Transportation Engineering

Traffic flow and capacity

Traffic flow is described by three macroscopic variables: flow q in veh/h, density k in veh/km and space-mean speed u in km/h. They are linked by the fundamental relation q = uk. Flow counts vehicles crossing a section per unit time, density counts vehicles per unit length of road, and every speed-flow and speed-density problem is built on this relation.

The linear Greenshields model assumes that speed falls linearly as density rises: u = u_f\left(1 - \dfrac{k}{k_j}\right), where u_f is the free-flow speed and k_j is the jam density. Substituting into q = uk gives a parabola that is zero at k = 0 and at k = k_j. Setting dq/dk = 0 places the maximum flow, or capacity, at k = k_j/2 and at speed u = u_f/2, so the capacity is C = u_f k_j/4. The speed at capacity is exactly half the free-flow speed. On the speed-flow curve, the characteristic points correspond to zero density at free flow, density k_j at zero speed, and density k_j/2 at capacity. Points on the uncongested branch have density below k_j/2, and points on the congested branch have density above it.

Since C = u_f k_j/4, the capacities of two roads are in the ratio C_A/C_B = (u_A k_A)/(u_B k_B), so both free-flow speed and jam density matter. Jam density is the reciprocal of the average space headway at standstill, so a spacing of s metres per vehicle in one lane gives k_j = 1000/s veh/km per lane. Together with u_f, this fixes the lane capacity. Average time headway is the reciprocal of flow, h = 3600/q seconds when q is in veh/h, so the headway at capacity is the smallest headway the stream can sustain.

For a flow below capacity, the linear model gives a quadratic in k with two roots, one on the uncongested and one on the congested branch. The same flow can therefore occur at two different speeds, each given by u = q/k.

The Greenberg logarithmic model, u = u_o \ln(k_j/k), moves capacity to k_m = k_j/e, with q_{max} = u_o k_j/e where e \approx 2.718. Setting the derivative of q = u_o k \ln(k_j/k) to zero gives this result, and the density at capacity is no longer half the jam density. When a speed-density relation is piecewise, the maximum flow must be found on each branch separately and the larger value taken. On a constant-speed branch flow is proportional to density, so its maximum lies at the upper end, while on a linear branch the maximum lies at the midpoint of the density range.

Two mean speeds must be distinguished. The time-mean speed is the arithmetic mean of spot speeds recorded at a point, whereas the space-mean speed is the harmonic mean, v_s = n/\sum(1/v_i), which equals total distance divided by total travel time. A harmonic mean never exceeds the arithmetic mean, so space-mean speed is less than or equal to time-mean speed for the same data. In a time-space diagram, each vehicle trajectory is a straight line whose slope is its speed, and the mean speed over the whole domain is the total distance travelled divided by the total time spent. This is the Edie definition of space-mean speed.

In signal design, Webster's optimum cycle length is C_0 = (1.5L + 5)/(1 - Y), where L is the total lost time per cycle (lost time per phase times the number of phases) and Y is the sum of the critical flow ratios. Each flow ratio is approach flow divided by saturation flow, and Y must stay below 1 for a feasible cycle.

A passenger car unit (PCU) converts each vehicle type into an equivalent number of cars for capacity analysis. The vehicle damage factor converts axle loads into equivalent standard axles for flexible pavement design. The California Bearing Ratio measures the strength and stability of the subgrade soil. Perception-reaction time is a component of stopping sight distance.

When demand exceeds the capacity at a bottleneck, the queue at any instant equals cumulative arrivals minus cumulative departures. A lane that is queue-discharging operates below its normal capacity, so the discharge falls once a queue exists. With a constant excess of arrivals over discharge, the queue therefore grows linearly and dissipates only when discharge exceeds arrivals.

A capacity-constrained section limits the flow that can pass through it, so a wider section downstream cannot receive more than the bottleneck discharge. Its operating point is read from the appropriate branch of the flow-density curve at that flow.

Intermediate sight distance is roughly twice the stopping sight distance, so where a sight distance of ISD is specified, the corresponding SSD is half of it.

Highway horizontal and vertical curves

Highway vertical curves are designed as parabolas because a parabola has a constant rate of change of grade, r = (G_2 - G_1)/L. This gives smooth riding and simple elevation calculation along the curve.

The algebraic difference of signed grades, A = G_2 - G_1, classifies the curve. A > 0 gives a sag curve and A < 0 gives a crest curve. The signs must be kept with the grades. Two positive grades form a sag when the second is steeper. Two negative grades form a crest when the second is steeper, and a rise followed by a descent is always a crest.

For a summit curve with sight distance S shorter than the curve length, L = \dfrac{AS^2}{200(\sqrt{h_1} + \sqrt{h_2})^2}, where h_1 is the driver eye height and h_2 is the object height. For S longer than L, L = 2S - \dfrac{200(\sqrt{h_1} + \sqrt{h_2})^2}{A}. The stopping distance is S = 0.278Vt + V^2/(254f) with V in km/h and t in seconds, and when the sight distance is stated to be compensated for gradient, the level-road value is used.

For a vertical curve with a given rate of change of grade r in percent per unit length, the length is L = A/r, where A is the algebraic difference of grades in percent.

On horizontal curves, superelevation e and side friction f together counteract centrifugal force. The requirement is e + f = V^2/(127R) with V in km/h and R in m. Once a permissible design friction is fixed, the superelevation is what remains, and a design friction near 0.15 is typical at 80 km/h.

The radius of a horizontal curve, R = V^2/[127(e+f)], can be reduced by lowering the design speed or by raising the superelevation. Increasing the design speed raises the required radius, while restricting heavy vehicles does not change the geometric radius.

