{(16.09)^2 + 2(9.81)(0.52)(18)} = 21.03\ \text{m/s} \end{aligned}$$

Answer: original speed of vehicle $A$ = 83.8 km/h and of vehicle $B$ = 75.7 km/h (speeds at impact: 52.8 and 57.9 km/h).

Isolated two-phase signal with pedestrian indications for road H (12 m) and road F (9.6 m). The ambers are given (5 s and 4 s), so the approach speeds are not needed for them.

Design basis (IRC:93-1985, Guidelines on Design and Installation of Road Traffic Signals, with Webster's method): two-phase signal, Phase 1 = road H moves, Phase 2 = road F moves. Pedestrians cross a road while the traffic on the cross road has green. Assumed (not given): saturation flow 1800 PCU/h per lane, start-up lost time 2 s per phase, pedestrian walking speed 1.2 m/s and initial walk (pedestrian start-up) time 5 s.

Step 1: Amber (change) time

Amber times are given: road H = 5 s, road F = 4 s.

Step 2: Pedestrian crossing time

Time to cross = initial walk + width / walking speed:

$$\begin{aligned} t_{p,H} &= 5 + \frac{12}{1.2} = 15.0\ \text{s}\\ t_{p,F} &= 5 + \frac{9.6}{1.2} = 13.0\ \text{s} \end{aligned}$$

Pedestrians crossing road H walk during Phase 2, so Phase 2 green must be at least 15.0 s. Pedestrians crossing road F walk during Phase 1, so Phase 1 green must be at least 13.0 s.

Step 3: Webster cycle time from traffic

$$\begin{aligned} y_H &= \frac{475}{1800} = 0.264, \quad y_F = \frac{325}{1800} = 0.181, \quad Y = 0.444\\ L &= (2+5) + (2+4) = 13\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(13)+5}{1-0.444} = 44.1\ \text{s} \end{aligned}$$ $$\begin{aligned} g_H &= \frac{0.264}{0.444}(31.1) = 18.5\ \text{s}, \quad G_H = g_H + 2 = 20.5\ \text{s}\\ g_F &= \frac{0.181}{0.444}(31.1) = 12.6\ \text{s}, \quad G_F = g_F + 2 = 14.6\ \text{s} \end{aligned}$$

Step 4: Check against pedestrian needs

Adopted cycle length $C = 21 + 5 + 15 + 4 = 45$ s.

Final timings

Pedestrian signals: pedestrians crossing road F get WALK for 5 s then flashing (clearance) for 8 s during Phase 1 (total 13 s of 21 s green); pedestrians crossing road H get WALK for 5 s then flashing for 10 s during Phase 2 (total 15 s of 15 s green). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

Phase diagram

Cycle = 45 s
Road H: [GREEN 21][A 5][  RED 19  ]
Road F: [ RED 26 ][GREEN 15][A 4]
Ped crossing F: [WALK+FLASH 21][DON'T WALK 24]
Ped crossing H: [DON'T WALK 26][WALK+FLASH 15][DW]

General requirements of traffic control devices

1. Fulfil a need: each device must be justified by a real need based on a traffic study.
2. Command attention: be conspicuous (size, colour, shape, retro-reflectivity).
3. Convey a clear and simple meaning that can be understood at a glance.
4. Command respect: be used only where needed, not over-used, so that drivers obey.
5. Give adequate time for response: placed in advance of the hazard, with enough time to read and act.
6. Be uniform: the same type of device, shape and colour everywhere in the country.
7. Be legible and visible at night (reflectorised or illuminated).
8. Be maintained: clean, repainted, repaired.
9. Be properly located and enforced by traffic police.

Types of traffic signs

Subdivisions: informatory signs include direction signs, place identification, route markers, facility information and parking signs.

Placement: within the cone of clear vision, 3 m lateral clearance from the edge, with a mounting height of 2.0-2.5 m above the ground; legends in a size that can be read at the speed of the road.

 Regulatory      Warning        Informatory
  .-----.        /\            +---------+
 /  STOP \      /  \           | HOSPITAL|
 \  (red) /     / !  \          +---------+
  `-----'      /______\        blue/green
 circle/octagon  triangle       rectangle

(Jam density is a number of vehicles per kilometre, so "veh/hr" in the question is read as veh/km.)

(a) 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 $k_j$:

$$v = v_f\left(1-\frac{k}{k_j}\right) = 80\left(1-\frac{k}{120}\right) = 80 - 0.6667\,k$$

Since $q = v\,k$:

$$q = v_f\,k\left(1-\frac{k}{k_j}\right) = 80\,k - 0.6667\,k^2$$

(b) Diagrams and critical values

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

$$\begin{aligned} k_m &= 120/2 = 60\ \text{veh/km}\\ v_m &= 80/2 = 40\ \text{km/h}\\ q_{max} &= \frac{80 \times 120}{4} = 2400\ \text{veh/h} \end{aligned}$$

v-k diagram (straight line)
 v (km/h)
   80 |*
      |   *
   40 |- - - *   <- at q max
      |         *
      +-----------*--- k (veh/km)
      0    60     120

q-k diagram (parabola)
 q (veh/h)
 2400 |        _ * _   <- q max
      |     *         *
      |   *             *
      +--*-------|-------*-- k
      0        60       120

v-q diagram (half parabola)
 v (km/h)
   80 |*
      |  *
   40 |- - - *  <- q max (nose)
      |     *
      |   *   (congested branch)
      +--------|--------- q (veh/h)
      0       2400

(c) Speed and density for a flow of 800 veh/h

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

$$0.6667k^2 - 80k + 800 = 0 \quad\Rightarrow\quad k = \frac{80 \pm \sqrt{80^2 - 4(0.6667)(800)}}{2(0.6667)}$$ $$k_1 = 11.01\ \text{veh/km},\quad k_2 = 108.99\ \text{veh/km}$$

• Uncongested (stable) state: $k = 11.0$ veh/km and $v = 80(1-11.01/120) = 72.7$ km/h.
• Congested (forced) state: $k = 109.0$ veh/km and $v = 7.3$ km/h.

Answer: for 800 veh/h, speed = 72.7 km/h at density = 11.0 veh/km (stable flow); the congested alternative is 7.3 km/h at 109.0 veh/km.

