IRC directs that superelevation be designed for 75% of the design speed with the friction term neglected. The reason is that

the governing case is the slow vehicle, which a curve banked for the full design speed would push towards the inner edge

friction is unreliable at high speed and cannot be counted on at all

75% of the design speed is the average speed observed on Indian highways

it makes the superelevation come out below the 7% maximum in all cases

Which of the following statements about transition curves are correct?

A transition curve introduces curvature gradually, so the centripetal acceleration builds at a controlled rate

It provides the length over which the superelevation and the extra widening can be run out

Its length is taken as the largest of the values from the three governing criteria

A transition curve introduces curvature gradually, so the centripetal acceleration builds at a controlled rate

It provides the length over which the superelevation and the extra widening can be run out

Its length is taken as the largest of the values from the three governing criteria

It reduces the radius of the circular curve it leads into, and so reduces the superelevation needed

A valley curve must be checked for headlight sight distance even when it satisfies the comfort criterion. The reason is that

at night the driver sees only as far as the headlight beam reaches, and the beam pitches down into the rising road

the centrifugal force reverses direction at night as the pavement cools

comfort is assessed for passengers and headlight distance for the driver, and the two use different speeds

the object height used for a valley curve is greater than for a summit curve

Geometric Design: Sight Distance, Horizontal and Vertical Alignment, Cant and Runway Length

Transportation Engineering is Section 6 of the Civil Engineering paper, and geometric design is the part of that paper where the design speed is not a limit but a multiplier, and almost every question in the chapter is a consequence of that. Raise the design speed of a road and its length does not change, but the stopping sight distance grows with the square of speed, the minimum radius grows with the square, the length of a summit curve grows with the square of the sight distance — which is itself already quadratic in speed — and the earthwork to deliver all of it grows faster still. The worked numbers below make the size of this concrete: doubling the design speed from 50 to 100 km/h multiplies the required sight distance by 2.9, not by 2, and the summit curve that provides it grows by a factor near eight. That is why a design speed is a decision about money as much as about travel time, and why an alignment is fixed early and changed reluctantly. The second thread is that every geometric element exists to serve one of two things — the driver’s eye or the vehicle’s tyres. Sight distances and vertical curves serve the eye: they ask whether a driver can see far enough to stop or overtake. Superelevation, radius and transition curves serve the tyres: they ask whether the friction available can supply the centripetal force the curve demands. Sorting each element into one of those two boxes before reaching for a formula is the fastest way to avoid using the wrong one, and it works for railways and runways as well as for roads.

Cross-sectional elements

The cross-section is where the standards look most arbitrary and are least so. Camber — the cross-slope from crown to edge — exists only to drain the surface, so its value is set by the surfacing: about 1.7 to 2% for a bituminous or cement concrete surface, 2.5 to 3% for a water-bound macadam, and up to 4% for earth, because a rougher, more permeable surface needs a steeper slope to shed water before it soaks in. Too little camber ponds water and destroys the pavement from above; too much makes the vehicle drift towards the edge and is uncomfortable. Carriageway width is a vehicle width of 2.44 m plus clearances: 3.75 m for a single lane, 7.0 m for two lanes, and 3.5 m per lane on a multi-lane road, where the shared clearance between lanes is counted once rather than twice. Kerbs, shoulders, the median and the right of way then complete the section, and kerb-to-kerb width is not carriageway width — a distinction that is worth checking in any question that mixes the terms.

Sight distance, and the term that dominates at speed

Stopping sight distance is the distance a driver needs to see in order to stop before an object in the road, and it has two parts: the lag distance covered while the driver perceives and reacts, and the braking distance covered once the brakes are on. SSD = 0.278Vt + V²/(254f) with V in km/h, t the reaction time — 2.5 s in IRC practice — and f the longitudinal friction, which is taken as 0.40 at 20 km/h falling to 0.35 at 80 and above, because friction itself degrades with speed. The first term is linear in speed and the second is quadratic, and the whole behaviour of the design follows from that difference.

