Geometric Design of Highway

Extra widening is the additional width of carriageway provided on a horizontal curve, over and above the normal width of the straight road.

Need (reasons)

1. Off-tracking: the rear wheels of a long vehicle do not follow the front wheels; they track inside, so the vehicle takes more width.
2. Difficulty of steering: the driver tends to keep greater clearance from the pavement edge on the curve (psychological).
3. Wider swing path of the front overhang, and the vehicle may encroach on the adjacent lane.
4. Greater clearance is needed between two vehicles that pass or overtake on a curve because the lateral position varies with speed.
5. Therefore: Extra widening $W_e = W_m + W_{ps}$ (mechanical + psychological).

Mechanical widening ($W_m$)

Let $l$ = length of wheelbase (rear axle to front axle), $R$ = mean radius of curve (for the outer rear wheel), $n$ = number of traffic lanes.

          O  (centre of curve)
          |\ 
          | \ R2
      R1  |  \
          |   \
          +----+ rear wheel
               l

In the right triangle formed by the centre $O$, the rear wheel track and the front wheel track:

$$\begin{aligned} R_2^2 &= R_1^2 - l^2 \\ R_1 - R_2 &= R_1 - \sqrt{R_1^2 - l^2} \approx \frac{l^2}{2R} \end{aligned}$$

For $n$ lanes, each vehicle gives the same off-tracking, so

$$W_m = \frac{n\, l^2}{2R}$$

Psychological widening ($W_{ps}$)

Empirical formula (IRC):

$$W_{ps} = \frac{V}{9.5\sqrt{R}}$$

where $V$ is design speed in km/h and $R$ is the radius of the curve in m.

Total extra widening

$$W_e = \frac{n\, l^2}{2R} + \frac{V}{9.5\sqrt{R}}$$

IRC takes $l = 6.1$ m for the design vehicle. The widening is generally provided for $R < 300$ m, since for larger radius it is negligible (NRS 2070/IRC practice).

Methods of providing

1. Widen on the inside edge (single curve) - mostly for the extra width to the inner side.
2. Half on each side (inner and outer edge) - for a large widening or when the road is on a hill with both sides in the cut.
3. Provide the widening gradually along the transition curve (or 1/3 length on the tangent in case of a simple circular curve) so that the full width is obtained at the beginning of the circular curve, with a smooth edge.
4. Add the widening in the pavement and keep the shoulders as usual.

Example: for $V = 60$ km/h, $R = 100$ m, 2 lanes, $l = 6.1$ m: $W_e \approx 2(6.1)^2/200 + 60/(9.5\times10) = 0.37 + 0.63 \approx 1.0$ m.

A transition curve is introduced between a straight and a circular curve so that centrifugal acceleration and superelevation are introduced gradually. The length is fixed from the following criteria, and the largest length is adopted.

1. Rate of change of centrifugal acceleration

When a vehicle moves at speed $v$ on a circular curve of radius $R$, centrifugal acceleration is $v^2/R$. On the transition curve it grows from 0 to $v^2/R$ in time $t = L_s/v$. If the allowed rate of change of acceleration is $C$ (m/s³):

$$C = \frac{v^2/R}{L_s/v} = \frac{v^3}{L_s R} \quad\Rightarrow\quad L_s = \frac{v^3}{C R}$$

With $v = V/3.6$ (V in km/h): $L_s = 0.0215\,\dfrac{V^3}{C R}$.

IRC: $C = \dfrac{80}{75+V}$, limited to $0.5 \le C \le 0.8$ m/s³.

2. Rate of introduction of superelevation

The pavement is raised from the normal camber to full superelevation $e$ over the transition length at a rate of 1 in $N$. Let $E$ = total rise of outer edge relative to inner edge = $e\,(W + W_e)$ when rotated about the inner edge.

$$L_s = N\,e\,(W+W_e)$$

(for rotation about the centre line, the rise of the outer edge is $\dfrac{(e+n)(W+W_e)}{2}$ and $L_s = N\dfrac{(e+n)(W+W_e)}{2}$ with $n$ = camber).

Typical $N$: 1 in 150 for plain and rolling, 1 in 100 for mountainous and steep terrain (IRC; NRS 2070 uses similar values).

3. Empirical formula (IRC)

$$L_s = \frac{2.7\,V^2}{R}\ \text{(plain and rolling)},\qquad L_s = \frac{V^2}{R}\ \text{(mountainous and steep)}$$

Design procedure

1. Find the three lengths by the three methods.
2. Adopt the largest value, rounded up to a multiple of 5 m.
3. Check minimum: not less than about 30 m or the distance travelled in 2-3 s; not more than the length that makes the circular curve negative (shift check).

Shape of transition

The most common type is the spiral (clothoid), in which the radius $r$ at distance $l$ along the curve satisfies $r\,l = R\,L_s$ (radius varies inversely with length). It is set out by the cubic parabola $y = \dfrac{x^3}{6RL_s}$. The shift of the circular curve is $s = \dfrac{L_s^2}{24R}$.

Geometric design fixes the visible dimensions of a highway - cross-section, horizontal and vertical alignment, sight distance and intersections - so that traffic is safe, comfortable and economical. The following factors control it.

1. Design speed

The most important control. It depends on road class, terrain and traffic. All elements (curve radius, sight distance, superelevation, gradient) are linked to it. NRS 2070 gives design speed by road class and terrain.

2. Topography (terrain)

Plain, rolling, hilly (mountainous) and steep terrain are classified by cross-slope. In steep terrain, speed and standards are lower.

3. Traffic factors

• Volume (AADT, design hour volume such as the 30th hour) decides the number of lanes and widths.
• Composition (cars, buses, trucks, non-motorised vehicles) and the capacity in PCU.
• Future traffic growth over the design life.

4. Design vehicle

Dimensions and turning radius of the largest vehicle (width, length, wheelbase, height) fix lane width, radius, extra widening and clearance. Vehicle performance (acceleration, braking) decides gradient and sight distance.

5. Human factors

Driver's reaction time (PIEV), eye height, visual ability, behaviour and pedestrian characteristics.

6. Environmental and aesthetic factors

Noise, pollution, scenery, land use; harmony of the road with the landscape.

7. Economy

Cost of construction, land, maintenance; benefit-cost basis. Standards must be balanced against funds.

8. Safety and capacity

Geometry should minimise accidents, provide sufficient capacity and level of service.