(d) Headway, spacing at maximum flow

At $q_{max} = 2400$ veh/h, $k_m = 60$ veh/km, $v_m = 40$ km/h:

$$\begin{aligned} \text{Average headway}\ h &= \frac{3600}{q_{max}} = \frac{3600}{2400} = 1.50\ \text{s}\\ \text{Average spacing}\ s &= \frac{1000}{k_m} = \frac{1000}{60} = 16.67\ \text{m} \end{aligned}$$

Answer: headway = 1.50 s, spacing = 16.67 m.

Isolated two-phase signal with pedestrian indications for road A (18 m) and road B (12 m).

Design basis (IRC:93-1985, Guidelines on Design and Installation of Road Traffic Signals, with Webster's method): two-phase signal, Phase 1 = road A moves, Phase 2 = road B moves. Pedestrians cross a road while the traffic on the cross road has green. Assumed (not given): saturation flow 1800 PCU/h per lane, start-up lost time 2 s per phase, pedestrian walking speed 1.2 m/s and initial walk (pedestrian start-up) time 5 s.

Step 1: Amber (change) time

Amber $I = t_r + \dfrac{v}{2a}$ with reaction time $t_r = 1$ s and comfortable deceleration $a = 3\ \text{m/s}^2$, kept within the IRC:93 range of 2-4 s:

$$\begin{aligned} I_A &= 1 + \frac{55/3.6}{2(3)} = 3.55 \to 4\ \text{s}\\ I_B &= 1 + \frac{40/3.6}{2(3)} = 2.85 \to 3\ \text{s} \end{aligned}$$

Step 2: Pedestrian crossing time

Time to cross = initial walk + width / walking speed:

$$\begin{aligned} t_{p,A} &= 5 + \frac{18}{1.2} = 20.0\ \text{s}\\ t_{p,B} &= 5 + \frac{12}{1.2} = 15.0\ \text{s} \end{aligned}$$

Pedestrians crossing road A walk during Phase 2, so Phase 2 green must be at least 20.0 s. Pedestrians crossing road B walk during Phase 1, so Phase 1 green must be at least 15.0 s.

Step 3: Webster cycle time from traffic

$$\begin{aligned} y_A &= \frac{300}{1800} = 0.167, \quad y_B = \frac{250}{1800} = 0.139, \quad Y = 0.306\\ L &= (2+4) + (2+3) = 11\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(11)+5}{1-0.306} = 31.0\ \text{s} \end{aligned}$$ $$\begin{aligned} g_A &= \frac{0.167}{0.306}(20.0) = 10.9\ \text{s}, \quad G_A = g_A + 2 = 12.9\ \text{s}\\ g_B &= \frac{0.139}{0.306}(20.0) = 9.1\ \text{s}, \quad G_B = g_B + 2 = 11.1\ \text{s} \end{aligned}$$

Step 4: Check against pedestrian needs

Adopted cycle length $C = 15 + 4 + 20 + 3 = 42$ s.

Final timings

Pedestrian signals: pedestrians crossing road B get WALK for 5 s then flashing (clearance) for 10 s during Phase 1 (total 15 s of 15 s green); pedestrians crossing road A get WALK for 5 s then flashing for 15 s during Phase 2 (total 20 s of 20 s green). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

Phase diagram

Cycle = 42 s
Road A: [GREEN 15][A 4][  RED 23  ]
Road B: [ RED 19 ][GREEN 20][A 3]
Ped crossing B: [WALK+FLASH 15][DON'T WALK 27]
Ped crossing A: [DON'T WALK 19][WALK+FLASH 20][DW]

A collision on a flat road is solved by (i) finding each vehicle's speed just after impact from its post-collision skid, (ii) applying conservation of linear momentum as a vector equation to get the speeds just before impact, and (iii) adding back the energy lost in the pre-collision skid.

Given: $W_A = 3.5$ t, $W_B = 6$ t, $f = 0.55$, $g = 9.81\ \text{m/s}^2$. Pre-collision skids: $d_A = 35$ m, $d_B = 25$ m. Post-collision skids: $d_A' = 30$ m, $d_B' = 10$ m.

Vehicle A comes from the West (moving East). Vehicle B comes from the South (moving North). After impact A slides 45 degrees North of West and B slides 50 degrees East of North.

where $\theta$ is the bearing from North, clockwise: $\theta_A = 90^\circ$, $\theta_B = 0^\circ$ (travel directions) and $\theta_A' = 315^\circ$, $\theta_B' = 50^\circ$ (skid directions after impact).

   before impact          after impact
   A: bearing 90 deg       A': bearing 315 deg
   B: bearing 0 deg       B': bearing 50 deg
   (bearing measured clockwise from North)

Step 1: Speeds just after collision

From work-energy, $v' = \sqrt{2 g f d'}$:

$$\begin{aligned} v_A' &= \sqrt{2(9.81)(0.55)(30)} = 17.99\ \text{m/s}\\ v_B' &= \sqrt{2(9.81)(0.55)(10)} = 10.39\ \text{m/s} \end{aligned}$$

Step 2: Momentum balance (vector form)

Take $x$ = East and $y$ = North, and let $u_A, u_B$ be the speeds just before impact. Weights are proportional to masses, so they can be used directly.

$$\begin{aligned} W_A u_A \sin\theta_A + W_B u_B \sin\theta_B &= W_A v_A' \sin\theta_A' + W_B v_B' \sin\theta_B' \\ W_A u_A \cos\theta_A + W_B u_B \cos\theta_B &= W_A v_A' \cos\theta_A' + W_B v_B' \cos\theta_B' \end{aligned}$$

Substituting the numbers:

$$\begin{aligned} 3.5\,u_A(1.0000) + 6\,u_B(0.0000) &= 3.5(17.992)(-0.7071) + 6(10.388)(0.7660) = 3.217 \\ 3.5\,u_A(0.0000) + 6\,u_B(1.0000) &= 3.5(17.992)(0.7071) + 6(10.388)(0.6428) = 84.593 \end{aligned}$$

Solving: $u_A = 0.92\ \text{m/s} = 3.3\ \text{km/h}$ and $u_B = 14.10\ \text{m/s} = 50.8\ \text{km/h}$ at the instant of impact.

