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GELE Surveying (Geomatics) — Horizontal Curves (Simple, Compound, Reverse)Detailed Explanation

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This is the "office hours" version of Horizontal Curves (Simple, Compound, Reverse) for the GELE 2026. No shortcuts, no hand-waving — just a full unpacking of why Professional Regulation Commission (PRC) — Board of Geodetic Engineering cares about each concept and how the Surveying (Geomatics) section items tend to play out on exam day. Read this once, then hit the practice questions with real understanding.

Exam context

For the Geodetic Engineer Licensure Examination, Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests Surveying (Geomatics) under a "Core" label, with Horizontal Curves (Simple, Compound, Reverse) in the 5th slot across 9 chapters. GELE candidates must clear the 70% weighted average, no sub-test below 50% cut on the 2026 paper, which draws about a meaningful share of Surveying (Geomatics) questions. Date to watch: September 2026.

Horizontal Curves (Simple, Compound, Reverse) - Detailed Explanation

Horizontal curves are fundamental elements of route surveying and highway/railway design. They connect two straight tangent lines (called 'back tangent' and 'forward tangent') in the horizontal plane, allowing vehicles to transition smoothly from one direction to another. For the PRC Civil Engineer Licensure Examination, horizontal curves consistently appear in the Surveying (Geomatics) portion, requiring you to compute curve elements, determine stationing of key points, and apply degree-of-curve relationships. This chapter covers three types: (1) Simple Circular Curves — a single arc of constant radius; (2) Compound Curves — two or more circular arcs of different radii turning in the same direction; and (3) Reverse Curves — two circular arcs turning in opposite directions, forming an S-shape. Mastery of the five standard curve elements (T, Lc, LC, E, M) and their derivations, together with correct stationing procedure, is the non-negotiable foundation for any CE board examinee.

Concepts

Simple Circular Curve — Geometry and Key Points

A simple circular curve is a single arc of radius R connecting two straight tangents that meet at the Point of Intersection (PI). The geometry is entirely defined by two parameters: the radius R and the intersection (deflection) angle I, which is the exterior angle between the two tangent directions. Key control points: • PC (Point of Curvature) — where the curve begins; located a tangent distance T back from the PI along the back tangent. • PI (Point of Intersection) — where the two tangents, if extended, would meet. • PT (Point of Tangency) — where the curve ends; located T ahead of the PI along the forward tangent. • Center (O) — the geometric center of the circular arc, equidistant (radius R) from every point on the curve. The angle at the center subtended by the full curve equals the intersection angle I (this is a direct consequence of the theorem that an inscribed angle equals half the central angle — but for tangent-chord geometry, the central angle equals I exactly). The five standard curve elements are derived purely from right-triangle trigonometry applied to the triangle formed by the PI, the center O, and either PC or PT: 1. Tangent Distance: T = R · tan(I/2) — Distance from PC to PI (or from PI to PT). 2. Length of Curve: Lc = (π · R · I) / 180 [I in degrees] — Arc length from PC to PT along the curve. Equivalently, Lc = R · I_rad where I_rad = I × π/180. 3. Long Chord: LC = 2R · sin(I/2) — Straight-line distance from PC to PT. 4. External Distance: E = R · (sec(I/2) − 1) — Distance from PI to the midpoint of the curve (measured toward the center). 5. Middle Ordinate: M = R · (1 − cos(I/2)) — Perpendicular distance from the midpoint of the long chord to the midpoint of the arc. Note the relationship: E > M always (except when I = 0). Also, LC < Lc always (chord is shorter than arc).

Examples

Each element follows directly from right-triangle trigonometry. The triangle formed by PC–PI–O (center) has a right angle at PC, hypotenuse PI–O = R/cos(I/2) = R·sec(I/2), and the leg PI–O minus R gives E. The half-chord from PC to mid-arc is R·sin(I/2), doubled for LC. Always verify consistency: chord must be less than arc, and external must exceed middle ordinate.

Scenario

A simple curve has R = 300 m and I = 40°. Compute all five curve elements: T, Lc, LC, E, and M.

Solution

Step 1 — Tangent distance: T = R tan(I/2) = 300 × tan(20°) tan(20°) = 0.36397 T = 300 × 0.36397 = 109.19 m Step 2 — Length of curve: Lc = πRI/180 = π × 300 × 40 / 180 Lc = 37,699.11 / 180 = 209.44 m Step 3 — Long chord: LC = 2R sin(I/2) = 2 × 300 × sin(20°) sin(20°) = 0.34202 LC = 600 × 0.34202 = 205.21 m Step 4 — External distance: E = R(sec(I/2) − 1) = 300 × (sec(20°) − 1) sec(20°) = 1/cos(20°) = 1/0.93969 = 1.06418 E = 300 × (1.06418 − 1) = 300 × 0.06418 = 19.25 m Step 5 — Middle ordinate: M = R(1 − cos(I/2)) = 300 × (1 − cos(20°)) = 300 × (1 − 0.93969) = 300 × 0.06031 = 18.09 m Check: LC < Lc → 205.21 < 209.44 ✓ Check: E > M → 19.25 > 18.09 ✓

Stationing is a chainaged distance measured continuously along the alignment. The curve replaces the straight-line path from PC to PT along the arc. Therefore, the station of PT equals the station of PC plus the arc length Lc. The tangent length 2T represents the straight-line distance PC–PI–PT and is irrelevant to stationing once inside the curve.

