CELE Surveying (Geomatics) — Spiral (Transition) CurvesExam Answer Templates

Marks

1

Topic

Definition and Purpose of Spiral Curves

Difficulty

easy

Template Id

T1

Examiner Tip

The examiner is checking for two ideas in one sentence: (1) curvature changes from 0 to 1/R, and (2) the purpose (smooth steering/superelevation). Hit both to guarantee the mark.

Model Answer

A spiral (transition) curve is a curve whose radius decreases linearly from infinity at the tangent point to the radius R of the circular curve at the spiral-to-curve point, allowing a gradual change in curvature and smooth superelevation runoff.

Question Type

very_short_answer

Answer Structure

• One concise sentence: define the curve by stating (a) radius varies from ∞ to R, and (b) its purpose — smooth steering and superelevation transition. [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing only 'a curve between a tangent and a circle' without mentioning the variable/changing radius loses the mark.
• Confusing transition curve with vertical curve — entirely wrong concept.

Key Phrases To Include

• radius decreases from infinity
• circular curve radius R
• gradual change in curvature
• superelevation runoff

Marks

1

Topic

Spiral Angle Formula

Difficulty

easy

Template Id

T2

Examiner Tip

Memorize the factor of 2 — it is the single most common error on this formula. The factor comes from the Euler spiral relationship: the curvature increases linearly, so the average curvature over Ls is 1/(2R).

Model Answer

θs = Ls / (2R) [radians] where: Ls = length of spiral (m), R = radius of the connecting circular curve (m), θs = spiral angle (rad).

Question Type

very_short_answer

Answer Structure

• Write the formula clearly. [0.5 mark]
• Define all variables with units. [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing θs = Ls/R (missing the factor of 2) — loses the mark entirely.
• Omitting units (radians) when the question explicitly asks for variables.

Key Phrases To Include

• θs = Ls / (2R)
• radians
• Ls = length of spiral
• R = radius of circular curve

Marks

1

Topic

Superelevation and Design Speed

Difficulty

medium

Template Id

T3

Examiner Tip

Board exams occasionally ask this as a conceptual trap. The derivation is: centripetal acceleration = V²/R; convert V from km/h to m/s by dividing by 3.6, so V²(km/h)/(3.6²) = V²/R·g → rearranges to e+f = V²/(g·3.6²·R) = V²/(127R).

Model Answer

The constant 127 = g × (3.6)² / 10 ≈ 9.81 × 12.96 ≈ 127.1 is a unit-conversion factor that arises when V is expressed in km/h and R in metres. It allows direct use of field units without manual conversion to m/s.

Question Type

very_short_answer

Answer Structure

• State what the constant represents and how it originates from unit conversion of V from km/h to m/s. [1 mark]

Scoring Breakdown

Common Mark Deductions

• Simply writing '127 is a constant' without explaining its origin loses the mark.
• Saying '127 = gravity' is incorrect and earns zero.

Key Phrases To Include

• unit conversion
• V in km/h
• R in metres
• g × 3.6²

Marks

1

Topic

Angle at Any Point on the Spiral

Difficulty

easy

Template Id

T4

Examiner Tip

The quadratic nature is the key differentiator between a spiral and a circular curve. Circular curves have constant curvature; spirals have linearly increasing curvature, which gives the squared relationship for the angle.

Model Answer

θ = θs × (ℓ/Ls)² The angle at any point varies as the square of the distance ℓ from the spiral's starting point (TS), not linearly.

Question Type

very_short_answer

Answer Structure

• State the formula θ = θs(ℓ/Ls)² and note the quadratic (not linear) variation. [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing θ = θs(ℓ/Ls) — linear relationship — loses the mark.
• Confusing ℓ (distance along spiral) with Ls (total spiral length).

Key Phrases To Include

• θ = θs(ℓ/Ls)²
• varies as the square
• quadratic
• distance from TS

Marks

2

Topic

Spiral Angle Computation

Difficulty

easy

Template Id

T5

Examiner Tip

Always show the conversion step explicitly, even if your calculator has a RAD→DEG button. The examiner must see '× (180/π)' to award the conversion mark.

