Horizontal Curves (Simple, Compound, Reverse) - Revision Notes
Horizontal curves are geometric elements used to transition a road, railway, or any linear infrastructure alignment from one straight tangent direction to another in the horizontal plane. Mastery of horizontal curve geometry is consistently tested in the PRC Civil Engineer Licensure Examination under Surveying (Geomatics). This chapter covers three curve types: the simple circular curve (most common), the compound curve (two arcs of different radii, same direction), and the reverse curve (two arcs of opposite direction forming an S-shape). All formulas are derived from basic circular geometry and trigonometry. A firm grasp of the five principal elements — Tangent (T), Curve Length (Lc), Long Chord (LC), External Distance (E), and Middle Ordinate (M) — plus correct stationing procedure, is essential for solving board examination problems efficiently.
1. Geometry of the Simple Circular Curve
Formulas
T = R × tan(I/2)
Example
Given R = 300 m, I = 40°: T = 300 × tan(20°) = 300 × 0.36397 = 109.19 m
Variables
T = tangent distance (m); R = radius of curve (m); I = intersection angle (degrees)
Application
Compute the distance from PC to PI (or from PI to PT). Used to locate the PC and PT from the PI station.
Lc = (π × R × I) / 180 OR Lc = R × I_rad
Example
R = 300 m, I = 40°: Lc = (π × 300 × 40) / 180 = 37699.11 / 180 = 209.44 m
Variables
Lc = arc length of curve (m); R = radius (m); I = central angle in degrees; I_rad = I in radians = I × π/180
Application
Gives the actual road distance along the curve from PC to PT. Used in stationing and earthwork volume calculations.
LC = 2R × sin(I/2)
Example
R = 300 m, I = 40°: LC = 2 × 300 × sin(20°) = 600 × 0.34202 = 205.21 m
Variables
LC = long chord (m); R = radius (m); I = central angle (degrees)
Application
Straight-line distance from PC to PT. Used in layout checks and chord-definition problems.
E = R × (sec(I/2) − 1) = R × (1/cos(I/2) − 1)
Example
R = 300 m, I = 40°: E = 300 × (sec 20° − 1) = 300 × (1.06418 − 1) = 300 × 0.06418 = 19.25 m
Variables
E = external distance (m); R = radius (m); I = central angle (degrees); sec = 1/cos
Application
Distance from PI to the midpoint of the curve (the point on the curve closest to the PI). Used in sight-distance and clearance checks.
M = R × (1 − cos(I/2))
Example
R = 300 m, I = 40°: M = 300 × (1 − cos 20°) = 300 × (1 − 0.93969) = 300 × 0.06031 = 18.09 m
Variables
M = middle ordinate (m); R = radius (m); I = central angle (degrees)
Application
Perpendicular distance from the midpoint of the long chord LC to the midpoint of the arc. Used in sight-distance checks for horizontal curves.
Exam tips
- Memorize the five formulas as a set: T(tan), Lc(pi), LC(sin), E(sec-1), M(1-cos). The pattern tan/sin/sec-1/1-cos at half-angle is easy to recall.
- In board exams, always box given values (R, I, Sta PI) and list unknowns before choosing a formula.
- E and M are related: E = M + M²/(2R) approximately, but use exact formulas in board exams.
- Check: E > M always (the PI is farther than the midchord point). If your E < M, you made an error.
- Quick sanity check for Lc: It must be greater than LC and less than πR (the full semicircle arc).
Key points
- A simple circular curve has a single, constant radius R throughout its length.
- The angle I (also called the intersection angle or deflection angle) is the angle between the two tangent lines meeting at the PI. It equals the central angle subtended by the curve.
- The PC (Point of Curvature) is where the alignment leaves the first tangent and enters the curve.
- The PT (Point of Tangency) is where the curve ends and the second tangent begins.
- The PI (Point of Intersection) is where the two tangents, if extended, meet — the driver never actually passes through the PI on a curved road.
- The five curve elements (T, Lc, LC, E, M) are all functions of R and I only.