Step 3: Original speeds (before the pre-collision skid)

$v_0 = \sqrt{u^2 + 2 g f d}$:

$$\begin{aligned} v_A &= \sqrt{(0.92)^2 + 2(9.81)(0.55)(35)} = 19.46\ \text{m/s}\\ v_B &= \sqrt{(14.10)^2 + 2(9.81)(0.55)(25)} = 21.65\ \text{m/s} \end{aligned}$$

Answer: original speed of vehicle $A$ = 70.0 km/h and of vehicle $B$ = 77.9 km/h (speeds at impact: 3.3 and 50.8 km/h).

(Jam density is a number of vehicles per kilometre, so "veh/hr" in the question is read as veh/km.)

(a) 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 $k_j$:

$$v = v_f\left(1-\frac{k}{k_j}\right) = 90\left(1-\frac{k}{150}\right) = 90 - 0.6\,k$$

Since $q = v\,k$:

$$q = v_f\,k\left(1-\frac{k}{k_j}\right) = 90\,k - 0.6\,k^2$$

(b) Diagrams and critical values

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

$$\begin{aligned} k_m &= 150/2 = 75\ \text{veh/km}\\ v_m &= 90/2 = 45\ \text{km/h}\\ q_{max} &= \frac{90 \times 150}{4} = 3375\ \text{veh/h} \end{aligned}$$

v-k diagram (straight line)
 v (km/h)
   90 |*
      |   *
   45 |- - - *   <- at q max
      |         *
      +-----------*--- k (veh/km)
      0    75     150

q-k diagram (parabola)
 q (veh/h)
 3375 |        _ * _   <- q max
      |     *         *
      |   *             *
      +--*-------|-------*-- k
      0        75       150

v-q diagram (half parabola)
 v (km/h)
   90 |*
      |  *
   45 |- - - *  <- q max (nose)
      |     *
      |   *   (congested branch)
      +--------|--------- q (veh/h)
      0       3375

(c) Speed and density for a flow of 800 veh/h

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

$$0.6k^2 - 90k + 800 = 0 \quad\Rightarrow\quad k = \frac{90 \pm \sqrt{90^2 - 4(0.6)(800)}}{2(0.6)}$$ $$k_1 = 9.49\ \text{veh/km},\quad k_2 = 140.51\ \text{veh/km}$$

• Uncongested (stable) state: $k = 9.5$ veh/km and $v = 90(1-9.49/150) = 84.3$ km/h.
• Congested (forced) state: $k = 140.5$ veh/km and $v = 5.7$ km/h.

Answer: for 800 veh/h, speed = 84.3 km/h at density = 9.5 veh/km (stable flow); the congested alternative is 5.7 km/h at 140.5 veh/km.

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

At $q_{max} = 3375$ veh/h, $k_m = 75$ veh/km, $v_m = 45$ km/h:

$$\begin{aligned} \text{Average headway}\ h &= \frac{3600}{q_{max}} = \frac{3600}{3375} = 1.07\ \text{s}\\ \text{Average spacing}\ s &= \frac{1000}{k_m} = \frac{1000}{75} = 13.33\ \text{m}\\ \text{Clearance}\ &= s - L = 13.33 - 6.1 = 7.23\ \text{m}\\ \text{Gap (time)}\ &= h - \frac{L}{v_m} = 1.07 - \frac{6.1}{12.500} = 0.58\ \text{s} \end{aligned}$$

Answer: headway = 1.07 s, spacing = 13.33 m, clearance = 7.23 m, gap = 0.58 s.

A collision on a flat road is solved by (i) finding each vehicle's speed just after impact from its post-collision skid, (ii) applying conservation of linear momentum as a vector equation to get the speeds just before impact, and (iii) adding back the energy lost in the pre-collision skid.

Given: $W_A = 4.4$ t, $W_B = 6$ t, $f = 0.55$, $g = 9.81\ \text{m/s}^2$. Pre-collision skids: $d_A = 39$ m, $d_B = 21$ m. Post-collision skids: $d_A' = 15$ m, $d_B' = 36$ m.

Vehicle A comes from the West (moving East); its weight is 4.4 t. Vehicle B comes from the South (moving North); its weight is 6 t. After impact A slides 51 degrees North of West and B slides 61 degrees East of North.

where $\theta$ is the bearing from North, clockwise: $\theta_A = 90^\circ$, $\theta_B = 0^\circ$ (travel directions) and $\theta_A' = 321^\circ$, $\theta_B' = 61^\circ$ (skid directions after impact).

   before impact          after impact
   A: bearing 90 deg       A': bearing 321 deg
   B: bearing 0 deg       B': bearing 61 deg
   (bearing measured clockwise from North)

Step 1: Speeds just after collision

From work-energy, $v' = \sqrt{2 g f d'}$:

$$\begin{aligned} v_A' &= \sqrt{2(9.81)(0.55)(15)} = 12.72\ \text{m/s}\\ v_B' &= \sqrt{2(9.81)(0.55)(36)} = 19.71\ \text{m/s} \end{aligned}$$

Step 2: Momentum balance (vector form)

Take $x$ = East and $y$ = North, and let $u_A, u_B$ be the speeds just before impact. Weights are proportional to masses, so they can be used directly.

$$\begin{aligned} W_A u_A \sin\theta_A + W_B u_B \sin\theta_B &= W_A v_A' \sin\theta_A' + W_B v_B' \sin\theta_B' \\ W_A u_A \cos\theta_A + W_B u_B \cos\theta_B &= W_A v_A' \cos\theta_A' + W_B v_B' \cos\theta_B' \end{aligned}$$

Substituting the numbers:

$$\begin{aligned} 4.4\,u_A(1.0000) + 6\,u_B(0.0000) &= 4.4(12.723)(-0.6293) + 6(19.710)(0.8746) = 68.202 \\ 4.4\,u_A(0.0000) + 6\,u_B(1.0000) &= 4.4(12.723)(0.7771) + 6(19.710)(0.4848) = 100.837 \end{aligned}$$

Solving: $u_A = 15.50\ \text{m/s} = 55.8\ \text{km/h}$ and $u_B = 16.81\ \text{m/s} = 60.5\ \text{km/h}$ at the instant of impact.

Step 3: Original speeds (before the pre-collision skid)

$v_0 = \sqrt{u^2 + 2 g f d}$:

$$\begin{aligned} v_A &= \sqrt{(15.50)^2 + 2(9.81)(0.55)(39)} = 25.71\ \text{m/s}\\ v_B &= \sqrt{(16.81)^2 + 2(9.81)(0.55)(21)} = 22.56\ \text{m/s} \end{aligned}$$

Answer: original speed of vehicle $A$ = 92.6 km/h and of vehicle $B$ = 81.2 km/h (speeds at impact: 55.8 and 60.5 km/h).