Scenario

Using the same curve (R = 300 m, I = 40°), the PI is at station 10+120 (i.e., 10,120 m). Find the stations of the PC and PT.

Solution

Step 1 — Station of PC: Sta PC = Sta PI − T = 10,120 − 109.19 = 10,010.81 m → 10+010.81 Step 2 — Station of PT: Sta PT = Sta PC + Lc = 10,010.81 + 209.44 = 10,220.25 m → 10+220.25 VERIFICATION (common mistake check): If you incorrectly add 2T to PI: 10,120 + 109.19 = 10,229.19 m — THIS IS WRONG. The correct approach always adds Lc to PC, not 2T to PI.

Applications

Determining the layout of highway alignments and locating stake positions for construction.
Computing earthwork volumes along curved sections of road.
Setting out curves in the field using deflection angles or chord-offset methods.
Checking sight distances on horizontal curves for traffic safety (DPWH standards).
Railway track alignment and cant (superelevation) computation.

Misconceptions

WRONG: Sta PT = Sta PI + T. CORRECT: Sta PT = Sta PC + Lc. The curve length, not the tangent, is added.
WRONG: Using I (not I/2) in the trig functions for T, LC, E, M. All five formulas use I/2.
WRONG: Using degrees directly in Lc = R·I without the π/180 conversion. Use Lc = πRI/180 or convert I to radians first.
WRONG: Confusing E (external) and M (middle ordinate). E uses secant (measured from PI inward); M uses cosine (measured from LC midpoint to arc midpoint). E > M always.
WRONG: Assuming LC = Lc. The long chord is always shorter than the arc length.

Related Concepts

Degree of Curve (arc and chord definition)
Stationing and chainage
Deflection angle method for curve layout
Compound and Reverse Curves
Vertical Curves (separate chapter)

Common Exam Questions

Example

Given R = 200 m and I = 60°, find the external distance E. E = 200(sec 30° − 1) = 200(1.15470 − 1) = 200(0.15470) = 30.94 m

Approach

Identify which formula applies (T, Lc, LC, E, or M). Substitute directly. Watch the angle: always use I/2 in the trig functions. Keep I in degrees for the Lc formula (use π/180 factor) or convert to radians first.

Question Type

Compute a specific curve element given R and I

Example

PI at 3+450, R = 500 m, I = 30°. T = 500 tan 15° = 500(0.26795) = 133.97 m Lc = π(500)(30)/180 = 261.80 m Sta PC = 3450 − 133.97 = 3316.03 m → 3+316.03 Sta PT = 3316.03 + 261.80 = 3577.83 m → 3+577.83

Approach

Compute T first. Sta PC = Sta PI − T. Then Sta PT = Sta PC + Lc. Do NOT add 2T to PI.

Question Type

Find stations of PC and PT given PI station and curve data

Example

For R = 300, I = 40°: M should equal E·cos(20°) = 19.25 × 0.93969 = 18.09 m ✓

Approach

Use relationships: E = T·tan(I/4) or E = (LC/2)·tan(I/4) as alternate checks. Also M = E·cos(I/2) is a useful cross-check.

Question Type

Verify or cross-check curve elements

Key Points To Remember

The intersection angle I equals the central angle subtended by the curve.
T = R tan(I/2); both PC and PT are equidistant from PI.
Lc = πRI/180 uses I in DEGREES; or Lc = R·I_rad with I in radians.
LC = 2R sin(I/2) — the straight chord, always shorter than Lc.
E uses the SECANT function: E = R(sec(I/2) − 1).
M uses the COSINE function: M = R(1 − cos(I/2)).
Both E and M are measured from the midpoint of LC to the midpoint of the arc.
The relationship E = M + M²/(2R) approximately holds for small angles.
Sta PT = Sta PC + Lc (add the ARC length, NOT 2T).
Sta PC = Sta PI − T.

