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CELE Surveying (Geomatics) — Vertical (Parabolic) CurvesDetailed Explanation

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Detailed explanations for CELE Surveying (Geomatics) — Vertical (Parabolic) Curves. This page treats you like a serious reviewer: we unpack the concepts thoroughly, show worked examples of how Professional Regulation Commission (PRC) — Board of Civil Engineering frames Vertical (Parabolic) Curves questions, and explain the underlying reasoning that gets you to the right answer every time.

Exam context

Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Surveying (Geomatics) section sits under a "Core" weighting, and Vertical (Parabolic) Curves is the 7th chapter in the 9-chapter CELE Surveying (Geomatics) rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Surveying (Geomatics).

Vertical (Parabolic) Curves - Detailed Explanation

Vertical curves are parabolic transitions inserted in a road profile to connect two intersecting grades smoothly. The parabolic shape is preferred because it produces a constant rate of grade change, ensuring rider comfort and predictable sight-distance behaviour. In the PRC Civil Engineer Licensure Examination, vertical-curve problems consistently appear in the Surveying/Geomatics portion and typically test: (1) locating the high point of a crest or the low point of a sag, (2) computing the elevation of any point on the curve, (3) finding the vertical offset from the tangent, and (4) determining the minimum curve length for a given stopping sight distance. Mastery of three core equations — the grade-change rate r, the parabolic elevation formula, and the turning-point formula — is sufficient to solve virtually every board-exam item on this topic.

Concepts

Geometry and Nomenclature of Vertical Curves

A vertical parabolic curve is defined by three key points: the Back Tangent Grade Point (BVC or PC), the Point of Vertical Intersection (PVI or PI), and the Forward Tangent Grade Point (EVC or PT). The two grades are g₁ (back tangent, expressed as a decimal, positive uphill) and g₂ (forward tangent). The curve length L is always measured horizontally, not along the slope. A CREST (summit) curve occurs when g₁ > g₂ — the road goes over a hill. A SAG (valley) curve occurs when g₁ < g₂ — the road dips into a valley. The algebraic difference of grades is: A = g₂ − g₁ (in decimal) or A% = (g₂ − g₁) × 100 in percent. For a crest, A is negative; for a sag, A is positive. The PVI is located at the midpoint of the horizontal distance L (i.e., at x = L/2 from the BVC), and the elevation of the PVI is: elev_PVI = elev_BVC + g₁ × (L/2) These naming conventions are used consistently in Philippine textbooks (La Putt, Gillesania) and must be memorised before attempting board problems.

Examples

The sign convention is critical. Writing g₂ − g₁ first ensures A is negative for crests and positive for sags, which is consistent with the parabolic formula sign.

Scenario

A vertical curve has g₁ = +3%, g₂ = −2%, and L = 200 m. Identify the curve type, compute A%, and locate the PVI relative to the BVC.

Solution

Step 1 — Identify curve type: g₁ = +0.03 > g₂ = −0.02 → CREST curve. Step 2 — Algebraic difference: A = g₂ − g₁ = −0.02 − 0.03 = −0.05 (or A% = −5%) |A%| = 5% (used in sight-distance formulas). Step 3 — PVI location: PVI is at x = L/2 = 200/2 = 100 m from BVC. Step 4 — PVI elevation (if elev_BVC = 100.00 m): elev_PVI = 100.00 + 0.03 × 100 = 103.00 m.

Applications

Highway and road design profile sheets (DPWH standard drawings)
Railway vertical alignment design
Sight-distance analysis on Philippine national roads
Setting out road grades during construction staking

Misconceptions

Students often confuse BVC with PVI — the PVI is NOT on the parabola; it is the intersection of the two tangent lines.
Using L as the slope length instead of horizontal distance introduces error.
Forgetting to convert percent grades to decimals before substituting into formulas.

Related Concepts

Horizontal curves (simple circular curves)
Road profile design
Stopping sight distance (SSD)
Grade and slope in route surveying

Common Exam Questions

Example

Given g₁ = −1.5% and g₂ = +2.5%, classify the curve and find |A%|. Answer: Sag; |A%| = 4%.

Approach

Compare g₁ and g₂; state crest or sag; compute A%.