9. Road classification

National highway, feeder road, district, urban and village road have different standards (NRS 2070).

Importance of geometric design

• Safe, smooth and efficient movement; lower accident rate.
• Lower vehicle operating cost and travel time.
• Proper use of land and funds, with room for future improvement.

Superelevation (cant) is the transverse slope given to the carriageway at a horizontal curve by raising the outer edge above the inner edge, so that a component of the vehicle weight balances the centrifugal force.

Derivation

Consider a vehicle of weight $W$ moving at speed $v$ on a curve of radius $R$, on a pavement banked at angle $\theta$ ($\tan\theta = e$ = superelevation).

              P = Wv^2/gR  (centrifugal, horizontal)
        ----->  o
              /|\  W (down)
   F_A+F_B  / | \
   friction/  |   \  R_A + R_B (normal)
   ______/___\|/___\_____ banked pavement (angle theta)

Forces along the inclined plane (taking the vehicle as a point and the friction at both axles $F = f\,(R_A+R_B)$ where $R_A+R_B$ is the total normal reaction):

$$\begin{aligned} P\cos\theta &= W\sin\theta + F \\ R_A+R_B &= W\cos\theta + P\sin\theta \\ F &= f\,(R_A+R_B) \end{aligned}$$

So

$$P\cos\theta = W\sin\theta + f\,(W\cos\theta + P\sin\theta)$$

Dividing by $W\cos\theta$, using $P/W = v^2/(gR)$ and $\tan\theta = e$:

$$\frac{v^2}{gR} = e + f + f\,e\,\frac{v^2}{gR}$$

The term $f\,e\,v^2/(gR)$ is very small and neglected:

$$\boxed{\,e + f = \frac{v^2}{gR}\,}$$

With $V$ in km/h and $R$ in m (g = 9.81):

$$e + f = \frac{V^2}{127\,R}$$

where $e$ = rate of superelevation ($\tan\theta$) and $f$ = design coefficient of lateral friction (IRC/NRS take 0.15).

Use

• For given $V$, $R$ and $f = 0.15$: $e = \dfrac{V^2}{127R} - 0.15$. Maximum $e$ = 7% (plain/rolling, IRC) or 10% (hills); in snowy areas 7%. If $e$ computed exceeds the maximum, adopt the maximum and check $f$.
• IRC adopts $e = V^2/(225R)$, which is the superelevation for 75% of the design speed with zero friction; the check for full speed is then made with $f$.

Stopping Sight Distance (SSD) is the minimum distance ahead along the road that a driver must be able to see an object (height 0.15 m) on the carriageway so that the vehicle running at the design speed can be stopped safely before reaching it. It is the sum of the lag distance and the braking distance.

Factors affecting SSD

1. Speed of the vehicle (increases SSD as $V^2$).
2. Total reaction time of the driver (PIEV, 2.5 s as per IRC/NRS).
3. Coefficient of longitudinal friction $f$ between tyre and pavement (0.35-0.40; depends on surface, tyre and weather).
4. Gradient of the road (down-grade increases SSD).
5. Brake efficiency and tyre condition; weight of the vehicle.
6. Visibility conditions such as night, fog, and the height of eye and object.

Derivation (level road)

Let $v$ = speed (m/s), $t$ = reaction time (s), $f$ = coefficient of friction, $l$ = lag distance, $l_b$ = braking distance.

Lag distance (vehicle moves at constant speed during reaction time):

$$l = v\,t$$

Braking distance: work done by friction = loss of kinetic energy

$$\begin{aligned} f\,W\,l_b &= \frac{1}{2}\frac{W}{g}v^2 \\ l_b &= \frac{v^2}{2gf} \end{aligned}$$

Therefore

$$SSD = v\,t + \frac{v^2}{2gf}$$

In terms of $V$ (km/h): $v = 0.278V$

$$\boxed{SSD = 0.278\,V\,t + \frac{V^2}{254\,f}}$$

For a gradient $n\%$ the braking term becomes $\dfrac{V^2}{254(f \pm 0.01n)}$ (+ upgrade, - downgrade).

Example: $V = 60$ km/h, $t = 2.5$ s, $f = 0.35$: $SSD = 41.7 + 40.5 = 82.2$ m.

Overtaking Sight Distance (OSD) is the minimum distance open to the vision of the driver of a vehicle intending to overtake a slow vehicle ahead, so that he can overtake safely, with no risk of collision from the opposite vehicle coming from the opposite direction.

Factors affecting OSD

1. Speeds of the overtaking, overtaken and opposite vehicles.
2. Spacing between vehicles (depends on speed).
3. Skill and reaction time of the driver.
4. Rate of acceleration of the overtaking vehicle.
5. Gradient of the road, and the road surface condition.
6. Width of the road (two-lane/multi-lane) and the traffic volume.

Derivation (two-lane two-way road)

Let $A$ = overtaking vehicle, $B$ = overtaken vehicle (speed $v_b$ m/s), $C$ = oncoming vehicle; $V$ = design speed of $A$ and $C$ (m/s), $a$ = acceleration of $A$, $t$ = reaction time (2 s).

 A  <-d1->|<-------- d2 -------->|<---- d3 ---->  C
 [A]..[B] |  [A]  ........ [B] [A]|   <- [C]
 start    |  overtaking path      | opposite car

$d_1$ - distance travelled during reaction time (A travels at speed $v_b$ behind B):

$$d_1 = v_b\,t$$

$d_2$ - distance during the overtaking operation. A starts at spacing $s$ behind B and finishes at spacing $s$ ahead of B, so A gains $2s$ on B in time $T$. Relative to B, A starts with zero relative speed and accelerates at $a$:

$$\begin{aligned} 2s &= \tfrac12 a T^2 \ \Rightarrow\ T = \sqrt{\frac{4s}{a}} \\ d_2 &= v_b\,T + 2s \end{aligned}$$

where $s = 0.7v_b + 6$ (m, $v_b$ in m/s).

$d_3$ - distance travelled by oncoming vehicle C during the overtaking time $T$ (C runs at design speed $v$):

$$d_3 = v\,T$$

Total

$$\boxed{OSD = d_1 + d_2 + d_3 = v_b t + (v_bT + 2s) + vT}$$

Where $v_b = V - 16$ km/h (speed of the overtaken vehicle) as per IRC.