Model Answer

Given: Ls = 60 m, R = 250 m Formula: θs = Ls / (2R) [radians] Substitute: θs = 60 / (2 × 250) θs = 60 / 500 θs = 0.1200 rad Convert to degrees: θs = 0.1200 × (180/π) θs = 0.1200 × 57.2958 θs = 6.875° ∴ Spiral angle θs ≈ 6.88°

Question Type

numerical

Answer Structure

• Line 1–2: Write given data and formula. [0.5 mark — method mark]
• Line 3–4: Correct substitution and computation in radians. [0.5 mark]
• Line 5–6: Correct conversion to degrees using ×(180/π). [0.5 mark]
• Line 7: Correct final answer with degree unit. [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using θs = Ls/R (omitting factor of 2) gives 0.24 rad = 13.75° — loses first mark.
• Forgetting to convert radians to degrees — loses second mark.
• Rounding error beyond ±0.05° — typically no deduction unless it propagates to a wrong answer.

Key Phrases To Include

• θs = Ls/(2R)
• radians
• × (180/π)
• 6.88° or 6.875°

Marks

2

Topic

Superelevation and Side Friction

Difficulty

easy

Template Id

T6

Examiner Tip

Part (a) and (b) are independent marks. Even if you make an arithmetic error in (a), you can still earn the mark in (b) by following through consistently — examiners call this 'error carried forward' (ECF).

Model Answer

Given: V = 80 km/h, R = 300 m, f = 0.14 (a) Required e + f: Formula: e + f = V² / (127R) e + f = (80)² / (127 × 300) e + f = 6400 / 38100 e + f = 0.168 (b) Design superelevation: e = (e + f) − f e = 0.168 − 0.14 e = 0.028 (or 2.8%) ∴ Required e + f = 0.168; design superelevation e = 0.028 (2.8%)

Question Type

numerical

Answer Structure

• Part (a): Formula + substitution + correct e+f value. [1 mark]
• Part (b): Subtract f from total to isolate e. [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using V in m/s instead of km/h (e.g., 22.22 m/s) and applying V²/(gR) — valid physics but misuses the 127 formula, leading to a different numerical result and possible mark deduction.
• Forgetting to subtract f to find e — answers only part (a).
• Reporting e without a percentage or decimal form clarification.

Key Phrases To Include

• e + f = V²/(127R)
• 6400/38100
• 0.168
• e = 0.028
• 2.8%

Marks

2

Topic

Angle at Any Point on the Spiral

Difficulty

medium

Template Id

T7

Examiner Tip

Always emphasize the squared exponent when substituting — write (ℓ/Ls)² and then evaluate step by step. This prevents the very common error of applying a linear ratio.

Model Answer

Given: Ls = 60 m, R = 250 m, ℓ = 30 m From T5: θs = 6.875° Formula: θ = θs × (ℓ/Ls)² Substitute: θ = 6.875° × (30/60)² θ = 6.875° × (0.5)² θ = 6.875° × 0.25 θ = 1.719° ∴ Spiral angle at ℓ = 30 m is θ ≈ 1.72°

Question Type

numerical

Answer Structure

• State formula θ = θs(ℓ/Ls)². [0.5 mark]
• Correctly compute (ℓ/Ls)² = (30/60)² = 0.25. [0.5 mark]
• Multiply by θs = 6.875° to get 1.719°. [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using linear interpolation: θ = θs × (ℓ/Ls) = 6.875 × 0.5 = 3.44° — wrong (linear, not quadratic), loses both marks.
• Using θs in radians but forgetting to apply the formula before converting — unit confusion.

Key Phrases To Include

• θ = θs(ℓ/Ls)²
• (30/60)² = 0.25
• quadratic variation
• 1.72°

Marks

3

Topic

Spiral Elements: Throw and Shift

Difficulty

medium

Template Id

T8

Examiner Tip

The relationship p = Throw/4 is a powerful self-check. Always verify this on the exam — if your two answers do not satisfy this ratio, you have made an error somewhere.

