Vertical (Parabolic) Curves - Exam Answer Templates
Proper answer writing is the single most controllable factor in your PRC board exam score. In Surveying problems on Vertical Parabolic Curves, examiners award marks for correct formula identification, proper sign convention, logical step-by-step solutions, and accurate final answers with correct units. A student who understands the concept but writes a disorganized solution will consistently lose 1–2 marks per problem — enough to push a passing grade into failure. These templates show you the exact structure, key phrases, and computational flow that earn full marks on every question type, from 1-mark recall to 5-mark design problems. Study the model answers as blueprints: replicate their format, not just their content.
Answer templates
Define a vertical parabolic curve and state why the parabola is preferred over other curve types in road design.
Examiner tip
Even for a 1-mark question, one mathematical expression (r = (g₂ − g₁)/L) shows mastery and guards against partial deductions.
Model answer
A vertical parabolic curve is a second-degree parabola used in highway profile design to connect two intersecting grades, providing a constant rate of grade change (r = (g₂ − g₁)/L). The parabola is preferred because it produces a uniform rate of change of slope, ensuring riding comfort and consistent sight distances along the curve.
Answer structure
- Line 1: State definition — second-degree parabola connecting two grades [½ mark]
- Line 2: State key property — constant rate of grade change, and practical reason (comfort/sight distance) [½ mark]
Scoring breakdown
Criteria
Correct definition identifying it as a parabola with constant grade-change rate, plus one practical reason (riding comfort or sight distance)
Common mark deductions
- Saying 'circular curve' instead of parabola — zero marks for concept error
- Omitting the reason why parabola is used — loses ½ mark
- Writing 'grade change is uniform' without defining what that means mathematically
Key phrases to include
- second-degree parabola
- constant rate of grade change
- g₂ − g₁ / L
- riding comfort
- sight distance
Distinguish between a crest (summit) curve and a sag curve. Give the grade condition for each.
Examiner tip
Always pair the grade condition with the controlling design criterion. A two-part answer that states both earns full marks consistently.
Model answer
A CREST (summit) curve occurs when g₁ > g₂ (the incoming grade is steeper upward than the outgoing grade), forming a hump — sight distance is the controlling design criterion. A SAG curve occurs when g₁ < g₂ (the road dips then rises), forming a valley — headlight sight distance and riding comfort govern its length.
Answer structure
- Part A: Crest — g₁ > g₂, hump shape, governed by stopping sight distance [½ mark]
- Part B: Sag — g₁ < g₂, valley shape, governed by headlight/comfort [½ mark]
Scoring breakdown
Criteria
Both types correctly identified with their grade conditions (g₁ > g₂ for crest; g₁ < g₂ for sag) and at least one design constraint each
Common mark deductions
- Reversing the grade conditions for crest and sag
- Omitting the design criterion that governs each type
- Not using grade notation (g₁, g₂) — vague descriptions lose marks
Key phrases to include
- g₁ > g₂
- g₁ < g₂
- crest
- sag
- stopping sight distance
- headlight
State the formula for the rate of grade change r in a parabolic vertical curve and identify each variable.
Examiner tip
Variable identification is worth half the marks on formula-recall questions. Never skip it, even under time pressure.
Model answer
r = (g₂ − g₁) / L
Where:
- g₁ = incoming (back) grade (expressed as a decimal, positive if ascending)
- g₂ = outgoing (forward) grade (expressed as a decimal, positive if ascending)
- L = length of the vertical curve (m)
- r = rate of grade change per unit length (m/m per m, or simply m⁻¹)
Answer structure
- Line 1: Write the formula r = (g₂ − g₁)/L [½ mark]
- Lines 2–5: Define each variable with correct sign convention [½ mark]
Scoring breakdown
Criteria
Correct formula with all four variables defined including sign convention for grades
Common mark deductions
- Writing r = (g₁ − g₂)/L (reversed) — wrong formula, lose entire mark
- Not defining variables — a formula alone earns only ½ mark
- Giving grades in percent without converting to decimal
Key phrases to include
- r = (g₂ − g₁)/L
- incoming grade
- outgoing grade
- decimal form
- positive if ascending
A parabolic vertical curve has an incoming grade of g₁ = +4% and an outgoing grade of g₂ = −2%. The curve length is L = 180 m. Compute: (a) the rate of grade change r, and (b) identify the type of curve.
Examiner tip
The grade comparison g₁ vs g₂ must be explicitly written. Examiners cannot give marks for conclusions not stated on paper.
Model answer
Given:
g₁ = +4% = +0.04 (ascending)
g₂ = −2% = −0.02 (descending)
L = 180 m
(a) Rate of grade change:
r = (g₂ − g₁) / L
r = (−0.02 − 0.04) / 180
r = −0.06 / 180
r = −0.000333 m⁻¹ (or −0.000333 per metre)
(b) Type of curve:
Since g₁ = +0.04 > g₂ = −0.02, this is a CREST (summit) curve.
The negative r confirms the grade decreases from PC to PT.
∴ r = −3.33 × 10⁻⁴ m⁻¹ and the curve is a CREST curve.
