CELE Surveying (Geomatics) — Vertical (Parabolic) CurvesStudy Notes
Thorough study notes for Vertical (Parabolic) Curves — the fastest path from zero to ready for CELE Surveying (Geomatics). Structured for self-study reviewers who cannot attend a review centre, these notes cover the full concept library plus the CELE-specific twists Professional Regulation Commission (PRC) — Board of Civil Engineering adds to its questions.
Exam context
On the CELE 2026, the Surveying (Geomatics) subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Civil Engineering's pattern. Vertical (Parabolic) Curves lands at position 7th out of 9 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Surveying (Geomatics) on a typical CELE paper.
Vertical (Parabolic) Curves - Study Notes
Vertical curves are essential geometric elements in highway design that smoothly connect two different road grades in profile. Unlike sharp tangent transitions that create discomfort and sight-distance restrictions, parabolic curves provide a constant rate of grade change, ensuring passenger comfort, structural durability, and adequate sight distance for safe vehicle operation. This study module covers the mathematical principles of parabolic curves, design formulas, location of critical points (summits and sags), elevation calculations, and sight-distance requirements as applied in Philippine road projects. All content aligns with NSCP 2015 standards and professional board-examination expectations.
Summary
Vertical parabolic curves are fundamental to highway design in the Philippines. They smoothly connect different road grades with a constant rate of change, ensuring comfort, safety, and adequate sight distance. Key concepts include: 1. **Mathematical Foundation:** The parabolic equation y = elev_PC + g₁·x + (r/2)·x² with constant rate r = (g₂ − g₁)/L governs all elevations on the curve. 2. **Critical Points:** The turning point (summit on crest curves, low point on sag curves) occurs at x = g₁·L/(g₁ − g₂) from the PC. This location must lie within the curve bounds [0, L]. 3. **Curve Classification:** Crest curves (g₁ > g₂) have r < 0 and require checking sight distance over the hump. Sag curves (g₁ < g₂) have r > 0 and are limited by headlight sight distance and drainage. 4. **Sight Distance Design:** Minimum curve lengths are calculated using A·S²/[200(√h₁ + √h₂)²] for crests and A·S²/400 for sags, where A is the algebraic grade difference in percent and S is the stopping sight distance. 5. **Practical Design:** The NSCP 2015 provides tables for quick reference. Designers must also consider comfort criteria (vertical acceleration ≤ 0.33 m/s²) and drainage requirements, especially in high-rainfall areas like the Philippines. 6. **Field Implementation:** Vertical curves are staked using calculated parabolic offsets from the initial tangent. Modern projects employ GPS and grade-control systems. 7. **Common Pitfalls:** Using percentages instead of decimals, measuring from the wrong reference point, and confusing crest/sag sight-distance formulas are frequent errors on board examinations. **For PRC Licensure Review:** Expect problems that require calculating elevations at arbitrary points, locating summits/sags, determining minimum curve lengths for sight distance, and verifying compliance with design standards. Strong understanding of grade sign conventions and systematic problem-solving approach are essential.
Sections
A vertical curve is a parabolic arc that connects two tangent grades (g₁ and g₂) smoothly without a sharp break. The curve is designed such that the rate of grade change is constant throughout its length. This constant rate of change distinguishes a parabolic vertical curve from a circular arc and ensures that vehicles experience uniform vertical acceleration. **Why Parabolic?** The parabola is chosen because: - It provides uniform rate of grade change (constant acceleration), which is comfortable for passengers - It simplifies elevation calculations using a quadratic equation - It matches common vehicle response characteristics - It is mathematically convenient for design and staking in the field **Grade Definition:** Grades are expressed as percentages or decimals: - Positive grade (+) indicates upslope (ascending) - Negative grade (−) indicates downslope (descending) - Example: g₁ = +3% means 3 m rise per 100 m horizontal distance **Curve Classification:** - **Crest (Summit) Curve:** g₁ > g₂ (grade changes from uphill to downhill or less uphill). The curve bulges upward. - **Sag (Valley) Curve:** g₁ < g₂ (grade changes from downhill to uphill or less downhill). The curve sags downward. Understanding these classifications is critical because they affect sight-distance requirements and design lengths differently.
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1. Fundamental Concept of Vertical Parabolic Curves
Examples
Crest curve; r = −0.00016/m
Problem
A crest vertical curve connects grade g₁ = +2.5% and grade g₂ = −1.5% over a curve length L = 250 m. Classify the curve and determine the rate of grade change per meter.
Solution
**Step 1: Classify the curve** Since g₁ (+2.5%) > g₂ (−1.5%), this is a CREST (summit) curve. **Step 2: Calculate rate of grade change** r = (g₂ − g₁)/L r = (−0.015 − 0.025)/250 r = (−0.040)/250 r = −0.00016/m (or −0.016%/m) The negative sign confirms the curve is a crest. The grade decreases by 0.016% for every meter of curve length.
Sag curve; r = +0.0001667/m
Problem
For a sag curve with g₁ = −3%, g₂ = +2%, and L = 300 m, determine if this is sag or crest, and find the rate of change.