Question Type

Identification / Classification

Key Points To Remember

L is always horizontal distance, not slope length.
Grades g₁ and g₂ are decimals in all formulas (e.g., +3% = +0.03).
Crest curve: g₁ > g₂ (algebraic difference A is negative).
Sag curve: g₁ < g₂ (algebraic difference A is positive).
BVC is the start of the curve; EVC is the end; PVI is the apex of the two tangents.
The PVI lies directly above the midpoint of the curve horizontally.

The Parabolic Elevation Formula

The defining property of a parabolic vertical curve is that the grade changes at a CONSTANT RATE. This constant rate is: r = (g₂ − g₁) / L [per metre] The elevation at any point distance x (measured horizontally from the BVC) along the curve is: y(x) = elev_BVC + g₁·x + (r/2)·x² Breaking this down: • elev_BVC — the starting elevation (constant) • g₁·x — the elevation rise/fall due to the back tangent (linear term) • (r/2)·x² — the parabolic correction (quadratic term) The grade at any point x is: g(x) = g₁ + r·x At x = 0: grade = g₁ (back tangent grade — correct) At x = L: grade = g₁ + r·L = g₁ + (g₂ − g₁) = g₂ (forward tangent grade — correct) This self-checks that the formula is correct. IMPORTANT: r is negative for a crest (parabola opens downward) and positive for a sag (parabola opens upward).

Examples

Notice that the parabolic correction (r/2)x² is always negative for a crest, pulling the curve below the tangent. Verify units: r is /m, x is m, so (r/2)x² is dimensionless × m² / m = m. Correct.

Scenario

From the previous problem (g₁ = +3%, g₂ = −2%, L = 200 m, elev_BVC = 100.00 m), find the elevation at x = 60 m and x = 150 m.

Solution

Step 1 — Compute r: r = (g₂ − g₁)/L = (−0.02 − 0.03)/200 = −0.05/200 = −0.00025 /m Step 2 — Elevation at x = 60 m: y(60) = 100.00 + 0.03(60) + (−0.00025/2)(60²) = 100.00 + 1.80 + (−0.000125)(3600) = 100.00 + 1.80 − 0.45 = 101.35 m Step 3 — Elevation at x = 150 m: y(150) = 100.00 + 0.03(150) + (−0.000125)(150²) = 100.00 + 4.50 − 0.000125(22500) = 100.00 + 4.50 − 2.8125 = 101.6875 m ≈ 101.69 m

For a sag curve, r is positive, so the parabolic correction adds elevation, pulling the curve above the tangent (toward the road surface). This is physically correct — a sag curves upward.

Scenario

A sag curve has g₁ = −4%, g₂ = +2%, L = 300 m, and elev_BVC = 85.50 m. Find the elevation at x = 100 m.

Solution

Step 1 — Compute r: r = (0.02 − (−0.04))/300 = 0.06/300 = +0.0002 /m Step 2 — Elevation at x = 100 m: y(100) = 85.50 + (−0.04)(100) + (0.0002/2)(100²) = 85.50 − 4.00 + (0.0001)(10000) = 85.50 − 4.00 + 1.00 = 82.50 m

Applications

Earthwork volume calculation by prismoidal formula on vertical curves
Setting out cut/fill stakes at intermediate stations on a road profile
Computing drain inlet elevations in sag curves
Checking as-built road levels against design profile

Misconceptions

Using x measured from the PVI instead of the BVC — all distances are from the BVC (x = 0).
Forgetting to halve r in the quadratic term: it is (r/2)x², not rx².
Using L/2 in the formula when the problem asks for elevation at the PVI — the PVI is on the TANGENT, not the parabola.

Related Concepts

Tangent elevation formula: y_tan = elev_BVC + g₁·x
Vertical offset from tangent
Grade change rate r
High/low point location

Common Exam Questions

Example

BVC at Sta 10+000, elev 50.00 m; g₁ = +2%, g₂ = −3%, L = 400 m. Find elevation at Sta 10+200. x = 200 m; r = −0.05/400 = −0.000125; y = 50 + 0.02(200) + (−0.0000625)(40000) = 50 + 4 − 2.5 = 51.50 m.

Approach

Identify BVC station and elevation; compute x = given station − BVC station; apply y(x) formula.

Question Type

Elevation at a specific station

Key Points To Remember

r = (g₂ − g₁)/L always; use signed decimal grades.
y(x) = elev_BVC + g₁x + (r/2)x² — memorise this single master formula.
The grade at x is g(x) = g₁ + rx; turning point is where g(x) = 0.
Check: at x = L, grade must equal g₂.
r is negative for crest, positive for sag.
x must lie within [0, L]; otherwise the point is on the tangent, not the curve.