SSD at t = 2.5 s, f = 0.35 — and why the second term wins
Design speed	Lag (linear)	Braking (quadratic)	SSD
40 km/h	27.80 m	18.00 m	45.80 m
50 km/h	34.75 m	28.12 m	62.87 m
80 km/h	55.60 m	71.99 m	127.59 m
100 km/h	69.50 m	112.49 m	181.99 m

🎯 Doubling the speed multiplies the sight distance by 2.9

Read the table across rather than down. At 40 km/h the lag distance is larger than the braking distance; by 80 km/h the braking distance has overtaken it, and by 100 it is 60 per cent larger. That crossover is the reason a design speed is expensive: going from 50 to 100 km/h multiplies SSD by 2.895, because the quadratic term grows fourfold while the linear term only doubles. And the consequence compounds down the design. The length of a summit curve goes as NS²/4.4, so it grows with the square of the sight distance — meaning that same doubling of design speed multiplies the summit curve length by about 8.4, and the earthwork under it by more still. It also cuts the other way, which is worth saying because it is the practical remedy: reducing a design speed by 20 per cent cuts SSD by nearly 30 per cent (80 → 64 km/h takes it from 127.59 m to 90.55 m), so where a hill or a town makes the geometry impossible, a lower posted speed with proper warning is a legitimate engineering answer rather than an admission of failure. Overtaking sight distance behaves the same way but larger — it is several times SSD, and providing it continuously on a two-lane road is usually impossible, which is why OSD is provided in stretches and marked, and why the marking is what makes the road safe rather than the geometry alone.

Horizontal alignment: superelevation, radius and the transition

A vehicle on a curve needs a centripetal force mv²/R, and only two things can supply it: the lateral friction between tyre and road, and the component of weight along a banked surface — the superelevation. Setting the two against the demand gives the governing equation e + f = V²/(127R) with V in km/h and R in metres. Everything in horizontal design is a way of splitting that sum. Both terms are capped: e is limited to 0.07 on plain and rolling terrain (10% in hills without snow), because a steeply banked curve is dangerous for a vehicle that has to stop on it or is moving slowly, and f is taken as 0.15, well below what a tyre can actually deliver, because the margin is what absorbs a wet surface, a worn tyre and a driver going faster than the design speed.

Compute e for 75% of the design speed with friction neglected: e = (0.75V)²/(127R). IRC does this so that a vehicle at three-quarters of the design speed needs no friction at all — the slow vehicle, not the fast one, is the governing case for the banking.
If that value exceeds e_max = 0.07, cap it at 0.07 and check that friction can carry the balance at the full design speed: is V²/(127R) − 0.07 ≤ 0.15?
If it cannot, the radius is inadequate. The ruling minimum radius is R = V²/(127(e + f)), which at 80 km/h with e = 0.07 and f = 0.15 is 229.06 m — so any curve tighter than that at 80 km/h needs a speed restriction, not a design.

The transition curve — a spiral, in Indian practice — exists because none of the above can be delivered instantaneously. A vehicle passing from a straight to a circular curve would need its full centripetal acceleration applied at a point, its superelevation to appear as a step, and its steering to turn discontinuously. The spiral introduces curvature linearly with distance, so the centripetal acceleration builds at a constant rate, the superelevation can be run out over its length, and the extra width for the curve can be introduced gradually. Its length is taken as the largest of three requirements: the rate of change of centripetal acceleration, the rate of introducing the superelevation, and the empirical IRC minimum — and taking the largest, rather than the one you calculated first, is where the mark is.

Vertical curves, and the branch that must be checked

A summit curve is designed for sight distance and nothing else — comfort is not the constraint, because centrifugal force on a summit acts upward and relieves the suspension rather than loading it. With driver eye height 1.2 m and object height 0.15 m, the standard result for the case L > S is L = NS²/4.4, where N is the algebraic deviation angle — the difference of the grades, counting downgrades as negative, so a +3% meeting a −5% gives N = 0.08 and not 0.02. A valley curve is designed for the opposite pair of reasons: comfort by day, because centrifugal force acts downward and adds to the passenger’s weight, and headlight sight distance by night, because the headlight beam pitches down into the road rather than along it, and a valley curve that is comfortable can still be too short to see over at night.

⚠️ Both branches must be tested, and one of them will contradict itself

Every vertical curve formula comes in two forms, one derived assuming the curve is longer than the sight distance and one assuming it is shorter, and each is valid only where its own assumption holds. The procedure is therefore: compute with one branch, then check the assumption you made. Take N = 0.08 and S = 128 m. The L > S branch gives L = NS²/4.4 = 0.08 × 16384/4.4 = 297.89 m; since 297.89 > 128, the assumption holds and the answer stands. Now test the other: L = 2S − 4.4/N = 256 − 55 = 201.00 m, which was derived on the assumption L < S, and 201.00 is not less than 128 — the branch contradicts itself and must be discarded. Candidates lose this mark in two ways, both avoidable. Some substitute into whichever formula they remember and never check, which gives a wrong answer that looks entirely reasonable. Others get a self-consistent result from both branches and panic; that cannot happen if the arithmetic is right, and if it appears to, the deviation angle has usually been computed as a difference of magnitudes rather than an algebraic difference. Grades meeting at a summit have opposite signs, so N is a sum of magnitudes, and getting 0.02 where 0.08 was intended will quietly make the curve four times too short.