Model Answer

Given: Ls = 70 m, R = 350 m (a) Tangent Offset (Throw): Formula: Throw = Ls² / (6R) Throw = (70)² / (6 × 350) Throw = 4900 / 2100 Throw = 2.333 m (b) Shift p: Formula: p = Ls² / (24R) p = (70)² / (24 × 350) p = 4900 / 8400 p = 0.583 m Check: p = Throw/4 = 2.333/4 = 0.583 m ✓ ∴ Tangent offset (throw) = 2.33 m; Shift p = 0.583 m

Question Type

numerical

Answer Structure

• Part (a): Formula for throw = Ls²/(6R) stated and substituted correctly. [1 mark]
• Part (a): Correct numerical answer: 2.33 m. [0.5 mark]
• Part (b): Formula for shift p = Ls²/(24R) stated and substituted correctly. [1 mark]
• Part (b): Correct numerical answer: 0.583 m. [0.5 mark — can also be awarded if verified as Throw/4]

Scoring Breakdown

Common Mark Deductions

• Confusing throw formula Ls²/(6R) with shift formula Ls²/(24R) — swapping them loses both part marks.
• Not showing the relationship p = Throw/4 when asked to find both — missing a verification opportunity.
• Using incorrect Ls² calculation (e.g., 70² = 4,900 vs. computing 70 × 70 = 4,900 — usually fine, but careless errors here cost marks).

Key Phrases To Include

• Throw = Ls²/(6R)
• p = Ls²/(24R)
• shift
• p = Throw/4
• 2.33 m
• 0.583 m

Marks

3

Topic

Spiral Angle and Intermediate Angle

Difficulty

medium

Template Id

T9

Examiner Tip

This is a classic 3-mark board-exam question. Allocate roughly 1 minute per mark. Part (b) depends on (a), so verify (a) first. The answer 1.91° = θs/4 (since 7.64°/4 = 1.91°) provides a quick sanity check at the midpoint.

Model Answer

Given: Ls = 80 m, R = 300 m (a) Spiral Angle θs: θs = Ls / (2R) [radians] θs = 80 / (2 × 300) = 80/600 = 0.13333 rad Convert to degrees: θs = 0.13333 × (180/π) = 0.13333 × 57.296 = 7.639° ∴ θs ≈ 7.64° (b) Angle at midpoint (ℓ = 40 m): θ = θs × (ℓ/Ls)² θ = 7.64° × (40/80)² θ = 7.64° × (0.5)² θ = 7.64° × 0.25 θ = 1.910° ∴ θ at midpoint ≈ 1.91°

Question Type

numerical

Answer Structure

• Part (a): Formula θs = Ls/(2R) and substitution. [0.5 mark]
• Part (a): Correct computation in radians = 0.1333 rad. [0.5 mark]
• Part (a): Correct conversion: 7.64°. [0.5 mark]
• Part (b): Formula θ = θs(ℓ/Ls)² and correct substitution. [0.5 mark]
• Part (b): Correct answer: 1.91°. [1 mark]

Scoring Breakdown

Common Mark Deductions

• In part (b), using ℓ/Ls = 40/80 = 0.5 without squaring it — gives 3.82° instead of 1.91°.
• Carrying forward an incorrect θs from part (a) into part (b) without flagging the error.

Key Phrases To Include

• θs = Ls/(2R)
• 0.1333 rad
• × (180/π)
• 7.64°
• θ = θs(ℓ/Ls)²
• 1.91°

Marks

3

Topic

Superelevation Check and Adequacy

Difficulty

medium

Template Id

T10

Examiner Tip

Adequacy problems are really three-step problems: compute demand, subtract friction, compare with limit. The third step earns its own mark, so always write a clear, explicit conclusion sentence — never leave it implied.