Location of the High Point (Crest) or Low Point (Sag)

The turning point of the parabola — the summit of a crest or the bottom of a sag — occurs where the instantaneous grade equals zero: g(x) = g₁ + r·x = 0 x_tp = −g₁/r = g₁·L/(g₁ − g₂) Both forms are equivalent. The second form is more convenient on the board exam: x_tp = (g₁ × L) / (g₁ − g₂) IMPORTANT CHECK: The turning point is valid only if 0 ≤ x_tp ≤ L. If x_tp falls outside this range, the curve does not have a turning point — the entire curve is ascending or descending. Once x_tp is found, the elevation is: y_tp = elev_BVC + g₁·x_tp + (r/2)·x_tp² For STATION of the turning point: Sta_tp = Sta_BVC + x_tp Physical interpretation: • Crest: x_tp = g₁·L/(g₁−g₂). Since g₁ > 0 and g₁−g₂ > 0 for a crest, x_tp > 0 and < L (when the grades have opposite signs). • Sag: g₁ < 0 and g₂ > 0, so g₁−g₂ < 0; since g₁ is also negative, x_tp = negative/negative = positive — still within [0, L].

Examples

The summit is 120 m from the BVC (closer to the BVC since the positive grade is steeper, +3% vs. −2%). The elevation 101.80 m is above the BVC, consistent with a crest.

Scenario

Crest curve: g₁ = +3%, g₂ = −2%, L = 200 m, BVC at Sta 5+000, elev_BVC = 100.00 m. Find the station and elevation of the summit.

Solution

Step 1 — Distance to summit: x_tp = g₁L/(g₁ − g₂) = (0.03 × 200)/(0.03 − (−0.02)) = 6.00/0.05 = 120 m from BVC Step 2 — Check: 0 ≤ 120 ≤ 200 ✓ Step 3 — Station of summit: Sta_tp = 5+000 + 120 = Sta 5+120 Step 4 — r: r = (−0.02 − 0.03)/200 = −0.00025 /m Step 5 — Elevation of summit: y_tp = 100.00 + 0.03(120) + (−0.000125)(120²) = 100.00 + 3.60 − 0.000125(14400) = 100.00 + 3.60 − 1.80 = 101.80 m

The low point falls 116.36 m from the BVC. The negative incoming grade (−4%) is steeper than the outgoing (+1.5%), so the low point is past the midpoint of the curve — consistent with the formula.

Scenario

Sag curve: g₁ = −4%, g₂ = +1.5%, L = 160 m, BVC at Sta 2+400, elev_BVC = 92.00 m. Find the station and elevation of the low point.

Solution

Step 1 — Distance to low point: x_tp = g₁L/(g₁ − g₂) = (−0.04 × 160)/(−0.04 − 0.015) = (−6.4)/(−0.055) = 116.36 m from BVC Step 2 — Check: 0 ≤ 116.36 ≤ 160 ✓ Step 3 — Station: Sta_tp = 2+400 + 116.36 = Sta 2+516.36 Step 4 — r: r = (0.015 − (−0.04))/160 = 0.055/160 = +0.0003438 /m Step 5 — Elevation: y_tp = 92.00 + (−0.04)(116.36) + (0.0003438/2)(116.36²) = 92.00 − 4.6544 + (0.0001719)(13539.65) = 92.00 − 4.6544 + 2.3284 = 89.674 m ≈ 89.67 m

Applications

Locating the highest point on a crest for drainage (water does not pond there)
Locating the low point of a sag for catch basin placement
Checking sight distance at the summit of a crest curve
Computing maximum road elevation for clearance analysis

Misconceptions

Using |g₁| − |g₂| instead of the signed difference g₁ − g₂ in the denominator.
Measuring x from the PVI (midpoint) instead of the BVC.
Forgetting to check that 0 ≤ x_tp ≤ L — a turning point outside this range does not exist on the curve.

Related Concepts

Derivative of the parabola equals zero at the turning point
Drainage design at sag low points
Sight distance over crest curves

Common Exam Questions

Example

g₁ = +5%, g₂ = −1%, L = 300 m, BVC Sta 3+000, elev 45.00 m. Summit: x = 0.05×300/(0.05+0.01) = 15/0.06 = 250 m; Sta 3+250; elev = 45 + 0.05(250) + (−0.00010/2)(250²) ... board answer ≈ 51.25 m.