Railway cant, and airport runway length

The geometric design of a railway track is the road problem with the friction term deleted. A flanged steel wheel on a steel rail has no useful lateral friction to call on, so the cant — the amount by which the outer rail is raised — must supply the whole centripetal force at the equilibrium speed: e = GV²/(127R) with G the dynamic gauge, 1.676 m on broad gauge. But a railway carries many speeds on one track, and no single cant can be right for all of them: too much and a slow goods train overloads the inner rail, too little and a fast passenger train overloads the outer. The resolution is to cant for a weighted equilibrium speed and permit a cant deficiency for faster trains, capped at 100 mm on BG, with the cant itself capped at 165 mm. That cap is what makes the arithmetic bite: at 100 km/h on a 600 m radius the equilibrium cant works out to 219.95 mm, which cannot be provided, so the curve must either be eased or the speed restricted.

Runway length: a 2000 m basic length at 600 m elevation, 35 °C, 0.5% effective gradient
Correction	Rule	Applied to	Running length
Basic length	At MSL, 15 °C, level	—	2000.00 m
Elevation	+7% per 300 m	The basic length	2280.00 m
Temperature	+1% per °C above the standard temperature at that elevation (15 − 0.0065 × 600 = 11.1 °C)	The elevation-corrected length	2824.92 m
Gradient	+20% per 1% effective gradient	The elevation- and temperature-corrected length	3107.41 m

The runway is only half of an airfield’s geometry. Taxiway design governs how an aircraft gets to and from it, and its constraints are gentler than the runway’s in every respect but one: taxiway width, shoulder width and the clearance to obstacles are all set by wingspan and undercarriage track rather than by speed, and taxiway gradients and sight distances are far less demanding because the aircraft is slow. The exception is the exit taxiway, and it is the one piece of taxiway geometry that is genuinely designed rather than dimensioned. An aircraft cannot leave the runway until it has slowed to the turnoff speed the exit allows, and the runway is blocked for every second it is still on it — so the exit angle is a capacity decision. A conventional 90° exit forces a near-stop, around 25 km/h, needing a turn radius of only R = V²/(125f) = 25²/(125 × 0.13) = 38.5 m. A rapid-exit taxiway at 30° lets the aircraft leave at about 65 km/h, which needs R = 260 m — nearly seven times the radius and far more pavement. What it buys is time on the runway: clearing a 2000 m runway at 25 km/h takes 288 seconds, and at 65 km/h 111 seconds, so the rapid exit returns about 177 seconds per movement. At a busy airport that is the difference between a runway that can take forty movements an hour and one that cannot, which is why rapid-exit taxiways are built at capacity-constrained airfields and 90° exits everywhere else.

⚠️ The corrections are sequential, not additive — and there is a limit on two of them

Two mistakes lose most of the marks on this topic. The first is applying all three percentages to the basic length and adding: 2000 × (1 + 0.14 + 0.239 + 0.10) = 2958 m, which is 149 m short of the right answer, because each correction is applied to the length as already corrected. Work them in order — elevation, then temperature, then gradient — and each multiplies what the previous one produced. The second is using 15 °C as the reference for the temperature correction. The standard atmospheric temperature falls with height at 6.5 °C per km, so at 600 m it is 15 − 0.0065 × 600 = 11.1 °C, and the correction is measured from that, not from sea level: using 15 °C would give a 20 °C excess instead of 23.9 °C and understate the length by 90 m. There is also a check the corrections themselves do not contain. If the elevation and temperature corrections together exceed 35 per cent, ICAO requires the runway length to be established by a specific study rather than by these rules. Here they come to 2824.92/2000 − 1 = 41.25%, comfortably over — so the honest answer to this question is 3107 m and a note that at this site the standard corrections have been pushed past the range they were fitted for. All of this exists because a hot, high airfield gives an aircraft thin air: less lift for the same wing and less thrust from the same engine, so it must run further to fly.