Saturation flow from the saturation headway: road H $s = 3600/3 = 1200$ PCU/h; road F $s = 3600/4 = 900$ PCU/h.

Step 1: Flow ratios

Flow ratio $y = q/s$ for each approach; the larger value in a phase is the critical one.

$$Y = 0.333 + 0.278 = 0.611$$

Step 2: Lost time

Lost time is not given; it is taken as 2 s per phase (including the 2 s amber), plus the all-red period for pedestrians $R$ = 15 s per cycle.

$$L = (0.0+2.0) + (0.0+2.0) + 15 = 19.0\ \text{s}$$

Step 3: Optimum cycle time (Webster)

$$C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(19.0) + 5}{1 - 0.611} = 86.1\ \text{s}$$

Adopt C = 90 s.

Step 4: Green times

Total effective green $= C - L = 90 - 19.0 = 71.0$ s, divided in proportion to $y$:

$$g_{1} = \frac{y}{Y}(C-L) = \frac{0.333}{0.611}(71.0) = 38.7\ \text{s}$$ $$g_{2} = \frac{y}{Y}(C-L) = \frac{0.278}{0.611}(71.0) = 32.3\ \text{s}$$

Displayed green $G = g +$ start-up loss (so that $\sum G + \sum I + R = C$):

Check: 39 + 2 + 32 + 2 + 15 = 90 s.

Phase diagram

|     P1 green 39      | A2|   P2 green 32    | A2|  R 15  |

Isolated two-phase signal with pedestrian indications for road A (15 m) and road B (12 m). Approach speeds are not needed because the ambers are given (3 s and 2 s).

Design basis (IRC:93-1985, Guidelines on Design and Installation of Road Traffic Signals, with Webster's method): two-phase signal, Phase 1 = road A moves, Phase 2 = road B moves. Pedestrians cross a road while the traffic on the cross road has green. Assumed (not given): saturation flow 1800 PCU/h per lane, start-up lost time 2 s per phase, pedestrian walking speed 1.2 m/s and initial walk (pedestrian start-up) time 5 s.

Step 1: Amber (change) time

Amber times are given: road A = 3 s, road B = 2 s.

Step 2: Pedestrian crossing time

Time to cross = initial walk + width / walking speed:

$$\begin{aligned} t_{p,A} &= 5 + \frac{15}{1.2} = 17.5\ \text{s}\\ t_{p,B} &= 5 + \frac{12}{1.2} = 15.0\ \text{s} \end{aligned}$$

Pedestrians crossing road A walk during Phase 2, so Phase 2 green must be at least 17.5 s. Pedestrians crossing road B walk during Phase 1, so Phase 1 green must be at least 15.0 s.

Step 3: Webster cycle time from traffic

$$\begin{aligned} y_A &= \frac{300}{1800} = 0.167, \quad y_B = \frac{250}{1800} = 0.139, \quad Y = 0.306\\ L &= (2+3) + (2+2) = 9\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(9)+5}{1-0.306} = 26.6\ \text{s} \end{aligned}$$ $$\begin{aligned} g_A &= \frac{0.167}{0.306}(17.6) = 9.6\ \text{s}, \quad G_A = g_A + 2 = 11.6\ \text{s}\\ g_B &= \frac{0.139}{0.306}(17.6) = 8.0\ \text{s}, \quad G_B = g_B + 2 = 10.0\ \text{s} \end{aligned}$$

Step 4: Check against pedestrian needs

Adopted cycle length $C = 15 + 3 + 18 + 2 = 38$ s.

Final timings

Pedestrian signals: pedestrians crossing road B get WALK for 5 s then flashing (clearance) for 10 s during Phase 1 (total 15 s of 15 s green); pedestrians crossing road A get WALK for 5 s then flashing for 12 s during Phase 2 (total 18 s of 18 s green). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

Phase diagram

Cycle = 38 s
Road A: [GREEN 15][A 3][  RED 20  ]
Road B: [ RED 18 ][GREEN 18][A 2]
Ped crossing B: [WALK+FLASH 15][DON'T WALK 23]
Ped crossing A: [DON'T WALK 18][WALK+FLASH 18][DW]

Methods of traffic control at an intersection

1. Uncontrolled (priority by rules): drivers give way to the vehicle on the right or on the main road; suitable for very low volume.
2. Traffic signs and markings: STOP and GIVE WAY signs, with stop lines; minor road gives way to major road (IRC:67, IRC:35).
3. Channelisation: islands and markings guide the movements (IRC:SP:41).
4. Rotary (roundabout): one-way circulation, no signals (IRC:65).
5. Traffic signals: fixed-time, vehicle-actuated or area-coordinated (IRC:93).
6. Police control by hand signals at peak times or special events.
7. Grade separation: flyover, underpass or interchange for very high volume.

Design of a two-phase signal by Webster's method

Data: roads A and B carry 400 and 300 PCU/h; saturation flows 1450 and 1150 PCU/h; all-red time for pedestrians $R$ = 12 s per cycle.

Step 1: Flow ratios

Flow ratio $y = q/s$ for each approach; the larger value in a phase is the critical one.

$$Y = 0.276 + 0.261 = 0.537$$

Step 2: Lost time

Lost time per phase is taken as 2 s (including the amber of 2 s), plus the all-red time $R$ = 12 s.

$$L = (0.0+2.0) + (0.0+2.0) + 12 = 16.0\ \text{s}$$

Step 3: Optimum cycle time (Webster)

$$C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(16.0) + 5}{1 - 0.537} = 62.6\ \text{s}$$

Adopt C = 65 s.

Step 4: Green times

Total effective green $= C - L = 65 - 16.0 = 49.0$ s, divided in proportion to $y$:

$$g_{1} = \frac{y}{Y}(C-L) = \frac{0.276}{0.537}(49.0) = 25.2\ \text{s}$$ $$g_{2} = \frac{y}{Y}(C-L) = \frac{0.261}{0.537}(49.0) = 23.8\ \text{s}$$

Displayed green $G = g +$ start-up loss (so that $\sum G + \sum I + R = C$):

Check: 25 + 2 + 24 + 2 + 12 = 65 s.