Approach

Apply x_tp = g₁L/(g₁−g₂); convert to station; substitute into parabolic elevation formula.

Question Type

Find station and elevation of high/low point

Example

g₁ = +3%, g₂ = +1%, L = 200 m: x_tp = 0.03×200/(0.03−0.01) = 6/0.02 = 300 m > L = 200 m. No high point within the curve — the curve is entirely ascending.

Approach

Compute x_tp; if x_tp > L or x_tp < 0, state 'no turning point within the curve' and the grade changes monotonically.

Question Type

Verify if turning point exists within the curve

Key Points To Remember

x_tp = g₁L/(g₁ − g₂) — memorise this; it is the most-tested formula.
Both g₁ and g₂ must have OPPOSITE signs for the turning point to fall within the curve.
Always verify 0 ≤ x_tp ≤ L before computing the turning point elevation.
Sta_tp = Sta_BVC + x_tp.
The turning point elevation uses the same parabolic formula — no special formula needed.
On a crest, the turning point is the HIGHEST elevation on the curve (drainage concern).
On a sag, it is the LOWEST elevation (drainage and sight-distance concern).

Vertical Offset from the Tangent

The vertical distance between the parabola and the back tangent at any point x is called the VERTICAL OFFSET (or tangent offset): offset(x) = y_parabola(x) − y_tangent(x) = [elev_BVC + g₁x + (r/2)x²] − [elev_BVC + g₁x] = (r/2)x² = [(g₂ − g₁)/(2L)] · x² This confirms the parabola property: offsets from a tangent vary as the SQUARE of the distance from the tangent point. IMPORTANT SIGN: • Crest: r < 0 → offset is negative → parabola is BELOW the tangent. • Sag: r > 0 → offset is positive → parabola is ABOVE the tangent. MAXIMUM OFFSET (at mid-curve, x = L/2): offset_max = (r/2)(L/2)² = [(g₂−g₁)/(2L)] × (L²/4) = (g₂−g₁)L/8 In absolute value: |offset_max| = |A| × L / 8 (A in decimal) or = |A%| × L / 800 (A% in percent) This maximum offset also equals the vertical distance from the PVI to the midpoint of the curve (sometimes called the 'throw' of the curve), expressed as: m = (g₁−g₂)L/8 for a crest (positive value, since g₁ > g₂)

Examples

The two methods agree, confirming both formulas. The '−1.25 m' means the road surface at mid-curve is 1.25 m lower than the straight-line projection of the back tangent — the curve dips below the tangent, as expected for a crest.

Scenario

For the crest curve (g₁ = +3%, g₂ = −2%, L = 200 m), find the maximum vertical offset from the tangent and the elevation at mid-curve.

Solution

Step 1 — Maximum offset: offset_max = (g₂−g₁)L/8 = (−0.02−0.03)(200)/8 = (−0.05)(200)/8 = −10/8 = −1.25 m Magnitude: 1.25 m; the parabola is 1.25 m BELOW the tangent at mid-curve. Step 2 — Elevation at mid-curve (x = 100 m) from the parabolic formula: r = −0.00025 /m y(100) = 100.00 + 0.03(100) + (−0.000125)(100²) = 100.00 + 3.00 − 1.25 = 101.75 m Step 3 — Cross-check via PVI: elev_PVI = 100.00 + 0.03 × 100 = 103.00 m Mid-curve elev = elev_PVI + offset_max = 103.00 + (−1.25) = 101.75 m ✓

Applications

Checking clearance under bridges at sag curves
Computing the 'throw' when setting out the curve by offset method in the field
Verifying as-built levels vs. design profile
Earthwork quantities estimation

Misconceptions

Confusing the direction of the offset — crest curves sit BELOW the tangent, sag curves ABOVE.
Computing offset at a point other than x = L/2 and calling it the maximum offset.
Using the formula m = (g₁+g₂)L/8 (incorrect) instead of m = (g₂−g₁)L/8.

Related Concepts

Mid-ordinate of a curve
Long chord vs. curve
Parabolic approximation for short curves

Common Exam Questions

Example

g₁ = −2%, g₂ = +4%, L = 400 m, elev_PVI = 30.00 m. offset_max = (+0.06)(400)/8 = +3.00 m (sag, above tangent). Mid-curve elev = 30.00 + 3.00 = 33.00 m.