Key takeaways

SSD = 0.278Vt + V²/(254f). The braking term is quadratic, so 50 → 100 km/h multiplies SSD by 2.895, not 2.
e + f = V²/(127R), with e ≤ 0.07 and f = 0.15; design e for 75% of the design speed, then cap it.
Summit curves are designed for sight distance, valley curves for comfort by day and headlight distance by night.
Both vertical-curve branches must be tested; the invalid one contradicts its own L > S or L < S assumption.
Railway cant e = GV²/(127R) has no friction term and is capped at 165 mm on BG — 100 km/h on R = 600 m needs 220 mm and fails.
Runway corrections are sequential, and the temperature datum is the standard temperature at that elevation, not 15 °C.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A highway is designed for 80 km/h. Taking a reaction time of 2.5 s, a longitudinal friction coefficient of 0.35 and a level road, the stopping sight distance in metres is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 127.59
127.59 m. SSD = 0.278Vt + V²/(254f) = 0.278 × 80 × 2.5 + 80²/(254 × 0.35) = 55.60 + 71.99 = 127.59 m. The 0.278 converts km/h to m/s (1/3.6), and the 254 in the second term is 2 × 9.81 × 3.6², so both terms take V in km/h — mixing units here is the commonest arithmetic slip. What the question is really probing is the balance between the two terms, because it changes with speed and drives the whole cost of a design. At 80 km/h the braking distance (71.99 m) already exceeds the lag distance (55.60 m); at 40 km/h it is the other way round, 18.00 against 27.80. The lag term is linear in speed, the braking term quadratic, so raising the design speed from 50 to 100 km/h raises SSD from 62.87 to 181.99 m — a factor of 2.895, not 2. And the consequence multiplies downstream: a summit curve is NS²/4.4, quadratic in the sight distance, so that same doubling of design speed lengthens the summit curve by about 8.4 times, and the earthwork under it by more still. Note also that this is a level road. On a gradient, f is replaced by (f ± n) — plus for an upgrade, which helps the vehicle stop, minus for a downgrade, which does not — and a two-way road on a slope must be designed for the downgrade direction.
A summit curve joins a +3% grade with a −5% grade. The stopping sight distance required is 128 m, and the design constant is 4.4 (eye height 1.2 m, object height 0.15 m). The required length of the summit curve in metres is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 297.89
297.89 m. First get N right, because this is where the question is set. The grades are +0.03 and −0.05, so the algebraic deviation is N = 0.03 − (−0.05) = 0.08. Taking it as a difference of magnitudes gives 0.02 and a curve four times too short — a silent, plausible-looking error. Now assume L > S and use L = NS²/4.4 = 0.08 × 128²/4.4 = 0.08 × 16384/4.4 = 297.89 m. Then check the assumption, which is the step that makes the answer defensible: is 297.89 > 128? Yes, so the branch is valid and this is the answer. For completeness, test the other branch: L = 2S − 4.4/N = 256 − 4.4/0.08 = 256 − 55 = 201.00 m. That formula was derived assuming L < S, and 201.00 is not less than 128, so it contradicts its own premise and is discarded. Exactly one branch survives, always. Two notes worth carrying. A summit curve is governed by sight distance alone — comfort is not a constraint on a summit, because the centrifugal force acts upward and unloads the suspension rather than loading it. And the sensitivity is severe: had the design speed been raised so that S became 182 m, the required length would go to 602 m, more than doubling for a 42% rise in sight distance, because L is quadratic in S and S is itself quadratic in speed.
IRC directs that superelevation be designed for 75% of the design speed with the friction term neglected. The reason is that
1. the governing case is the slow vehicle, which a curve banked for the full design speed would push towards the inner edge
2. friction is unreliable at high speed and cannot be counted on at all
3. 75% of the design speed is the average speed observed on Indian highways
4. it makes the superelevation come out below the 7% maximum in all cases
Show answer
Answer: A — the governing case is the slow vehicle, which a curve banked for the full design speed would push towards the inner edge
The slow vehicle governs. This is the most-missed idea in horizontal design, because the intuition points the wrong way: it feels as though the fast vehicle must be the critical case, since it is the one that could skid outward. But look at what superelevation does. Banking supplies centripetal force whether or not the vehicle needs it. A curve banked for the full design speed is banked steeply, and a vehicle crawling round it at 20 km/h — a loaded truck on a gradient, or anything in a queue — has far more banking than its speed calls for, so the surplus component of its weight pushes it down the slope towards the inner edge, and friction must now act outward to hold it. On a wet surface, with a high-sided vehicle, that is a real overturning and sliding risk, and it is worst exactly where traffic is slowest. Designing e for three-quarters of the design speed places the equilibrium point where most traffic actually is, and leaves the faster vehicles to make up the difference with friction — which they can, because they have good tyre loading and the design reserves f = 0.15 for exactly this. So the rule is: e = (0.75V)²/(127R), capped at 0.07, then check that V²/(127R) − e ≤ 0.15 at the full speed. Option (b) is wrong — friction is explicitly relied on for the fast case. Option (c) invents a statistic. Option (d) is false: at V = 80 and R = 250 the 75% rule still gives e = 0.1134, above 0.07, so the cap still has to be applied.
