Why Locate the Highest or Lowest Point on a Vertical Curve?
On a summit curve, the highest point is where a driver travelling along the road experiences the worst (most restricted) sight distance — because the road ahead dips away on both sides from this peak, hiding whatever lies beyond. Similarly, on a valley curve, the lowest point is where the road reaches its minimum elevation, affecting drainage design (water accumulates here) and where night visibility using headlights is most critical.
These points are also important for: setting drainage inlets at the correct location, calculating elevations along the vertical curve profile, and ensuring sight distance requirements are actually met at the most critical section of the curve.
Highest Point on Summit Curve
For a summit curve designed as a square parabola (y = ax²), the highest point is where the gradient of the parabolic curve becomes zero — i.e., where the tangent to the parabola is perfectly horizontal. At this point, the road is neither rising nor falling, and the design vehicle has its most restricted forward visibility.
Derivation
For the parabola y = ax² + bx + c, the gradient dy/dx = 2ax + b = 0 at the extreme point.
At x = 0 (VPC), the initial slope = n₁ (first gradient). As we move along the parabola, the slope decreases at a constant rate (2a per unit length). The slope becomes zero at:
2ax₀ + n₁ = 0 → But since the total change in slope over L is N = n₁ + n₂, we get a = N/2L. Therefore:
x₀ = n₁L / N
Where x₀ is measured from the VPC (Vertical Point of Curvature) — the starting point of the summit curve.
Important Observations
- If n₁ = n₂ (equal gradients): x₀ = n₁L/(n₁+n₂) = L/2 — highest point at the midpoint of the curve
- If n₁ > n₂: x₀ > L/2 — highest point shifts toward the flatter (second) gradient side
- If n₁ < n₂: x₀ < L/2 — highest point is closer to VPC
- Always verify: x₀ must be less than L (the point must fall within the curve, not on the tangent extension)
Lowest Point on Valley Curve
The lowest point on a valley curve is located where the gradient of the vertical profile becomes zero — the bottom of the dip. Its location depends on whether we model the valley curve as a square parabola or cubic parabola.
Using Square Parabola (for Objective/Quick Problems)
By the same derivation as the summit curve highest point:
x₀ = n₁L / N
Where n₁ = first (approaching) gradient (take its absolute value), L = length of valley curve, N = deviation angle = |n₁ − n₂|. Measured from the first tangent point (VPC) of the valley curve.
Using Cubic Parabola (for Subjective/Detailed Analysis)
Since IRC recommends a cubic parabola for valley curves (fully transitional), the lowest point location is:
x₀ = L × √(n / 2N)
Where n = the flatter of the two gradients meeting at the valley curve, N = total deviation angle = |n₁ − n₂|, L = total length of valley curve.
Summary Comparison
| Property | Highest Point (Summit) | Lowest Point (Valley) |
|---|---|---|
| Formula (square parabola) | x₀ = n₁L/N | x₀ = n₁L/N |
| Formula (cubic parabola) | Not applicable | x₀ = L√(n/2N) |
| Measured from | VPC (start of summit curve) | VPC (start of valley curve) |
| IRC curve type | Square parabola | Cubic parabola |
| Engineering significance | Worst sight distance location | Drainage inlet location + HSD check |
| If gradients equal | At midpoint (L/2) | At midpoint (L/2) |
| If unequal | Shifts to flatter gradient side | Shifts to flatter gradient side |
Solved Example 1 — Highest Point on Summit Curve
Problem: A summit curve is formed at the junction of +1% and −2% grades. The length of curve = 153.41 m. Find the location of the highest point from VPC.
Solution:
n₁ = 0.01 (first/approaching gradient)
n₂ = −0.02 (second/leaving gradient)
N = |n₁ − n₂| = |0.01 − (−0.02)| = 0.03
L = 153.41 m
x₀ = n₁L/N = (0.01 × 153.41)/0.03 = 1.5341/0.03 = 51.14 m from VPC
Check: 51.14 m < 153.41 m ✔ (point falls within curve)
Solved Example 2 — Lowest Point on Valley Curve
Problem: A valley curve connects −2% and +1% grades. Length of curve = 75 m. Locate the lowest point (i) using square parabola and (ii) using cubic parabola.
Solution:
n₁ = 0.02 (first/approaching gradient, absolute value)
N = |−0.02 − 0.01| = 0.03
n = flatter gradient = 0.01
L = 75 m
(i) Square Parabola: x₀ = n₁L/N = (0.02 × 75)/0.03 = 1.5/0.03 = 50 m from VPC
(ii) Cubic Parabola: x₀ = L√(n/2N) = 75 × √(0.01/(2 × 0.03)) = 75 × √(0.1667) = 75 × 0.4082 = 30.6 m from VPC
The cubic parabola formula gives a different (generally smaller) value as it accounts for the non-uniform curvature of the transition curve. IRC recommends using the cubic parabola result for detailed/subjective design calculations.
Key Summary
- Highest point on summit = where gradient = 0 = x₀ = n₁L/N from VPC
- Lowest point on valley (square par.) = x₀ = n₁L/N from VPC
- Lowest point on valley (cubic par.) = x₀ = L√(n/2N) from VPC
- For equal gradients: extreme point at midpoint L/2
- For unequal gradients: extreme point shifts toward flatter gradient side
- Always verify x₀ < L (must fall within the curve)
- Summit: square parabola (IRC) | Valley: cubic parabola (IRC)