Solution
**Step 1: Classify** Since g₁ (−3%) < g₂ (+2%), this is a SAG (valley) curve. **Step 2: Calculate r** r = (g₂ − g₁)/L r = (0.02 − (−0.03))/300 r = 0.05/300 r = 0.0001667/m (positive, as expected for sag) The grade increases (becomes less negative, then positive) by 0.01667% per meter.
Key Points
- Parabolic curves provide constant rate of grade change: r = (g₂ − g₁)/L per unit length
- Elevation at any point: y = elev_PC + g₁·x + (r/2)·x²
- Crest curves require sight-distance checks over the hump
- Sag curves require headlight sight-distance and comfort checks
- The turning point (summit or sag) occurs at: x = g₁·L/(g₁ − g₂) from the point of curvature (PC)
The elevation at any point along a parabolic vertical curve is calculated using the fundamental parabolic equation. This equation is derived from the constant-rate-of-change principle and is essential for field staking and design verification. **The Parabolic Elevation Equation:** y = elev_PC + g₁·x + (r/2)·x² Where: - y = elevation at distance x from the point of curvature (PC) - elev_PC = elevation at the beginning of the curve (in meters) - g₁ = initial grade (as a decimal; positive uphill, negative downhill) - x = distance along the curve from the PC (in meters, 0 ≤ x ≤ L) - r = rate of grade change = (g₂ − g₁)/L **Alternative Form Using Tangent Offset:** The elevation can also be expressed as: y = elev_PC + g₁·x + (A/200L)·x² Where A is the algebraic grade difference in percent: A = (g₂ − g₁) × 100 This form is convenient when grades are given as percentages rather than decimals. **Physical Interpretation:** The first term (elev_PC) is the starting elevation. The second term (g₁·x) is the elevation gain along the initial tangent. The third term ((r/2)·x²) is the parabolic offset from the initial tangent, which accounts for the curvature. For crest curves, this offset is negative (the curve sags below the tangent); for sag curves, it is positive (the curve rises above the tangent). **Step-by-Step Calculation Procedure:** 1. Convert grades to decimals (divide percentages by 100) 2. Calculate r = (g₂ − g₁)/L 3. Verify the value of x (must be 0 ≤ x ≤ L) 4. Substitute into the parabolic equation 5. Double-check sign conventions and arithmetic
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2. Elevation Calculations Along the Curve
Examples
At x = 50 m: 151.19 m; at x = 100 m: 151.75 m; at x = 150 m: 151.69 m
Problem
A vertical crest curve has the following data: PC elevation = 150.00 m, g₁ = +3%, g₂ = −2%, L = 200 m. Find the elevation at x = 50 m, x = 100 m (mid-curve), and x = 150 m from the PC.
Solution
**Step 1: Calculate rate of grade change** r = (g₂ − g₁)/L = (−0.02 − 0.03)/200 = −0.05/200 = −0.00025/m **Step 2: Calculate elevation at x = 50 m** y₅₀ = 150.00 + 0.03(50) + (−0.00025/2)(50)² y₅₀ = 150.00 + 1.50 − 0.3125 y₅₀ = 151.1875 m ≈ 151.19 m **Step 3: Calculate elevation at x = 100 m (mid-curve)** y₁₀₀ = 150.00 + 0.03(100) + (−0.00025/2)(100)² y₁₀₀ = 150.00 + 3.00 − 1.25 y₁₀₀ = 151.75 m **Step 4: Calculate elevation at x = 150 m** y₁₅₀ = 150.00 + 0.03(150) + (−0.00025/2)(150)² y₁₅₀ = 150.00 + 4.50 − 2.8125 y₁₅₀ = 151.6875 m ≈ 151.69 m **Observation:** The highest elevation is at the summit (which we'll calculate in Section 3), not necessarily at the mid-point.
At x = 75 m: 93.44 m
Problem
For a sag curve, elev_PC = 95.50 m, g₁ = −4%, g₂ = +1%, L = 150 m. Calculate the elevation at x = 75 m (mid-curve).
Solution
**Step 1: Convert and calculate r** g₁ = −0.04, g₂ = +0.01 r = (0.01 − (−0.04))/150 = 0.05/150 = 0.0003333/m **Step 2: Apply parabolic equation at x = 75 m** y₇₅ = 95.50 + (−0.04)(75) + (0.0003333/2)(75)² y₇₅ = 95.50 − 3.00 + (0.00016667)(5625) y₇₅ = 95.50 − 3.00 + 0.9375 y₇₅ = 93.4375 m ≈ 93.44 m **Note:** The elevation first decreases (following g₁ = −4%) then increases as the sag curve transitions to g₂ = +1%. The minimum elevation will be at the low point (calculated in Section 3).