- The long chord LC is always less than Lc (arc length) for any curve with I > 0.
- For very flat curves (small I or large R), LC ≈ Lc, but for sharp curves the difference is significant.
Definitions
PC — Point of Curvature
The point at which the alignment transitions from the back tangent to the circular curve. The beginning of the curve.
Importance
PC station is the reference from which PT is computed by adding Lc.
PI — Point of Intersection
The point where the two tangent lines of the alignment meet when extended. Also called the vertex or intersection point.
Importance
The PI station is typically given in exam problems; T is subtracted to find PC station.
PT — Point of Tangency
The point at which the circular curve ends and the alignment returns to a straight tangent. The end of the curve.
Importance
Sta PT = Sta PC + Lc. This is the most common stationing error in board exams.
Intersection Angle I (or Delta, Δ)
The deflection angle between the two tangents, measured at the PI. Equal to the central angle of the curve.
Importance
I is used in every curve element formula. Ensure it is in degrees for T, Lc (÷180), LC, E, and M.
External Distance E
The distance from the PI to the nearest point on the curve (midpoint of curve arc), measured along the bisector of the intersection angle.
Importance
Uses secant function — a common source of confusion with M which uses cosine.
Middle Ordinate M
The perpendicular distance from the midpoint of the long chord to the midpoint of the arc.
Importance
Used in AASHTO/DPWH sight-distance calculations for horizontal curves on roads.
Common mistakes
- Using I instead of I/2 in sin, cos, tan functions — always halve the central angle before applying trigonometry.
- Forgetting to convert I to radians when using Lc = R × I_rad (or equivalently, using Lc = π R I / 180 to keep I in degrees).
- Confusing E (uses secant/reciprocal of cosine) with M (uses cosine directly). Remember: E is farther from the curve (at PI), so E > M always.
- Adding 2T to PI station to get PT — WRONG. Sta PT = Sta PC + Lc. The road goes around the arc, not along the tangents.
- Treating LC (long chord) and Lc (arc length) as the same — they are different. LC < Lc always.
- Ignoring units: if I is in degrees, use the π/180 factor in Lc; if I is already in radians, do not multiply again.
2. Degree of Curve — Arc and Chord Definitions
Formulas
R = 1145.916 / D (Arc Definition, 20-m arc)
Example
D = 4°: R = 1145.916 / 4 = 286.479 m ≈ 286.48 m
Variables
R = radius in meters; D = degree of curve in degrees (central angle per 20-m arc)
Application
Convert between degree of curve and radius for arc-definition problems. Most common in Philippine board exams.
R = 10 / sin(D/2) (Chord Definition, 20-m chord)
Example
D = 4°: R = 10 / sin(2°) = 10 / 0.034899 = 286.56 m (slightly different from arc definition)
Variables
R = radius in meters; D = degree of curve in degrees (central angle per 20-m chord)
Application
Used when the problem explicitly states chord definition. Less common than arc definition in PH board exams.
Derivation of R = 1145.916/D: Set arc = 20 m = R × D_rad = R × (D × π/180). Solve: R = 20 × 180 / (π × D) = 3600/(π × D) = 1145.916/D
Example
For D = 2°: R = 1145.916/2 = 572.96 m. Check: arc = R × D_rad = 572.96 × (2π/180) = 572.96 × 0.03491 = 20.0 m ✓
Variables
Derivation only — shows where the constant 1145.916 comes from
Application
Understanding the derivation prevents formula memorization errors and helps reconstruct it during exams.
Exam tips
- The constant 1145.916 = 20 × 180 / π. You can re-derive it in seconds if you forget it during the exam.
- If D is given (e.g., '3-degree curve, arc definition'), immediately compute R = 1145.916/3 = 381.97 m, then proceed with standard curve element formulas.
- For compound or reverse curves, each curve has its own D (and thus its own R). Treat them separately.
Key points
- The degree of curve D is the central angle (in degrees) subtended by a standard length of arc or chord.