Degree of Curve — Arc and Chord Definitions

The degree of curve D is a measure of the sharpness of a curve. It is defined as the central angle (in degrees) subtended by a standard chord or arc length. There are two definitions in use: 1. ARC DEFINITION (standard in Philippine practice and most SI systems): D is the central angle subtended by a 20-m arc. Since the arc length formula gives: 20 = (π R D) / 180 Solving for R: R = (20 × 180) / (π × D) = 3600 / (π D) Numerically: R = 1145.916 / D [R in metres, D in degrees] This is the formula you will use in 99% of board exam problems in the Philippines. 2. CHORD DEFINITION (used in US practice, occasionally appears in exams): D is the central angle subtended by a 20-m chord. Since the chord formula gives: 20 = 2R sin(D/2) Solving: sin(D/2) = 10/R → R = 10 / sin(D/2) For small D (gentle curves), sin(D/2) ≈ D·π/360, so both definitions give nearly the same R. For sharp curves (D > 5°), the difference becomes significant. Practical implication: • A 1° curve has R ≈ 1145.92 m — very gentle (expressways). • A 6° curve has R ≈ 191.0 m — moderately sharp (national roads). • A 10° curve has R ≈ 114.6 m — sharp (local roads). Relationship between D and curve elements: Once you find R from D, all five elements follow as before: T = R tan(I/2), Lc = πRI/180 = (20/D)×I, etc. Note that Lc = (I/D) × 20 m is a very quick formula: the curve length is simply the ratio of the total central angle to the degree of curve, multiplied by the standard arc length (20 m).

Examples

Method B works because the arc definition is built on the 20-m arc unit: every degree of central angle corresponds to exactly 20/D metres of arc. So for I degrees of total central angle and D degrees per 20-m arc, the total arc = (I/D) × 20. This is the fastest way to compute Lc when D is given.

Scenario

A highway curve has a degree of curve D = 4° (arc definition) and an intersection angle I = 50°. Find R, T, and Lc.

Solution

Step 1 — Radius from arc definition: R = 1145.916 / D = 1145.916 / 4 = 286.48 m Step 2 — Tangent distance: T = R tan(I/2) = 286.48 × tan(25°) tan(25°) = 0.46631 T = 286.48 × 0.46631 = 133.58 m Step 3 — Length of curve (two methods): Method A: Lc = πRI/180 = π(286.48)(50)/180 = 250.00 m Method B (quick): Lc = (I/D) × 20 = (50/4) × 20 = 12.5 × 20 = 250.00 m ✓ Both methods agree perfectly — use Method B as a quick check!

The chord and arc definitions converge for small central angles because sin(D/2) ≈ D·π/360 for small D. For sharp curves (D > 6°), always identify which definition the problem uses — the exam will often state it explicitly.

Scenario

A curve is described as having D = 5° using the chord definition. Find the radius.

Solution

Using chord definition: sin(D/2) = 10/R sin(2.5°) = 10/R 0.04362 = 10/R R = 10/0.04362 = 229.25 m Comparison with arc definition: R_arc = 1145.916/5 = 229.18 m Difference = 229.25 − 229.18 = 0.07 m (very small for D = 5°) Conclusion: For D ≤ 5°, the two definitions are practically interchangeable.

Applications

Specifying road geometry in DPWH highway design standards.
Calculating curve lengths directly from chainage stations and D.
Field layout using the sub-arc or sub-chord intervals.
Comparing sharpness of different curves in route selection.

Misconceptions

WRONG: Using R = 1145.916/D for the chord definition. This formula is ONLY for the arc definition.
WRONG: Thinking a higher D means a larger radius. Higher D = sharper curve = smaller R.
WRONG: Using a 100-ft (30.48-m) standard without converting when working in SI units.

Related Concepts

Simple Circular Curve Elements
Curve length and stationing
Sub-chord and sub-arc layout in the field

Common Exam Questions

Example

D = 3°, I = 45°, arc definition. R = 1145.916/3 = 381.97 m T = 381.97 tan(22.5°) = 381.97 × 0.41421 = 158.22 m Lc = (45/3) × 20 = 300.00 m

Approach

Convert D to R using R = 1145.916/D (arc definition) or R = 10/sin(D/2) (chord definition). Then apply T, Lc, LC, E, M formulas. Use Lc = (I/D)×20 as a quick Lc calculation.

Question Type

Given D, find R and compute all curve elements

Example

D = 10°: R_arc = 114.59 m; R_chord = 10/sin(5°) = 10/0.08716 = 114.73 m. Difference = 0.14 m.

Approach

Compute R from both definitions and note the difference. Show that for small D, differences are negligible.

Question Type

Compare arc vs chord definition for a given D

Key Points To Remember

Arc definition (Philippine standard): R = 1145.916 / D
Chord definition: sin(D/2) = 10/R, so R = 10/sin(D/2)
Quick Lc formula with arc definition: Lc = (I/D) × 20 m
Higher D means sharper curve and smaller radius.
For D ≤ 2°, arc and chord definitions give virtually identical results.
Always verify which definition is used in the problem statement.
The 20-m standard arc corresponds to SI practice; US practice uses 100 ft.