Answer structure
- Block 1: List all given values in decimal form with units [prerequisite — no marks if skipped]
- Block 2 (a): Write formula r = (g₂ − g₁)/L, substitute, compute r with correct sign [1 mark]
- Block 3 (b): State the grade comparison g₁ vs g₂ and name the curve type [1 mark]
Scoring breakdown
Criteria
Correct computation of r = −0.000333 m⁻¹ with proper negative sign
Criteria
Correct identification as CREST curve with grade comparison stated
Common mark deductions
- Using percent grades directly without converting to decimals (e.g., computing with 4 and −2 instead of 0.04 and −0.02)
- Dropping the negative sign from r — loses the (a) mark
- Calling it a 'summit' without stating the grade comparison — loses the (b) mark
Key phrases to include
- r = (g₂ − g₁)/L
- g₁ > g₂
- CREST curve
- negative r
- decimal conversion
For the crest curve in T4 (g₁ = +4%, g₂ = −2%, L = 180 m), the elevation of the PC is 210.50 m. Find the elevation at a point 90 m from the PC.
Examiner tip
The ½ in (r/2)x² is the most common arithmetic mistake in PRC board exams. Highlight it in your formula.
Model answer
Given:
g₁ = +0.04, g₂ = −0.02, L = 180 m
elev_PC = 210.50 m
x = 90 m (from PC)
Step 1 — Compute r:
r = (g₂ − g₁)/L = (−0.02 − 0.04)/180 = −0.000333 m⁻¹
Step 2 — Apply the elevation equation:
y = elev_PC + g₁x + (r/2)x²
y = 210.50 + (0.04)(90) + (−0.000333/2)(90)²
y = 210.50 + 3.60 + (−0.0001667)(8100)
y = 210.50 + 3.60 − 1.350
y = 212.75 m
∴ Elevation at x = 90 m from PC = 212.75 m
Answer structure
- Block 1: List given data, compute r [background, no separate mark but required]
- Block 2: Write y = elev_PC + g₁x + (r/2)x² [formula statement — ½ mark]
- Block 3: Substitute all values correctly [½ mark]
- Block 4: Correct final elevation with unit [1 mark]
Scoring breakdown
Criteria
Correct formula stated and values substituted with correct signs
Criteria
Correct final elevation y = 212.75 m
Common mark deductions
- Writing y = elev_PC + g₁x + rx² (missing the ½ factor) — arithmetic error leading to wrong answer
- Using x = 90 m but measuring from PI instead of PC
- Not showing intermediate steps — loses process marks even if final answer is correct
Key phrases to include
- y = elev_PC + g₁x + (r/2)x²
- r/2
- x² term
- 210.50 + 3.60 − 1.350
Derive the formula for the location of the high point (or low point) of a parabolic vertical curve from the general elevation equation, and compute the distance from the PC to the summit for: g₁ = +3%, g₂ = −2%, L = 200 m.
Examiner tip
When a question says 'derive', it explicitly requires calculus steps. Jumping to the formula earns at most 1 of 3 marks. Always differentiate.
Model answer
DERIVATION:
The elevation at distance x from the PC is:
y = elev_PC + g₁x + (r/2)x² … (1)
The slope (grade) at any x is dy/dx:
dy/dx = g₁ + rx … (2)
At the high/low point, dy/dx = 0:
g₁ + rx = 0
x = −g₁/r
Substituting r = (g₂ − g₁)/L:
x_turning = −g₁ / [(g₂ − g₁)/L] = g₁L / (g₁ − g₂) … (Formula)
NUMERICAL APPLICATION:
Given: g₁ = +0.03, g₂ = −0.02, L = 200 m
x = g₁L / (g₁ − g₂)
x = (0.03)(200) / (0.03 − (−0.02))
x = 6 / 0.05
x = 120 m from PC
VERIFICATION CHECK:
0 ≤ 120 ≤ 200 m ✓ (turning point lies within the curve)
∴ The summit is located 120 m from the PC.
Answer structure
- Part 1 (Derivation): State y equation → differentiate → set to zero → solve for x [1.5 marks]
- Part 2 (Formula): Write x = g₁L/(g₁ − g₂) clearly labeled [0.5 mark]
- Part 3 (Numerical): Substitute values and compute x = 120 m [0.5 mark]
- Part 4 (Verification): Check 0 ≤ x ≤ L [0.5 mark]
Scoring breakdown
Criteria
Correct differentiation dy/dx = g₁ + rx and setting equal to zero
Criteria
Correct derivation of formula x = g₁L/(g₁ − g₂) or x = −g₁/r
Criteria
Correct numerical answer x = 120 m with verification check stated
Common mark deductions
- Skipping the derivation and jumping straight to the formula — loses 2 marks for a 'derive' question
- Not verifying that x falls within the curve length
- Sign error: using g₁ − g₂ = 0.03 − 0.02 = 0.01 instead of 0.03 − (−0.02) = 0.05
Key phrases to include
- dy/dx = g₁ + rx
- dy/dx = 0
- x = −g₁/r
- x = g₁L/(g₁ − g₂)
- 0 ≤ x ≤ L
For the crest curve: g₁ = +3%, g₂ = −2%, L = 200 m, elev_PC = 100.00 m. Compute (a) the elevation of the summit, and (b) the maximum vertical offset from the back tangent.