Model Answer

Given: V = 100 km/h, R = 400 m, f = 0.12, e_max = 0.08 Step 1 — Compute required e + f: e + f = V² / (127R) e + f = (100)² / (127 × 400) e + f = 10000 / 50800 e + f = 0.1969 Step 2 — Compute required superelevation: e = (e + f) − f e = 0.1969 − 0.12 e = 0.0769 (or 7.69%) Step 3 — Adequacy check: Required e = 0.0769 Allowable e_max = 0.08 Since 0.0769 < 0.08, the curve is ADEQUATE. ∴ Required e + f = 0.197; design e = 7.69% < 8% limit — curve is adequate.

Question Type

numerical

Answer Structure

• Step 1: Formula + substitution + e+f = 0.197. [1 mark]
• Step 2: Subtract f: e = 0.197 − 0.12 = 0.0769 (7.69%). [1 mark]
• Step 3: Compare with e_max and state adequacy conclusion. [1 mark]

Scoring Breakdown

Common Mark Deductions

• Failing to perform the adequacy check (Step 3) — the question explicitly asks for it, so omitting this loses 1 mark.
• Using f = 0 and equating e = 0.197 — ignoring given friction factor.
• Stating 'not adequate' when e = 7.69% < 8% — wrong conclusion due to direction of inequality.

Key Phrases To Include

• e + f = V²/(127R)
• 0.1969
• e = 0.0769
• 7.69%
• adequate
• e < e_max

Marks

5

Topic

Spiral Curve Theory, Elements, and Derivation

Difficulty

hard

Template Id

T11

Examiner Tip

For a 5-mark long answer, use numbered sections with clear headings. Examiners scan for structure first. The derivation is the core of the marks — set up the integral explicitly, even if you know the result. Showing dθ = dℓ/ρ and the curvature equation earns method marks even if you make an arithmetic slip.

Model Answer

1. PURPOSE OF A SPIRAL (TRANSITION) CURVE A spiral (transition) curve is inserted between a straight tangent and a circular curve to provide: (a) Gradual steering adjustment — curvature increases from 0 (at the tangent) to 1/R (at the circular curve), letting the driver steer smoothly without abrupt changes. (b) Progressive superelevation runoff — the pavement cross-fall changes from normal camber to full design superelevation over the spiral length Ls, preventing abrupt tilting of the road surface. 2. KEY GEOMETRIC ELEMENTS • Ls = total length of spiral (m) • R = radius of the connecting circular curve (m) • θs = spiral angle — angle turned through from TS to SC • TS = Tangent-to-Spiral point (start of spiral) • SC = Spiral-to-Curve point (end of spiral / start of circle) • Throw = Ls²/(6R) — tangent offset at SC (m) • Shift p = Ls²/(24R) — inward shift of circular arc (m); note p = Throw/4 3. DERIVATION OF THE SPIRAL ANGLE A spiral is an Euler (Cornu) spiral where curvature (1/ρ) increases linearly with arc length ℓ: 1/ρ = ℓ / (Ls × R) ... (i) At any point, the infinitesimal angle turned is dθ = dℓ/ρ. Substituting (i): dθ = ℓ dℓ / (Ls × R) Integrate from 0 to Ls: θs = ∫₀^Ls [ℓ/(Ls·R)] dℓ = 1/(Ls·R) × [ℓ²/2]₀^Ls = 1/(Ls·R) × Ls²/2 = Ls / (2R) [radians] 4. DEGREE EQUIVALENT θs (degrees) = Ls × 90 / (π × R) 5. INTERMEDIATE ANGLE Following the same integration from 0 to ℓ: θ = ℓ² / (2·Ls·R) = θs × (ℓ/Ls)² This shows the angle grows quadratically with distance — a unique and testable property of the spiral. ∴ The spiral angle θs = Ls/(2R) (rad) provides the total angular rotation of the spiral; the angle at any intermediate point varies as the square of the fractional distance from TS.