Phase diagram

|    P1 green 25    | A2|   P2 green 24    | A2|   R 12  |

Critical road user characteristics

1. Physical: vision (acuity, field of vision, colour vision, glare), hearing, strength and agility. They fix sign sizes, letter heights, lighting, sight distances and footpath gradients.
2. Mental and psychological: attention, judgement, emotional stability and attitude. They influence overtaking decisions, gap acceptance and compliance with signals.
3. Reaction time (PIEV): about 2.5 s is used for stopping sight distance, headlight sight distance and intersection sight distance (IRC:SP:23, IRC:66, Nepal Road Standard 2070).
4. Pedestrian characteristics: walking speed about 1.2 m/s (IRC:103 and IRC:93 use 1.2 m/s; slower, about 0.9 m/s, for children and the elderly); space needed about 0.5-0.75 m² per person; they need footpaths, crossings, refuge islands and signal time.
5. Cyclists: speed 10-25 km/h, need separate lanes and safe turning at intersections.
6. Age and gender, fatigue, intoxication and disability: these add to variation, so design uses the 85th-percentile or a conservative value (for example the 15th-percentile pedestrian speed for crossing time).
7. Behaviour and compliance: lane discipline, helmet and seat-belt use, jaywalking and parking habits vary by country and must be reflected in the design and enforcement plan (in Nepal mixed traffic is common).
8. Mixed traffic: in Nepal slow and fast vehicles share the carriageway, so wider shoulders and segregation are required.

How static characteristics of a vehicle affect road design

Examples: the 2.5 m width and 11 m length of a bus fix the minimum bay size; the 4.75 m height fixes bridge clearance; the large turning radius of trucks fixes the radius of intersection corners; the axle load fixes pavement thickness.

Basic requirements of an intersection at grade

1. Safety: few and simple conflict points, good sight triangles (IRC:66, IRC:SP:41), and low approach speeds.
2. Capacity: enough lanes and green time or rotary width for the design hour volume.
3. Clear, simple layout: roads meet at about 90 degrees (not less than 60 degrees); avoid more than four arms; flat approach gradient.
4. Channelisation and control: islands, markings, signs, signals or a rotary where needed; provision for pedestrians and cyclists.
5. Drainage, lighting and economy.

Factors affecting night visibility

• Headlight intensity, beam type (dipped or high) and alignment.
• Glare from opposing headlights, and the driver's glare recovery time.
• Street lighting level and uniformity.
• Contrast between object and background; colour and reflectance of the pavement (a wet or dark surface reflects little).
• Speed of the vehicle (less time to see and react) and driver age and eyesight.
• Weather (rain, fog, dust), cleanliness of windscreen, and retro-reflective signs and markings.

Lateral offset of lighting poles

Poles are placed at least 0.6 m (minimum 0.3 m on very restricted kerbed streets) behind the kerb face on urban roads, and about 1.5 m or more from the carriageway edge (outside the shoulder) on high-speed roads without kerbs, as in IS 1944 / IRC practice.

Reasoning: (1) a pole set too close is hit by vehicles whose wheels or mirrors overhang; (2) a larger offset gives the driver recovery space and reduces the severity of impact (frangible or breakaway poles are used where offset is small); (3) it keeps the footpath and sight lines clear; (4) too large an offset reduces the light falling on the carriageway and increases the bracket length, so the offset is a balance between safety and lighting efficiency.

Design the vehicular and pedestrian timings of an isolated two-phase signal for roads X (15 m) and Y (10.5 m).

Design basis (IRC:93-1985, Guidelines on Design and Installation of Road Traffic Signals, with Webster's method): two-phase signal, Phase 1 = road X moves, Phase 2 = road Y moves. Pedestrians cross a road while the traffic on the cross road has green. Assumed (not given): saturation flow 1800 PCU/h per lane, start-up lost time 2 s per phase, pedestrian walking speed 1.2 m/s and initial walk (pedestrian start-up) time 5 s.

Step 1: Amber (change) time

Amber $I = t_r + \dfrac{v}{2a}$ with reaction time $t_r = 1$ s and comfortable deceleration $a = 3\ \text{m/s}^2$, kept within the IRC:93 range of 2-4 s:

$$\begin{aligned} I_X &= 1 + \frac{50/3.6}{2(3)} = 3.31 \to 4\ \text{s}\\ I_Y &= 1 + \frac{35/3.6}{2(3)} = 2.62 \to 3\ \text{s} \end{aligned}$$

Step 2: Pedestrian crossing time

Time to cross = initial walk + width / walking speed:

$$\begin{aligned} t_{p,X} &= 5 + \frac{15}{1.2} = 17.5\ \text{s}\\ t_{p,Y} &= 5 + \frac{10.5}{1.2} = 13.8\ \text{s} \end{aligned}$$

Pedestrians crossing road X walk during Phase 2, so Phase 2 green must be at least 17.5 s. Pedestrians crossing road Y walk during Phase 1, so Phase 1 green must be at least 13.8 s.

Step 3: Webster cycle time from traffic

$$\begin{aligned} y_X &= \frac{250}{1800} = 0.139, \quad y_Y = \frac{200}{1800} = 0.111, \quad Y = 0.250\\ L &= (2+4) + (2+3) = 11\ \text{s}\\ C_0 &= \frac{1.5L+5}{1-Y} = \frac{1.5(11)+5}{1-0.250} = 28.7\ \text{s} \end{aligned}$$ $$\begin{aligned} g_X &= \frac{0.139}{0.250}(17.7) = 9.8\ \text{s}, \quad G_X = g_X + 2 = 11.8\ \text{s}\\ g_Y &= \frac{0.111}{0.250}(17.7) = 7.9\ \text{s}, \quad G_Y = g_Y + 2 = 9.9\ \text{s} \end{aligned}$$

Step 4: Check against pedestrian needs

Adopted cycle length $C = 14 + 4 + 18 + 3 = 39$ s.

Final timings

Pedestrian signals: pedestrians crossing road Y get WALK for 5 s then flashing (clearance) for 9 s during Phase 1 (total 14 s of 14 s green); pedestrians crossing road X get WALK for 5 s then flashing for 12 s during Phase 2 (total 18 s of 18 s green). All pedestrian indications show DON'T WALK (red man) while their cross traffic moves.