IRC gives about 470 m for $V = 80$ km/h on a two-way road (about 300 m for 60 km/h). On one-way roads $d_3$ is not needed, so $OSD = d_1 + d_2$.

A transition curve is a curve of gradually changing radius, provided between a straight and a circular curve (or between two circular curves) so that the radius changes from infinity at the tangent to $R$ at the circular curve.

Objectives and need

1. To introduce the centrifugal force gradually from zero to full value, avoiding a sudden jerk to the passengers and the load.
2. To enable gradual introduction of superelevation over the length of the curve.
3. To provide a way for the gradual introduction of extra widening.
4. To improve the appearance of the road, avoiding the visible kink at the junction of the straight and curve.
5. To reduce the chance of skidding, overturning and lateral shift of the vehicle, so that speed is more uniform.
6. To keep the vehicle within its lane (a vehicle naturally follows a transition path while entering a curve).

Types of transition curve

1. Spiral (clothoid / Euler spiral) - radius varies inversely with the length: IRC recommends this.
2. Cubic parabola - $y = x^3 /(6RL_s)$; practically the same as the spiral for small deflection.
3. Bernoulli's lemniscate - used mostly for urban roads and railways at large deflection.

Superelevation is provided by rotating the pavement cross-section over the transition length (or on the tangent before the curve if there is no transition). The pavement is first changed from a cambered section to a flat or one-way section, and then to the full superelevation. The change should be gradual (rate 1 in 150 to 1 in 60 as per terrain).

1. Eliminating the crown (removing the adverse camber)

• First the outer half is rotated about the crown until it becomes level with the inner half (normal camber is removed), then the whole pavement is rotated to give the required superelevation.
• Done over the first part of the transition length.

 normal:   /\        flat:  ____    full e:     ___/
          /  \              ----           ___/

2. Rotating about the centre line

• The pavement is turned about the centre line: outer edge is raised and the inner edge lowered by an equal amount, each by half of the total difference.
• Advantage: keeps the centre-line level fixed (profile unchanged), half of the earthwork on each side. Mostly used on plains.
• Disadvantage: the inner edge goes below the ground profile, possible drainage problem.

   outer edge up   ____.-----   
   centre fixed ---+
   inner edge down ----'

3. Rotating about the inner edge

• The inner edge is the fixed (pivot) and the outer edge is raised; the centre line also rises.
• Used where the drainage is critical or the inner edge level can't be changed (e.g. in cuttings, hill roads).
• Disadvantage: the profile of the centre line is changed and the outer edge is raised more, which affects levels and appearance.

   outer edge up  ______----.
   centre line up    .--'
   inner edge fixed ---

4. Rotating about the outer edge

• Outer edge level fixed and the inner edge lowered; used where the outer edge profile must be kept (for appearance or a curve on a hillside).

Remarks

• The pavement is generally rotated about the centre line on plain roads and about the inner edge in drainage-sensitive cases.
• If no transition curve is provided, about two-thirds of the superelevation run-off is placed on the tangent and one-third inside the circular curve (IRC). Extra widening is introduced over the same length.

The PIEV theory explains the driver's total reaction time. It divides the time between the instant the driver sees an object and the instant the brake begins to act into four parts.

1. P - Perception: the time taken to see and receive the signal/object through the senses.
2. I - Intellection: the time to understand and interpret what the object means (is it dangerous?).
3. E - Emotion: the time to decide on a reaction (to brake, steer or ignore), affected by fear, fatigue or alcohol.
4. V - Volition: the time to carry out the decision, i.e. the muscles act to apply the brake.

$$t = t_P + t_I + t_E + t_V$$

The total time ranges from 0.5 s to over 4 s; the design value is 2.5 s (IRC, NRS 2070) used for SSD, which covers about 85-90% of drivers.

Example

A driver driving at 60 km/h sees a child run onto the road: he sees the child (P), recognises the danger (I), decides to brake (E), and moves the foot to the brake pedal (V). During 2.5 s the vehicle travels $0.278\times60\times2.5 \approx 42$ m (the lag distance) before braking starts.

Use: PIEV time decides the lag distance in the sight distance, and the location of signs and signals.

Design steps of superelevation (IRC practice, followed in NRS 2070 designs):

1. Fix the design speed $V$ (km/h) and the radius $R$ of the curve from the road class and terrain.
2. Compute the superelevation for 75% of the design speed with no friction: $$e = \frac{V^2}{225\,R}$$
3. Compare with the maximum value: 7% for plain and rolling terrain, 10% for hilly terrain (IRC). If $e$ is more than the maximum, adopt $e_{max}$.
4. Minimum value: not less than the camber (normally 2.5%); if the computed $e$ is smaller than the camber, provide the camber value (or a one-way cross fall).
5. Check friction at the full design speed: $$f = \frac{V^2}{127\,R} - e \le 0.15$$ If $f > 0.15$, either reduce the speed (restrict) or increase $R$ or $e$.
6. Decide the rate of introduction of superelevation (1 in 150 plain/rolling, 1 in 60 to 1 in 100 in hills) and find the length of the transition.
7. Choose the method of attaining superelevation (about the centre line or inner edge) and prepare the diagram of edge levels.
8. Provide extra widening and introduce it along the same transition length.

Extra widening is the additional carriageway width given on a horizontal curve beyond the normal width of the straight section.

Reasons

1. Off-tracking: the rear wheels follow a smaller radius path than the front wheels, so a long vehicle needs more width (mechanical widening).
2. Wider swing of the front end: the front overhang sweeps a larger arc outside the track of the front wheels.
3. Driver's difficulty: on curves drivers find it harder to keep to the lane; they tend to drive away from the pavement edges and need extra clearance (psychological widening).
4. Vehicle crossing: a safer clearance between two vehicles passing or overtaking on a curve is needed, as lateral clearances become uneven with speed.
5. Higher speed vehicles need more width for steering.

The extra width is $W_e = \dfrac{n l^2}{2R} + \dfrac{V}{9.5\sqrt{R}}$ and is generally provided when $R < 300$ m; it is introduced gradually along the transition curve.

Curve resistance

Curve resistance is the additional tractive resistance a vehicle has to overcome when it moves along a horizontal curve. It arises from the sideways slip of the tyres, the lateral friction needed to resist the centrifugal force, the steering action and the off-tracking of the rear wheels. It increases as the radius becomes smaller. A gradient plus a sharp curve at the same place can overload the vehicle.