Extra widening compensates for the additional lateral space a vehicle needs on a curve. The IRC expression is W_e = \dfrac{n l^2}{2R} + \dfrac{V}{9.5\sqrt{R}}. The mechanical term depends on the number of lanes n, the longest wheelbase l and the radius R, and the psychological term depends on V and R. Superelevation and longitudinal gradient do not appear. Both terms are computed separately and added to give the total widening in metres.

Superelevation is provided on highways and railways because vehicles and trains travel fast on curves. Taxiways are normally laid without superelevation since aircraft taxi at low speed.

Crowning raises the middle of a straight pavement above its edges for drainage, whereas centrifugal force on a curve is resisted by superelevation, which raises the outer edge. The lateral friction factor is kept low on curves for passenger comfort, which makes it lower than the longitudinal friction used for braking in sight distance computations, about 0.35 to 0.40. Grade compensation reduces the gradient on a curve to offset the extra resistance of curvature, and on two-way roads a longer sight distance such as ISD governs, so summit curves are longer than on one-way roads.

The point of curve (PC) is where the tangent ends and the circular curve begins, and the point of tangency (PT) marks the reverse transition.

For a simple circular curve with deflection angle \Delta and radius R, the tangent length is T = R\tan(\Delta/2). The central angle equals the deflection angle, so \Delta can be found from a measured tangent length.

Railway curve cant and equilibrium speed

On a railway curve, the outer rail is raised above the inner rail by cant C, so that a component of train weight balances centrifugal force at the equilibrium speed. For Indian broad gauge the cant formula is C = \dfrac{GV^2}{127R}, with G the gauge in mm, V in km/h and R in m. With dynamic gauge G = 1750 mm, the constant G/127 is about 13.8, so C \approx 13.8\,V^2/R. Each alternative constant corresponds to a different gauge value, so only the one matching the dynamic gauge is correct.

The degree of curve D is the angle subtended at the centre by a 30.5 m arc, so the radius is R = \dfrac{30.5 \times 180}{\pi D} m.

Cant deficiency C_d is the difference between the cant required for the maximum sanctioned speed and the cant actually provided, C_d = \dfrac{G(V_{max}^2 - V_e^2)}{127R}, where V_e is the equilibrium speed. It arises when faster trains run on a curve designed for a lower equilibrium speed. Cant excess is the reverse case, in which a slow train receives more cant than its equilibrium requirement.

Where several train speeds share one curve, the equilibrium speed is taken as the weighted mean of the speeds, with weights equal to the number of trains at each speed. The equilibrium cant e = GV^2/(127R) is then computed from that mean speed.

When a curve already has a cant provided, C_p, and an allowable cant deficiency C_d is permitted, the maximum speed satisfies V_{max} = \sqrt{\dfrac{127R(C_p + C_d)}{G}}.

Worked example: for R = 500 m, G = 1750 mm and V = 60 km/h, the equilibrium cant is C = \dfrac{1750 \times 60^2}{127 \times 500} \approx 99.2 mm.

Stopping sight distance

Stopping sight distance (SSD) is the minimum length of road ahead that must be visible to a driver so that the vehicle can come to rest before reaching a stationary obstacle. It is the sum of two parts: the distance covered during the perception-reaction time, d_1 = vt, and the braking distance, d_2 = \frac{v^2}{2a}, so that SSD = vt + \frac{v^2}{2a}. Subtracting the braking distance from the reaction distance, or using either part alone, does not give the stopping sight distance.

When the speed, deceleration and SSD are given, the reaction time is found from t = \frac{SSD - v^2/(2a)}{v}. For example, at v = 20 m/s with a = 3 m/s^2 and SSD = 100 m, the braking distance is 66.7 m, the reaction distance is 33.3 m, and t = 1.67 s.

On a gradient G, positive for an ascent and negative for a descent, the deceleration becomes a = g(\eta f \pm G), where \eta is the brake efficiency and f is the coefficient of friction. The product \eta f is the effective friction, so brake efficiency 60% and f = 0.7 give \eta f = 0.42. On a descent the gravity component assists the vehicle, so the minus sign applies and the braking distance increases. For example, at v = 15 m/s with \eta f = 0.40 and a 3% descent, a = 9.81(0.40 - 0.03) = 3.63 m/s^2 and the braking distance is about 31.0 m.

At a vertical curve, the design length is set by sight distance. A descending grade followed by an ascending grade forms a sag curve, and the algebraic grade difference A (in percent) is used with the headlight sight distance D = h + S\tan\alpha, where h is the headlight height and \alpha is the beam angle. When S is greater than the curve length L, the curve length is L = 2S - \frac{2D}{A}.

Runway effective gradient

The effective gradient of a runway is the difference between the highest and the lowest elevation on the runway centre line divided by the total runway length, expressed as a percentage. The profile must be built up from the given gradients, so each section's level change is the gradient multiplied by the section length, and the levels are then accumulated from a starting reduced level.

When the highest point comes before the lowest point, the difference is simply the change in level between those two points. The net change over the full length is not the effective gradient, since it can be much smaller than the actual range of levels along the runway.

• effective gradient — maximum level difference along the runway divided by its length. Example: levels ranging from 98 m to 103.6 m over 2000 m give \frac{5.6}{2000} \times 100 = 0.28\%.

The runway length is also corrected for gradient. The ICAO correction increases the length by 20 percent for every 1 percent of effective gradient, so L_c = L(1 + 0.20 g) with g in percent. For example, a length of 700 m with g = 1\% becomes 840 m.