- Philippine and metric surveying practice uses a 20-m standard length (either arc or chord).
- US practice uses a 100-ft (≈ 30.48 m) chord or arc; do not mix systems in board problems.
- Arc Definition (most common in Philippine boards): D is the central angle for a 20-m arc. Gives R = 1145.916 / D.
- Chord Definition: D is the central angle for a 20-m chord. Gives sin(D/2) = 10/R, so R = 10 / sin(D/2).
- For flat curves (small D, large R), arc and chord definitions give nearly identical R values. For sharp curves (large D), the difference is notable.
- Knowing which definition is being used is critical — the problem will specify or it can be inferred from context.
Definitions
Degree of Curve D
The central angle, in degrees, subtended at the center of a circular curve by a standard arc (arc definition) or chord (chord definition) length of 20 m in metric practice.
Importance
D characterizes curve sharpness: higher D = sharper curve = smaller R. Board exams frequently give D instead of R.
Arc Definition
The degree of curve is defined as the central angle subtended by a 20-m arc. Leads to R = 1145.916/D. Standard in Philippine metric surveying.
Importance
Most PRC board exam problems use this definition unless explicitly stated otherwise.
Chord Definition
The degree of curve is defined as the central angle subtended by a 20-m chord. Leads to R = 10/sin(D/2). More common in older US-based references.
Importance
Knowing both definitions prevents formula confusion when old references or specific problem statements are used.
Common mistakes
- Dividing 1145.916 by the wrong quantity — the formula is R = 1145.916/D, not D = 1145.916/R (though that rearrangement is also valid).
- Using the chord definition formula when the problem implies arc definition (or vice versa) — read the problem carefully.
- Forgetting that 1145.916 is derived from a 20-m station length. In older US problems with 100-ft stations, the constant is 5729.58 ft/degree (or 1745.33 m/degree) — do not mix.
3. Stationing Along the Alignment
Formulas
Sta PC = Sta PI − T
Example
Sta PI = 10+120 (= 10120 m), T = 109.19 m: Sta PC = 10120 − 109.19 = 10010.81 m = 10+010.81
Variables
Sta PC = chainage of Point of Curvature (m); Sta PI = chainage of Point of Intersection (m); T = tangent distance (m)
Application
Locates the beginning of the curve on the alignment. Always performed first in stationing problems.
Sta PT = Sta PC + Lc
Example
Sta PC = 10+010.81, Lc = 209.44 m: Sta PT = 10010.81 + 209.44 = 10220.25 m = 10+220.25
Variables
Sta PT = chainage of Point of Tangency (m); Sta PC = chainage of Point of Curvature (m); Lc = arc length of curve (m)
Application
Locates the end of the curve. Critical: use arc length Lc, NOT 2T. This is the most tested stationing formula.
Exam tips
- The golden rule: Sta PC = PI − T; Sta PT = PC + Lc. Write these two lines first in any stationing problem.
- If the answer choices cluster near a value obtained by adding 2T to PI, recognize this as the distractor for the classic 'add arc, not tangents' error.
- Always verify: Sta PT > Sta PC (curve runs forward); Sta PT > Sta PI if T < Lc/2 (common for large I).
Key points
- Stations are continuous chainage distances (in meters) measured along the road centerline from a reference origin (usually Sta 0+000).
- Notation: Sta 10+120 means 10,120 m from the origin. The '+' separates kilometers from meters.
- The PI station is usually given. From it, subtract T to get the PC station.
- Add the arc length Lc (NOT 2T, NOT the long chord LC) to the PC station to get the PT station.
- Stationing follows the actual path of travel — along the curve, not across chords or back along tangents.
- If multiple curves are involved (compound or reverse), stationing is continuous: the PT of curve 1 is the PCC (Point of Compound Curvature) or PRC (Point of Reverse Curvature) and starting station for curve 2.
Definitions
Station (Chainage)
A point on the alignment identified by its cumulative distance from the reference origin, expressed in the format km+m (e.g., 5+240 = 5240 m from origin).