Grade compensation

Grade compensation (grade reduction) is the reduction of the design gradient on a horizontal curve so that the total resistance (grade + curve) at the curve is not more than the resistance at the straight section.

IRC formula (R in metres):

$$\text{Grade compensation} = \frac{30 + R}{R}\ \%\ \ \text{(max } \frac{75}{R}\%)$$

• It is not needed when the gradient is flatter than 4%, and in practice matters mainly for sharp curves (small $R$).
• Example: $R = 60$ m: $(30+60)/60 = 1.5\%$, maximum $75/60 = 1.25\%$, so adopt 1.25%. If the ruling gradient is 6%, use $6 - 1.25 = 4.75\%$ on the curve.
• Used chiefly in hill roads.

Grade compensation is the reduction of gradient on a sharp horizontal curve. The reasons are:

1. Curve resistance: on a curve the vehicle meets extra tractive resistance due to lateral friction, tyre slip and steering. If the full ruling gradient also exists there, the total resistance may be more than the engine can overcome, causing loss of speed or stalling.
2. Equal total resistance: the gradient is reduced by an amount equal to the curve resistance so that the total tractive effort on the curve is the same as on the straight with the ruling gradient.
3. Safety and speed uniformity: vehicles keep a uniform speed, driving is easier and the accident risk reduced, especially for heavy trucks.
4. Hill roads: where hairpin bends and sharp curves are on long steep grades, the heavy vehicles can not climb without compensation.

IRC rule: compensation $= \dfrac{30+R}{R}\%$, not exceeding $\dfrac{75}{R}\%$ (R in m); not needed for gradients flatter than 4%.

The factors affecting stopping sight distance are:

1. Speed of the vehicle: SSD increases roughly with the square of speed.
2. Total reaction time of the driver (PIEV time): longer time increases the lag distance; 2.5 s adopted in design.
3. Coefficient of longitudinal friction $f$ between tyre and road, which depends on the road surface (smooth, wet, rough) and tyre condition.
4. Gradient of the road: descending grade increases the braking distance, ascending grade decreases it.
5. Efficiency of brakes and weight (loading) of the vehicle.
6. Weather and visibility: rain, fog, night; height of the driver's eye and object height.
7. Driver characteristics: age, fatigue, alertness, alcohol.
8. Tyre type and condition.

Given: $g_1 = +1/60 = +1.667\%$, $g_2 = -1/45 = -2.222\%$; SSD $= 210$ m, OSD $= 600$ m; $L_{max} = 750$ m.

$$N = \frac{1}{60} + \frac{1}{45} = 0.03889$$

Sight line heights (IRC): eye height $H = 1.2$ m; object height for SSD $h = 0.15$ m, for OSD and ISD $h = 1.2$ m.

(i) Stopping sight distance ($S = 210$ m)

Assuming $S < L$:

$$L = \frac{N S^2}{\left(\sqrt{2H}+\sqrt{2h}\right)^2} = \frac{0.03889\,(210)^2}{(\sqrt{2.4}+\sqrt{0.3})^2} = \frac{1715.0}{4.397} = 390.0\ \text{m}$$

$S = 210 < 390$ so the assumption holds. $L = 390$ m < 750 m: OK.

(ii) Overtaking sight distance ($S = 600$ m)

$$L = \frac{N S^2}{(\sqrt{2.4}+\sqrt{2.4})^2} = \frac{0.03889\,(600)^2}{9.6} = 1458.3\ \text{m}$$

$S < L$, valid. But 1458 m is more than the limit of 750 m - not feasible.

Intermediate sight distance $= 2\times SSD = 420$ m:

$$L = \frac{0.03889\,(420)^2}{9.6} = 714.6\ \text{m} \ (< 750\ \text{m})$$

Discussion

The summit curve of 750 m can not give full overtaking sight distance. Providing OSD would need 1458 m, which is too costly. So adopt $L \approx 715$-$750$ m, which gives the intermediate sight distance (at least 420 m) and more than the SSD everywhere; overtaking is allowed only where there is a clear view, and "no overtaking" signs and markings are provided on the curve. A length of 750 m gives ISD of about 430 m.

Stopping distance is the total distance travelled by a vehicle from the instant the driver sights an object or hazard ahead until the vehicle comes to a complete stop. It is the sum of the lag distance (travelled during the reaction time) and the braking distance:

$$SD = v\,t + \frac{v^2}{2g\,(f \pm G)}$$

Factors affecting stopping distance

1. Speed of the vehicle.
2. Reaction time (PIEV) of the driver.
3. Coefficient of longitudinal friction (type and condition of road surface and tyres, wet or dry).
4. Gradient of the road.
5. Brake efficiency and the weight/loading of the vehicle.
6. Weather and visibility, driver's alertness (also fatigue, alcohol).

Relation between stopping distance and stopping sight distance

• The stopping sight distance (SSD) is the clear length of road ahead that the road must give so that the driver can see an object and stop in time.
• It is equal to the stopping distance computed for the design speed with the design values of $t = 2.5$ s and $f$: so the design SSD is simply the stopping distance: $$SSD = 0.278\,V\,t + \frac{V^2}{254\,(f \pm 0.01n)}$$
• The available sight distance at any point of a horizontal or vertical curve (measured with a driver eye height 1.2 m and an object height 0.15 m) must be not less than the stopping distance.
• If the available sight distance is less than the stopping distance, the vehicle cannot stop before the object, so the road geometry (curve length, set-back, speed limit) is revised.

$g_1=+4\%,\ g_2=-4\%$, so $N=0.08$. This is a summit curve (IRC/NRS 2070: eye height 1.2 m; object height 0.15 m for SSD and 1.2 m for OSD).

(a) Stopping sight distance S = 150 m

Assuming $L>S$:

$$L=\frac{NS^2}{4.4}=\frac{0.08(150)^2}{4.4}=409.1\ \text{m}$$

$L>S$, so valid. L ≈ 410 m, which is less than 800 m, so it is possible.

(b) Overtaking sight distance S = 500 m

For $h_1=h_2=1.2$ m:

$$L=\frac{NS^2}{9.6}=\frac{0.08(500)^2}{9.6}=2083.3\ \text{m}$$

This exceeds the 800 m limit. With L = 800 m the sight distance actually available is

$$S=\sqrt{\frac{9.6L}{N}}=\sqrt{\frac{9.6(800)}{0.08}}=309.8\ \text{m}$$

which is much less than the 500 m OSD.

