6.3 Sag Vertical Curve Geometry and Sight Distance

6.3 Sag Vertical Curve Geometry and Sight Distance

Geometry of Sag Curves

A sag vertical curve is a concave vertical transition between two tangent grades. Sag curves are used when connecting a downgrade to an upgrade, a downgrade to a gentler downgrade, or an upgrade to a steeper upgrade. Like crest curves, they are modeled as equal-tangent parabolas, splitting the horizontal curve length ($L$) equally on either side of the Point of Vertical Intersection (PVI).

The rate of vertical curvature ($K = L/A$) remains a core design parameter. However, unlike crest curves, where design is almost exclusively governed by stopping sight distance, sag vertical curve design is controlled by four separate criteria:

1. Headlight Sight Distance (Safety): The primary night safety driver.
2. Passenger Comfort: Preventing excessive vertical centripetal force.
3. Drainage Control: Preventing water pooling at the bottom of curbed sections.
4. General Appearance: Aesthetics and driver anticipation.

---

Headlight Sight Distance Criteria

During daylight hours, a driver's sight distance on a sag curve is generally unrestricted unless blocked by an overhead structure (such as a bridge). However, at night, the sight distance is limited by the area of the roadway illuminated by the vehicle's headlights.

Mathematical Formulation

To model night sight distance, AASHTO assumes:

• The height of the headlight above the roadway surface is $h_3 = 2.0\text{ ft}$ ($0.60\text{ m}$).
• The headlight beam has an upward divergence angle (spread) of $\theta = 1.0^\circ$ above the longitudinal axis of the vehicle.

Case 1: Sight Distance is Less Than or Equal to Curve Length ($S \le L$)

The minimum curve length is given by:

$L = \frac{A S^2}{200 \left( h_3 + S \tan \theta \right)}$

Substituting $h_3 = 2.0\text{ ft}$ and $\theta = 1.0^\circ$:

• US Customary Units: $L = \frac{A S^2}{400 + 3.5S}$
• Metric Units: $L = \frac{A S^2}{120 + 3.5S}$

Case 2: Sight Distance is Greater Than Curve Length ($S > L$)

If the required sight distance exceeds the curve length:

$L = 2S - \frac{200 \left( h_3 + S \tan \theta \right)}{A}$

Substituting standard design values:

• US Customary Units: $L = 2S - \frac{400 + 3.5S}{A}$
• Metric Units: $L = 2S - \frac{120 + 3.5S}{A}$

---

Passenger Comfort Criteria

When a vehicle travels through a sag vertical curve, centripetal acceleration acts downward, adding to the force of gravity. If this vertical acceleration is too high, passenger discomfort occurs (a feeling of being "pushed" into the seat).

To ensure comfort, AASHTO limits vertical centripetal acceleration to $1.0\text{ ft/s}^2$ ($0.3\text{ m/s}^2$). The minimum curve length required to satisfy comfort is:

• US Customary Units: $L = \frac{A V^2}{46.5}$
• Metric Units: $L = \frac{A v^2}{395}$

Where:

• $V$ = Design speed in mph
• $v$ = Design speed in km/h
• $A$ = Algebraic difference in grades (in percent)

---

Drainage Criteria

On curbed roadways, sag vertical curves present a drainage challenge. At the very bottom of the curve, the grade passes through $0.0%$. If the curve is too long (too flat near the bottom), stormwater will not drain quickly to the gutters, leading to water accumulation and hydroplaning risks.

To maintain adequate drainage on curbed roads, a minimum longitudinal slope of $0.3%$ must be maintained within 50 feet of the low point. This requirement is met if the rate of vertical curvature ($K$) satisfies:

$K \le 167\text{ ft/\% grade}$

Thus, for drainage, we have a maximum length limit ($L_{\max} = 167 \cdot A$) rather than a minimum length limit. If the calculated design curve length exceeds this maximum, designers must specify special gutter profiles or closely spaced catch basins.

---

Low Point Location and Elevation

Locating the lowest point of a sag vertical curve is vital for placing catch basins to collect runoff.

Station of the Low Point

The horizontal distance ($x_m$) from the PVC to the low point on an equal-tangent parabolic curve is:

$x_m = \frac{|G_1| L}{|G_2 - G_1|} = \frac{|G_1| L}{A}$

Where:

• $G_1$ = Slope of the entering tangent (in percent)
• $L$ = Total length of the curve
• $A$ = Algebraic difference in grades ($|G_2 - G_1|$)

Note: The low point only lies within the curve if the tangents slope in opposite directions (one negative, one positive). If both are positive, the low point is at the PVC ($x_m = 0$). If both are negative, the low point is at the PVT ($x_m = L$).

The station of the low point is:

$\text{Sta}_{LP} = \text{Sta}_{PVC} + x_m$

Elevation of the Low Point

The elevation of the low point ($E_{LP}$) is calculated by applying the parabolic elevation equation:

$E_{LP} = E_{PVC} + \left( \frac{G_1}{100} \right) x_m + \frac{A x_m^2}{200 L}$

Alternatively, using the absolute value of $G_1$ in percent, the shortcut formula is:

$E_{LP} = E_{PVC} - \frac{G_1^2 L}{200 A}$

Step-by-Step Design Example

A sag vertical curve connects a $-3.0%$ grade to a $+2.0%$ grade. The design speed is 50 mph, requiring a stopping sight distance of 425 feet. The PVC is at station $10+00$ and elevation $100.00\text{ ft}$.

1. Calculate Grade Difference (A): $A = |2.0 - (-3.0)| = 5.0\%$
2. Verify Length for Headlight Sight Distance: Assume $S \le L$: $L = \frac{A S^2}{400 + 3.5S} = \frac{5.0 \cdot 425^2}{400 + 3.5(425)} = \frac{903,125}{1887.5} \approx 478.5\text{ ft}$ Since $425\text{ ft} \le 478.5\text{ ft}$, the assumption is correct. The minimum curve length for headlight sight distance is 479 feet.
3. Verify Length for Comfort: $L_c = \frac{A V^2}{46.5} = \frac{5.0 \cdot 50^2}{46.5} = \frac{12,500}{46.5} \approx 268.8\text{ ft}$ Since headlight control requires a longer curve ($479\text{ ft} > 269\text{ ft}$), headlight control governs. Let's round the design curve length to $500\text{ ft}$.
4. Locate the Low Point: $x_m = \frac{|G_1| L}{A} = \frac{3.0 \cdot 500}{5.0} = 300\text{ ft}$ $\text{Sta}_{LP} = (10+00) + 300 = 13+00$
5. Calculate Low Point Elevation: $E_{LP} = E_{PVC} - \frac{G_1^2 L}{200 A} = 100.00 - \frac{3.0^2 \cdot 500}{200 \cdot 5.0} = 100.00 - 4.50 = 95.50\text{ ft}$