Examiner tip
Maximum offset is ALWAYS at the mid-point (x = L/2), not at the summit. Many students confuse these two points — the examiner knows this and tests it deliberately.
Model answer
Given:
g₁ = +0.03, g₂ = −0.02, L = 200 m, elev_PC = 100.00 m
From T6: x_summit = 120 m
(a) ELEVATION OF SUMMIT:
r = (g₂ − g₁)/L = (−0.02 − 0.03)/200 = −0.00025 m⁻¹
y = elev_PC + g₁x + (r/2)x²
y = 100.00 + (0.03)(120) + (−0.00025/2)(120)²
y = 100.00 + 3.60 + (−0.000125)(14400)
y = 100.00 + 3.60 − 1.80
y_summit = 101.80 m ← ANSWER (a)
(b) MAXIMUM VERTICAL OFFSET (mid-curve, x = L/2 = 100 m from PC):
Offset formula: Δ = [(g₂ − g₁)/(2L)] × x²
At x = L/2 = 100 m:
Δ = (−0.05 / 400)(100)² = (−0.000125)(10000) = −1.25 m
|Δ_max| = 1.25 m BELOW the back tangent ← ANSWER (b)
Alternate direct formula: Δ_max = (g₂ − g₁)L/8
Δ_max = (−0.05)(200)/8 = −1.25 m ✓
Answer structure
- Block 1: Restate r and x_summit from prior work [no new marks, but required]
- Block 2 (a): Apply y equation at x = 120 m, show all arithmetic, state summit elev = 101.80 m [1.5 marks]
- Block 3 (b): State offset formula, substitute x = L/2 = 100 m, compute −1.25 m and state direction [1.5 marks]
Scoring breakdown
Criteria
Correct r computed and elevation equation applied at x = 120 m
Criteria
Correct summit elevation = 101.80 m
Criteria
Correct maximum offset = 1.25 m stated as 'below the back tangent' with correct formula shown
Common mark deductions
- Computing offset at x = 120 m (the summit) instead of x = L/2 = 100 m (mid-curve)
- Stating '1.25 m offset' without specifying direction (below or above tangent)
- Using Δ = (g₁ − g₂)L/8 with wrong sign, getting +1.25 and calling it 'above tangent'
Key phrases to include
- r = −0.00025
- 101.80 m
- Δ_max = (g₂ − g₁)L/8
- −1.25 m
- below the back tangent
- x = L/2
A sag vertical curve joins grades g₁ = −4% and g₂ = +1.5% over a length of L = 160 m. The elevation of the PC is 85.20 m. Find: (a) the location of the low point from the PC, and (b) the elevation of the low point.
Examiner tip
Sag problems have negative g₁ values. Write it as −0.04 and keep the negative through every formula. The formula x = g₁L/(g₁ − g₂) naturally gives a positive x because both numerator and denominator are negative for a sag curve.
Model answer
Given:
g₁ = −0.04, g₂ = +0.015, L = 160 m, elev_PC = 85.20 m
Curve type check: g₁ < g₂ → SAG curve ✓
(a) LOCATION OF LOW POINT:
x = g₁L / (g₁ − g₂)
x = (−0.04)(160) / (−0.04 − 0.015)
x = −6.40 / (−0.055)
x = 116.36 m from PC
Verification: 0 ≤ 116.36 ≤ 160 ✓ (low point within curve)
(b) ELEVATION OF LOW POINT:
r = (g₂ − g₁)/L = (0.015 − (−0.04))/160 = 0.055/160 = +0.0003438 m⁻¹
y = elev_PC + g₁x + (r/2)x²
y = 85.20 + (−0.04)(116.36) + (0.0003438/2)(116.36)²
y = 85.20 − 4.6544 + (0.0001719)(13539.6)
y = 85.20 − 4.6544 + 2.3274
y = 82.873 m ≈ 82.87 m
∴ Low point at x = 116.36 m from PC; elevation = 82.87 m
Answer structure
- Block 1: Identify curve type (SAG), state g₁ and g₂ as decimals [prerequisite]
- Block 2 (a): Apply x = g₁L/(g₁ − g₂), show arithmetic, verify within L [1 mark]
- Block 3 (b): Compute r, apply elevation equation at x = 116.36 m [1 mark]
- Block 4: Correct final elevation 82.87 m [1 mark]
Scoring breakdown
Criteria
Correct low-point location x = 116.36 m with verification check
Criteria
Correct r = +0.0003438 m⁻¹ and proper substitution in elevation equation
Criteria
Correct low-point elevation = 82.87 m
Common mark deductions
- Using g₁ = +0.04 (dropping the negative sign) — everything downstream is wrong
- Not checking that x falls within the curve length
- Using r = (g₁ − g₂)/L instead of (g₂ − g₁)/L
Key phrases to include
- SAG curve
- g₁ < g₂
- x = g₁L/(g₁ − g₂)
- 116.36 m
- r = +0.0003438
- 82.87 m
A 300-m crest vertical curve has g₁ = +2% and g₂ = −3%. The elevation at PC (Sta. 10+000) is 150.00 m. Determine the station and elevation of the summit, and the elevation at Sta. 10+150 (mid-curve).