Phase diagram

Cycle = 39 s
Road X: [GREEN 14][A 4][  RED 21  ]
Road Y: [ RED 18 ][GREEN 18][A 3]
Ped crossing Y: [WALK+FLASH 14][DON'T WALK 25]
Ped crossing X: [DON'T WALK 18][WALK+FLASH 18][DW]

Spacing of street lights

The spacing is found from the luminous flux that must reach the road surface (lumen method):

$$S = \frac{F \times U_f \times M}{E \times W}$$

where $S$ = spacing between lamps (m), $F$ = luminous flux of the lamp (lumen), $U_f$ = utilisation factor (fraction of lamp flux that reaches the road; from the luminaire curves for the given mounting height, width and arrangement), $M$ = maintenance factor (0.7-0.8, allows for dust and lamp ageing), $E$ = required average illuminance (lux) and $W$ = width of the road to be lit (m).

Steps:

1. Fix the required illuminance $E$ for the road class from IS 1944/CIE/Nepal standards.
2. Select the mounting height $H$, the arrangement (single side, staggered, opposite or central) and the luminaire type.
3. Choose the lamp (wattage and lumen output $F$).
4. Read the utilisation factor $U_f$ from the manufacturer's curve for $W/H$ and the overhang.
5. Take the maintenance factor $M$.
6. Compute $S$ from the formula; for staggered or opposite arrangement use the full width $W$ and divide $F$ appropriately (for opposite arrangement use half width per row).
7. Check that $S/H$ lies within 3 to 5 (cut-off luminaires: 3, semi-cut-off: 4-5), and that the uniformity ratio (min/avg) is at least 0.3. Adjust the lamp or height if necessary.

Example: $F = 15000$ lm, $U_f = 0.4$, $M = 0.75$, $E = 20$ lux, $W = 10$ m: $S = \dfrac{15000 \times 0.4 \times 0.75}{20 \times 10} = 22.5$ m. For a mounting height of 7.5 m, $S/H = 3$, which is acceptable.

Types of parking

Parking is classified as on-street (on the carriageway edge) and off-street (away from the carriageway).

A. On-street parking (vehicles park along the kerb):

 Parallel            45 degree           90 degree
 ______________     \ \ \ \ \          | | | | | |
 [car] [car] [car]   \_\_\_\_\_         |_|_|_|_|_|
 ========= kerb      ===== kerb          ===== kerb

B. Off-street parking:

• Surface parking lots: open areas near markets or terminals; low cost, require land.
• Multi-storey (multi-level) parking: several floors; best use of costly land, high construction cost; ramp or mechanical access.
• Underground parking: below buildings or public squares; keeps surface free, expensive.
• Mechanical or automated parking: lifts and stackers, used where land is very scarce.
• Park-and-ride facilities: at city edges, linked to public transport.

Nepal practice: on-street parking is regulated by the Metropolitan/Municipal offices and Traffic Police (parallel parking on one side, paid parking zones), and off-street parking is required in new buildings by the Nepal National Building Code (NBC 206 / municipal building bye-laws).

The acceleration falls linearly with speed, so $\dfrac{dV}{dt} = 2.01 - 0.05V$ is solved by separating variables. Assumption: the car starts at the truck's speed, $V_0 = 35$ km/h $= 9.722$ m/s at $t = 0$.

Speed as a function of time

$$\int \frac{dV}{2.01-0.05V} = \int dt \;\Rightarrow\; -20\ln(2.01-0.05V) = t + C$$

At $t=0$, $V=V_0$. This gives

$$V(t) = 40.2 - (40.2 - 9.722)\,e^{-0.05t} = 40.2 - 30.478\,e^{-0.05t}\ \text{m/s}$$

and the acceleration is $a = 2.01 - 0.05V = 0.05(40.2 - V_0)e^{-0.05t} = 1.5239\,e^{-0.05t}$.

(i) Acceleration after 20 s and 40 s

$$\begin{aligned} V(20) &= 40.2 - 30.478e^{-1} = 28.988\ \text{m/s}, & a(20) &= 2.01 - 0.05(28.988) = 0.561\ \text{m/s}^2\\ V(40) &= 40.2 - 30.478e^{-2} = 36.075\ \text{m/s}, & a(40) &= 2.01 - 0.05(36.075) = 0.206\ \text{m/s}^2 \end{aligned}$$

(ii) Maximum speed attainable

Speed is maximum when $dV/dt = 0$:

$$V_{max} = \frac{2.01}{0.05} = 40.2\ \text{m/s} = 144.7\ \text{km/h}$$

(This is approached asymptotically; a real car would be limited by the speed limit and road.)

(iii) Distance travelled in 260 s

$$\begin{aligned} x &= \int_0^{260} V\,dt = 40.2t - \frac{30.478}{0.05}\left(1 - e^{-0.05t}\right)\\ &= 40.2(260) - 609.56\left(1 - e^{-13}\right) = 9842.4\ \text{m} \end{aligned}$$

Answer: (i) $a_{20} = 0.561$ m/s$^2$, $a_{40} = 0.206$ m/s$^2$; (ii) $V_{max} = 40.2$ m/s (144.7 km/h); (iii) distance $\approx$ 9842 m (9.84 km).

The moving observer method gives the flow $q$ and mean stream speed from counts made by a test car running with and against the stream.

Reading of the table: the vehicles she meets while driving southward are the northbound stream she is measuring ($N_s$); while driving northward, $N_o$ = cars that overtake her and $N_p$ = cars she overtakes. Section length 800 m, test speed 20 km/h on both runs.

Average counts (4 runs)

$$\begin{aligned} \bar N_s &= \frac{107+113+30+79}{4} = 82.25\\ \bar N_o &= \frac{10+25+15+18}{4} = 17.00, \quad \bar N_p = \frac{74+41+5+9}{4} = 32.25\\ y &= \bar N_o - \bar N_p = -15.25 \end{aligned}$$

Travel times

$$t_w = t_a = \frac{0.8}{20} = 0.04\ \text{h} = 2.4\ \text{min}$$

Traffic flow

$$q = \frac{N_s + y}{t_w + t_a} = \frac{82.25 + (-15.25)}{0.04 + 0.04} = 837.5\ \text{veh/h}$$

Journey time and speed of the stream

$$\bar t = t_w - \frac{y}{q} = 0.04 - \frac{(-15.25)}{837.5} = 0.0582\ \text{h} = 3.49\ \text{min}$$ $$V_j = \frac{L}{\bar t} = \frac{0.8}{0.0582} = 13.74\ \text{km/h}$$

Answer: average flow = 837.5 veh/h; journey speed = 13.7 km/h (journey time 3.49 min). The stream is slower than the test car because more vehicles were overtaken by her than overtook her.