Key Points
- Elevation equation: y = elev_PC + g₁·x + (r/2)·x²
- Use decimal grades, not percentages, in the parabolic equation
- x is measured from the PC (start of curve), not from the PI (intersection of tangents)
- The term (r/2)·x² is the vertical offset from the initial tangent line
- For crest curves, r is negative, so the offset (r/2)·x² is negative
- For sag curves, r is positive, so the offset (r/2)·x² is positive
- The equation applies only within the curve boundaries: 0 ≤ x ≤ L
The turning point of a vertical curve is the location where the tangent slope becomes zero. This point is a maximum (summit) on a crest curve and a minimum (sag) on a sag curve. Finding this point is critical for design checks and field work. **Mathematical Derivation:** The grade (slope) at any point is the derivative of the elevation equation: Slope = dy/dx = g₁ + r·x Setting the slope to zero: g₁ + r·x = 0 x = −g₁/r = −g₁/[(g₂ − g₁)/L] = g₁·L/(g₁ − g₂) This can be rearranged as: x = |g₁|·L/A Where A is the absolute value of the algebraic grade difference: A = |g₂ − g₁| **Critical Observation:** The turning point is measured from the PC (point of curvature), not from the PI (point of intersection). Always verify that 0 < x < L; if x falls outside this range, the turning point is outside the curve. **Elevation at the Turning Point:** Once x is determined, substitute it back into the parabolic equation: y_turning = elev_PC + g₁·x + (r/2)·x² **Alternative Compact Formula:** For a crest curve, the maximum elevation can be calculated as: y_max = elev_PC + (g₁²·L)/(2(g₁ − g₂)) = elev_PC + (g₁²·L·100)/(2A) (using A in percent) For a sag curve, the minimum elevation is: y_min = elev_PC + (g₁²·L)/(2(g₁ − g₂)) **Field Implications:** - For crest curves, the summit location determines the highest point where drainage or sight-distance obstructions might occur - For sag curves, the low point determines where water collects and where comfort criteria are most critical - On horizontal curves combined with vertical curves, the critical point helps in evaluating combined safety
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3. Location and Elevation of Critical Points (Summit and Sag)
Examples
Summit at x = 120 m from PC; elevation = 101.80 m
Problem
A crest curve has g₁ = +3%, g₂ = −2%, L = 200 m, and elev_PC = 100.00 m. Find the station (distance from PC) and elevation of the summit.
Solution
**Step 1: Check if turning point is within the curve** Since g₁ = +3% > g₂ = −2%, this is a crest curve. The turning point should be between 0 and L. **Step 2: Calculate distance to summit from PC** x_summit = g₁·L/(g₁ − g₂) x_summit = 0.03(200)/(0.03 − (−0.02)) x_summit = 6/0.05 = 120 m Verification: 0 < 120 < 200 ✓ (turning point is within the curve) **Step 3: Calculate r** r = (g₂ − g₁)/L = (−0.02 − 0.03)/200 = −0.00025/m **Step 4: Calculate summit elevation** y_summit = 100.00 + 0.03(120) + (−0.00025/2)(120)² y_summit = 100.00 + 3.60 − 1.80 y_summit = 101.80 m **Answer:** The summit is located 120 m from the PC, at elevation 101.80 m. If the PC station is 5+200, then the summit is at station 5+320.
Low point at x = 120 m from PC; elevation = 78.70 m
Problem
A sag curve has g₁ = −3%, g₂ = +1.5%, L = 180 m, and elev_PC = 80.50 m. Find the station and elevation of the low point.
Solution
**Step 1: Verify it's a sag curve** Since g₁ = −3% < g₂ = +1.5%, this is a sag (valley) curve. ✓ **Step 2: Calculate distance to low point from PC** x_low = g₁·L/(g₁ − g₂) x_low = (−0.03)(180)/(−0.03 − 0.015) x_low = −5.4/(−0.045) x_low = 120 m Verification: 0 < 120 < 180 ✓ **Step 3: Calculate r** r = (0.015 − (−0.03))/180 = 0.045/180 = 0.00025/m **Step 4: Calculate low-point elevation** y_low = 80.50 + (−0.03)(120) + (0.00025/2)(120)² y_low = 80.50 − 3.60 + 1.80 y_low = 78.70 m **Answer:** The low point is 120 m from the PC, at elevation 78.70 m. This is where water would accumulate and where vertical acceleration is maximum (affects comfort/drainage requirements).
Summit at x = 100 m; elevation = 111.00 m (1.00 m above PC)
Problem
For a crest curve with g₁ = +2%, g₂ = −3%, L = 250 m, verify that the summit lies within the curve and find its elevation relative to the PC (elev_PC = 110.00 m).
Solution
**Step 1: Distance to summit** x_summit = g₁·L/(g₁ − g₂) = 0.02(250)/(0.02 − (−0.03)) = 5/0.05 = 100 m Check: 0 < 100 < 250 ✓ Summit is within the curve. **Step 2: Rate of change** r = (−0.03 − 0.02)/250 = −0.0002/m **Step 3: Summit elevation** y_summit = 110.00 + 0.02(100) + (−0.0002/2)(100)² y_summit = 110.00 + 2.00 − 1.00 y_summit = 111.00 m **Elevation gain from PC to summit:** 111.00 − 110.00 = 1.00 m This can be verified by calculating how far the initial tangent (at +2%) would rise: 0.02(100) = 2.00 m, but the curve rises only 1.00 m, meaning the curve is 1.00 m below the initial tangent at the summit location.