Approach

Compute offset_max = (g₂−g₁)L/8; then mid-curve elevation = elev_PVI + offset_max.

Question Type

Maximum offset / mid-curve elevation

Key Points To Remember

offset(x) = [(g₂ − g₁)/(2L)] × x² — parabolic, proportional to x².
Maximum offset at x = L/2 (mid-curve): m = (g₂−g₁)L/8.
Negative offset means the curve is below the tangent (crest).
Positive offset means the curve is above the tangent (sag).
The maximum offset equals the vertical distance from the PVI to the curve midpoint.
Useful shortcut: elevation at mid-curve = elev_PVI + offset_max (with correct sign).

Sight Distance and Minimum Curve Length

Road design codes (AASHTO, DPWH Highway Design Guidelines) require that vertical curves be long enough to provide adequate STOPPING SIGHT DISTANCE (SSD) — the distance a driver needs to see an object on the road and stop safely. For CREST curves, the sight line passes over the hump. The minimum curve length L is related to the stopping sight distance S and the algebraic grade difference A = |g₁ − g₂| (in %). Two cases arise: CASE 1 — S < L (sight distance is less than the curve length): L = A·S² / [200(√h₁ + √h₂)²] CASE 2 — S > L (sight distance exceeds the curve length): L = 2S − 200(√h₁ + √h₂)² / A With: h₁ = driver eye height (typically 1.08 m per DPWH/AASHTO) h₂ = object height (typically 0.60 m for a passenger car tail-light) For h₁ = 1.08 m, h₂ = 0.60 m: (√h₁ + √h₂)² = (√1.08 + √0.60)² ≈ (1.0392 + 0.7746)² ≈ (1.8138)² ≈ 3.29 m So the simplified crest formula (Case 1) becomes: L = A·S² / 658 (with L and S in metres, A in %) For SAG curves, sight distance is limited by the reach of headlights at night: Case 1 (S < L): L = A·S² / [200(H + S·tan β)] where H = headlight height ≈ 0.60 m, β = headlight upward angle ≈ 1° Simplified: L ≈ A·S² / 120 (common approximation used in Philippine board reviews) PRACTICAL RULE: If the problem does not specify which case, assume S < L (Case 1) first, compute L, then verify S < L. If not satisfied, use Case 2.

Examples

In this case S > L, so Case 2 governs. Many board-exam problems accept either approach; always verify your assumption to avoid errors.

Scenario

Find the minimum length of a crest curve for SSD = 120 m, g₁ = +3%, g₂ = −2% (use h₁ = 1.08 m, h₂ = 0.60 m).

Solution

Step 1 — A%: A% = |g₂ − g₁| × 100 = |−2 − 3| = 5% Step 2 — (√h₁ + √h₂)²: = (√1.08 + √0.60)² = (1.03923 + 0.77460)² = (1.81383)² = 3.290 m Step 3 — L (assuming S < L): L = A·S² / [200(√h₁+√h₂)²] = 5 × (120)² / (200 × 3.290) = 5 × 14400 / 658 = 72000 / 658 = 109.4 m Step 4 — Check: S = 120 m > L = 109.4 m → assumption S < L is VIOLATED. Step 5 — Use Case 2: L = 2S − 200(√h₁+√h₂)²/A = 2(120) − 200(3.290)/5 = 240 − 131.6 = 108.4 m Step 6 — Verify: L = 108.4 m < S = 120 m ✓ (consistent with S > L assumption) Minimum curve length = 108.4 m ≈ 109 m (round up to next design increment).

The board exam often uses the simplified formula L = AS²/658 for crests. Know when to check the S vs. L condition in the longer, computation-heavy items.

Scenario

(Simplified — common board-exam shortcut) A crest curve, A% = 5%, SSD = 120 m, using L = AS²/658.

Solution

L = 5 × (120²)/658 = 72000/658 = 109.4 m Since L = 109.4 m < S = 120 m, the Case 1 assumption is violated. Many Philippine board-exam problems use the simplified formula and accept L ≈ 109 m as the answer without checking the assumption. Be prepared for both approaches.

Applications

DPWH highway design — minimum vertical curve lengths for national roads
Design speed selection and its influence on minimum L
Safety audit of existing vertical curves on Philippine highways
Bridge approach design requiring adequate sight distance

Misconceptions

Using A as a decimal instead of percent in sight-distance formulas — the formula is calibrated for A in %.
Skipping the Case 1 vs. Case 2 check — this is a frequent error source.
Using h₁ = 1.20 m (truck height) instead of h₁ = 1.08 m for passenger car design.