Importance
Correct stationing is essential for earthwork, drainage structure, and pavement quantity computations.
PCC — Point of Compound Curvature
The common point where two curves of a compound curve meet. The PT of the first curve equals the PC of the second curve.
Importance
Station of PCC = Sta PC1 + Lc1 (same principle as PT stationing).
PRC — Point of Reverse Curvature
The common point where two curves of a reverse curve meet, curving in opposite directions.
Importance
Station of PRC = Sta PC1 + Lc1. A tangent section may be inserted between curves for superelevation transition.
Common mistakes
- Using Sta PT = Sta PI + T (WRONG — this gives a point on the forward tangent, not the PT of the curve).
- Using Sta PT = Sta PI − T + 2T = Sta PI + T (same wrong result approached differently).
- Using LC (chord length) instead of Lc (arc length) to add to PC — always use the arc.
- Writing stations incorrectly: 10+010.81 ≠ 10+010.8 when precision matters in computation checks.
4. Compound Curves
Formulas
T1 = R1 × tan(I1/2); T2 = R2 × tan(I2/2)
Example
R1 = 200 m, I1 = 30°, R2 = 300 m, I2 = 25°: T1 = 200 tan 15° = 53.59 m; T2 = 300 tan 12.5° = 66.49 m
Variables
T1, T2 = tangent distances for curves 1 and 2; R1, R2 = radii; I1, I2 = central angles of each curve
Application
Compute the tangent of each individual curve. Used to locate PC, PCC, and PT along the alignment.
Lc1 = πR1I1/180; Lc2 = πR2I2/180
Example
R1 = 200 m, I1 = 30°: Lc1 = π(200)(30)/180 = 104.72 m. R2 = 300 m, I2 = 25°: Lc2 = π(300)(25)/180 = 130.90 m
Variables
Lc1, Lc2 = arc lengths of curves 1 and 2; R1, R2 = radii; I1, I2 = central angles (degrees)
Application
Compute arc lengths for stationing: Sta PCC = Sta PC + Lc1; Sta PT = Sta PCC + Lc2.
t1 + t2 = Length of common tangent between back-tangent PI and forward-tangent PI
Example
If the full tangent lengths (from back-tangent PC to PI and from PI to PT) are given, set up: Full back tangent = T1 + (t1 part of common tangent); use triangle geometry.
Variables
t1 = T1 = tangent from PC to PI (measured on back tangent extended); t2 = T2 = tangent from PI to PT
Application
In the compound curve triangle (formed by the two tangents and the common tangent), use sine rule to find unknown angles or distances when full geometry is given.
Exam tips
- In compound curve problems, always identify and list: R1, R2, I1, I2 separately before computing. Label a diagram.
- Sta PT = Sta PC + Lc1 + Lc2 — simply add both arc lengths to the PC station.
- If only the total I and one R are given, you likely need additional constraints (e.g., equal tangents or a given PC-PT distance) to solve — look for these in the problem stem.
Key points
- A compound curve consists of two (or more) simple circular arcs of different radii (R1, R2) turning in the same direction, joined at a common tangent point called the PCC.
- The PCC is simultaneously the PT of the first curve and the PC of the second curve.
- The overall deflection angle I = I1 + I2, where I1 and I2 are the central angles of the first and second curves respectively.
- Each curve has its own T, Lc, LC, E, M computed independently using its own R and I.
- The common (or 'middle') tangent at the PCC is tangent to both curves simultaneously.
- Compound curves are used where terrain, right-of-way, or alignment controls prevent a single radius — common in mountain roads like those in the Benguet or Mountain Province highways in the Philippines.
- The tangent lengths from PI to PC and from PI to PT are not equal unless R1 = R2 (simple curve case).
Definitions
PCC — Point of Compound Curvature
The point at which a curve of one radius ends and a curve of a different radius begins, both turning in the same direction. The common tangent at this point is tangent to both arcs.