A broad gauge railway curve of radius 600 m carries trains at 100 km/h. Taking the dynamic gauge as 1.676 m, the equilibrium cant in mm is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 219.95
219.95 mm. e = GV²/(127R) = 1.676 × 100²/(127 × 600) = 16,760/76,200 = 0.21995 m = 219.95 mm. The formula is the road equation with the friction term deleted, and that deletion is the whole difference between the two subjects: a flanged steel wheel on a steel rail has almost no lateral friction to call on, so the banking must supply the entire centripetal force at the equilibrium speed. Having got the number, the question is what to do with it — and the answer is that it cannot be provided. The maximum permissible cant on Indian broad gauge is 165 mm, and 219.95 exceeds it by a third. So the design has three honest options and no fourth: ease the curve to a larger radius (at 165 mm, the permissible radius at 100 km/h is 800 m), restrict the speed on this curve (at 600 m and 165 mm the equilibrium speed is 86.6 km/h), or run at 165 mm and accept a cant deficiency of 55 mm, which is within the 100 mm normally allowed on BG. That third option is what usually happens, and it explains why a railway curve is canted at all rather than perfectly: a track carries goods trains at 60 and passenger trains at 100 on the same rails, no single cant is right for both, and the system is designed as a compromise with an explicit deficiency allowance rather than as an equilibrium. Cant excess for the slow train and cant deficiency for the fast one are two names for the same unavoidable fact.
A runway has a basic length of 2000 m. The site is at an elevation of 600 m with an airport reference temperature of 35 °C, and the effective gradient is 0.5%. Applying the standard corrections in sequence (7% per 300 m elevation, 1% per °C above the standard temperature at that elevation, 20% per 1% gradient), the corrected runway length in metres is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 3107.41
3107.41 m. Apply the corrections in sequence, each to the length already corrected. Elevation: 7% × 600/300 = 14%, so 2000 × 1.14 = 2280.00 m. Temperature: the datum is the standard atmospheric temperature at that elevation, 15 − 0.0065 × 600 = 11.1 °C, so the excess is 35 − 11.1 = 23.9 °C and the correction is 23.9% × 2280 = 544.92, giving 2824.92 m. Gradient: 20% × 0.5 = 10% of that, 282.49, giving 3107.41 m. Two errors account for most lost marks here. Applying all three percentages to the basic length and adding gives 2000 × 1.479 = 2958 m, short by 149 m, because the corrections compound rather than sum. And using 15 °C as the temperature datum gives a 20 °C excess instead of 23.9 °C and understates the answer by about 90 m — the standard atmosphere cools at 6.5 °C/km, so a site at 600 m has a standard temperature of 11.1 °C, not 15. There is also a check the corrections do not contain, and a good answer mentions it: when the elevation and temperature corrections together exceed 35%, ICAO requires a specific study rather than these rules. Here they reach 2824.92/2000 − 1 = 41.25%, so this site is past the range the rules were fitted for. All of it comes from one physical fact — a hot, high airfield gives thin air, so the same wing makes less lift and the same engine less thrust, and the aircraft must run further to fly. That is why runway length, alone among the geometric quantities in this chapter, depends on the weather.
Which of the following statements about transition curves are correct?
1. A transition curve introduces curvature gradually, so the centripetal acceleration builds at a controlled rate
2. It provides the length over which the superelevation and the extra widening can be run out
3. Its length is taken as the largest of the values from the three governing criteria
4. It reduces the radius of the circular curve it leads into, and so reduces the superelevation needed
Show answer
Answer: A — A transition curve introduces curvature gradually, so the centripetal acceleration builds at a controlled rate; B — It provides the length over which the superelevation and the extra widening can be run out; C — Its length is taken as the largest of the values from the three governing criteria
