CELE Surveying (Geomatics) — Vertical (Parabolic) CurvesExam Answer Templates

Marks

1

Topic

Definition and Properties of Parabolic Curves

Difficulty

easy

Template Id

T1

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.

Question Type

very_short_answer

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

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

Marks

1

Topic

Crest vs. Sag Curves

Difficulty

easy

Template Id

T2

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.

Question Type

very_short_answer

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

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

Marks

1

Topic

Rate of Grade Change

Difficulty

easy

Template Id

T3

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)

Question Type

very_short_answer

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

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

Marks

2

Topic

Rate of Grade Change and Curve Classification

Difficulty

easy

Template Id

T4

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.

Question Type

numerical

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

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

Marks

2

Topic

Elevation Along the Curve

Difficulty

medium

Template Id

T5

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

Question Type

numerical

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

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

Marks

3

Topic

Location of High/Low Point

Difficulty

medium

Template Id

T6

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.

Question Type

numerical

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

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

Marks

3

Topic

Summit Elevation and Vertical Offsets

Difficulty

medium

Template Id

T7

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 ✓

Question Type

numerical

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

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

Marks

3

Topic

Low Point Location — Sag Curves

Difficulty

medium

Template Id

T8

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

Question Type

numerical

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

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

Marks

5

Topic

Full Station and Elevation Computation

Difficulty

hard

Template Id

T9

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

Question Type

numerical

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

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

Marks

5

Topic

Sight Distance Design — Crest Curves

Difficulty

hard

Template Id

T10

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)

Question Type

numerical

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

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

Marks

2

Topic

Vertical Offsets from Tangent

Difficulty

medium

Template Id

T11

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

Question Type

numerical

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

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

Marks

5

Topic

Turning Point Outside the Curve

Difficulty

hard

Template Id

T12

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

Question Type

numerical

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

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

Marks

2

Topic

Common Pitfalls and Sign Conventions

Difficulty

easy

Template Id

T13

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: 1. 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. 2. 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). 3. 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. 4. 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.

Question Type

short_answer

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

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

Marks

5

Topic

Full Curve Computation from PI Data

Difficulty

hard

Template Id

T14

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

Question Type

numerical

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

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

Marks

2

Topic

Minimum Curve Length from Rate of Grade Change

Difficulty

medium

Template Id

T15

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)

Question Type

numerical

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

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