Key Points
- Distance to turning point from PC: x = g₁·L/(g₁ − g₂)
- Alternatively: x = |g₁|·L/|g₂ − g₁|
- Always verify that 0 < x < L; if not, the turning point is outside the curve
- For crest curves (g₁ > g₂), r is negative, and x is positive (turning point within curve)
- For sag curves (g₁ < g₂), r is positive, and x is positive (turning point within curve)
- Elevation at turning point uses the parabolic equation with the calculated x value
- The grade changes sign at the turning point (from positive to negative for crest, vice versa for sag)
Understanding the relationship between the parabolic curve and its tangent lines is essential for design calculations, particularly for sight distance and comfort analysis. The vertical offset from the tangent to the curve at any point characterizes the parabolic shape. **Vertical Offset from Initial Tangent:** At any distance x from the PC, the elevation of the initial tangent line (at grade g₁) would be: y_tangent = elev_PC + g₁·x The actual elevation on the parabolic curve is: y_curve = elev_PC + g₁·x + (r/2)·x² The vertical offset (difference between curve and tangent) is: Offset = y_curve − y_tangent = (r/2)·x² For a crest curve (r < 0), this offset is negative (curve below tangent). For a sag curve (r > 0), the offset is positive (curve above tangent). **Maximum Offset at Mid-Curve:** The maximum vertical separation between the curve and the initial tangent occurs at x = L/2 (mid-point of the curve): Offset_max = (r/2)·(L/2)² = (r·L²)/8 = (g₂ − g₁)·L/8 This is also equal to A·L/800 when A is the algebraic grade difference in percent. **Important Property for Design:** The parabolic property that "offsets are proportional to the square of the distance" is used extensively in: - Sight-distance calculations (line-of-sight corrections over humps) - Comfort checks (evaluating vertical acceleration) - Staking procedures (using parabolic offset tables) **Practical Calculation Example:** For the crest curve in Example 1 (g₁ = +3%, g₂ = −2%, L = 200 m): - r = −0.00025/m - Maximum offset at mid-curve = (−0.00025/2)·(100)² = −1.25 m - This means the curve at mid-length is 1.25 m BELOW the initial tangent line **Application in Field Staking:** When staking a vertical curve, field crews often use offset tables. If the tangent elevation at x = 50 m is 151.50 m and the parabolic offset is −0.3125 m, the actual curve elevation is 151.50 − 0.3125 = 151.1875 m.
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4. Vertical Offset and Chord Properties
Examples
At x = 50 m: −0.3125 m; at x = 100 m: −1.25 m; at x = 150 m: −2.8125 m; maximum offset: −1.25 m
Problem
For the crest curve with g₁ = +3%, g₂ = −2%, L = 200 m, calculate the vertical offset at x = 50 m, x = 100 m, and x = 150 m. Also calculate the maximum offset.
Solution
**Step 1: Calculate r** r = (−0.02 − 0.03)/200 = −0.00025/m **Step 2: Calculate offsets using Offset = (r/2)·x²** At x = 50 m: Offset₅₀ = (−0.00025/2)·(50)² = −0.3125 m At x = 100 m: Offset₁₀₀ = (−0.00025/2)·(100)² = −1.25 m At x = 150 m: Offset₁₅₀ = (−0.00025/2)·(150)² = −2.8125 m **Step 3: Calculate maximum offset at mid-curve (x = L/2 = 100 m)** Offset_max = (−0.00025/2)·(100)² = −1.25 m Alternatively: |Offset_max| = |A|·L/800 = 5 × 200/800 = 1.25 m ✓ **Interpretation:** The parabolic curve sags increasingly below the initial tangent as you move along the curve. At the mid-point, the maximum drop is 1.25 m. This is typical of crest curves used in highway design.
Maximum offset = 1.80 m (positive, sag curve)
Problem
A sag curve has g₁ = −4%, g₂ = +2%, L = 240 m. Find the maximum vertical offset at mid-curve and verify it using the formula A·L/800.
Solution
**Step 1: Calculate r** r = (0.02 − (−0.04))/240 = 0.06/240 = 0.00025/m **Step 2: Maximum offset at x = L/2 = 120 m** Offset_max = (0.00025/2)·(120)² = (0.000125)·(14400) = 1.80 m **Step 3: Verify using A·L/800** A = |g₂ − g₁|·100 = |0.02 − (−0.04)|·100 = 6% Offset_max = 6 × 240/800 = 1440/800 = 1.80 m ✓ **Interpretation:** For a sag curve, the curve rises 1.80 m ABOVE the initial tangent line at the mid-point. This increased elevation is needed to maintain sight distance at night (headlight beam angle) and affects drainage at the low point.