Intermediate sight distance

ISD $=2\times SSD=300$ m:

$$L=\frac{0.08(300)^2}{9.6}=750\ \text{m}$$

This is within 800 m, so ISD can be provided.

Discussion

Answer: adopt L = 800 m (or at least 750 m). It satisfies SSD (210 m available) and ISD (310 m), but full OSD cannot be provided within 800 m. Overtaking must be prohibited on this summit (no-overtaking signs and continuous centre line).

The cross-section of a highway shows the elements in a section cut at right angle to the centre line. The main elements are:

|<-------------- Right of way ------------------>|
|<--------- Formation width -------->|
 _     shoulder   carriageway   shoulder     _
  \  ____________ ___/^\___ ____________  /
   \/ kerb/edge   camber crown            \/
 drain     ===========================   drain

1. Carriageway (pavement): the portion of the road meant for vehicles; consists of lanes. Lane width 3.0-3.75 m (3.5 m typical for NH) depending on class, speed and terrain (NRS 2070/IRC).
2. Shoulders: strips on both sides of the carriageway (1.5-2.5 m); used for emergency stopping, lateral support to the pavement and pedestrians.
3. Camber (cross fall): the transverse slope of the carriageway (2-3% for paved roads) to drain water.
4. Kerbs: raised edges between carriageway and footpath/median (urban roads and bridges).
5. Median (central reserve): on divided highways, a strip between opposite directions with minimum width 1.2-5 m.
6. Side drains: open drains beside the road, usually trapezoidal or triangular.
7. Side slopes: slopes of embankment (2H:1V, i.e. 2 horizontal to 1 vertical) or cutting.
8. Formation (roadway) width: carriageway + shoulders (+ median + drains in some definitions); the width at subgrade level.
9. Right of way (ROW): the total land width acquired for the road with allowance for widening, utilities and services (e.g. 25-50 m for NH as per class).
10. Footpath, cycle track and service road: in urban stretches.
11. Roadside planting, guard rails, signs, and boundary stones.
12. Building line and control line: the lines beyond which construction is restricted.

An urban road has more elements than a rural road because of pedestrians, parking, drainage and utility services.

|<--------------- Right of Way --------------->|
|footpath|green|  carriageway  |median| carriageway |green|footpath|
|  1.5-2m|    | L1  L2  parking|      | L1  L2      |     |        |
 =kerb==         ______/\______  (camber)           ==kerb==
 | drain below footpath / utilities                          |

Elements

1. Carriageway: lanes for vehicles; each lane 3.0-3.5 m (bus lane 3.5 m); number of lanes depends on the capacity needed. Camber 2-2.5%.
2. Median (divider): separates opposing flows in multi-lane roads; may have trees and street lights; minimum 1.2 m (preferably more).
3. Kerb: raised edge (100-150 mm) separating the carriageway from the footpath; also guides the drainage to the gully.
4. Footpath (sidewalk): for pedestrians, minimum 1.5 m (more in busy areas); a slight slope outward.
5. Parking lane: 2.0-2.5 m wide parallel parking along the kerb.
6. Cycle track: separate path, usually 2.0 m for two-way; protects the cyclists.
7. Drainage: storm sewers, gully pits, or covered drains under the footpath.
8. Utilities: water, sewer, electric, telecom lines under footpath or green strip.
9. Street furniture: street lights, signs, bus stops, benches, trees and green verge.
10. Right of way and building line: the total corridor; buildings are set back from the building line for light, safety and future widening.
11. Service roads / frontage roads: parallel to the main road for local access.

When the inner edge can not be lowered (drainage problem), the pavement is rotated about the inner edge and the outer edge is raised. The two-way (cambered) section of slope $n$ on each side is changed to a one-way section of slope $e$ over the transition length $L$ in three steps.

Notation

$W$ = pavement width, $n$ = camber, $e$ = full superelevation ($e>n$), $L$ = length of transition. Inner edge level is held fixed. The outer edge must rise by $eW$ relative to the inner edge over $L$ (uniform rate $eW/L$).

Stages (cross-sections)

 A (normal)      B (outer flat)  C (one-way n)   D (one-way e)
   /\            ____/            ___.-'          ___.---'
  /  \          /                ./              ./   
 inner fixed at all sections --> ^ (same level)

• Section A (start): normal crown, both halves slope $n$.
• Stage 1 (A to B): outer half is rotated about the crown until it is level (slope $0$). The outer edge rises $nW/2$, so B is at a distance $$x_B = \frac{nW/2}{eW/L} = \frac{nL}{2e}$$
• Stage 2 (B to C): the outer half is rotated further until the whole pavement has a uniform one-way slope $n$. The outer edge has risen $nW$ in all: $$x_C = \frac{nW}{eW/L} = \frac{nL}{e}$$
• Stage 3 (C to D): the whole pavement is rotated about the inner edge, increasing the one-way slope from $n$ to $e$ at the end of the transition, $x_D = L$. The outer edge has risen $eW$.

Level of the outer edge

At distance $x$ from the start, the outer edge is above its original level by $\dfrac{eW}{L}x$ (relative to the inner edge). At D: outer edge $= eW$ above the inner edge; the centre line has risen by $\dfrac{(e-n)W}{2}$ above its original level.

The extra widening, if any, is provided during the same length $L$.

When a vehicle moves along a horizontal curve of radius $R$ at speed $v$, a centrifugal force $P = \dfrac{Wv^2}{gR}$ acts outwards (horizontal, through the centre of gravity). The ratio $P/W = v^2/(gR)$ is called the centrifugal ratio. It has the following effects:

1. Tendency to overturn the vehicle about the outer wheels, because $P$ acts at the height $h$ of the centre of gravity and the weight $W$ resists with the lever arm $b/2$ ($b$ = wheel track). Overturning begins when $$\frac{v^2}{gR} \ge \frac{b}{2h}$$
2. Tendency to skid sideways (lateral skidding) outwards when $P$ is more than the lateral friction $fW$: $$\frac{v^2}{gR} \ge f$$
3. Discomfort and strain to the passengers and the load; shift of the load in goods vehicles.
4. Reduced steering control, driver drifts to the outer edge and may encroach to the opposite lane.
5. Uneven tyre wear and higher fuel use due to lateral slip.
6. Higher chance of accident, especially for vehicles with high centre of gravity (trucks, buses), or in wet roads.