Stationing of Curve Points

Stationing (chainage) is the system of continuously measuring distances along a road or railway alignment from a reference point (usually station 0+000). Stations are written as kilometer + meters, e.g., 5+240 means 5,240 m from the origin. Fundamental stationing rules for horizontal curves: 1. Sta PC = Sta PI − T The PC is located T metres back from the PI along the back tangent. 2. Sta PT = Sta PC + Lc The PT is located Lc metres ahead of the PC, measured ALONG THE ARC. CRITICAL: You do NOT add or subtract T from the PI station to get the PT. The alignment passes along the curve from PC to PT, so the chainaged distance is Lc, the arc length. Stationing of intermediate points on the curve: For any point P on the curve at a central angle θ from the PC (θ ≤ I): Sta P = Sta PC + (πRθ/180) [θ in degrees] Staking the curve in the field: Conventionally, the first full station after the PC and all subsequent full stations up to the PT are staked. If the PC is at 10+010.81 and stations are every 20 m, the first full stake is at 10+020, then 10+040, etc. Sub-arc to first full station: l₁ = (first full station after PC) − Sta PC Sub-arc to last sub-station before PT: l₂ = Sta PT − (last full station before PT) These sub-arcs correspond to central angles: θ₁ = (l₁ × D) / 20° [arc definition] θ₂ = (l₂ × D) / 20°

Examples

The alignment follows the arc, not the two tangent segments. So the 'savings' relative to going around via the PI = 2T − Lc = 162.46 − 157.08 = 5.38 m. This is why adding T to PI gives a station that is 5.38 m beyond the actual PT station.

Scenario

A simple curve has R = 250 m, I = 36°, and the PI is at station 5+240. Find Sta PC and Sta PT.

Solution

Step 1 — Compute T: T = R tan(I/2) = 250 × tan(18°) tan(18°) = 0.32492 T = 250 × 0.32492 = 81.23 m Step 2 — Compute Lc: Lc = πRI/180 = π(250)(36)/180 = π(250)(0.2) × π ... let's compute directly: = (π × 250 × 36) / 180 = 28274.33 / 180 = 157.08 m Step 3 — Sta PC: Sta PC = 5240 − 81.23 = 5158.77 m → 5+158.77 Step 4 — Sta PT: Sta PT = 5158.77 + 157.08 = 5315.85 m → 5+315.85 Verification: PI − T + Lc = 5240 − 81.23 + 157.08 ≠ PI + T (5321.23) The PT station (5315.85) is less than PI+T (5321.23), confirming Lc < 2T for this case. Note: Lc vs 2T comparison: 157.08 vs 162.46 → Lc < 2T for small I. ✓

Any point on the curve can be stationed simply by adding its arc distance from the PC to the PC station. The central angle is found by the inverse of the arc-length formula.

Scenario

For the above curve (Sta PC = 5+158.77, Sta PT = 5+315.85, D via arc def.), find the station of a point P on the curve that is 60 m along the arc from the PC.

Solution

Sta P = Sta PC + arc distance = 5158.77 + 60.00 = 5218.77 m → 5+218.77 Central angle to P: θ = (l/R) × (180/π) = (60/250) × (180/π) = 0.24 × 57.2958° = 13.75° This point subtends 13.75° of central angle from the PC.

Applications

Field staking of roads and railways using chainage pegs.
Quantity estimation: earthwork volumes at given stations.
Locating drainage structures, culverts, and bridges along the alignment.
As-built surveys to verify constructed alignment matches design.

Misconceptions

WRONG: Sta PT = Sta PI + T. This is the most common board-exam mistake.
WRONG: Forgetting that the alignment follows the arc, not the tangent segments.
WRONG: Using the long chord LC instead of arc Lc for computing Sta PT.

Related Concepts

Simple Curve Elements (T and Lc)
Degree of Curve
Compound Curve stationing
Sub-arc layout in field surveying

Common Exam Questions

Example

PI at 8+350, R = 400 m, I = 24°. T = 400 tan 12° = 400(0.21256) = 85.02 m. Lc = π(400)(24)/180 = 167.55 m. Sta PC = 8350 − 85.02 = 8264.98 = 8+264.98. Sta PT = 8264.98 + 167.55 = 8432.53 = 8+432.53.

Approach

Always compute T first. Sta PC = PI − T. Compute Lc. Sta PT = PC + Lc. This is the most frequently tested stationing problem.

Question Type

Find Sta PC and Sta PT from given PI station and curve data

Example

Point at central angle 15° from PC; R = 400 m. Arc = π(400)(15)/180 = 104.72 m. Sta = 8264.98 + 104.72 = 8369.70 = 8+369.70.

Approach

Determine the arc length from PC to the point (given directly or computed from central angle). Add to Sta PC.

