GELE Surveying (Geomatics) — Spiral (Transition) CurvesRevision Notes
Final-week revision notes for Spiral (Transition) Curves. If you have already studied the full chapter, this page is your go-to refresher before sitting the GELE. Compact, high-yield, and aligned with what Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests in the Surveying (Geomatics) subtest.
Exam context
On the GELE 2026, the Surveying (Geomatics) subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Geodetic Engineering's pattern. Spiral (Transition) Curves lands at position 6th out of 9 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Surveying (Geomatics) on a typical GELE paper.
Spiral (Transition) Curves - Revision Notes
A spiral (transition) curve is inserted between a straight tangent and a circular curve to provide a gradual change in curvature from infinity (tangent) to a finite value (1/R at the circular arc). This prevents the abrupt steering demand and sudden centrifugal force that would occur if the alignment shifted directly from tangent to circle. On Philippine highways, spiral curves are designed following DPWH Highway Design Guidelines and the principles embodied in AASHTO standards, which Philippine practice adopts. This chapter is a consistent source of 2–4 items on the PRC Civil Engineer Licensure Examination (Surveying/Geomatics component). Mastery of the three core relationships — the spiral angle formula, the quadratic angle variation, and the superelevation–friction equation — is sufficient to answer virtually every board question on this topic.
Sections
Exam Tips
- Draw the TS–SC–CS–ST sequence on your scratch paper before solving any spiral problem.
- Remember: at TS, R = ∞ (curvature = 0); at SC, R = design radius (curvature = 1/R).
Key Points
- A spiral curve transitions the alignment from a tangent (R = ∞) to a circular curve (R = finite) over a length Ls.
- The defining property: curvature (1/R) increases linearly with distance along the spiral, so R × ℓ = R × Ls = constant (called the 'clothoid' or 'Euler spiral' property).
- This linear curvature variation allows a driver to steer smoothly and allows superelevation (banking) to be developed gradually.
- On the board exam, spirals appear in three question types: (1) computing the spiral angle, (2) finding the angle or offset at an intermediate point, and (3) computing superelevation or design speed.
Definitions
Term
Spiral (Transition) Curve
Definition
A curve whose radius decreases linearly from infinity at the tangent end (TS) to the design radius R at the spiral-to-curve point (SC). Also called a clothoid or Euler spiral.
Importance
Fundamental concept — examined directly on the PRC board exam.
Term
TS (Tangent-to-Spiral)
Definition
The point where the straight tangent ends and the spiral begins.
Importance
Used as the reference origin for distance ℓ measured along the spiral.
Term
SC (Spiral-to-Curve)
Definition
The point where the spiral ends and the circular arc begins. At SC, the radius equals R.
Importance
The spiral angle θs is measured at this point.
Term
CS (Curve-to-Spiral)
Definition
The point where the circular arc ends and the exit spiral begins (symmetric layout).
Importance
For symmetric spiralized curves, CS mirrors the SC point.
Term
ST (Spiral-to-Tangent)
Definition
The point where the exit spiral ends and the straight tangent resumes.
Importance
Completes the spiralized curve alignment.
Section Title
1. Purpose and Concept of the Spiral Curve
Common Mistakes
- Confusing TS, SC, CS, ST labels — memorize the sequence: Tangent → Spiral → Curve → Spiral → Tangent.
- Thinking curvature increases linearly means radius increases linearly — it is the curvature (1/R) that is linear, not the radius itself.
Formulas
Example
Ls = 80 m, R = 300 m → θs = 80/(2×300) = 0.1333 rad = 0.1333 × (180/π) = 7.64°
Formula
θs = Ls / (2R) [radians]
Variables
θs = spiral angle (rad or °); Ls = full length of spiral (m); R = radius of the circular curve (m).
Application
Compute the total angle subtended by the spiral at its far end (SC point). Convert to degrees: θs° = (Ls × 90) / (πR) = (Ls / 2R) × (180/π).
Example
For θs = 7.64° and ℓ = 40 m (midpoint of Ls = 80 m): θ = 7.64 × (40/80)² = 7.64 × 0.25 = 1.91°
Formula
θ = θs × (ℓ / Ls)²
Variables
θ = angle at point distance ℓ from TS (same units as θs); ℓ = distance from TS to the point (m); Ls = total spiral length (m); θs = spiral angle.
Application
Find the spiral angle at any intermediate point. The angle grows with the square of the fractional distance — at the midpoint, θ = θs/4, not θs/2.
Exam Tips
- Memorize: θs [rad] = Ls/(2R). Memorize: θ at intermediate point uses (ℓ/Ls) SQUARED.