Related Concepts

Design speed and stopping sight distance tables (DPWH HDG)
Passing sight distance (PSD) for two-lane highways
Horizontal curve sight distance
Sag curve headlight sight distance

Common Exam Questions

Example

A% = 4%, SSD = 150 m, h₁ = 1.08 m, h₂ = 0.60 m. L = 4×150²/658 = 136.8 m. Check: L = 136.8 m < S = 150 m — Case 2: L = 300 − 200(3.29)/4 = 300 − 164.5 = 135.5 m ≈ 136 m.

Approach

Compute A%; apply L = AS²/[200(√h₁+√h₂)²]; check S vs. L; use Case 2 if needed.

Question Type

Minimum crest curve length for given SSD

Key Points To Remember

A% = |g₂ − g₁| × 100 — use percent in sight-distance formulas.
Crest: L = A·S²/[200(√h₁+√h₂)²]; assume S < L unless told otherwise.
Sag: L ≈ A·S²/120 (simplified Philippine board-exam form).
Check assumption: if computed L ≤ S, use Case 1; if L > S, re-examine.
Standard heights: h₁ = 1.08 m (eye), h₂ = 0.60 m (object) for passenger cars.
Longer curves improve safety but increase earthwork cost — a design trade-off.

Practice Problems

Notice the summit falls exactly at x = 250 m (the mid-curve in this case is at x = 200 m, but the summit is at x = 250 m because g₁ is steeper). The maximum offset from the tangent (−4.00 m) is always at mid-curve x = L/2, NOT at the summit. These are two different points — a common confusion in the board exam.

Problem

PROBLEM 1 (Board-style). A symmetrical parabolic curve is to connect two grades: g₁ = +5% and g₂ = −3%. The BVC is at Sta 10+000 with elevation 210.50 m and the curve length L = 400 m. (a) Find the station and elevation of the summit. (b) Find the elevation at Sta 10+250. (c) Find the maximum offset from the back tangent.

Solution

(a) Summit location: r = (g₂−g₁)/L = (−0.03−0.05)/400 = −0.08/400 = −0.0002 /m x_tp = g₁L/(g₁−g₂) = 0.05×400/(0.05−(−0.03)) = 20/0.08 = 250 m Check: 0 ≤ 250 ≤ 400 ✓ Sta_summit = 10+000 + 250 = Sta 10+250 y_summit = 210.50 + 0.05(250) + (−0.0001)(250²) = 210.50 + 12.50 − 0.0001(62500) = 210.50 + 12.50 − 6.25 = 216.75 m (b) Elevation at Sta 10+250: x = 10+250 − 10+000 = 250 m This IS the summit → elev = 216.75 m (same as above) Alternative for non-summit point (Sta 10+100, x = 100 m): y(100) = 210.50 + 0.05(100) + (−0.0001)(10000) = 210.50 + 5.00 − 1.00 = 214.50 m (c) Maximum offset (at x = L/2 = 200 m): offset_max = (g₂−g₁)L/8 = (−0.08)(400)/8 = −4.00 m The parabola is 4.00 m BELOW the back tangent at mid-curve. Cross-check: elev_PVI = 210.50 + 0.05×200 = 220.50 m Mid-curve elev = y(200) = 210.50 + 0.05(200) + (−0.0001)(40000) = 210.50 + 10.00 − 4.00 = 216.50 m offset = 216.50 − 220.50 = −4.00 m ✓

The low point at Sta 4+940 is past the midpoint (which would be at x = 120 m, Sta 4+920) because g₁ = −3.5% is steeper than g₂ = +2.5%. A heavier incoming downgrade pushes the low point farther along the curve — this matches the formula result and should be physically intuitive.

Problem

PROBLEM 2 (Board-style). A sag vertical curve joins g₁ = −3.5% and g₂ = +2.5% with L = 240 m. The BVC is at Sta 4+800, elevation 55.20 m. (a) Find the station and elevation of the low point. (b) Find the elevation at Sta 5+020.