Examiner tip
On 5-mark problems, organize your work in clearly labeled STEPS. Examiners are instructed to award marks per step — a well-labeled solution earns partial marks even if the final answer is wrong.
Model answer
GIVEN DATA:
g₁ = +0.02, g₂ = −0.03, L = 300 m
Sta. PC = 10+000, elev_PC = 150.00 m
Sta. PT = 10+000 + 300 = 10+300
Curve type: g₁ = 0.02 > g₂ = −0.03 → CREST curve ✓
STEP 1 — Rate of grade change:
r = (g₂ − g₁)/L = (−0.03 − 0.02)/300 = −0.05/300 = −0.0001667 m⁻¹
STEP 2 — Location of summit (x from PC):
x = g₁L/(g₁ − g₂) = (0.02)(300)/(0.02 − (−0.03))
x = 6.0/0.05 = 120 m from PC
Station of summit = 10+000 + 120 = Sta. 10+120
Verification: 0 ≤ 120 ≤ 300 ✓
STEP 3 — Elevation of summit (x = 120 m):
y = elev_PC + g₁x + (r/2)x²
y = 150.00 + (0.02)(120) + (−0.0001667/2)(120)²
y = 150.00 + 2.40 + (−0.00008333)(14400)
y = 150.00 + 2.40 − 1.20
y_summit = 151.20 m ← Elevation at Sta. 10+120
STEP 4 — Elevation at Sta. 10+150 (x = 150 m from PC):
y = 150.00 + (0.02)(150) + (−0.0001667/2)(150)²
y = 150.00 + 3.00 + (−0.00008333)(22500)
y = 150.00 + 3.00 − 1.875
y = 151.125 m ≈ 151.13 m ← Elevation at Sta. 10+150
SUMMARY:
- Summit: Sta. 10+120, Elevation = 151.20 m
- Sta. 10+150: Elevation = 151.13 m
Answer structure
- Block 1: Data organization — list all given values in decimal form, identify PC and PT stations [0.5 mark]
- Block 2: Compute r = −0.0001667 m⁻¹ [0.5 mark]
- Block 3: Compute summit location x = 120 m → Sta. 10+120 with verification [1 mark]
- Block 4: Compute summit elevation = 151.20 m using y equation [1.5 marks]
- Block 5: Compute elevation at Sta. 10+150 (x = 150 m) = 151.13 m [1.5 marks]
Scoring breakdown
Criteria
Correct r = −0.0001667 m⁻¹ and correct station of PC and PT
Criteria
Correct summit station Sta. 10+120 with verification that x = 120 m is within curve
Criteria
Correct summit elevation 151.20 m with full working shown
Criteria
Correct identification of x = 150 m for Sta. 10+150 and correct elevation computation
Criteria
Correct final elevation at Sta. 10+150 = 151.13 m with summary provided
Common mark deductions
- Converting station to x incorrectly — e.g., using x = 150 for the summit instead of computing x = g₁L/(g₁ − g₂)
- Not showing the (r/2) factor — automatic deduction if formula is incorrectly written
- Leaving out the summary box — 0.5-mark deduction for poor answer organization on 5-mark items
- Computing Sta. PT = 10+300 incorrectly or omitting it entirely
Key phrases to include
- r = −0.0001667
- x = 120 m
- Sta. 10+120
- 151.20 m
- x = 150 m
- 151.13 m
- verification: 0 ≤ 120 ≤ 300
A crest vertical curve is designed for a stopping sight distance S = 120 m. The algebraic difference of grades A = |g₁ − g₂| = 5%. Using the AASHTO/DPWH formula for the case S < L, compute the minimum curve length L. Assume eye height h₁ = 1.08 m and object height h₂ = 0.60 m.
Examiner tip
This is a classic PRC trap: always verify the S < L or S > L assumption after computing L. If the assumption fails, use the other formula and verify again. Showing both checks earns full marks even if you made a minor arithmetic error.
Model answer
GIVEN DATA:
S = 120 m (stopping sight distance)
A = 5% = 0.05 (algebraic grade difference)
h₁ = 1.08 m (driver eye height)
h₂ = 0.60 m (object height — DPWH standard)
FORMULA (Crest curve, assuming S < L):
L = A·S² / [200(√h₁ + √h₂)²]
STEP 1 — Compute the denominator term:
√h₁ = √1.08 = 1.0392 m^0.5
√h₂ = √0.60 = 0.7746 m^0.5
√h₁ + √h₂ = 1.0392 + 0.7746 = 1.8138 m^0.5
(√h₁ + √h₂)² = (1.8138)² = 3.2899 m
STEP 2 — Compute 200(√h₁ + √h₂)²:
200 × 3.2899 = 657.98 m (≈ 658)
STEP 3 — Compute L:
L = A·S² / 657.98
Note: A must be in fractional (decimal) form: A = 0.05
L = (0.05)(120)² / 657.98
L = (0.05)(14400) / 657.98
L = 720 / 657.98
L = 1.094 m ← This result is unreasonably small!