Remedies: superelevation, adequate radius, a transition curve, restricting speed, and the use of warning signs. In a well-designed curve, skidding happens before overturning because normally $f < b/2h$.

Consider a vehicle of weight $W$ moving at speed $v$ on a horizontal curve of radius $R$ on a pavement with superelevation $e$ (neglect the small term $f\,e$). Wheel track $= b$, height of centre of gravity above the road $= h$, coefficient of lateral friction $= f$. Centrifugal force $P = \dfrac{Wv^2}{gR}$.

        P ---> o  (c.g. at height h)
               |
   <- b/2 ->   |
   inner wheel   outer wheel (overturning about this)

1. Condition for lateral skidding

Skidding occurs when the centrifugal force along the pavement exceeds the sum of the friction and the weight component:

$$\begin{aligned} P\cos\theta &\ge W\sin\theta + F \\ \frac{v^2}{gR} &\ge e + f \end{aligned}$$

So the maximum safe speed against skidding is $v = \sqrt{gR(e+f)}$ or $V = \sqrt{127R(e+f)}$ km/h. For a flat road ($e=0$): $\dfrac{v^2}{gR} \ge f$.

2. Condition for overturning

Take moments about the outer wheel contact point. The overturning moment $= P\,h$ and the restoring moment $= W\,b/2$ (flat road). Overturning starts when the reaction at the inner wheel becomes zero:

$$P\,h \ge W\frac{b}{2} \ \Rightarrow\ \frac{v^2}{gR} \ge \frac{b}{2h}$$

With superelevation, $\dfrac{v^2}{gR} \ge \dfrac{b}{2h} + e$.

Result

• Safe against skidding if $\dfrac{v^2}{gR} < e + f$.
• Safe against overturning if $\dfrac{v^2}{gR} < \dfrac{b}{2h} + e$.
• For usual vehicles $b/2h \approx 0.45$-$0.6$, which is greater than $f$ (0.15), so a vehicle will skid before it overturns.

Types of highway curves

Horizontal curves

• Simple circular curve - a single arc of constant radius joining two tangents.
• Compound curve - two or more arcs of different radii in the same direction with a common tangent at the junction.
• Reverse curve - two arcs bending in opposite directions.
• Transition (spiral) curve - varying radius between a tangent and a circular curve.

Vertical curves

• Summit (crest) curve - formed when an upgrade meets a downgrade or a lesser upgrade.
• Valley (sag) curve - when a downgrade meets an upgrade or lesser downgrade.

Effects on a vehicle on a horizontal curve without superelevation

When the pavement is flat (no superelevation), the whole centrifugal force $P = Wv^2/(gR)$ must be resisted by lateral friction alone.

1. Lateral skidding: if $v^2/(gR) > f$ the tyres slip sideways and the vehicle leaves the curve; a high speed or low $f$ (wet road) makes it likely.
2. Overturning: if $v^2/(gR) > b/(2h)$ the vehicle overturns about the outer wheels, a risk for high vehicles such as buses and trucks.
3. Load transfer: the outer wheels carry more load than the inner wheels, so the outer tyres wear more and the load shifts.
4. Passenger discomfort: passengers are thrown outward.
5. Driver difficulty: to prevent skidding the driver must slow down, so the speed and capacity decrease; or, if he does not, he runs on the outer lane and may meet opposing vehicles.
6. Higher fuel use and tyre wear.

Reason: the centrifugal force has no counterbalance except friction, and the friction available ($f = 0.15$ at design) is usually not enough for the design speed. A superelevation $e$ (and an adequate radius) provides a component of weight against $P$ so that $e + f = V^2/(127R)$ is satisfied.

Gradient (grade) is the rate of rise or fall of the road along its length, expressed as a percentage (or 1 in n).

Types of gradient

1. Ruling (design) gradient: the maximum gradient that is used in the design of a road. Heavy vehicles can pull up the grade without loss of speed. IRC: plain/rolling 3.3% (1 in 30), mountainous and steep 5%.
2. Limiting gradient: a steeper value used where the topography makes the ruling gradient costly. IRC: 5% (1 in 20) in plain/rolling, 6% in hills. Not long continuous.
3. Exceptional gradient: the steepest value allowed in short stretches (not more than 100 m) in unavoidable conditions. IRC: 6.7% (plain/rolling), 7% (hills). Nepal hill roads allow higher values as per the class in NRS 2070.
4. Minimum gradient: for drainage on a kerbed road or a flat terrain, at least 0.5% for lined drains, 1% for unlined drains (IRC).
5. Average gradient: the ratio of the total rise or fall to the length of a section; used for assessing the hill road.
6. Momentum gradient: a steep grade after a downgrade where the vehicle uses its momentum to climb.

(Figures follow IRC; NRS 2070 should be read for the exact table by road class and terrain.)

Factors in selecting gradients

1. Terrain - plain, rolling, hilly, or mountainous.
2. Design speed and the class of road.
3. Traffic volume and composition - high proportion of trucks needs flatter grades.
4. Construction and maintenance cost - flatter grades require more earthwork.
5. Vehicle characteristics - power-to-weight ratio, tractive resistance, brake performance.
6. Drainage needs and climatic conditions (snow, ice, rain).
7. Length of grade - long grades need flatter gradient; use rest levels.
8. Horizontal curves - provide grade compensation.
9. Safety: sight distance on vertical curves, braking on long downgrades.

Sight distance is the length of the road ahead visible to the driver. Adequate sight distance is necessary for safe operation.

1. Stopping Sight Distance (SSD)

The minimum distance needed to see an object on the road and stop before hitting it (eye height 1.2 m and object height 0.15 m, IRC). It is the sum of the lag distance and the braking distance:

$$SSD = 0.278\,V\,t + \frac{V^2}{254\,f}$$

($V$ in km/h, $t = 2.5$ s, $f \approx 0.35$-$0.40$). It must be provided on every point of the road. Example: $V = 60$ km/h gives about 82 m (IRC table: 80 m).