Solution

(a) Low point: x_tp = g₁L/(g₁−g₂) = (−0.035×240)/(−0.035−0.025) = (−8.40)/(−0.060) = 140 m from BVC Check: 0 ≤ 140 ≤ 240 ✓ Sta_low = 4+800 + 140 = Sta 4+940 r = (g₂−g₁)/L = (0.025−(−0.035))/240 = 0.060/240 = +0.00025 /m y_low = 55.20 + (−0.035)(140) + (0.00025/2)(140²) = 55.20 − 4.90 + (0.000125)(19600) = 55.20 − 4.90 + 2.45 = 52.75 m (b) Elevation at Sta 5+020: x = 5+020 − 4+800 = 220 m Check: 220 ≤ 240 ✓ (within curve) y(220) = 55.20 + (−0.035)(220) + (0.000125)(220²) = 55.20 − 7.70 + 0.000125(48400) = 55.20 − 7.70 + 6.05 = 53.55 m

Always round UP to the next practical design increment; never round down, as this would reduce safety. In the board exam, carry at least 2 decimal places for intermediate calculations, but the final answer is usually an integer or 1 decimal place.

Problem

PROBLEM 3 (Board-style). Determine the minimum length of a crest vertical curve if the algebraic difference of grades is A = 6% and the stopping sight distance is SSD = 100 m. Use h₁ = 1.08 m and h₂ = 0.60 m.

Solution

Step 1 — Compute (√h₁ + √h₂)²: √1.08 = 1.03923 m^0.5 √0.60 = 0.77460 m^0.5 Sum = 1.81383 Sum² = 3.290 m Step 2 — Assume S < L (Case 1): L = A·S²/[200·(√h₁+√h₂)²] = 6×(100)²/[200×3.290] = 60000/658.0 = 91.19 m Step 3 — Check: S = 100 m > L = 91.19 m → Case 1 assumption is VIOLATED (S > L). Step 4 — Use Case 2: L = 2S − 200(√h₁+√h₂)²/A = 2(100) − 200(3.290)/6 = 200 − 109.67 = 90.33 m Step 5 — Verify: L = 90.33 m < S = 100 m ✓ Minimum curve length ≈ 90.3 m → design value: use L = 91 m or round to the next 10-m increment: L = 100 m.

This is a very common board-exam trap: both grades have the same sign. The turning-point formula gives x_tp > L, confirming no true summit within the curve. Always perform this bounds check before computing the high/low point elevation.

Problem

PROBLEM 4 (Board-style). Two grades g₁ = +4% and g₂ = +1% are connected by a 300-m parabolic vertical curve. BVC at Sta 7+200, elevation 88.00 m. Does a crest or high point occur within the curve? If so, find its station and elevation. If not, explain why.

Solution

Step 1 — Check curve type: g₁ = +4%, g₂ = +1%. Both positive → the road is always ascending, only the slope is decreasing. This is a CREST curve but without a true summit (no grade reversal). Step 2 — Compute x_tp: x_tp = g₁L/(g₁−g₂) = (0.04×300)/(0.04−0.01) = 12/0.03 = 400 m Step 3 — Check bounds: x_tp = 400 m > L = 300 m → the theoretical turning point is BEYOND the curve. Conclusion: No high point exists within the curve. The curve rises throughout its entire length. The highest elevation is at the EVC (x = L = 300 m). EVC elevation: r = (0.01−0.04)/300 = −0.01/100 = −0.0001/m (wait: −0.03/300 = −0.0001) Actually: r = (g₂−g₁)/L = (0.01−0.04)/300 = −0.03/300 = −0.0001 /m y(300) = 88.00 + 0.04(300) + (−0.0001/2)(300²) = 88.00 + 12.00 + (−0.00005)(90000) = 88.00 + 12.00 − 4.50 = 95.50 m

Note that the summit (x = 312.5 m) and the mid-curve (x = 250 m) are different points. The maximum offset is always at mid-curve; the maximum elevation is at the summit. Both appear in board exams — do not confuse them.

Problem

PROBLEM 5 (Board-style — multi-part). A 500-m parabolic curve has BVC at Sta 12+000, elevation 300.00 m; g₁ = +2.5%; g₂ = −1.5%. (a) Find r. (b) Find the summit station and elevation. (c) Find the elevation at x = 300 m. (d) Find the elevation at the EVC. (e) Find the maximum vertical offset from the tangent.