- Quick check: at ℓ = Ls/2, θ should be θs/4 — if you get θs/2, you used linear variation (wrong).
- To convert: θs [deg] = θs [rad] × (180/π) ≈ θs [rad] × 57.296.
Key Points
- The total angle the spiral turns through from TS to SC is the spiral angle θs.
- θs = Ls / (2R) in RADIANS. Convert to degrees using 180/π.
- At any intermediate point at distance ℓ from TS, the angle θ varies as the SQUARE of the distance ratio: θ = θs × (ℓ/Ls)².
- This quadratic variation is the most frequently tested relationship on the board — distinguish it clearly from linear variation.
Definitions
Term
Spiral Angle (θs)
Definition
The total central angle through which the spiral deflects from the TS to the SC. Numerically equal to the central angle of a circular arc of length Ls and radius R.
Importance
Primary formula tested on the board — θs = Ls/(2R) in radians.
Term
Clothoid / Euler Spiral
Definition
The mathematical curve where curvature varies linearly with arc length. The spiral used in highway design is a clothoid.
Importance
Understanding this property justifies the quadratic angle relationship.
Section Title
2. Spiral Angle and Its Variation Along the Spiral
Common Mistakes
- Using θs = Ls/R instead of Ls/(2R) — the factor of 2 in the denominator is critical and frequently missed.
- Applying LINEAR variation for the intermediate angle (θ = θs × ℓ/Ls) instead of QUADRATIC (θ = θs × (ℓ/Ls)²).
- Forgetting to convert radians to degrees when the answer is required in degrees.
- Using degrees directly in θs = Ls/(2R) — the formula gives radians; conversion is required.
Formulas
Example
Ls = 80 m, R = 300 m → Throw = 80²/(6×300) = 6400/1800 = 3.56 m
Formula
Throw (tangent offset at SC) ≈ Ls² / (6R)
Variables
Ls = spiral length (m); R = circular curve radius (m). Result is in meters.
Application
Estimates how far the SC point is offset perpendicularly from the original tangent line. Used to check clearance.
Example
Ls = 80 m, R = 300 m → p = 80²/(24×300) = 6400/7200 = 0.889 m (= Throw/4 = 3.56/4 ✓)
Formula
Shift (p) ≈ Ls² / (24R)
Variables
p = shift or throw-in of the circular arc (m); Ls = spiral length (m); R = circular curve radius (m).
Application
The circular arc is moved inward by p to allow the spiral to fit tangentially. Also used to compute the shifted tangent length.
Exam Tips
- Memorize both: throw = Ls²/(6R); shift = Ls²/(24R). The denominator 24 = 4 × 6.
- If you only remember one, derive the other: shift = throw / 4.
Key Points
- The tangent offset (perpendicular distance from the initial tangent to the SC point) is called the 'throw': x ≈ Ls²/(6R).
- The shift p is the inward offset of the circular arc center needed to accommodate the spiral: p ≈ Ls²/(24R).
- Note the relationship: shift p = throw/4 (one-quarter of the throw). Board exams test these separately, so do not confuse them.
- The horizontal distance along the tangent from TS to the foot of the throw is approximately: y ≈ Ls (for small spiral angles, Ls ≈ long chord).
Definitions
Term
Throw
Definition
The perpendicular distance from the initial tangent to the SC point (end of the spiral). Equals Ls²/(6R) approximately.
Importance
Directly tested in board problems; do not confuse with shift.
Term
Shift (p)
Definition
The distance the circular arc must be moved inward (toward the center) to allow the spiral to connect tangentially from the TS. Equals Ls²/(24R) ≈ Throw/4.
Importance
Used in computing the spiral tangent length; tested separately from throw.
Section Title
3. Offsets, Throw, and Shift
Common Mistakes
- Using Ls²/(24R) when the question asks for the throw (should be Ls²/6R) and vice versa.
- Not recognizing that shift = throw/4 — this relationship can serve as a quick verification.
Formulas
Example
V = 80 km/h, R = 300 m → e + f = 80²/(127×300) = 6400/38100 = 0.168. If f_allowable = 0.15, then e = 0.168 − 0.15 = 0.018 (1.8% superelevation).
Formula
e + f = V² / (127R)
Variables
e = superelevation rate (dimensionless, e.g., 0.08 for 8%); f = side-friction factor (dimensionless); V = design speed (km/h); R = radius of circular curve (m). The constant 127 = g × (1000/3600)² × (1/g) corrected = 9.81 × (1/3.6)² ÷ 9.81 in consistent SI conversion.