Importance
Critical for stationing: Sta PCC = Sta PC + Lc1 (arc of first curve). Then Sta PT = Sta PCC + Lc2.
Common (Middle) Tangent
The tangent line at the PCC, common to both curves of the compound curve. It lies between the back tangent and the forward tangent.
Importance
Used in geometric constructions to find T1 and T2 when the compound curve triangle is analyzed.
Common mistakes
- Treating the compound curve as a single curve and using a combined I with a single R — each curve must be handled separately.
- Stationing error: Sta PT = Sta PC + Lc1 + Lc2 (correct), not Sta PC + T1 + T2.
- Confusing the sum I = I1 + I2 with a single curve's I when computing elements.
5. Reverse Curves
Formulas
For parallel tangents with R1 ≠ R2: d = (R1 + R2)(1 − cosI) where I1 = I2 = I (equal central angles for parallel tangent case)
Example
d = 12 m, R1 = R2 = R (equal radii), I1 = I2 = I: 12 = 2R(1−cosI). If I = 20°: 12 = 2R(1−cos20°) = 2R(0.06031); R = 12/(2×0.06031) = 99.48 m
Variables
d = perpendicular offset between parallel tangents (m); R1, R2 = radii of the two curves (m); I = central angle of each curve (degrees) — equal for parallel tangents
Application
Relates the geometry of a reverse curve with parallel tangents to the radii and central angle. Used to find unknown R or d.
For equal radii and parallel tangents: d = 2R(1 − cosI)
Example
R = 150 m, I = 25°: d = 2(150)(1−cos25°) = 300(1−0.9063) = 300(0.0937) = 28.11 m
Variables
d = perpendicular offset (m); R = radius (m, both curves equal); I = central angle of each curve (degrees)
Application
Simplified formula for the most common board exam reverse curve scenario.
L = 2R sinI (Length of the chord between PC and PT of the full reverse curve, equal radii, parallel tangents)
Example
R = 150 m, I = 25°: L = 2(150)sin25° = 300(0.4226) = 126.79 m
Variables
L = distance between PC1 and PT2 measured along the common tangent direction; R = radius; I = central angle of each curve
Application
Finds the straight-line distance between the start and end of the full reverse curve.
Exam tips
- When the problem says 'parallel tangents' + 'reverse curve', immediately write: I1 = I2 = I and use d = (R1+R2)(1−cosI) [or 2R(1−cosI) if equal radii].
- Draw the reverse curve S-shape on scratch paper: label PC1, PRC, PT2, the two centers O1 and O2 on opposite sides, and the offset d. This prevents sign and direction errors.
- For the chord from PC1 to PT2 with equal radii and parallel tangents: L = 2R sinI (derived from the isoceles triangle formed by O1, PC1, and PRC).
Key points
- A reverse curve consists of two circular arcs turning in opposite directions (one left, one right), joined at a common point called the PRC (Point of Reverse Curvature).
- The overall alignment forms an S-shape in plan view.
- At the PRC, the two curves share a common tangent but curve away from each other — creating an abrupt change in direction of centrifugal force.
- On high-speed roads, a tangent section is inserted between the two reverse curves to allow for superelevation runoff and driver reaction — a 'tangent-separated reverse curve.'
- For parallel tangents (a common board exam scenario), a geometric relationship between R1, R2, and the perpendicular offset d between tangents is used: d = Lc1×sinI1 / (2) type relationships derived from geometry.
- The most common special case: parallel tangents with equal radii (R1 = R2 = R), giving d = 2R(1−cosI) where I is the central angle of each curve.
- Reverse curves are prohibited on high-speed highways (design speed > 80 km/h per DPWH Design Guidelines) without a tangent between them.
- Stationing: Sta PRC = Sta PC1 + Lc1; Sta PT = Sta PRC + Lc2.
Definitions
PRC — Point of Reverse Curvature
The common point at which the first curve (curving, say, to the right) ends and the second curve (curving to the left) begins. The common tangent at this point is shared by both arcs.