Vision supplies about 90 percent of the information a driver uses, so its characteristics control how a road must be designed.

• Visual acuity: the ability to see detail. Static acuity is needed to read signs; dynamic acuity (seeing moving objects) is poorer and falls as speed rises. A sign letter height is chosen so that a driver of normal acuity can read it in the time available.
• Cone of vision: clear vision is within about 3 degrees; fairly clear vision within about 10 degrees; useful peripheral vision up to about 120 degrees (about 160-180 degrees for movement). Signs must be placed within the clear cone.
• Peripheral vision: detects vehicles and pedestrians at the sides; it narrows with speed (about 100 degrees at 30 km/h to about 40 degrees at 100 km/h) and with age and alcohol.
• Colour vision: red, green and amber must be recognised. About 8 percent of males are colour blind; signals therefore also use fixed positions (red on top) and shapes.
• Glare vision and recovery: headlight glare at night reduces vision for several seconds. Recovery time is about 3-6 s from dark to bright, longer in old age. Hence anti-glare screens, median planting and low-mounted headlights are used.
• Depth perception: judging distance and relative speed is vital in overtaking and gap acceptance; it is poor at night and in fog.
• Night vision and adaptation: the eye takes time to adapt to darkness; street lighting at tunnel exits and underpasses must provide a gradual transition.

The design consequences are: sign size and placement, retro-reflective markings, lighting levels, sight-distance lengths, and signal lens colours and positions.

Spot speed

Spot speed is the instantaneous speed of a vehicle at a specified point (spot) of the road. It is used for speed limits, design speed, sight distance, signal timing, crash analysis and checking road geometry (85th-percentile speed is the usual regulatory speed).

Methods of spot speed study

1. Stopwatch (enoscope-based) method: a short base length (about 30 m for speeds below 40 km/h, 50-100 m for higher speeds) is marked; the time a vehicle takes to cross it is measured with a stopwatch, and $v = d/t$. Simple and cheap but less accurate; the enoscope (mirror box) reduces parallax error.
2. Pavement contact devices: pneumatic road tubes or piezo/induction loops across the lane record the passing of axles and give speed from the time between two sensors; automatic and accurate, good for long counts.
3. Radar (Doppler) speed meter / lidar gun: a hand-held or tripod unit measures speed directly using the Doppler effect; quick and accurate, but drivers may notice and slow.
4. Video and image processing: vehicles are recorded and speeds computed from frames and a known base length; gives permanent records and classification.
5. Automatic number-plate / GPS / floating car methods are used for long sections and for travel time.

Sample size is at least 30-50 vehicles, chosen at random; observations are made in off-peak free-flow periods under good weather (IRC:SP:19).

Advantages and limitations of a rotary

Advantages

• No signals needed; traffic flows continuously, so delay is low at moderate volumes.
• Crossing conflicts are replaced by weaving, merging and diverging, so crashes are fewer and less severe.
• Low speeds and low cost of operation and maintenance; no power needed.
• Suitable for junctions of four or more arms with balanced flows.

Limitations

• Needs a large area of land, so it is unsuitable in built-up areas.
• Not suitable for very high traffic (above about 3,000 PCU/h) or a very high share of weaving traffic.
• Pedestrians cross with difficulty and the detour adds distance to through and turning vehicles.
• Not suitable where one road is far more important than the other, or where slow vehicles are many.
• Congestion on one arm can lock up the whole rotary.

Road 1 and Road 2 each run in one phase, so this is a two-phase signal. The design hour flows of the two approaches of each road are given and the saturation flow is the same for both directions of a road (3675 and 1890 PCU/h).

Step 1: Flow ratios

Flow ratio $y = q/s$ for each approach; the larger value in a phase is the critical one.

$$Y = 0.231 + 0.132 = 0.364$$

Step 2: Lost time

Lost time per phase = start-up lost time 2 s + amber 3 s = 5 s; no all-red time is given.

$$L = (2.0+3.0) + (2.0+3.0) = 10.0\ \text{s}$$

Step 3: Optimum cycle time (Webster)

$$C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(10.0) + 5}{1 - 0.364} = 31.4\ \text{s}$$

Adopt C = 35 s.

Step 4: Green times

Total effective green $= C - L = 35 - 10.0 = 25.0$ s, divided in proportion to $y$:

$$g_{1} = \frac{y}{Y}(C-L) = \frac{0.231}{0.364}(25.0) = 15.9\ \text{s}$$ $$g_{2} = \frac{y}{Y}(C-L) = \frac{0.132}{0.364}(25.0) = 9.1\ \text{s}$$

Displayed green $G = g +$ start-up loss (so that $\sum G + \sum I + R = C$):

Check: 18 + 3 + 11 + 3 = 35 s.

Phase diagram

|      P1 green 18       |   A3  | P2 green 11  |   A3  |

Parking terms

• Parking accumulation: the number of vehicles parked in a study area at a given instant. Plotting it against time gives the accumulation curve; its peak gives the demand.
• Parking volume: total number of vehicles parked in an area during a given period (usually a day); it counts each vehicle once.
• Parking load: the area under the accumulation curve, in vehicle-hours (or vehicle-minutes) in a given period; it shows the total parking demand in time.
• Parking duration: the length of time a vehicle stays parked.
• Parking turnover: the rate of use of a parking space, the number of vehicles parked per bay in a given time.

$$\text{Turnover} = \frac{\text{Number of vehicles parked in the period}}{\text{Number of parking bays available}}$$

• Parking index (occupancy): the ratio of the number of bays occupied to the number of bays available, as a percentage. It measures how well the supply is used.

$$\text{Parking index} = \frac{\text{Parking load (vehicle-hours)}}{\text{Parking capacity (bay-hours)}} \times 100$$

where parking capacity = number of bays x hours of study. An index near 100 percent means the supply is exhausted; a very low value means over-supply.

