How is Valley Curve Length Determined?
Unlike summit curves where sight distance alone governs the design, valley curve length must satisfy two independent design criteria simultaneously — and the larger of the two governs the final design. These are the comfort criterion (limiting passenger discomfort from centrifugal force acting downward) and the safety criterion (ensuring adequate visibility at night under headlight illumination).
Criterion 1 — Comfort (Rate of Centrifugal Acceleration)
At a valley curve, the centrifugal force acts downward, increasing the effective weight on the vehicle suspension. If the rate of change of centrifugal acceleration is too rapid, passengers experience a jarring sensation. The comfort criterion limits this rate to C = 0.6 m/s³ as per IRC:
Ls = √(Nv³/C)
Where N = deviation angle = |n₁ − n₂|, v = speed in m/s, C = 0.6 m/s³ (IRC).
Criterion 2 — Safety (Head Light Sight Distance)
At night, visibility is restricted to the area illuminated by the vehicle’s headlights. The valley curve must be designed so that headlights (height h = 0.75 m, beam angle β = 1° as per IRC) illuminate the road ahead for at least the stopping sight distance:
Case 1: When L > S (Curve Longer than HSD)
L = NS² / (2h + 2S tan β) = NS² / (1.5 + 0.035S)
Case 2: When L < S (HSD Longer than Curve)
L = 2S − (2h + 2S tan β)/N = 2S − (1.5 + 0.035S)/N
Final design length = Maximum of comfort criterion (Ls) and HSD criterion (Lv)
Solved Example — Valley Curve Length
Problem: A sag (valley) curve connects gradients n₁ = −2% and n₂ = +1%. Head light sight distance = 150 m. Find length of valley curve.
Step 1: N = |n₁ − n₂| = |−0.02 − 0.01| = 0.03
Step 2: Assume L > S: L = NS²/(1.5 + 0.035S) = (0.03 × 150²)/(1.5 + 5.25) = 675/6.75 = 100 m
Check: 100 m < 150 m — assumption L > S is wrong. Switch to L < S case.
Step 3: L = 2S − (1.5 + 0.035S)/N = 2×150 − (1.5 + 5.25)/0.03 = 300 − 225 = 75 m
Check: 75 m < 150 m ✔ (L < S assumption correct)
∴ Length of Valley Curve = 75 m
Location of Highest/Lowest Point on Vertical Curves
Highest Point on Summit Curve
The highest point is where the gradient along the parabola is zero (tangent is horizontal). Its distance from the VPC (first tangent point) is:
x₀ = n₁ × L / N (for square parabola)
Lowest Point on Valley Curve
Similarly, for a valley curve the lowest point distance from the first tangent point:
x₀ = n₁ × L / N (for square parabola — IRC objective problems)
x₀ = L × √(n/2N) (for cubic parabola — IRC subjective problems)
Where n = the flatter of the two gradients meeting at the curve.
Complete Formula Reference Table
| Parameter | Summit Curve (SSD) | Summit Curve (OSD) | Valley Curve (HSD) |
|---|---|---|---|
| L > S | NS²/4.4 | NS²/9.6 | NS²/(1.5+0.035S) |
| L < S | 2S − 4.4/N | 2S − 9.6/N | 2S−(1.5+0.035S)/N |
| Highest/Lowest point | n₁L/N | n₁L/N | n₁L/N (sq.par.) |
| Curve type | Square parabola | Square parabola | Cubic parabola |
| Key IRC values | h₁=1.2m, h₂=0.15m | h₁=h₂=1.2m | h=0.75m, β=1° |
Key Summary Points
- Valley curve length = max of comfort criteria and HSD criteria
- Comfort: Ls = √(Nv³/C), C = 0.6 m/s³ (IRC)
- HSD L>S: NS²/(1.5+0.035S) | L<S: 2S−(1.5+0.035S)/N
- Highest/lowest point from VPC: x₀ = n₁L/N
- Gradient effect neglected in vertical curve sight distance calculations (IRC)
- Serviceability indicator (IRC 32-2015): Good = <1800 mm/km | Fair = 2400 mm/km | Poor = 3200 mm/km