Solution

(a) r = (g₂−g₁)/L = (−0.015−0.025)/500 = −0.04/500 = −0.00008 /m (b) Summit: x_tp = g₁L/(g₁−g₂) = (0.025×500)/(0.025−(−0.015)) = 12.5/0.04 = 312.5 m Sta_summit = 12+000 + 312.5 = Sta 12+312.5 y_summit = 300.00 + 0.025(312.5) + (−0.00004)(312.5²) = 300.00 + 7.8125 − 0.00004(97656.25) = 300.00 + 7.8125 − 3.9063 = 303.906 m ≈ 303.91 m (c) x = 300 m: y(300) = 300.00 + 0.025(300) + (−0.00004)(300²) = 300.00 + 7.50 − 0.00004(90000) = 300.00 + 7.50 − 3.60 = 303.90 m (d) EVC (x = L = 500 m): y(500) = 300.00 + 0.025(500) + (−0.00004)(500²) = 300.00 + 12.50 − 0.00004(250000) = 300.00 + 12.50 − 10.00 = 302.50 m Cross-check: elev_EVC = elev_BVC + g₁L + r/2·L² = same calculation ✓ (e) Maximum offset (x = L/2 = 250 m): offset_max = (g₂−g₁)L/8 = (−0.04×500)/8 = −20/8 = −2.50 m (2.50 m below the back tangent at mid-curve) Cross-check: tangent elev at x=250: 300+0.025×250 = 306.25 m parabola at x=250: y = 300+0.025(250)+(−0.00004)(62500) = 300+6.25−2.50 = 303.75 m offset = 303.75 − 306.25 = −2.50 m ✓

Exam Preparation Tips

MEMORISE THREE MASTER FORMULAS: (1) r = (g₂−g₁)/L, (2) y(x) = elev_BVC + g₁x + (r/2)x², (3) x_tp = g₁L/(g₁−g₂). These three solve 90% of board-exam vertical curve items.
ALWAYS CONVERT GRADES TO DECIMALS before substituting into formulas. Writing +3% as 0.03 and −2% as −0.02 is non-negotiable.
CHECK THE TURNING-POINT BOUNDS: After computing x_tp, verify 0 ≤ x_tp ≤ L. Many exam problems deliberately use grades of the same sign to trap students.
DISTINGUISH MID-CURVE FROM SUMMIT/LOW POINT: Maximum offset always at x = L/2; turning point at x = g₁L/(g₁−g₂). These are rarely the same station.
SIGN DISCIPLINE: For a crest, r < 0 → the parabola is below the tangent. For a sag, r > 0 → the parabola is above the tangent. Draw a sketch to confirm the physics.
STATION ARITHMETIC: Keep station notation consistent (e.g., Sta 5+120 means 5120 m from datum). Compute x = Sta_point − Sta_BVC; never use station numerals directly in the formula.
FOR SIGHT-DISTANCE ITEMS: Use A in percent (not decimal). Remember L = AS²/658 as the quick crest formula (with h₁=1.08m, h₂=0.60m). Always verify the S vs. L assumption.
TWO-MINUTE CHECK METHOD: After getting an elevation, verify it makes physical sense. For a crest, the summit should be higher than both the BVC and EVC. For a sag, the low point should be lower than both ends.
REVIEW PAST BOARD QUESTIONS: Vertical curve problems appear in virtually every CE board exam cycle. Focus on La Putt (Surveying textbook) chapters and Gillesania review books for Filipino board-exam-specific examples.
PRACTICE UNIT CONSISTENCY: r has units of per metre (/m); x is in metres; (r/2)x² comes out in metres — always verify dimensional homogeneity as a quick check.

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In summary

Vertical parabolic curves are among the most consistently tested topics in the PRC Civil Engineer Licensure Examination Surveying component. The entire subject reduces to three master equations: r = (g₂−g₁)/L, the parabolic elevation formula y(x) = elev_BVC + g₁x + (r/2)x², and the turning-point formula x_tp = g₁L/(g₁−g₂). Rigorous sign discipline — keeping grades as signed decimals and interpreting r as negative for crests and positive for sags — prevents the majority of errors. The mid-curve maximum offset formula m = (g₂−g₁)L/8 and the crest sight-distance formula L = AS²/658 (simplified) complete the examination toolkit. Always verify the turning-point bounds (0 ≤ x_tp ≤ L) and the Case 1 vs. Case 2 condition in sight-distance problems — these two checks separate high scorers from average examinees. With consistent practice on board-style numerical problems using SI units and the step-by-step flowcharts provided in this chapter, a CE reviewee can confidently approach any vertical curve item in the licensure examination.

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