CORRECTION — A is expressed in PERCENT form in this formula:
L = A(%)·S² / [200(√h₁ + √h₂)²]
L = (5)(120)² / 657.98
L = (5)(14400) / 657.98
L = 72000 / 657.98
L = 109.43 m
STEP 4 — VERIFY ASSUMPTION (S < L):
S = 120 m vs L = 109.43 m → S > L ✗ (assumption is NOT satisfied!)
USE ALTERNATE FORMULA (S > L):
L = 2S − [200(√h₁ + √h₂)²/A(%)]
L = 2(120) − (657.98/5)
L = 240 − 131.60
L = 108.40 m
Re-check: S = 120 > L = 108.40 ✓ (assumption confirmed — use this formula)
∴ Minimum crest curve length L = 108.40 m ≈ 109 m (round up for safety)
Answer structure
- Block 1: List all given data including units; state design formula for S < L [0.5 mark]
- Block 2: Compute √h₁ + √h₂ and its square; compute denominator 200(...)² [1 mark]
- Block 3: Substitute into L formula with A in percent — compute L = 109.43 m [1 mark]
- Block 4: Check assumption S < L — find it is violated; switch to S > L formula [1.5 marks]
- Block 5: Compute L = 108.40 m using alternate formula, re-verify assumption, state final answer [1 mark]
Scoring breakdown
Criteria
Correct formula stated with proper variables (h₁, h₂, A, S, L) and correct denominator computation
Criteria
Correct computation of 200(√h₁ + √h₂)² = 657.98 m
Criteria
Correct L from S < L formula = 109.43 m, with recognition that A is in percent
Criteria
Correct checking that S > L (assumption fails) and switching to S > L formula
Criteria
Correct final L = 108.40 m using S > L formula, with assumption re-verified
Common mark deductions
- Using A as decimal (0.05) in the L formula when A must be in percent (5) — gives nonsensical L ≈ 1 m
- Not verifying the S vs L assumption and not switching formulas when violated
- Rounding √h₁ and √h₂ too early, causing significant error in denominator
Key phrases to include
- L = A·S²/[200(√h₁+√h₂)²]
- A in percent
- √h₁ + √h₂ = 1.8138
- 657.98
- assumption S < L
- S > L formula: L = 2S − [...]
- 108.40 m
What is the vertical offset (in metres) from the entry tangent line to the curve at the mid-point of a vertical curve with g₁ = +3%, g₂ = −2%, and L = 200 m?
Examiner tip
Maximum offset ≠ summit location. The maximum offset from the tangent is always at x = L/2. Practice distinguishing between these two special points.
Model answer
Given:
g₁ = +0.03, g₂ = −0.02, L = 200 m
Mid-point: x = L/2 = 100 m
Method 1 — Direct offset formula:
Offset at mid-curve = (g₂ − g₁)L / 8
= (−0.02 − 0.03)(200) / 8
= (−0.05)(200) / 8
= −10/8 = −1.25 m
Method 2 — Offset formula at any x:
Δ = [(g₂ − g₁)/(2L)] × x²
Δ = [(−0.05)/(2×200)] × (100)²
Δ = (−0.000125)(10000) = −1.25 m
Sign interpretation: Δ = −1.25 m means the curve is 1.25 m BELOW the entry tangent at mid-curve (expected for a crest curve).
∴ Vertical offset at mid-curve = 1.25 m below the entry tangent
Answer structure
- Line 1: Identify mid-point x = L/2 = 100 m [prerequisite]
- Line 2: Apply Δ = (g₂ − g₁)L/8 or Δ = [(g₂−g₁)/2L]x² [1 mark]
- Line 3: Compute Δ = −1.25 m and state direction (below tangent) [1 mark]
Scoring breakdown
Criteria
Correct formula applied at x = L/2 with correct sign for (g₂ − g₁)
Criteria
Correct answer |Δ| = 1.25 m with direction stated as 'below the entry tangent'
Common mark deductions
- Reporting +1.25 m (wrong sign/direction) — loses direction mark
- Using x = 120 m (the summit location) instead of x = 100 m (mid-curve)
- Using the offset formula without the ½ factor: Δ = [(g₂−g₁)/L]x² — gives double the correct answer
Key phrases to include
- x = L/2 = 100 m
- Δ = (g₂ − g₁)L/8
- −1.25 m
- below the entry tangent
- crest curve
A symmetrical parabolic vertical curve has a back tangent grade of +1.5% and a forward tangent grade of +4.0%. The curve length is 240 m and the elevation of the PC is 72.00 m. Classify the curve, find the location and elevation of the low point (if any), and find the elevation at Sta. 60 m from PC.
Examiner tip
Checking whether the turning point is inside the curve is worth 1.5 marks on its own. This is a classic examiner trap — when x < 0 or x > L, you must recognize there is no turning point in the curve and explain why.
Model answer
GIVEN DATA:
g₁ = +0.015 (ascending), g₂ = +0.040 (ascending)
L = 240 m, elev_PC = 72.00 m
STEP 1 — CURVE CLASSIFICATION:
Since g₁ = +0.015 < g₂ = +0.040, this is a SAG curve.