Application
Determines the total centripetal demand. Given V and R, find the combined requirement, then allocate between superelevation (design choice) and friction (material/tire limit).
Example
e = 0.06, w = 3.65 m (one lane), Δ = 1/200 (0.005 m/m) → Ls = 0.06 × 3.65 / 0.005 = 43.8 m, round up to nearest design value.
Formula
Ls (min for superelevation runoff) = (e × w) / (Δ)
Variables
Ls = minimum spiral length (m); e = design superelevation rate; w = width of rotating lane(s) (m); Δ = rate of superelevation change per unit length (e.g., 0.005 m/m per lane-width change, per AASHTO).
Application
Ensures superelevation is developed gradually enough to be comfortable and safe.
Exam Tips
- Memorize 127 as the magic constant for V in km/h and R in meters.
- If V is given in m/s: use e + f = V²/(9.81R) instead (no unit conversion needed in SI).
- On the board exam, typical values: e_max = 0.08 to 0.10; f ranges from 0.10 (high speed) to 0.17 (low speed).
Key Points
- On a curved road, centripetal force is resisted by both the superelevation (banking) e and the side friction factor f.
- The fundamental equation is: e + f = V²/(127R), where V is in km/h and R is in meters.
- The constant 127 comes from unit conversion: g = 9.81 m/s² and the conversion factor (3.6)² = 12.96, giving g × (3.6)² ≈ 127.
- Design procedure: select e within code limits (DPWH/AASHTO typically limits e to 0.08 or 0.10), then check that friction demand f = V²/(127R) − e does not exceed the allowable coefficient.
- The spiral length Ls is often set so the full superelevation e is developed over the spiral: Ls = (e × w) / (runoff rate), where w is the lane width.
Definitions
Term
Superelevation (e)
Definition
The transverse slope of the road surface on a curve, expressed as a ratio (e.g., 0.08 = 8%). Acts as banking to partially counteract centrifugal tendency.
Importance
Central to highway curve design; the primary formula e + f = V²/(127R) is a board staple.
Term
Side-Friction Factor (f)
Definition
The coefficient of lateral friction between the tire and the pavement. Depends on speed and pavement condition; AASHTO provides maximum values by design speed.
Importance
Must not be exceeded in design — exceeding f means skidding on the curve.
Term
Superelevation Runoff
Definition
The length over which the full superelevation is developed (or removed) — typically coincides with the spiral length.
Importance
Sets the minimum spiral length in practice.
Section Title
4. Superelevation, Side Friction, and Design Speed
Common Mistakes
- Using V in m/s in the formula e + f = V²/(127R) — V must be in km/h. If V is given in m/s, convert first: V_kmh = V_ms × 3.6.
- Using R in km or any unit other than meters.
- Solving for e + f and reporting it as the superelevation e, ignoring the friction component f.
- Forgetting that the formula gives a MINIMUM R for a given e, f, and V — not a unique answer.
Formulas
Example
Ls = 60 m, R = 250 m → θs = 60×90/(π×250) = 5400/785.4 = 6.875° ≈ 6°52.5'
Formula
θs [deg] = (Ls × 90) / (π × R)
Variables
Direct degree version of θs = Ls/(2R) [rad] × (180/π), simplified to (Ls/2R) × (180/π) = Ls × 90/(πR).
Application
Use this form when the answer is needed directly in degrees, saving a conversion step.
Exam Tips
- Write formulas on your scratch paper before reading the choices — this prevents being misled by distractors.
- Check: for a well-designed highway, θs is typically 5° to 15°. If you get 45°, re-check your substitution.
- Unit trap: if Ls is given in kilometers (rare), convert to meters before substituting.
Key Points
- Three master formulas cover virtually all PRC board questions on spiral curves: (1) θs = Ls/(2R), (2) θ = θs(ℓ/Ls)², (3) e + f = V²/(127R).
- Supporting formulas: throw = Ls²/(6R); shift p = Ls²/(24R).
- Problem-solving sequence: identify what is given (Ls, R, V, e, f, ℓ), identify what is asked, select formula, substitute, convert units if needed.
- Always confirm units: Ls and R in meters; V in km/h; angles in radians or degrees as required.
Section Title
5. Board-Exam Formula Summary and Problem-Solving Strategy
Common Mistakes
- Mixing up Ls/2R (spiral angle formula) with Ls/R (full central angle of a circular arc of the same length — different quantity).
- Applying the throw formula Ls²/(6R) at an intermediate point — this formula gives the offset only at the SC end.