Importance
Sta PRC = Sta PC1 + Lc1 for stationing. The abrupt direction of curvature reversal at PRC is why tangent sections are needed on high-speed roads.
Parallel Tangents
A condition in reverse curve problems where the back tangent of curve 1 and the forward tangent of curve 2 are parallel (but offset by a perpendicular distance d).
Importance
This condition forces I1 = I2 (the two central angles must be equal), greatly simplifying the algebra. This is a very common board exam setup.
Common mistakes
- Forgetting that parallel tangents force I1 = I2 — students sometimes try to solve with different central angles and get an unsolvable system.
- Confusing d (perpendicular distance between parallel tangents) with the chord length or arc length.
- Using the same formula for compound and reverse curves — these are fundamentally different geometric configurations.
- Applying reverse curve formulas to equal-radii cases with the general (R1 ≠ R2) formula instead of the simplified one — check if R1 = R2 first.
Connections
- Vertical Curves (Parabolic): Horizontal curves deal with geometry in plan (top view); vertical curves deal with geometry in profile (side view). Both require stationing — same PC/PT stationing principle applies. Combined design (horizontal + vertical) forms the 3D road alignment.
- Route Surveying / Highway Engineering (DPWH Design Guidelines): Minimum radii for horizontal curves are set by design speed and superelevation limits. The relationship R_min = V²/(127(e+f)) connects horizontal curve radius to traffic engineering design — a common interdisciplinary board exam question.
- Sight Distance (Stopping Sight Distance on Horizontal Curves): The middle ordinate M determines the lateral clearance needed to provide stopping sight distance on the inside of a horizontal curve. S = √(8RM) gives the stopping sight distance S for a required lateral clearance M.
- Earthwork and Mass Diagram: Stations of PC, PT, PCC, and PRC define the limits of geometric sections used in cross-section earthwork computations and mass diagram analysis.
- Spirals / Transition Curves: A next-level topic where a clothoid (Euler spiral) transitions between a tangent (R=∞) and a circular arc, avoiding the abrupt change in curvature at PC. This is the 'missing link' that makes high-speed roads safer than simple circular curves alone.
- Traverse and Coordinate Geometry: The deflection angles (I) in horizontal curve layout correspond to turning angles in a traverse. Coordinates of PC, PT, and points on the curve can be computed using survey coordinate methods — frequently combined in advanced board exam problems.
- Laws: RA 544 (Republic Act 544, as amended by RA 1582) — the Civil Engineering Law of the Philippines — includes surveying within the scope of civil engineering practice. DPWH Department Order No. 52 (Design Guidelines, Criteria and Standards for Public Highways) governs minimum curve radii and superelevation on Philippine national roads.
Quick review questions
A simple curve has R = 250 m and I = 36°. What is the tangent distance T?
Answer
T = 250 × tan(18°) = 250 × 0.32492 = 81.23 m
Apply T = R tan(I/2). Half of 36° is 18°. tan(18°) = 0.32492. Multiply: 250 × 0.32492 = 81.23 m. Always halve I before taking the tangent.
For the same curve (R = 250 m, I = 36°), what is the arc length Lc?
Answer
Lc = π(250)(36)/180 = 28274.33/180 = 157.08 m
Lc = πRI/180. Plug in: π × 250 × 36 = 28274.33; divide by 180 = 157.08 m. Equivalently, Lc = R × I_rad = 250 × (36π/180) = 250 × 0.6283 = 157.08 m.
What is the external distance E for R = 250 m, I = 36°?
Answer
E = 250 × (sec18° − 1) = 250 × (1/cos18° − 1) = 250 × (1.05146 − 1) = 250 × 0.05146 = 12.86 m
E = R(sec(I/2) − 1). cos(18°) = 0.95106, so sec(18°) = 1/0.95106 = 1.05146. E = 250 × 0.05146 = 12.86 m. Note E > M (always true).
What is the middle ordinate M for R = 250 m, I = 36°?