Question Type

long_answer

Answer Structure

• Section 1: Purpose — at least 2 points (steering + superelevation). [1 mark]
• Section 2: Geometric elements — at least 5 named elements with symbols/formulas. [1 mark]
• Section 3: Derivation — set up curvature-linear relationship, show integration. [2 marks]
• Section 4/5: Final formula and intermediate angle formula with quadratic explanation. [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating θs = Ls/(2R) without showing the derivation in a 5-mark question — loses 2 derivation marks.
• Listing geometric elements without formulas for Throw and Shift — partial credit only.
• Writing purpose as only 'smooth driving' without distinguishing steering from superelevation.

Key Phrases To Include

• curvature increases linearly
• superelevation runoff
• Euler spiral
• dθ = dℓ/ρ
• θs = Ls/(2R)
• θ = θs(ℓ/Ls)²
• quadratic variation
• TS and SC

Marks

5

Topic

Superelevation Demand and Spiral Length for Runoff

Difficulty

hard

Template Id

T12

Examiner Tip

This type of compound problem tests both superelevation demand AND spiral length. Budget 5–6 minutes. Watch for the negative e scenario — it is an intentional trap testing whether you understand that friction alone can handle gentle curves. Always state your physical interpretation.

Model Answer

Given: V = 60 km/h, R = 180 m, f = 0.16, e_max = 0.10 Runoff rate = 1:150, half-width w = 3.5 m PART (a) — Superelevation demand: e + f = V² / (127R) e + f = (60)² / (127 × 180) e + f = 3600 / 22860 e + f = 0.1574 Required e = (e + f) − f = 0.1574 − 0.16 = −0.0026 Since the required e is negative (demand is less than friction alone), the friction is sufficient. Set e = 0 (no superelevation needed). However, check: if e must be at least 2% (normal crown) for drainage, e = 0.02 is applied as minimum. ∴ e + f = 0.157; required e = 0 (friction is sufficient; apply minimum drainage cross-fall of 2%). PART (b) — Minimum spiral length for superelevation runoff: The superelevation change = e × w (height change over width) where e = design superelevation = 0.02 (minimum, for computation; use actual design e) Using the general formula for superelevation runoff length: Ls = e × w × (runoff rate denominator) = e × w × 150 For e = 0.02 (minimum value applied): Ls = 0.02 × 3.5 × 150 = 10.5 m (very short — minimum only) For the case where the problem intends e from the demand formula (some PRC problems use e + f total = 0.157, assuming e absorbs all): Ls = 0.157 × 3.5 × 150 = 82.4 m Practical minimum spiral length (DPWH-based rule-of-thumb: Ls ≥ V/3.6 = 60/3.6 ≈ 17 m) — take the larger: Ls_min ≈ 82.4 m (governs when e = 0.157 is used) ∴ Minimum spiral length Ls ≈ 82.4 m SUMMARY: (a) e + f = 0.157; required e = 0 (or minimum 2% for drainage) (b) Ls_min ≈ 82.4 m (based on full e + f developed over spiral)

Question Type

numerical

Answer Structure

• Part (a), Step 1: Formula e+f = V²/(127R) and substitution. [1 mark]
• Part (a), Step 2: Correct computation e+f = 0.157 and conclusion on e. [1 mark]
• Part (b), Step 1: State runoff formula Ls = e × w × rate. [1 mark]
• Part (b), Step 2: Correct substitution. [1 mark]
• Part (b), Step 3: Correct Ls and practical check. [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not recognizing that required e is negative (friction exceeds demand) — blindly solving e = 0.157 − 0.16 = negative and failing to interpret it correctly.
• Using full road width (7 m) instead of superelevated half-width (3.5 m) in the runoff formula.
• Not applying the runoff formula at all — computing Ls from some other formula without justification.

Key Phrases To Include

• e + f = V²/(127R)
• 3600/22860 = 0.1574
• Ls = e × w × runoff rate
• runoff rate 1:150
• half-width 3.5 m
• 82.4 m

Marks

2

Topic

Spiral Elements: Throw and Shift

Difficulty

easy

Template Id

T13

Examiner Tip

Board exams love testing throw vs. shift because the formulas look similar (only the denominator constant differs). A reliable memory aid: 'Throw = 6, Shift = 24' — shift is 4× the denominator, so shift is ¼ the throw.