Speed-density relation

Speed falls as density rises. The simplest (Greenshields) model is linear:

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

where $v_f$ is the free-flow speed (at $k \to 0$) and $k_j$ is the jam density (where $v = 0$).

 v ^
vf |*
   |   *
   |      *
   |         *
   |            *
   +---------------*--> k
   0               kj

Speed-flow relation

Since $q = vk$, eliminating $k$ gives $q = k_j\left(v - v^2/v_f\right)$, a parabola with maximum flow $q_{max} = v_f k_j/4$ at $v = v_f/2$.

 v ^
vf |*
   |  *
   |    *   uncongested
vm |- - - -*----- (v = vf/2)
   |       *  congested
   |     *
   +-----|----> q
   0   qmax

The upper branch is stable (uncongested) flow, and the lower branch is forced (congested) flow. Flow rises with density up to $k_m = k_j/2$ (capacity) and then falls to zero at jam density.

After the collision the two vehicles move together, so the common post-impact speed comes from the skid distance, the impact speeds from momentum conservation in two directions, and the original speeds from the pre-collision skids.

Given: car $m_1 = 430$ kg moving East, truck $m_2 = 1299$ kg moving South, $f = 0.5$, pre-collision skids 15 m (car) and 10 m (truck), post-collision skid $d' = 8$ m at $65^\circ$ South of East.

        N
        ^
        |      car (East) ->
  W ----+----> E
        |\ 65 deg
        | \  locked pair
        v  v (S of E)
        S   truck comes moving South

Step 1: Speed of the locked vehicles after impact

$$v' = \sqrt{2gfd'} = \sqrt{2(9.81)(0.5)(8)} = 8.859\ \text{m/s}$$

Step 2: Momentum conservation

Total mass $M = 430 + 1299 = 1729$ kg. East component (only the car had eastward momentum) and South component (only the truck had southward momentum):

$$\begin{aligned} m_1 u_1 &= M v'\cos 65^\circ \Rightarrow u_1 = \frac{1729(8.859)(0.4226)}{430} = 15.054\ \text{m/s}\\ m_2 u_2 &= M v'\sin 65^\circ \Rightarrow u_2 = \frac{1729(8.859)(0.9063)}{1299} = 10.687\ \text{m/s} \end{aligned}$$

Step 3: Original speeds

$$\begin{aligned} v_1 &= \sqrt{u_1^2 + 2gf(15)} = \sqrt{(15.054)^2 + 2(9.81)(0.5)(15)} = 19.33\ \text{m/s}\\ v_2 &= \sqrt{u_2^2 + 2gf(10)} = \sqrt{(10.687)^2 + 2(9.81)(0.5)(10)} = 14.57\ \text{m/s} \end{aligned}$$

Answer: initial speed of the car = 69.6 km/h and of the truck = 52.5 km/h (impact speeds 54.2 and 38.5 km/h).

Advantages of one-way traffic

• Fewer conflict points (a four-arm junction of two one-way streets has far fewer than the 32 of two-way streets), so fewer and less severe crashes.
• Higher capacity (typically 20-50 percent more) as no opposing flow, no right-turn crossing and less need for signal phases.
• Easier signal coordination and progressive flow; fewer signal phases.
• Parking and loading can be allowed on both sides without blocking opposing flow.
• Higher speed and less delay; less glare at night; fewer head-on and sideswipe crashes.
• Pedestrians only look in one direction when crossing.
• No central median needed; easier lane use.

Limitations: longer travel distance for some trips, confusion for strangers, and need for a parallel return street.

Conflict points at a four-arm intersection (both roads two-way)

              N
          |  D | M  |
          |    |    |
  ------- X--X-X--X --------
   M D    X  X X  X     D M
  ------- X--X-X--X --------
   D M    X  X X  X     M D
  ------- X--X-X--X --------
          |    |    |
          |  M | D  |
              S
 X = crossing, M = merging, D = diverging
 (4 diverging + 4 merging near arm ends; 
  crossing points in the central area)

Each of the four approaches has through, left-turn and right-turn movements (12 movements in all).

Classification of conflicts

1. Diverging: one stream splits into two (a vehicle leaves the approach stream to turn). Least severe; 2 per approach = 8.
2. Merging: two streams join into one at an exit. Moderately severe; 2 per exit = 8.
3. Crossing: two streams cut across each other's path (right turners across opposing through traffic, and cross-road through/turning movements). Most severe and most numerous = 16.

Hence crossing conflicts (half of all) are the main target for control by signals, islands, rotary or grade separation.

Human-vehicle-environment (HVE) system

The traffic stream is the combined behaviour of three interacting elements: the Human (road user), the Vehicle and the Environment (road and surroundings). Safe and efficient flow needs all three to match.

        +--------------+
        |    HUMAN     |  driver, pedestrian,
        | (perception, |  cyclist, passenger
        |   reaction)  |
        +------+-------+
       controls|    ^ feedback
               v    |
        +------+----+--+      +--------------------+
        |   VEHICLE    |<---->|    ENVIRONMENT     |
        | size, power, |      | road geometry,     |
        | brakes, load |      | surface, signs,    |
        +--------------+      | weather, lighting  |
                              +--------------------+

• Human: receives information (mostly visual), decides and acts (steering, braking). Errors in perception, judgement or reaction are the main cause of crashes (about 80-90 percent involve human error).
• Vehicle: its dimensions, weight, power, braking and visibility set the road width, gradient, curve radius and pavement strength required.
• Environment: road geometry, surface friction, traffic control devices, lighting, weather and adjacent land use affect how the driver sees and responds.

A mismatch in any one (for example a fast vehicle, a sharp curve and a wet surface) causes crashes, so design must fit the "design driver" and "design vehicle" to the road.

Physical characteristics of the driver that affect traffic performance

• Vision: acuity (about 6/6 needed for a licence), the cone of clear vision (3-5 degrees), the field of vision (clear about 10 degrees, fairly clear about 12-14 degrees, peripheral up to about 120-180 degrees), colour perception (colour blindness for red/green is critical for signals), glare sensitivity, glare recovery and depth perception. Speed narrows the field of vision (at 100 km/h only about 40 degrees).
• Hearing: horn and siren detection; deafness is partly compensated by vision, but affects awareness of emergency vehicles.
• Reaction ability: muscle coordination, strength and speed of hand/foot response; fatigue, age and disability reduce it.
• Age and health: reaction slows and vision/hearing fade in old age; illness, epilepsy and heart conditions are hazards.
• Fatigue, alcohol and drugs: lengthen reaction time, impair judgement and cause drowsiness; Nepal's Motor Vehicles and Transport Management Act 2049 prohibits driving under the influence.