(Both grades are positive/ascending, but the forward grade is steeper — the road sags then climbs more steeply.)
r = (g₂ − g₁)/L = (0.040 − 0.015)/240 = 0.025/240 = +0.0001042 m⁻¹ (positive → sag confirmed)
STEP 2 — LOW POINT LOCATION:
x = g₁L/(g₁ − g₂)
x = (0.015)(240)/(0.015 − 0.040)
x = 3.60/(−0.025)
x = −144 m
INTERPRETATION: x = −144 m is NEGATIVE → the low point is NOT within the curve (it lies before the PC). Since the sag is caused by the grade change from +1.5% to +4.0%, the grade is increasing throughout and there is no turning point within the curve. The minimum elevation occurs at PC (x = 0).
STEP 3 — ELEVATION AT PC (minimum):
elev_PC = 72.00 m (given — this is the lowest point on the curve)
STEP 4 — ELEVATION AT x = 60 m FROM PC:
y = elev_PC + g₁x + (r/2)x²
y = 72.00 + (0.015)(60) + (0.0001042/2)(60)²
y = 72.00 + 0.90 + (0.00005208)(3600)
y = 72.00 + 0.90 + 0.1875
y = 73.09 m
SUMMARY:
- Curve type: SAG (both grades ascending but g₂ > g₁)
- Low point: NOT within the curve; minimum elevation at PC = 72.00 m
- Elevation at x = 60 m from PC = 73.09 m
Answer structure
- Block 1: Classify curve with grade comparison — SAG because g₁ < g₂ [0.5 mark]
- Block 2: Compute r = +0.0001042 m⁻¹ (positive confirms sag) [0.5 mark]
- Block 3: Compute x = −144 m and interpret — low point outside curve [1.5 marks]
- Block 4: State minimum elevation at PC = 72.00 m [0.5 mark]
- Block 5: Compute elevation at x = 60 m = 73.09 m [2 marks]
Scoring breakdown
Criteria
Correct curve classification as SAG with grade comparison stated
Criteria
Correct r = +0.0001042 m⁻¹ computed and sign interpreted
Criteria
x = −144 m computed and correctly interpreted as low point outside the curve
Criteria
Correct elevation at x = 60 m = 73.09 m with full working
Criteria
Summary and complete interpretation provided, minimum point correctly identified
Common mark deductions
- Not checking if x is within [0, L] and blindly reporting x = −144 m as the answer
- Forgetting that a sag can occur even when both grades are positive
- Rounding r too aggressively early, causing compound arithmetic error
Key phrases to include
- g₁ < g₂
- SAG curve
- r = +0.0001042
- x = −144 m
- outside the curve
- minimum at PC
- 73.09 m
Enumerate and explain the four most common sign-convention errors in vertical curve calculations.
Examiner tip
This type of enumeration question rewards students who have practiced enough problems to know the common mistakes. List specific formulas and their correct vs wrong versions to demonstrate mastery.
Model answer
Common sign-convention errors in parabolic vertical curve calculations:
- GRADE SIGN ERROR: Using g₁ = 3% instead of g₁ = +0.03 (decimal). Grades MUST be converted to decimals with correct sign. Positive = ascending, negative = descending.
- FORMULA INVERSION: Using r = (g₁ − g₂)/L instead of the correct r = (g₂ − g₁)/L. This reverses the sign of r and flips all conclusions (e.g., incorrectly identifying a crest as a sag).
- x-REFERENCE POINT ERROR: Measuring x from the PI (intersection point) instead of from the PC (beginning of curve / BVC). All distances x in the elevation equation are from PC.
- OFFSET DIRECTION ERROR: For a crest curve, the vertical offset from the tangent is negative (curve lies below tangent). Students frequently report this as positive without specifying direction, losing the direction mark.
Answer structure
- Error 1: Grade sign and decimal conversion [0.5 mark]
- Error 2: Formula inversion r = (g₁ − g₂)/L [0.5 mark]
- Error 3: Measuring x from PI not PC [0.5 mark]
- Error 4: Offset direction [0.5 mark]
Scoring breakdown
Criteria
4 distinct sign/convention errors clearly named and explained (0.5 mark each)
Common mark deductions
- Repeating the same error in different words — only one unique error counts
- Listing errors without explanation — half marks only
- Mentioning 'calculation errors' rather than conceptual sign errors
Key phrases to include
- positive = ascending
- g₂ − g₁
- from PC (BVC)
- below tangent for crest
- decimal form
A crest vertical curve has g₁ = +5%, g₂ = −3%, and L = 400 m. The PI is at Sta. 5+200 with elevation 180.00 m. Compute the elevation of the PC, the PT, and the summit.
Examiner tip
When the PI elevation is given instead of the PC elevation, you must work backward along the back tangent. Going from PI to PC means moving BACKWARD against g₁ direction — so elev_PC = elev_PI − g₁(L/2). This is the most common trap in PI-given problems.