(a), (b) and (c). (a) is the primary purpose. On a straight the radius is infinite and the centripetal acceleration zero; on the circular curve it is v²/R. Joining them directly would demand that full acceleration at a point, with the steering turned discontinuously — physically impossible for the vehicle and violently uncomfortable for the passenger. A spiral introduces curvature linearly with distance travelled, so 1/R grows at a constant rate and the acceleration builds smoothly. (b) is the practical reason it has the length it does: superelevation cannot appear as a step either, so it is run out along the transition, and so is the extra width a vehicle needs on a curve because its rear wheels track inside its front ones. (c) is the procedural point that carries the mark. Three criteria give three candidate lengths — the rate of change of centripetal acceleration (L = 0.0215V³/(CR)), the rate at which superelevation may be introduced (a limit on the longitudinal slope of the pavement edge), and the empirical IRC minimum for the terrain — and the design takes the largest, because each is a separate requirement that must independently be met. Computing one, or averaging them, is the common way to lose this. (d) is wrong, and it is wrong about geometry rather than about practice: the transition leads into the circular curve and does not alter it. The circular radius is fixed by e + f = V²/(127R) and is unchanged by anything the spiral does. What the transition does change is the alignment, because inserting a spiral shifts the circular curve inward by an amount called the shift — a real consequence that has to be set out on the ground, but not a reduction in radius.
A highway is designed for 80 km/h with e_max = 0.07 and f = 0.15. The ruling minimum radius of the horizontal curve in metres is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 229.06
229.06 m. Rearranging the governing equation e + f = V²/(127R) gives R = V²/(127(e + f)) = 6400/(127 × 0.22) = 6400/27.94 = 229.06 m. This is the tightest curve that can be designed at 80 km/h, using both the maximum banking permitted and the full lateral friction the code allows. Its value lies in what it rules out. A curve of R = 250 m at this speed demands e + f = 6400/(127 × 250) = 0.2016, which is under 0.22, so it is feasible — but only just, and every bit of the friction reserve is committed. A curve of R = 200 m demands 0.252, above what is available, so at 80 km/h it is not a design at all: it needs either a larger radius or a posted speed restriction, and no amount of extra banking will rescue it, because e is already at its cap. That cap is not conservatism for its own sake. Superelevation beyond about 7% is dangerous for the slow vehicle, which slides towards the inner edge, and the friction figure of 0.15 is well below what a good tyre delivers because the margin absorbs a wet surface, a worn tyre and a driver exceeding the design speed. Note also that the ruling minimum is not an absolute minimum: IRC also publishes an absolute minimum radius for use where terrain makes the ruling value impossible, at the cost of a lower assumed speed on that stretch — which is the same trade the runway and railway sections of this chapter make, and the honest way to design in hills.
A valley curve must be checked for headlight sight distance even when it satisfies the comfort criterion. The reason is that
1. at night the driver sees only as far as the headlight beam reaches, and the beam pitches down into the rising road
2. the centrifugal force reverses direction at night as the pavement cools
3. comfort is assessed for passengers and headlight distance for the driver, and the two use different speeds
4. the object height used for a valley curve is greater than for a summit curve
Show answer
Answer: A — at night the driver sees only as far as the headlight beam reaches, and the beam pitches down into the rising road
At night the sight distance is set by the headlight beam, not by the geometry of the road. On a summit curve the obstruction is the crest of the road itself, and the criterion is the same by day and by night: the driver’s eye at 1.2 m must see an object at 0.15 m over the hump. On a valley curve nothing obstructs the view at all in daylight — you can see right across the dip — so the daytime criterion is comfort, because centrifugal force on a sag acts downward and adds to the passengers’ weight rather than relieving it. At night the situation inverts completely. The driver sees only what the headlights illuminate; the beam leaves the vehicle at about 0.75 m above the road with a small upward divergence of around 1°, and on a sag the road ahead rises away from the beam, so the lit patch is short and can easily be shorter than the stopping sight distance. A valley curve that is perfectly comfortable can therefore be unsafe after dark, which is why both criteria are computed and the longer length governs — and in practice the headlight criterion usually does. Option (b) is physical nonsense. Option (c) invents a distinction: both criteria use the design speed. Option (d) is wrong in a way worth knowing — the headlight criterion uses no object height at all; it uses the beam height of 0.75 m and the beam’s upward divergence, which is a different geometry from the summit case rather than a different number in the same one.