Question Type

Find the station of a specific point on the curve

Key Points To Remember

Sta PC = Sta PI − T (subtract tangent from PI).
Sta PT = Sta PC + Lc (add arc length to PC).
NEVER compute Sta PT as Sta PI + T or Sta PI + (something involving T).
Stationing follows the actual alignment path: back tangent → arc → forward tangent.
For intermediate curve points, add the partial arc length to Sta PC.
Sub-chords and sub-arcs arise at the beginning and end of the curve where the spacing doesn't fall exactly on a full station.

Compound Curves

A compound curve consists of two or more simple circular curves of DIFFERENT radii that turn in the SAME direction and join at a common tangent point called the PCC (Point of Compound Curvature) or, in some references, the PRC. The two component curves share a common tangent at the PCC — meaning the center of each arc lies on the same line through the PCC, and there is no abrupt change in direction at that point (though there is a change in curvature). Notation for a two-arc compound curve: • R₁ = radius of the first (back) curve, I₁ = its central angle • R₂ = radius of the second (forward) curve, I₂ = its central angle • Total intersection angle: I = I₁ + I₂ • T₁ = R₁ tan(I₁/2) — tangent of first curve • T₂ = R₂ tan(I₂/2) — tangent of second curve The compound curve occupies a total tangent length from the PC (back) to the PT (forward). The key geometric problem is usually: Given the two radii, two central angles, and the distance between the PI₁ (if separate) — find the common PCC station and the overall layout. Typical configuration: The overall PI lies at the intersection of the two original tangents. The back tangent length from PC to PI and the forward tangent length from PI to PT are: Back tangent length = T₁ + (T₂·sin(I₁)/sin(I)) Forward tangent length = T₂ + (T₁·sin(I₂)/sin(I)) [These are derived from the sine rule on the triangle formed by PC, PI, and the intermediate point.] Stationing: Sta PC is found from the PI station as usual for the first curve: Sta PC = Sta PI − (back tangent length) Sta PCC = Sta PC + Lc₁ where Lc₁ = πR₁I₁/180 Sta PT = Sta PCC + Lc₂ where Lc₂ = πR₂I₂/180

Examples

The stationing follows the alignment continuously: back tangent → first arc (Lc₁) → second arc (Lc₂) → forward tangent. Each arc is computed independently using its own radius and central angle. The given back tangent length accounts for the compound geometry at the PI.

Scenario

A compound curve has R₁ = 400 m, I₁ = 25°, R₂ = 250 m, I₂ = 20°. The total intersection angle I = 45°. The PI is at station 12+000. The back tangent length (PC to PI) = 200.45 m (given). Find Sta PC, Sta PCC, and Sta PT.

Solution

Given: Back tangent = 200.45 m (includes effects of both curves on the PI position). Step 1 — Sta PC: Sta PC = Sta PI − back tangent = 12000 − 200.45 = 11799.55 m → 11+799.55 Step 2 — Length of first curve: Lc₁ = πR₁I₁/180 = π(400)(25)/180 = 31415.93/180 = 174.53 m Step 3 — Sta PCC: Sta PCC = Sta PC + Lc₁ = 11799.55 + 174.53 = 11974.08 m → 11+974.08 Step 4 — Length of second curve: Lc₂ = πR₂I₂/180 = π(250)(20)/180 = 15707.96/180 = 87.27 m Step 5 — Sta PT: Sta PT = Sta PCC + Lc₂ = 11974.08 + 87.27 = 12061.35 m → 12+061.35

Applications

Transitioning between high-speed and low-speed sections of a highway.
Matching existing road geometry where space constraints prevent a single-radius curve.
Railway track design where grade and curvature must be balanced.

Misconceptions

WRONG: Treating a compound curve as a single curve with average radius (R₁+R₂)/2.
WRONG: Assuming the PCC station is simply the midpoint between PC and PT.
WRONG: Using the total I with a single radius to compute Lc.

Related Concepts

Simple Circular Curve
Stationing
Reverse Curves
Transition (Spiral) Curves

Common Exam Questions

Example

R₁=300m, I₁=20°, R₂=200m, I₂=30°, Sta PC=4+100. Lc₁=π(300)(20)/180=104.72m. Lc₂=π(200)(30)/180=104.72m. PCC=4+204.72. PT=4+309.44.

Approach

Identify R₁, I₁, R₂, I₂. Compute Lc₁ and Lc₂ separately. Sta PCC = Sta PC + Lc₁. Sta PT = Sta PCC + Lc₂.

Question Type

Find PCC and PT stations in a compound curve

Key Points To Remember

Compound curves have two different radii (R₁ ≠ R₂) turning in the SAME direction.
The PCC (Point of Compound Curvature) is where the two curves meet.
At PCC, there is a common tangent — direction is continuous but curvature changes abruptly.
Total intersection angle: I = I₁ + I₂.
Each arc has its own T, Lc, computed separately using respective R and I.
Stationing: PC → (Lc₁) → PCC → (Lc₂) → PT.
Compound curves are used where different design speeds or constraints apply to different sections.