Key Points
- Vertical offset = (r/2)·x² = (parabolic term only, from elevation equation)
- Crest curves: offset is negative (curve sags below tangent)
- Sag curves: offset is positive (curve rises above tangent)
- Maximum offset at mid-curve: |offset_max| = |A|·L/800 (A in percent)
- Offset is zero at PC and PT (beginning and end of curve), maximum at center
- Field staking uses: actual elevation = tangent elevation + parabolic offset
- The parabolic property (offset ∝ x²) is exact and fundamental to curve design
Sight distance is one of the primary determinants of vertical curve length in modern road design. The parabolic shape of the curve affects the driver's ability to see obstacles ahead. Different sight-distance criteria apply to crest and sag curves. **Crest Curves — Stopping Sight Distance:** On a crest (summit) curve, the driver's line of sight is obstructed by the curve itself. The required curve length is determined by the need to see an object of height h₂ (typically 0.15 m for a low obstacle) from a driver's eye height h₁ (typically 1.07 m) at a distance equal to the stopping sight distance S. For S < L (stopping distance less than curve length): L = A·S²/[200(√h₁ + √h₂)²] For S ≥ L (stopping distance greater than curve length): L = 2S − 200(√h₁ + √h₂)²/A Where: - L = length of vertical curve (m) - A = algebraic grade difference = |g₂ − g₁|·100 (percent) - S = stopping sight distance (m) - h₁ = driver's eye height above road surface (typically 1.07 m) - h₂ = object height (typically 0.15 m for a vehicle tail light) **Typical Stopping Sight Distance (from AASHTO/NSCP):** - 40 km/h: S ≈ 40 m - 60 km/h: S ≈ 65 m - 80 km/h: S ≈ 100 m - 100 km/h: S ≈ 140 m - 120 km/h: S ≈ 180 m **Sag Curves — Sight Distance:** On a sag (valley) curve, the sight limitation is usually due to headlight beam throw at night rather than line-of-sight obstruction. The criterion is that the driver must see the beam reach the road surface at the stopping-sight-distance. For sag curves, the design equation is: L = A·S²/[200(h + S·tan(θ))] Where: - h = headlight height (approximately 0.6 m) - θ = headlight beam angle (approximately 1° above horizontal) - For practical design: L = A·S²/400 (simplified form) Alternatively, NSCP specifies minimum curve lengths based on algebraic grade difference and design speed. **Design Procedure:** 1. Determine design speed and corresponding stopping sight distance S 2. Calculate algebraic grade difference A = |g₂ − g₁|·100 3. Select curve type (crest or sag) 4. Apply appropriate sight-distance formula 5. Calculate minimum required L 6. Compare with other design criteria (comfort, drainage) and use the larger value **Practical Note for Philippine Design:** The NSCP 2015 (which governs Philippine road design) provides tables of minimum vertical curve lengths based on design speed and algebraic grade change. These tables are derived from the formulas above but provide quick reference values. Always verify local requirements and any project-specific criteria from the Department of Public Works and Highways (DPWH).
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5. Sight Distance on Vertical Curves
Examples
Minimum L = 194 m (use 195 m); summit at x ≈ 121 m
Problem
Design a crest vertical curve for a highway with design speed 100 km/h. The grades are g₁ = +2.5% and g₂ = −1.5%. The stopping sight distance for this speed is 140 m. Calculate the minimum required curve length and verify that the turning point lies within the curve.
Solution
**Step 1: Calculate algebraic grade difference** A = |g₂ − g₁|·100 = |−0.015 − 0.025|·100 = |−0.04|·100 = 4% **Step 2: Determine sight-distance parameters** For a crest curve: - h₁ = 1.07 m (driver eye height) - h₂ = 0.15 m (object height) - S = 140 m (stopping sight distance at 100 km/h) **Step 3: Calculate √h₁ + √h₂** √h₁ + √h₂ = √1.07 + √0.15 = 1.0344 + 0.3873 = 1.4217 (√h₁ + √h₂)² = (1.4217)² = 2.0211 **Step 4: Check if S < L** We don't know L yet, but we'll assume S < L first (most common case). **Step 5: Apply crest formula for S < L** L = A·S²/[200(√h₁ + √h₂)²] L = 4 × (140)²/(200 × 2.0211) L = 4 × 19600/404.22 L = 78400/404.22 L = 193.95 m Verification: S (140 m) < L (194 m) ✓ Our assumption was correct. **Step 6: Verify turning point is within curve** x = g₁·L/(g₁ − g₂) = 0.025 × 193.95/(0.025 − (−0.015)) x = 4.849/0.04 = 121.2 m Check: 0 < 121.2 < 193.95 ✓ Turning point is within the curve. **Answer:** Minimum required curve length L = 194 m (round to 195 m for construction). The summit occurs at approximately 121 m from the PC.
Minimum L = 180 m; low point at x ≈ 108 m
Problem
A sag vertical curve has g₁ = −3%, g₂ = +2%, and the stopping sight distance for the design speed is 120 m. Calculate the minimum curve length using the simplified sag formula.
Solution
**Step 1: Calculate algebraic grade difference** A = |g₂ − g₁|·100 = |0.02 − (−0.03)|·100 = 5% **Step 2: Apply simplified sag formula** L = A·S²/400 = 5 × (120)²/400 L = 5 × 14400/400 L = 72000/400 L = 180 m **Step 3: Verify by checking if turning point is within curve** x = g₁·L/(g₁ − g₂) = (−0.03) × 180/(−0.03 − 0.02) x = −5.4/(−0.05) = 108 m Check: 0 < 108 < 180 ✓ Low point is within the curve. **Answer:** Minimum curve length for the sag curve is L = 180 m. The low point occurs at 108 m from the PC, which is slightly offset from the geometric center due to the unequal grades.
Crest L ≈ 380 m; Sag L ≈ 384 m (similar for this case)
Problem
Compare the curve lengths for a crest and a sag curve with the same algebraic grade difference A = 6% and S = 160 m.