2. Overtaking Sight Distance (OSD)

The minimum distance for the safe overtaking of a slower vehicle on a two-lane road without colliding with the oncoming vehicle:

$$OSD = d_1 + d_2 + d_3 = v_b t + (v_b T + 2s) + vT$$

where $v_b$ = speed of overtaken vehicle, $T = \sqrt{4s/a}$, $s = 0.7v_b + 6$ (IRC). It is provided on straight sections; example 470 m at 80 km/h on a two-way road.

3. Intermediate Sight Distance (ISD)

Where OSD can not be provided, ISD $= 2\times SSD$ gives some chance to overtake.

4. Other sight distances

• Headlight sight distance for valley curves at night.
• Safe sight distance at intersections (approach sight triangle).
• Decision sight distance at complex locations.

A valley (sag) curve is a vertical curve with the centre of curvature above the road. There is no obstruction to vision during the day, so the length is designed from two criteria: comfort and headlight sight distance at night. The longest value is adopted. The curve is taken as a parabola of length $L$, with algebraic difference of grades $N$ (in ratio) and radius of curvature $R \approx L/N$.

1. Comfort criterion

A vehicle moving at speed $v$ (m/s) on the vertical curve feels a centripetal acceleration $v^2/R$ which rises gradually from 0 at the start. The allowable rate of change of this acceleration is $C = 0.6$ m/s³.

The valley curve can be treated as two transition curves (each of length $L/2$ and deflection $N/2$). For a spiral transition curve, the deflection angle is $\dfrac{L_s}{2R}$ and $L_sR = \dfrac{v^3}{C}$. Here, $L_s = L/2$:

$$\frac{N}{2} = \frac{L/2}{2R}\ \Rightarrow\ R = \frac{L}{2N}$$ $$\frac{L}{2}\cdot\frac{L}{2N} = \frac{v^3}{C} \ \Rightarrow\ L = 2\sqrt{\frac{N\,v^3}{C}}$$

The minimum radius from comfort:

$$R_{min} = \frac{L}{2N} = \sqrt{\frac{v^3}{C\,N}}$$

Here, $N$ = $|g_1 - g_2|$ as ratio, $v = V/3.6$.

2. Night visibility (headlight) criterion

Assume the headlight height $h$ above the road and a beam angle $\alpha$ upward (usually $h = 0.75$ m, $\alpha = 1^\circ$). The driver must see up to the stopping distance $S$.

Case (i): $L > S$. The beam touches the road at distance $S$ from the vehicle, at the height $y$ of the curve above the tangent:

$$h + S\tan\alpha = \frac{N S^2}{2L} \ \Rightarrow\ L = \frac{N S^2}{2\,(h + S\tan\alpha)}$$

Radius: $R = \dfrac{L}{N} = \dfrac{S^2}{2(h + S\tan\alpha)}$.

Case (ii): $L < S$.

$$L = 2S - \frac{2\,(h + S\tan\alpha)}{N}$$

Design

Find $L$ from comfort and from headlight criteria; adopt the greater. Then $R = L/N$ and check that $R$ is more than the minimum radius.

Given: $g_1 = +4\%$, $g_2 = -3.5\%$, posted speed $V = 70$ km/h, highest point at 120 m from the beginning of the vertical curve (BVC), eye height $h_1 = 1.08$ m, reaction time $t = 2.5$ s, $f = 0.35$.

Assumption: only the driver's eye height is given, so the object (the other car or an obstruction) is taken at road level, $h_2 = 0$. The sight distance to be checked is the stopping sight distance.

Step 1: Length of the vertical curve

$$N = g_1 - g_2 = 0.04 + 0.035 = 0.075$$

Highest point from BVC: $x = \dfrac{g_1 L}{N}$

$$120 = \frac{0.04\,L}{0.075}\ \Rightarrow\ L = 225\ \text{m}$$

Step 2: SSD required at the posted speed

$$\begin{aligned} SSD &= 0.278\,V\,t + \frac{V^2}{254 f} \\ &= 0.278(70)(2.5) + \frac{70^2}{254\times0.35} \\ &= 48.65 + 55.12 = 103.8\ \text{m} \end{aligned}$$

Step 3: Sight distance available on the curve

For a summit curve with $S < L$ and $h_2 = 0$, $H = 1.08$ m:

$$L = \frac{N S^2}{2H}\ \Rightarrow\ S = \sqrt{\frac{2HL}{N}} = \sqrt{\frac{2(1.08)(225)}{0.075}} = 80.5\ \text{m}$$

$S = 80.5\ \text{m} < L = 225$ m, so the assumption $S < L$ is valid.

Step 4: Check

Available sight distance 80.5 m $<$ required SSD 103.8 m. The sight distance is not sufficient at 70 km/h, so the posted speed sign is faulty.

Step 5: Safe posted speed

Find $V$ for which SSD = 80.5 m:

$$\frac{V^2}{88.9} + 0.695V - 80.5 = 0\ \Rightarrow\ V = 59.2\ \text{km/h}$$

Check: at 60 km/h, SSD = 82.2 m (slightly greater than 80.5 m), at 55 km/h SSD = 72.3 m (< 80.5 m).

Answer: The posted 70 km/h is unsafe (SSD 103.8 m > available 80.5 m). The posted speed should be 55 km/h (the calculated limit is about 59 km/h, rounded down to a practical value of 55 km/h).

Note: if the IRC object height of 0.15 m is used, the available distance becomes 110.5 m and 70 km/h would be acceptable (maximum 72.9 km/h); the answer above uses the data as given (no object height).

Given: $R = 400$ m, $V = 85$ km/h, $c = 0.52$ m/s³, intersection angle (interior) $= 125^\circ 25'$, PI chainage $= 1075.5$ m.

Reading: the "angle of intersection" is the interior angle between the straights, so the deflection angle $\Delta = 180^\circ - 125^\circ25' = 54^\circ35' = 54.583^\circ$.

(i) Length of transition curve

$v = 85/3.6 = 23.61$ m/s.

$$L_s = \frac{v^3}{cR} = \frac{(23.61)^3}{0.52\times400} = \frac{13162.6}{208} = 63.3\ \text{m}$$

Check with the IRC empirical formula (plain/rolling): $2.7V^2/R = 2.7(85)^2/400 = 48.8$ m $<63.3$ m. The rate of superelevation criterion is not requested (no width data). Adopt

$L_s = 65$ m.