Reverse Curves

A reverse curve consists of two circular arcs of equal or different radii that curve in OPPOSITE directions, meeting at a common point called the PRC (Point of Reverse Curvature). In plan view, the alignment traces an S-shape. Key features: • The two arcs turn in opposite directions (e.g., first arc turns left, second arc turns right). • At the PRC, the common tangent is shared, but the centers of the two arcs lie on opposite sides. • There is an abrupt change in the direction of superelevation at the PRC. Why reverse curves are restricted on high-speed roads: In modern highway design (DPWH standards, AASHTO), reverse curves without a tangent between them are generally prohibited on high-speed roads because: 1. There is insufficient distance to transition superelevation (banking) from one side to the other. 2. Drivers cannot comfortably and safely transition from one curve direction to the other at high speed. A minimum tangent length between the two curves is required for superelevation runoff. Special case — Reverse curve with parallel tangents: This is the most common board-exam type. If the two end tangents are parallel (offset by a perpendicular distance d), then: d = R₁(1 − cos I₁) + R₂(1 − cos I₂) For equal radii (R₁ = R₂ = R) and equal central angles (I₁ = I₂ = I): d = 2R(1 − cos I) And the horizontal distance between the PCs: L = R₁ sin I₁ + R₂ sin I₂ (For equal R and I: L = 2R sin I) Stationing for a reverse curve: Sta PC₁ → + Lc₁ → Sta PRC → + Lc₂ → Sta PT (same sequential approach as for compound curves)

Examples

The trigonometric identity (1 − cos I)/sin I = tan(I/2) is the key to solving this system without iteration. This type of problem — parallel tangents, equal radii — is extremely common on CE board exams. Memorize tan(I/2) = d/L as an equivalent quick formula: tan(I/2) = (d/2)/(L/2) = d/L = 12/80 = 0.15. ✓

Scenario

Two parallel tangents are offset by d = 12 m. A reverse curve with equal radii R₁ = R₂ = R and equal central angles I₁ = I₂ = I connects them. The distance between the PCs along the tangent direction is L = 80 m. Find R and I.

Solution

For a reverse curve with parallel tangents, equal R and equal I: Equation 1: d = 2R(1 − cos I) → 12 = 2R(1 − cos I) ... (1) Equation 2: L = 2R sin I → 80 = 2R sin I → R sin I = 40 ... (2) From (1): R(1 − cos I) = 6 ... (1') From (2): R sin I = 40 ... (2') Divide (1') by (2'): (1 − cos I) / sin I = 6/40 = 0.15 Using the identity: (1 − cos I)/sin I = tan(I/2) So: tan(I/2) = 0.15 I/2 = arctan(0.15) = 8.531° I = 17.06° From (2'): R = 40/sin(17.06°) = 40/0.29350 = 136.29 m Verification with (1'): R(1 − cos I) = 136.29(1 − cos 17.06°) = 136.29(1 − 0.95592) = 136.29(0.04408) = 6.00 m ✓

Applications

Urban road design where limited right-of-way forces S-curves.
Railway realignment to avoid obstacles.
Connecting parallel roads with different chainages.
Layout of interchange ramps (with adequate tangent lengths for superelevation).

Misconceptions

WRONG: Confusing reverse curve (S-shape, opposite directions) with compound curve (same direction, different radii).
WRONG: For parallel tangent problems, forgetting that the identity tan(I/2) = d/L applies only when R₁=R₂ and I₁=I₂.
WRONG: Assuming reverse curves are always permissible — they require a tangent insert on high-speed roads.

Related Concepts

Compound Curves
Simple Circular Curve Elements
Superelevation and transition spirals
DPWH highway design standards

Common Exam Questions

Example

d = 8 m, L = 60 m. tan(I/2) = 8/60 = 0.13333. I/2 = 7.595°. I = 15.19°. R = 30/sin(15.19°) = 30/0.26196 = 114.53 m.

Approach

Set up the two equations: d = 2R(1−cosI) and L = 2R sinI. Divide to get tan(I/2) = d/L. Solve for I, then find R from L = 2R sinI.

Question Type

Find R and I for a reverse curve with parallel tangents

Example

From above: Lc = 2 × π(136.29)(17.06)/180 = 2 × 40.59 = 81.18 m.

Approach

Compute each arc: Lc₁ = πR₁I₁/180, Lc₂ = πR₂I₂/180. Total length = Lc₁ + Lc₂.

Question Type

Find the length of the reverse curve

Key Points To Remember

Reverse curves turn in OPPOSITE directions — the S-shape alignment.
PRC = Point of Reverse Curvature (where the two arcs meet).
For parallel tangents with equal radii and equal I: d = 2R(1−cosI); L = 2R sinI.
Reverse curves without an intervening tangent are prohibited on high-speed roads (DPWH).
Stationing: Sta PT = Sta PC + Lc₁ + Lc₂.
The common tangent at PRC ensures directional continuity but curvature reverses.
Board exams often give parallel tangent conditions and ask for R or I.