Model answer
GIVEN DATA:
g₁ = +0.05, g₂ = −0.03, L = 400 m
Sta. PI = 5+200, elev. PI = 180.00 m
Curve type: g₁ > g₂ → CREST curve ✓
STEP 1 — LOCATE PC AND PT:
For symmetrical curve: PC is L/2 = 200 m BEFORE PI
Sta. PC = 5+200 − 200 = Sta. 5+000
Sta. PT = 5+200 + 200 = Sta. 5+400
STEP 2 — ELEVATION OF PC:
The PC is 200 m before the PI on the BACK tangent (grade g₁ = +0.05 going forward, so going backward from PI to PC the elevation decreases by g₁ × L/2).
elev. PC = elev. PI − g₁ × (L/2)
elev. PC = 180.00 − (0.05)(200)
elev. PC = 180.00 − 10.00 = 170.00 m
STEP 3 — ELEVATION OF PT:
PT is 200 m after the PI on the FORWARD tangent (grade g₂ = −0.03).
elev. PT = elev. PI + g₂ × (L/2)
elev. PT = 180.00 + (−0.03)(200)
elev. PT = 180.00 − 6.00 = 174.00 m
STEP 4 — RATE OF GRADE CHANGE:
r = (g₂ − g₁)/L = (−0.03 − 0.05)/400 = −0.08/400 = −0.0002 m⁻¹
STEP 5 — SUMMIT LOCATION:
x = g₁L/(g₁ − g₂) = (0.05)(400)/(0.05 − (−0.03)) = 20/0.08 = 250 m from PC
Verification: 0 ≤ 250 ≤ 400 ✓
Station of summit: Sta. 5+000 + 250 = Sta. 5+250
STEP 6 — ELEVATION OF SUMMIT:
y = elev_PC + g₁x + (r/2)x²
y = 170.00 + (0.05)(250) + (−0.0002/2)(250)²
y = 170.00 + 12.50 + (−0.0001)(62500)
y = 170.00 + 12.50 − 6.25
y_summit = 176.25 m at Sta. 5+250
SUMMARY:
- PC: Sta. 5+000, Elevation = 170.00 m
- PT: Sta. 5+400, Elevation = 174.00 m
- Summit: Sta. 5+250, Elevation = 176.25 m
Answer structure
- Block 1: Identify curve as crest; compute PC and PT stations [0.5 mark]
- Block 2: Compute elev. PC = 170.00 m from PI elevation using back tangent [1 mark]
- Block 3: Compute elev. PT = 174.00 m using forward tangent [0.5 mark]
- Block 4: Compute r = −0.0002 m⁻¹ [0.5 mark]
- Block 5: Compute summit x = 250 m → Sta. 5+250 with verification [1 mark]
- Block 6: Compute summit elevation = 176.25 m with full y-equation [1.5 marks]
Scoring breakdown
Criteria
Correct PC and PT stations with correct elevations (170.00 m and 174.00 m)
Criteria
Correct r = −0.0002 m⁻¹ and proper computation from PI elevation
Criteria
Correct summit station Sta. 5+250 with x = 250 m and verification
Criteria
Correct summit elevation = 176.25 m with elevation equation shown
Criteria
Organized summary with all three answers clearly labeled
Common mark deductions
- Adding g₁ × L/2 to PI elevation to get PC elevation instead of subtracting (direction error)
- Measuring x from the PI instead of from the PC when using the elevation equation
- Forgetting to add PC station to x to get summit station
Key phrases to include
- Sta. 5+000
- 170.00 m
- Sta. 5+400
- 174.00 m
- r = −0.0002
- Sta. 5+250
- 176.25 m
- L/2 = 200 m
Two grades, g₁ = −2% and g₂ = +3%, are connected by a vertical curve. What is the minimum length of curve required if the maximum rate of grade change must not exceed 0.3% per station (1 station = 20 m)?
Examiner tip
Rate-of-grade-change problems always require unit consistency. Convert everything to 'per metre' before computing L, or use the station-based approach end-to-end — never mix the two.
Model answer
Given:
g₁ = −2% = −0.02
g₂ = +3% = +0.03
Max. rate of grade change: r_max = 0.3% per 20-m station = 0.003/20 = 0.00015 per metre
Step 1 — Algebraic grade difference:
|g₂ − g₁| = |0.03 − (−0.02)| = |0.05| = 0.05 (or 5%)
Step 2 — Minimum L from rate of change criterion:
r = (g₂ − g₁)/L
Rearranging: L = (g₂ − g₁)/r
Using r in consistent units (per metre):
L = 0.05/0.00015 = 333.33 m
Alternatively using station-based calculation:
Number of stations = |g₂(%) − g₁(%)| / r_max(% per station)
= 5% / (0.3% per station) = 16.67 stations
L = 16.67 × 20 m = 333.33 m
∴ Minimum curve length L = 333.33 m (round up to 340 m or nearest even station)
Answer structure
- Block 1: Convert r_max to consistent units (per metre or per station) [0.5 mark]
- Block 2: Compute |g₂ − g₁| = 5% [0.5 mark]
- Block 3: Compute L = |g₂ − g₁|/r_max = 333.33 m [1 mark]
Scoring breakdown
Criteria
Correct algebraic grade difference = 5% and correct formula L = (g₂ − g₁)/r
Criteria
Correct minimum L = 333.33 m with consistent units throughout
Common mark deductions
- Using r_max = 0.3% per station = 0.003 per metre (forgetting to divide by 20) — gives wrong L = 16.67 m
- Using g₁ − g₂ instead of g₂ − g₁ in the numerator
- Not rounding up to a practical station value
Key phrases to include
- |g₂ − g₁| = 5%
- r_max = 0.00015 per metre
- L = (g₂ − g₁)/r
- 333.33 m
- round up
Mark wise strategy
Recall-and-state. These are definition, classification, or formula-identification questions. Write the answer immediately with no preamble. Include one mathematical expression to demonstrate conceptual mastery.