(ii) Shift

$$s = \frac{L_s^2}{24R} = \frac{65^2}{24\times400} = \frac{4225}{9600} = 0.44\ \text{m}$$

(iii) Chainages

Spiral angle $\phi_s = \dfrac{L_s}{2R} = 0.0813$ rad $= 4.66^\circ$.

$$\begin{aligned} T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} \\ &= (400.44)\tan(27.292^\circ) + 32.5 = 206.6 + 32.5 = 239.1\ \text{m} \end{aligned}$$

Length of circular curve:

$$L_c = R\left(\Delta - 2\phi_s\right) = 400\,(0.95269 - 0.16250) = 316.1\ \text{m}$$

Answer: $L_s = 65$ m; shift $= 0.44$ m; curve begins at chainage 836.4 m and ends at 1282.5 m (total length $316.1 + 130 = 446.1$ m).

If the 125°25' were taken as the deflection angle instead, $T_s = 808.6$ m, the curve would begin at 266.9 m and end at 1207.5 m.

Given: $V = 100$ km/h, $g_1 = -1/40 = -2.5\%$, $g_2 = +1/30 = +3.33\%$, BVC elevation $= 312.56$ m.

Assumptions (IRC/NRS): reaction time 2.5 s, $f = 0.35$, comfort rate $C = 0.6$ m/s³, headlight height 0.75 m, beam angle $1^\circ$.

$$N = 0.025 + 0.03333 = 0.05833$$

$v = 100/3.6 = 27.78$ m/s.

1. Comfort condition

$$L = 2\sqrt{\frac{N v^3}{C}} = 2\sqrt{\frac{0.05833\times 27.78^3}{0.6}} = 2\sqrt{2083.8} = 91.3\ \text{m}$$

2. Headlight sight distance

SSD:

$$SSD = 0.278(100)(2.5) + \frac{100^2}{254\times0.35} = 69.5 + 112.5 = 182.0\ \text{m}$$

Assuming $S < L$:

$$L = \frac{N S^2}{2\,(h + S\tan\alpha)} = \frac{0.05833\,(182.0)^2}{2\,(0.75 + 182.0\tan1^\circ)} = \frac{1932.2}{2(3.927)} = 246.0\ \text{m}$$

$S = 182 < 246$ so the assumption holds.

3. Design length

Adopt the larger: $L = 246$ m, rounded to $L = 250$ m.

4. Lowest point

Distance from BVC:

$$x = \frac{g_1 L}{N} = \frac{0.025\times250}{0.05833} = 107.1\ \text{m}$$

Elevation (curve equation $y = BVC + g_1x + \dfrac{N x^2}{2L}$ with $g_1 = -0.025$):

$$\begin{aligned} RL_{low} &= 312.56 - 0.025(107.14) + \frac{0.05833(107.14)^2}{2(250)} \\ &= 312.56 - 2.679 + 1.339 = 311.22\ \text{m} \end{aligned}$$

5. End of curve (EVC)

$$RL_{EVC} = 312.56 + \frac{(g_1+g_2)L}{2} = 312.56 + \frac{(-0.025+0.03333)(250)}{2} = 313.60\ \text{m}$$

The EVC is at 250 m from the BVC.

Answer: $L = 250$ m (comfort 91.3 m, headlight 246.0 m); lowest point is 107.1 m from BVC at RL 311.22 m; end of curve at 250 m from BVC with RL 313.60 m.

Data

$R=400$ m (centre line), $L=200$ m, $n=1.9$ m (centre line of road to centre of inner lane). The sight line is taken along the centre of the inner lane, so its radius is $R-n=398.1$ m. The setback $m$ is measured from the centre line of the pavement to the obstruction line.

(a) Stopping sight distance S = 90 m ($S<L$)

Half the angle subtended by the sight distance:

$$\frac{\alpha}{2}=\frac{S}{2(R-n)}=\frac{90}{2(398.1)}=0.11304\ \text{rad}=6.477^\circ$$ $$m=R-(R-n)\cos\frac{\alpha}{2}=400-398.1\cos(6.477^\circ)=4.441\ \text{m}$$

(b) Overtaking sight distance S = 300 m ($S>L$)

For $S>L$ the angle is based on the curve length:

$$\frac{\alpha}{2}=\frac{L}{2(R-n)}=\frac{200}{2(398.1)}=0.25119\ \text{rad}=14.392^\circ$$ $$m=R-(R-n)\cos\frac{\alpha}{2}+\frac{S-L}{2}\sin\frac{\alpha}{2}=14.394+50\sin(14.392^\circ)=26.822\ \text{m}$$

Answer: setback from the centre line = 4.44 m for SSD = 90 m and 26.82 m for OSD = 300 m.

Reading of the question: the first digit of the PVI elevation is unclear in the scan; I have taken the elevation as 150.5 m. The method is the same for any value (add or subtract the difference to every level).

A −3.5% grade meeting +4% is a valley curve; $N=0.035+0.04=0.075$.

Length (night visibility, $S=250$ m)

$$L=\frac{NS^2}{2(h+S\tan\alpha)}=\frac{0.075(250)^2}{2(0.6+250\tan1.5^\circ)}=328.0\ \text{m}$$

$L>S$, so the formula is valid. Adopt L = 330 m; half length = 165 m.

Levels at BVC and EVC (PVI = 150.50 m)

$$RL_{BVC}=150.50+0.035(165)=156.275\ \text{m},\qquad RL_{EVC}=150.50+0.04(165)=157.100\ \text{m}$$

Lowest point

$$x=\frac{g_1L}{N}=\frac{0.035(330)}{0.075}=154.00\ \text{m from BVC}$$

For a valley curve with $x$ from BVC: $y=RL_{BVC}-0.035x+\dfrac{0.075x^2}{2(330)}$.

$$RL_{lowest}=156.275-0.035(154.00)+\frac{0.075(154.00)^2}{2(330)}=153.580\ \text{m}$$

Levels 75 m left and 60 m right of PVI

• Left: $x=165-75=90$ m, $y=156.275-0.035(90)+\dfrac{0.075(90)^2}{2(330)}=\mathbf{154.045}$ m
• Right: $x=165+60=225$ m, $y=156.275-0.035(225)+\dfrac{0.075(225)^2}{2(330)}=\mathbf{154.153}$ m

Answer: BVC 156.28 m; lowest point 153.58 m (at 154.0 m from BVC); EVC 157.10 m; 75 m left 154.05 m; 60 m right 154.15 m.