Solution
**For the Crest Curve:** L_crest = A·S²/[200(√h₁ + √h₂)²] L_crest = 6 × (160)²/(200 × 2.0211) L_crest = 6 × 25600/404.22 L_crest = 153600/404.22 L_crest ≈ 379.7 m ≈ 380 m **For the Sag Curve:** L_sag = A·S²/400 = 6 × (160)²/400 L_sag = 6 × 25600/400 L_sag = 153600/400 L_sag = 384 m **Comparison:** - Crest curve (sight over hump): L ≈ 380 m - Sag curve (headlight sight): L ≈ 384 m Both are similar in this case, but crest curves generally require longer lengths when the algebraic grade difference and sight distance are large. The difference becomes more pronounced at higher speeds and steeper grades.
Key Points
- Crest curves: sight distance limited by line-of-sight over the hump
- Sag curves: sight distance limited by headlight beam throw at night
- For crest curves: L = A·S²/[200(√h₁ + √h₂)²] when S < L
- For sag curves: L = A·S²/400 (simplified common form)
- A = algebraic grade difference in percent: A = |g₂ − g₁|·100
- Typical heights: driver eye h₁ = 1.07 m, object height h₂ = 0.15 m
- Design curves using the larger of sight-distance and comfort requirements
- NSCP 2015 provides tables; formulas give precise calculations for specific sites
Beyond the fundamental mathematics, vertical curve design must consider multiple practical factors, especially in the Philippine context where terrain, climate, and safety standards all play critical roles. **Comfort Criteria (Vertical Acceleration):** Passengers experience vertical acceleration as the vehicle traverses the curve. The rate of change of grade (r in m/m²) directly relates to this acceleration. Maximum allowable vertical acceleration: a_max ≈ g/30 ≈ 0.33 m/s² (or roughly 3.3% of gravity) This translates to a minimum curve length: L_comfort = r·v²/a_max Where v is the design speed in m/s. For practical design, NSCP provides minimum lengths based on comfort. **Drainage and Surface Water Control:** On sag curves, water tends to collect at the low point. Minimum lengths are often specified to prevent ponding and ensure adequate longitudinal drainage slopes. Philippine standards typically require minimum L for sag curves to prevent standing water, especially in areas with heavy rainfall. **Grade Transition Issues:** When transitioning from one grade to another: - Smooth transitions are required for vehicle handling - Sudden grade changes cause structural stress on vehicles and pavements - Inadequate curve length leads to "bottoming out" on sag curves and excessive front-end dip on crest curves **Combined Horizontal and Vertical Alignment:** In real projects, horizontal (circular) curves are often combined with vertical curves. When both are present simultaneously: - Safety visibility is reduced further (cannot see around a horizontal curve over a vertical hump) - Design must consider the more restrictive criterion - Philippine design standards address combined horizontal-vertical alignment effects **Field Implementation (Staking):** Vertical curves are staked in the field using: 1. **Offset Method:** Calculate parabolic offsets from the tangent and set points accordingly 2. **Elevation Method:** Calculate elevations at key points (25 m or 50 m intervals) and set using level/GPS 3. **Profile Control:** Modern projects use GPS and grade control systems **Pavement Design Implications:** The rate of grade change affects pavement design: - Steeper curves (larger r values) cause greater structural loading - Sag curves experience more severe water infiltration due to compression stresses - Thermal stresses in concrete pavements are affected by vertical alignment **Environmental and Safety Considerations in Philippine Context:** Specific to Philippine highway design: - Typhoon/monsoon season drainage considerations - Landslide-prone areas require stable, well-drained grades - Urban areas often have tighter constraints on curve length - Rural or mountainous projects may have cost/constructability limits **Use of Design Tables:** The NSCP 2015 provides tabulated minimum vertical curve lengths for various design speeds and algebraic grade differences. These tables consider: - Stopping sight distance - Passing sight distance (on rare occasion for special cases) - Comfort criteria - Drainage requirements Designers select the larger of the calculated sight-distance length and the comfort/drainage minimum.
Heading
6. Design Considerations and Practical Applications
Examples
Sight-distance: 53 m; Comfort: 42 m; Drainage (governing): 100 m
Problem
A vertical sag curve in a low-lying area of Metro Manila experiences heavy rainfall. The algebraic grade difference is 5% and the design speed is 60 km/h (S = 65 m). Calculate both the sight-distance-based length and a reasonable drainage-based minimum length. Which governs?
Solution
**Step 1: Sight-distance-based length (sag curve)** L_sight = A·S²/400 = 5 × (65)²/400 L_sight = 5 × 4225/400 = 21125/400 = 52.8 m **Step 2: Comfort-based length** For a vertical curve, the rate of change is r = A/(100L) in terms of percentage per meter. To ensure a_vertical ≤ 0.33 m/s²: L_comfort = A·v²/(100·a_max) where v is in m/s v_60kmh = 60/3.6 = 16.67 m/s L_comfort = 5 × (16.67)²/(100 × 0.33) = 5 × 277.9/33 ≈ 42.1 m **Step 3: Drainage-based minimum** For a sag curve in a high-rainfall area, best practice is to use at least: L_drain ≥ 80–100 m (industry practice for proper sump drainage) **Step 4: Determine governing length** - Sight distance: 53 m - Comfort: 42 m - Drainage (recommended): 80–100 m For this high-rainfall location, **L_drain = 100 m GOVERNS**. This ensures adequate sump design and prevents standing water that could lead to pavement deterioration and hydroplaning risk. **Answer:** Use L = 100 m minimum (drainage is the governing criterion for this metro Manila project).