Model Answer

(a) Tangent offset (throw): Throw = Ls² / (6R) (b) Shift p (inward setback of the circular arc): p = Ls² / (24R) Relationship: p = Throw / 4 The shift is exactly one-quarter of the tangent offset. This relationship arises because the denominators 24 and 6 differ by a factor of 4.

Question Type

short_answer

Answer Structure

• Formula for Throw = Ls²/(6R). [0.5 mark]
• Formula for p = Ls²/(24R). [0.5 mark]
• Statement of relationship: p = Throw/4. [1 mark]

Scoring Breakdown

Common Mark Deductions

• Swapping the formulas — writing Throw = Ls²/(24R) — loses first mark.
• Not stating the ratio relationship between throw and shift — loses second mark.

Key Phrases To Include

• Ls²/(6R)
• Ls²/(24R)
• shift
• throw
• p = Throw/4
• one-quarter

Marks

2

Topic

Spiral Length from Spiral Angle

Difficulty

medium

Template Id

T14

Examiner Tip

When θs is given in degrees and the formula is in radians, the conversion step is a free mark — show it explicitly. This is one of the most reliable mark-grabbing steps in spiral problems.

Model Answer

Given: R = 200 m, θs = 9° Step 1 — Convert θs to radians: θs = 9° × (π/180) = 9 × 0.017453 = 0.15708 rad Step 2 — Solve for Ls from θs = Ls/(2R): Ls = 2R × θs Ls = 2 × 200 × 0.15708 Ls = 400 × 0.15708 Ls = 62.83 m ∴ Spiral length Ls ≈ 62.8 m

Question Type

numerical

Answer Structure

• Convert θs from degrees to radians. [0.5 mark]
• Rearrange formula: Ls = 2R × θs. [0.5 mark]
• Correct substitution and computation: 62.83 m. [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using θs = 9° directly in the formula without converting to radians — gives Ls = 2(200)(9) = 3600 m, a wildly wrong answer.
• Forgetting to multiply by 2 in Ls = 2Rθs — gives Ls = Rθs = 31.4 m.

Key Phrases To Include

• × (π/180)
• 0.15708 rad
• Ls = 2R × θs
• 62.83 m

Marks

3

Topic

Relationship Between Throw and Shift

Difficulty

medium

Template Id

T15

Examiner Tip

A verification question requires you to show BOTH computations AND the comparison step. Writing only the two numbers without the ratio check does not earn the verification mark. Write 'Throw/4 = ___ = p ✓' explicitly.

Model Answer

Given: Ls = 100 m, R = 500 m Step 1 — Compute Throw: Throw = Ls² / (6R) Throw = (100)² / (6 × 500) Throw = 10000 / 3000 Throw = 3.333 m Step 2 — Compute Shift p: p = Ls² / (24R) p = (100)² / (24 × 500) p = 10000 / 12000 p = 0.833 m Step 3 — Verify p = Throw/4: Throw/4 = 3.333 / 4 = 0.833 m Computed p = 0.833 m ✓ The two values match exactly, confirming p = Throw/4. ∴ Throw = 3.33 m; Shift p = 0.833 m; Verified: p = Throw/4 = 0.833 m ✓

Question Type

numerical

Answer Structure

• Correct computation of Throw = Ls²/(6R) = 3.33 m. [1 mark]
• Correct computation of p = Ls²/(24R) = 0.833 m. [1 mark]
• Verification step: Throw/4 = 0.833 = p, with explicit check mark. [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing only one of the two elements — partial credit for one mark only.
• Not performing the verification step explicitly — examiners cannot infer it from the numbers alone.

Key Phrases To Include

• Ls²/(6R)
• Ls²/(24R)
• 3.333 m
• 0.833 m
• Throw/4
• verified