Practice Problems

Notice that when R = 180 m, the formula Lc = πRI/180 simplifies to Lc = πI (in metres, with I in degrees). This is a useful mental shortcut. All five elements are computed from the half-angle I/2 = 26°, using tan, sin, cos, and secant respectively.

Problem

Problem 1 (Curve Elements): A simple circular curve has R = 180 m and I = 52°. Compute: (a) T, (b) Lc, (c) LC, (d) E, (e) M.

Solution

(a) T = R tan(I/2) = 180 tan(26°) = 180(0.48773) = 87.79 m (b) Lc = πRI/180 = π(180)(52)/180 = π(52) = 163.36 m [Note: π × 52 = 163.36 — a nice simplification when R = 180 makes the 180s cancel] (c) LC = 2R sin(I/2) = 2(180) sin(26°) = 360(0.43837) = 157.81 m (d) E = R(sec(I/2) − 1) = 180(sec 26° − 1) sec 26° = 1/cos 26° = 1/0.89879 = 1.11270 E = 180(1.11270 − 1) = 180(0.11270) = 20.29 m (e) M = R(1 − cos(I/2)) = 180(1 − cos 26°) = 180(1 − 0.89879) = 180(0.10121) = 18.22 m Consistency checks: LC < Lc → 157.81 < 163.36 ✓; E > M → 20.29 > 18.22 ✓

Always compute T and Lc first, then apply the two fundamental stationing formulas. Showing the incorrect alternative helps reinforce why the correct method matters — the error is small for gentle curves but grows as I increases.

Problem

Problem 2 (Stationing): A curve has R = 320 m and I = 30°. The PI is at station 7+480. Find the stations of PC and PT.

Solution

Step 1 — Compute T: T = R tan(I/2) = 320 tan(15°) = 320(0.26795) = 85.74 m Step 2 — Compute Lc: Lc = πRI/180 = π(320)(30)/180 = 167.55 m Step 3 — Stations: Sta PC = 7480 − 85.74 = 7394.26 m → 7+394.26 Sta PT = 7394.26 + 167.55 = 7561.81 m → 7+561.81 Wrong answer (common mistake): Sta PT = 7480 + 85.74 = 7565.74 m — this adds T to PI instead of Lc to PC. The difference = 7565.74 − 7561.81 = 3.93 m (the shortening of the arc vs 2T).

The quick formula Lc = (I/D)×20 is derived from the definition of D: one degree of D subtends exactly 20 m of arc. So I degrees of total curve subtends (I/D) groups of 20 m. This is computationally faster than using π explicitly and is very useful in multiple-choice exams.

Problem

Problem 3 (Degree of Curve): A road curve is specified as D = 5° using the arc definition. The intersection angle is I = 60°. Find: (a) R, (b) T, (c) Lc using the quick formula, (d) E.

Solution

(a) R = 1145.916/D = 1145.916/5 = 229.18 m (b) T = R tan(I/2) = 229.18 tan(30°) = 229.18(0.57735) = 132.33 m (c) Quick formula: Lc = (I/D) × 20 = (60/5) × 20 = 12 × 20 = 240.00 m Verify: Lc = πRI/180 = π(229.18)(60)/180 = π(229.18)(1/3) = 240.00 m ✓ (d) E = R(sec 30° − 1) = 229.18(1.15470 − 1) = 229.18(0.15470) = 35.45 m

The identity tan(I/2) = d/L (valid only for equal R and equal I) is the fastest entry point into reverse-curve-with-parallel-tangents problems. Once I is found, R follows from the L equation. Always verify with the d equation. Note that the total curve length (91.54 m) is slightly longer than L (90 m), which makes geometric sense.

Problem

Problem 4 (Reverse Curve — Parallel Tangents): Two parallel highway tangents are separated by a perpendicular distance of d = 15 m. A reverse curve with equal radii and equal central angles connects them. The distance along the tangent between the two PC points is L = 90 m. Find R, I, and the total length of the reverse curve.

Solution

Step 1 — Find I using tan(I/2) = d/L: tan(I/2) = 15/90 = 0.16667 I/2 = arctan(0.16667) = 9.462° I = 18.92° Step 2 — Find R from L = 2R sin I: 90 = 2R sin(18.92°) sin(18.92°) = 0.32444 90 = 2R(0.32444) R = 90/(2 × 0.32444) = 90/0.64888 = 138.70 m Verification: d = 2R(1 − cos I) = 2(138.70)(1 − cos 18.92°) cos(18.92°) = 0.94604 d = 277.40(1 − 0.94604) = 277.40(0.05396) = 14.97 m ≈ 15 m ✓ (rounding) Step 3 — Total curve length: Lc_each = πRI/180 = π(138.70)(18.92)/180 = 45.77 m Total Lc = 2 × 45.77 = 91.54 m

This five-part problem mirrors typical CE board exam multi-part questions. The midpoint of the curve is simply Sta PC + Lc/2, not the PI. The PI is the intersection of the tangents and lies outside the curve, while the midpoint lies on the arc itself.