Minimum L = 186 m; summit at x ≈ 99 m
Problem
Design a vertical curve for a highway in a mountainous region where g₁ = +4%, g₂ = −3.5%, design speed 80 km/h (S = 100 m). Calculate the minimum length and verify that the summit location is accessible (i.e., does not create a blind spot near a critical point).
Solution
**Step 1: Calculate A** A = |−0.035 − 0.04|·100 = 7.5% **Step 2: Calculate sight-distance length** For a crest curve with S = 100 m: L = A·S²/[200(√h₁ + √h₂)²] (√h₁ + √h₂)² ≈ 2.02 L = 7.5 × (100)²/(200 × 2.02) L = 7.5 × 10000/404 L ≈ 185.6 m ≈ 186 m **Step 3: Verify summit location** x_summit = g₁·L/(g₁ − g₂) = 0.04 × 186/(0.04 − (−0.035)) x_summit = 7.44/(0.075) = 99.2 m The summit is at 99.2 m from the PC, which is close to the mid-point (93 m). This indicates relatively balanced grades, which is good for sight distance. **Step 4: Check if this length is sufficient for terrain** In a mountainous region, the designer should verify: - Are there any critical sight points (intersections, rest areas) near x = 99 m? - Is 186 m feasible in the available terrain? - Are there other constraints (structures, environmental concerns)? If the terrain cannot accommodate 186 m, the road must be redesigned with different grades or lower design speed. **Answer:** Minimum L = 186 m; summit at x ≈ 99 m from PC. For mountainous terrain, verify constructability and absence of critical sight obstructions.
Key Points
- Vertical acceleration on curves should not exceed roughly 0.33 m/s² for comfort
- Sag curves require special attention to drainage, especially in high-rainfall areas
- Minimum curve lengths are governed by sight distance, comfort, AND drainage
- Combined horizontal-vertical curves are more restrictive than either alone
- Field staking requires accurate calculation of offsets or elevations at 25–50 m intervals
- Philippine standards (NSCP 2015) provide tables for quick reference
- Steeper vertical grades increase structural loads on pavements
- Drainage design is critical on sag curves to prevent standing water and pavement damage
This section consolidates all essential formulas and provides a systematic problem-solving approach for vertical curve questions on the PRC Civil Engineer Licensure Examination. **Core Formulas:** 1. **Rate of Grade Change:** r = (g₂ − g₁)/L 2. **Elevation Along Curve:** y = elev_PC + g₁·x + (r/2)·x² Alternative: y = elev_PC + g₁·x + [A/(200L)]·x² (A in percent) 3. **Distance to Turning Point:** x = g₁·L/(g₁ − g₂) = |g₁|·L/|g₁ − g₂| 4. **Vertical Offset from Tangent:** Offset = (r/2)·x² = [A/(200L)]·x² 5. **Maximum Offset at Mid-Curve:** |Offset_max| = |r|·L²/8 = A·L/800 (A in percent) 6. **Crest Curve Sight Distance (S < L):** L = A·S²/[200(√h₁ + √h₂)²] 7. **Sag Curve Sight Distance:** L = A·S²/400 (simplified) **Systematic Problem-Solving Approach:** **For Elevation/Offset Calculations:** 1. Identify g₁, g₂, L, and elev_PC 2. Convert grades to decimals (divide percent by 100) 3. Calculate r = (g₂ − g₁)/L 4. Identify the location (distance x from PC) 5. Apply: y = elev_PC + g₁·x + (r/2)·x² 6. Report elevation with proper units and precision (usually 2 decimal places for meters) **For Turning Point Calculations:** 1. Classify curve as crest (g₁ > g₂) or sag (g₁ < g₂) 2. Calculate x = g₁·L/(g₁ − g₂) 3. Verify: 0 < x < L (must be within curve) 4. If valid, substitute x into elevation equation 5. Report distance from PC and elevation **For Sight Distance Calculations:** 1. Determine design speed and corresponding S from tables 2. Calculate A = |g₂ − g₁|·100 (in percent) 3. Check if crest or sag curve 4. Apply appropriate formula (crest or sag) 5. Calculate minimum L 6. Verify that turning point is within the curve (optional but good practice) 7. Compare with comfort/drainage requirements and use the larger value **Common Exam Question Patterns:** - "Find the elevation at a given point on the curve" → Use elevation equation - "Find the location and elevation of the summit/sag" → Use turning-point formula - "What is the vertical offset at mid-curve?" → Use Offset_max formula - "Design a curve for a given sight distance and speed" → Use sight-distance formula - "Verify that the curve meets design standards" → Check that calculated L meets or exceeds minimum required **Critical Sign Conventions:** - Grades: positive (+) for uphill, negative (−) for downhill - r: negative for crest, positive for sag - Offset: negative (below tangent) for crest, positive (above tangent) for sag - Turning point location x: always positive, measured from PC **Common Mistakes to Avoid:** 1. **Using percentages instead of decimals** in the elevation equation 2. **Measuring x from the PI instead of the PC** (always from the point of curvature) 3. **Forgetting to check that x is within [0, L]** 4. **Incorrect grade signs** (especially when both grades are negative or positive) 5. **Confusing crest and sag sight-distance formulas** 6. **Rounding errors** (use enough decimal places in intermediate steps) 7. **Forgetting to convert km/h to m/s** when using velocity-based formulas **Example Problem-Solving Flow for Exam:** Question: "A vertical crest curve connects g₁ = +2.5% and g₂ = −1.5% over 150 m. The PC is at elevation 200 m. Find: (a) the summit elevation, (b) the distance to the summit from PC, (c) the minimum curve length for a stopping sight distance of 100 m." **Solution Approach:** a) Convert grades: g₁ = 0.025, g₂ = −0.015 Calculate r = (−0.015 − 0.025)/150 = −0.000267/m Find x = 0.025 × 150/(0.025 − (−0.015)) = 3.75/0.04 = 93.75 m Calculate y = 200 + 0.025(93.75) + (−0.000267/2)(93.75)² = 200 + 2.34375 − 1.17 = 201.17 m **Answer: 201.17 m** b) **Answer: 93.75 m from PC (or 93.8 m)** c) Calculate A = |−0.015 − 0.025|·100 = 4% Apply: L = 4 × (100)²/[200 × 2.02] ≈ 98.5 m **Answer: Minimum 98.5 m (round to 100 m); the 150 m curve exceeds this requirement** ✓
Heading
7. Summary of Key Formulas and Problem-Solving Strategy
Examples
See detailed table above. Both curves meet design requirements. Curve 1 summit: 120 m from PC at 51.8 m elev. Curve 2 low point: 102.86 m from PC at 46.97 m elev.