Connections
- Simple Circular Curves: The spiral connects the tangent to the circular arc; spiral design requires knowing R and the degree of curvature from circular curve theory.
- Route Surveying / Curve Staking: Field layout of spirals uses the quadratic angle formula (θ = θs(ℓ/Ls)²) to compute deflection angles at successive stations.
- Highway Design (DPWH Standards): Minimum spiral lengths are governed by superelevation runoff rates — connecting this chapter to pavement cross-section design.
- Centripetal Force / Vehicle Dynamics (Engineering Mechanics): The formula e + f = V²/(127R) derives from Newton's second law applied to a vehicle on a banked curve — connects Surveying to Engineering Mechanics.
- Superelevation Table Design (AASHTO/DPWH): Maximum e values (0.08 for areas with ice/snow risk, 0.10 for tropical countries like the Philippines) set the design boundary for this formula.
- Earthworks / Cross-Sections: Superelevation changes the cross-slope of the road, affecting cut-and-fill volume calculations in earthworks.
Exam Strategy
For PRC board spiral curve questions, execute this 4-step protocol: (1) Identify and list all given values with units — Ls, R, V, e, f, or ℓ. (2) Write the applicable formula from the master list: θs = Ls/(2R), θ = θs(ℓ/Ls)², e+f = V²/(127R), throw = Ls²/(6R), shift = Ls²/(24R). (3) Check units — V must be in km/h, all lengths in meters, angles in radians before conversion. (4) Substitute and solve, then sanity-check the answer: θs should be between 3° and 20° for typical highway spirals; e+f should be between 0.10 and 0.25 for typical speeds and radii. The most common exam traps are: (a) forgetting the factor of 2 in the denominator of θs = Ls/(2R), (b) using linear instead of quadratic angle variation at an intermediate point, and (c) substituting V in m/s into the 127-constant formula. Allocate no more than 3 minutes per spiral question — if you have the three master formulas memorized, each question is a direct substitution.
Quick Review Questions
A spiral curve has a length Ls = 60 m and connects to a circular curve of radius R = 250 m. What is the spiral angle in degrees?
θs = Ls/(2R) = 60/(2×250) = 60/500 = 0.12 rad. Convert: 0.12 × (180/π) = 0.12 × 57.296 = 6.875°. Alternatively, θs = 60×90/(π×250) = 5400/785.4 = 6.875°.
For the spiral in Question 1, what is the spiral angle at a point 30 m from the TS (midpoint)?
θ = θs × (ℓ/Ls)² = 6.875° × (30/60)² = 6.875° × 0.25 = 1.719°. Note: the angle at the midpoint is θs/4, not θs/2, due to the quadratic relationship.
A highway curve has R = 400 m and design speed V = 100 km/h. Compute the required (e + f).
e + f = V²/(127R) = 100²/(127×400) = 10000/50800 = 0.1969 ≈ 0.197. If the allowable friction factor is f = 0.12, the required superelevation would be e = 0.197 − 0.12 = 0.077 (7.7%).
Find the throw (tangent offset at the SC point) for a spiral with Ls = 70 m and R = 350 m.
Throw = Ls²/(6R) = 70²/(6×350) = 4900/2100 = 2.333 m.
What is the shift (p) for the same spiral (Ls = 70 m, R = 350 m)?
p = Ls²/(24R) = 70²/(24×350) = 4900/8400 = 0.583 m. Verify: shift = throw/4 = 2.333/4 = 0.583 m ✓
What is the minimum spiral length required to develop a superelevation of e = 0.06 over a single lane of width 3.6 m at a runoff rate of 1 in 200 (0.005 m/m)?
Ls = (e × w) / Δ = (0.06 × 3.6) / 0.005 = 0.216 / 0.005 = 43.2 m. This sets the minimum length for comfortable and safe superelevation development.
A spiral angle is given as θs = 0.20 rad. The radius of the circular curve is R = 200 m. What is the spiral length Ls?
From θs = Ls/(2R): Ls = 2R × θs = 2 × 200 × 0.20 = 80 m. This is the inverse application of the basic formula — given θs and R, find Ls.
On a curve of R = 300 m with design speed V = 80 km/h, if the superelevation is set to e = 0.08, what side-friction factor f is developed?
e + f = V²/(127R) = 80²/(127×300) = 6400/38100 = 0.168. Therefore f = 0.168 − 0.08 = 0.088. This is within acceptable limits (AASHTO allows f ≈ 0.14 at 80 km/h), confirming the design is safe.
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Horizontal Curves (Simple, Compound, Reverse)
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Vertical (Parabolic) Curves
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