Problem

Problem 5 (Board-Exam Style — Multiple Elements): A highway curve has a degree of curve D = 3° (arc definition). The PI is at station 6+500 and the deflection angle is I = 48°. A road sign is to be placed at the midpoint of the curve. Find: (a) R, (b) T, (c) Sta PC, (d) Sta PT, (e) Station of the midpoint of the curve.

Solution

(a) Radius: R = 1145.916/3 = 381.97 m (b) Tangent distance: T = 381.97 tan(24°) = 381.97(0.44523) = 170.06 m (c) Sta PC: Sta PC = 6500 − 170.06 = 6329.94 m → 6+329.94 (d) Length of curve and Sta PT: Lc = (I/D) × 20 = (48/3) × 20 = 16 × 20 = 320.00 m Sta PT = 6329.94 + 320.00 = 6649.94 m → 6+649.94 (e) Station of midpoint: Midpoint is at arc distance = Lc/2 = 160 m from PC. Sta midpoint = 6329.94 + 160.00 = 6489.94 m → 6+489.94 Note: The midpoint of the curve (6+489.94) is NOT the same as the PI station (6+500). They differ by 10.06 m — this is the external distance area effect.

Exam Preparation Tips

MEMORIZE the five formulas in one line: T=Rtan(I/2), Lc=πRI/180, LC=2Rsin(I/2), E=R(sec(I/2)−1), M=R(1−cos(I/2)). Write them out 10 times before the exam.
TATTOO this in your memory: Sta PT = Sta PC + Lc. NOT Sta PI + T. This single mistake is responsible for the most wrong answers in the CE board surveying portion.
For degree-of-curve problems using the arc definition, always start with R = 1145.916/D. Then use the quick Lc = (I/D)×20 to get curve length without computing π explicitly.
For reverse curves with parallel tangents, the shortcut is tan(I/2) = d/L (equal radii and equal I). Derive this from the two equations d=2R(1−cosI) and L=2R sinI by dividing and applying the half-angle identity.
In multiple-choice exams, compute ALL five curve elements even if only one is asked. The extra work takes 2 minutes and allows you to eliminate wrong choices using consistency checks (LC < Lc, E > M).
Always keep your calculator in DEGREE mode, not radian mode. The most insidious error is computing trig functions in the wrong mode — your answer will be completely wrong but you won't notice until you check.
When the problem says 'degree of curve' without specifying arc or chord definition, default to the ARC DEFINITION (R = 1145.916/D) in Philippine board exams. Specify otherwise if the chord definition formula is required.
Practice the E vs M distinction: E uses SECANT (extends to PI), M uses COSINE (stays within the arc). A memory trick: E is External = goes out to PI = uses sec. M is Middle = stays inside = uses cos.
For compound curve problems, treat each arc completely independently — compute T₁, Lc₁ for the first curve with R₁, I₁; and T₂, Lc₂ for the second with R₂, I₂. Station sequentially: PC → PCC → PT.
Time management: Horizontal curve problems in the CE board typically take 3–5 minutes each. If a problem requires more than 7 minutes, flag it and move on — return after completing faster problems.
Review the DPWH Highway Design Standards for minimum radius requirements by design speed. Board exams occasionally include policy questions: e.g., 'Why are reverse curves without tangent inserts prohibited on expressways?' Answer: inadequate superelevation transition distance.
Practice verifying answers: After computing Sta PT, check that the arc length between PC and PT (= Lc) is plausible relative to the geometry. For gentle curves, Lc is only slightly longer than LC (the chord). If Lc << LC, you made an error.

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In summary

Horizontal curves — simple, compound, and reverse — form one of the most reliably tested topics in the PRC Civil Engineer Licensure Examination Surveying (Geomatics) portion. The entire computational framework rests on five formulas (T, Lc, LC, E, M) derived from basic circular geometry, plus two stationing rules (Sta PC = PI − T; Sta PT = PC + Lc). The degree-of-curve relationship R = 1145.916/D (arc definition) and the quick formula Lc = (I/D)×20 dramatically speed up calculations on time-pressured board exams. For compound curves, treat each arc independently and station sequentially through the PCC. For reverse curves with parallel tangents, the elegant identity tan(I/2) = d/L reduces a two-equation system to a single-step angle computation. The most consequential mistake to avoid — and the one most frequently cited in post-exam reviews — is computing Sta PT by adding T to the PI instead of adding Lc to the PC. Internalize the physical meaning: the road follows the arc, and stationing measures distance along the road. Master these principles through repeated problem-solving with the worked examples in this chapter, and horizontal curves will become one of your most confident scoring areas on the CE board.

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