Problem
Complete Design Problem: A new highway section in Bulacan has two vertical curves. Curve 1 (crest): g₁ = +3%, g₂ = −2%, L = 200 m, elev_PC = 50 m. Curve 2 (sag): g₁ = −2%, g₂ = +1.5%, L = 180 m, elev_PC = 48 m. For both curves, calculate: (i) location and elevation of summit/sag, (ii) vertical offset at mid-curve, (iii) design speed requirement if S = 90 m.
Solution
**CURVE 1 (CREST):** **(i) Location and Elevation of Summit:** r₁ = (−0.02 − 0.03)/200 = −0.00025/m x₁ = 0.03 × 200/(0.03 − (−0.02)) = 6/0.05 = 120 m y₁ = 50 + 0.03(120) + (−0.00025/2)(120)² y₁ = 50 + 3.6 − 1.8 = 51.8 m **Summit at 120 m from PC; elevation 51.8 m** **(ii) Vertical Offset at Mid-Curve:** Offset = (−0.00025/2) × (100)² = −1.25 m Alternatively: A = 5%, L = 200 m Offset_max = 5 × 200/800 = 1.25 m (magnitude) **Offset = −1.25 m (curve sags 1.25 m below tangent)** **(iii) Minimum L for S = 90 m:** A₁ = 5% L_min = 5 × (90)²/[200 × 2.02] = 40500/404 ≈ 100.2 m Since the actual L = 200 m > 100.2 m, **the 200 m curve is adequate** ✓ --- **CURVE 2 (SAG):** **(i) Location and Elevation of Low Point:** r₂ = (0.015 − (−0.02))/180 = 0.035/180 = 0.0001944/m x₂ = (−0.02) × 180/(−0.02 − 0.015) = −3.6/(−0.035) = 102.86 m y₂ = 48 + (−0.02)(102.86) + (0.0001944/2)(102.86)² y₂ = 48 − 2.057 + 1.029 = 46.972 m ≈ 46.97 m **Low point at 102.86 m from PC; elevation 46.97 m** **(ii) Vertical Offset at Mid-Curve:** Offset = (0.0001944/2) × (90)² = 0.788 m Alternatively: A = |1.5 − (−2)|·100 = 3.5% Offset_max = 3.5 × 180/800 = 0.7875 m ≈ 0.79 m (magnitude) **Offset = +0.79 m (curve rises 0.79 m above tangent)** **(iii) Minimum L for S = 90 m:** A₂ = 3.5% L_min = 3.5 × (90)²/400 = 28350/400 = 70.875 m ≈ 71 m Since the actual L = 180 m > 71 m, **the 180 m curve is adequate** ✓ --- **Summary Table:** | Criterion | Curve 1 (Crest) | Curve 2 (Sag) | |-----------|-----------------|---------------| | Summit/Sag Location | 120 m from PC | 102.86 m from PC | | Elevation | 51.8 m | 46.97 m | | Max Offset (mid) | −1.25 m | +0.79 m | | Min L for S=90m | 100.2 m | 70.9 m | | Actual L | 200 m ✓ | 180 m ✓ | | Status | Adequate | Adequate |
Key Points
- Use decimals (not percentages) in the elevation equation y = elev_PC + g₁·x + (r/2)·x²
- Distance x is always measured from the PC (point of curvature), never from the PI
- Turning point must satisfy: 0 < x < L; otherwise it is outside the curve
- For crest curves: r < 0, offset < 0 (curve below tangent)
- For sag curves: r > 0, offset > 0 (curve above tangent)
- Sight-distance formulas differ significantly between crest and sag curves
- Always report final answers with appropriate units and precision
- Verify that calculated curve lengths meet design standards (sight distance, comfort, drainage)
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