Do
- State the formula or definition in the first line
- Include a relevant mathematical expression (e.g., r = (g₂ − g₁)/L)
- Use correct engineering terminology (BVC, EVC, grade, offset)
- Convert % grades to decimals if any numerical value is involved
Don't
- Do not write lengthy introductory sentences
- Do not leave out the unit for any numerical value
- Do not use layman terms (e.g., 'hill' instead of 'crest curve')
- Do not spend more than 2 minutes — move on
Show formula → substitute → compute → state result with unit. Two-mark questions typically have two distinct sub-parts or two sequential computation steps. Earn one mark per sub-part by clearly labeling your work.
Do
- Label each sub-part clearly: (a) and (b)
- Write the formula before substituting values
- Show at least one intermediate calculation step
- Box or underline the final answer with its unit
Don't
- Do not go straight from given data to final answer without showing formula
- Do not skip the grade-to-decimal conversion step
- Do not forget to state the direction for offset answers (above/below tangent)
- Do not round intermediate values — keep 4–5 significant figures until the final answer
Structure as: Given → Formula → Step-by-step computation → Verification → Answer. Three-mark questions usually have three distinct scoring points. Verification (checking x within curve, or checking sign of r) earns the third mark.
Do
- Write a GIVEN block listing all data in correct units
- Label computation steps clearly (Step 1, Step 2, Step 3)
- Include the verification check: 0 ≤ x ≤ L for turning point problems
- State the curve type (crest or sag) at the beginning
Don't
- Do not derive formulas unless specifically asked — state the formula and use it
- Do not use only one significant figure for intermediate results
- Do not omit the verification check — it costs 1 mark
- Do not write a wall of numbers without labels — examiners cannot follow it
Treat as a mini-engineering report. Use: Given → Solution (Steps 1–n) → Summary table. Five-mark problems usually have 5 distinct scoring points. Each step must be self-contained so the examiner can follow and award partial marks if a later step uses an incorrect earlier result but with correct method.
Do
- Write a complete GIVEN DATA block — label every variable with its unit
- Number each computational step sequentially
- Verify every assumption (S < L or S > L; x within curve length)
- Provide a SUMMARY box at the end listing all asked values
- Include a rough profile sketch labeling PC, PI, PT, g₁, g₂
Don't
- Do not skip steps even if they seem trivial — every step is a potential mark
- Do not use approximate values (e.g., π ≈ 3.14) unless specified
- Do not present only the final answer — it earns at most 1 of 5 marks
- Do not cross out work unless you have a replacement — crossed work may still be marked
- Do not leave the assumption unverified in sight-distance problems
General answer writing tips
- Always define variables with units before substituting numerical values — write 'g₁ = +3% = +0.03, g₂ = −2% = −0.02, L = 200 m' as a header block before any calculation.
- Use the correct sign convention consistently: grades going uphill are positive (+), downhill are negative (−). State this convention explicitly if the question is worth 3 marks or more.
- For any numerical problem, box or underline your final answer and always include the unit (m, m/m, %, etc.). An unboxed or unitless answer is an easy deduction.
- When finding the high/low point, always verify that the computed x-value lies within 0 ≤ x ≤ L. State this check explicitly — examiners reward it.
- Sketch a rough profile (PC, PI, PT, grade lines, and curve) even for numerical questions — this earns diagram credit and helps you avoid sign errors.
- Write the general elevation equation y = elev_PC + g₁x + (r/2)x² first, then substitute values — never go straight to substitution without stating the formula.
- For crest vs. sag identification, explicitly state 'Since g₁ > g₂, this is a CREST (summit) curve' or 'Since g₁ < g₂, this is a SAG curve' to signal conceptual understanding.
- In sight-distance design problems, state both the assumed condition (S < L or S > L) and verify it after computing L — an unverified assumption loses at least one mark.
Diagrams
Vertical Parabolic Curve Problem-Solving Decision Tree
A decision-tree flowchart guiding students through the correct formula selection and verification steps for any vertical parabolic curve problem type. Follow from START to identify which formula and verification check apply based on what the problem is asking.
Vertical Parabolic Curves — Formula and Concept Overview
A comprehensive mind map of all vertical parabolic curve concepts, formulas, and key points. Use this as a quick-reference overview before attempting problems. The central branches represent the major topic areas covered in the chapter.
Sight Distance Design: Which Formula to Use?
A flowchart showing the two-formula approach for sight-distance-based vertical curve length design. The critical decision point is verifying the S vs L assumption after computing L. This diagram directly supports Template T10 and the common board-exam trap of failing to check the assumption.
Distribution of PRC Board Exam Question Types — Vertical Curves
An estimated distribution of PRC Civil Engineer board exam question types within the Vertical Parabolic Curves topic. Elevation computation and turning-point location together account for 55% of questions — prioritize these topics. Sight distance design (20%) is the hardest question type and is worth practicing thoroughly.