Answer
M = 250 × (1 − cos18°) = 250 × (1 − 0.95106) = 250 × 0.04894 = 12.24 m
M = R(1 − cos(I/2)). cos(18°) = 0.95106. M = 250 × 0.04894 = 12.24 m. Confirm E (12.86 m) > M (12.24 m) — correct.
A 4° curve uses the arc definition (20-m arc). What is the radius R?
Answer
R = 1145.916 / 4 = 286.48 m
Arc definition formula: R = 1145.916/D. D = 4°, so R = 1145.916/4 = 286.479 m ≈ 286.48 m. The constant 1145.916 = 20 × 180/π.
The PI of a curve is at Sta 5+240. T = 81.23 m and Lc = 157.08 m. Find Sta PC and Sta PT.
Answer
Sta PC = 5+240 − 81.23 = 5+158.77; Sta PT = 5+158.77 + 157.08 = 5+315.85
Sta PC = Sta PI − T = 5240 − 81.23 = 5158.77 m (= 5+158.77). Sta PT = Sta PC + Lc = 5158.77 + 157.08 = 5315.85 m (= 5+315.85). Never add 2T to PI.
A compound curve has R1 = 200 m, I1 = 30°, R2 = 350 m, I2 = 20°. Sta PC = 2+000. Find Sta PT.
Answer
Lc1 = π(200)(30)/180 = 104.72 m; Lc2 = π(350)(20)/180 = 122.17 m. Sta PCC = 2+000 + 104.72 = 2+104.72. Sta PT = 2+104.72 + 122.17 = 2+226.89
Compute arc lengths separately. Lc1 = πR1I1/180 = 104.72 m; Lc2 = πR2I2/180 = 122.17 m. Stationing: Sta PCC = Sta PC + Lc1; Sta PT = Sta PCC + Lc2. Total arc = 104.72 + 122.17 = 226.89 m added to PC.
Two reverse curves with equal radii R and equal central angles I connect parallel tangents separated by d = 10 m. If I = 18°, find R.
Answer
d = 2R(1 − cosI): 10 = 2R(1 − cos18°) = 2R(0.04894); R = 10 / (2 × 0.04894) = 10 / 0.09789 = 102.16 m
For parallel tangents with equal radii: d = 2R(1 − cos I). Substituting: 10 = 2R(1 − 0.95106) = 2R(0.04894). Solving: R = 10/0.09789 = 102.16 m. Note: parallel tangents force I1 = I2 = I.
What is the long chord of a simple curve with R = 400 m and I = 50°?
Answer
LC = 2(400)sin(25°) = 800 × 0.42262 = 338.09 m
LC = 2R sin(I/2). Half of 50° is 25°. sin(25°) = 0.42262. LC = 2 × 400 × 0.42262 = 338.09 m. Check: LC must be less than Lc = π(400)(50)/180 = 349.07 m — yes, 338.09 < 349.07 ✓
Which is greater — the external distance E or the middle ordinate M — and why?
Answer
E > M always. E = R(sec(I/2)−1) is the distance from PI to the arc; M = R(1−cos(I/2)) is from the midchord to the arc. The PI is farther from the curve center than the midchord midpoint.
Geometrically, E is measured from the PI outward to the curve, while M is the shorter offset from the midpoint of the long chord to the arc. Since PI lies beyond the arc's midpoint, E > M for all values of I and R. In the specific example (R=250, I=36°): E=12.86 m > M=12.24 m.
Diagrams
Horizontal Curves — Complete Concept Overview
A comprehensive mind map showing the three types of horizontal curves (simple, compound, reverse) with their key elements, formulas, and concepts organized hierarchically. Use this for a quick overview before diving into individual sections.
Step-by-Step Problem-Solving Flowchart for Horizontal Curve Problems
A decision-tree flowchart guiding the student through the complete solution process for any horizontal curve problem. Starting from curve type identification, through radius computation, element formulas, and stationing — covering all three curve types.
Simple Curve Element Relationships — Which Formula Uses Which Trig Function
A visual reference showing which trigonometric function (tan, sin, sec, cos) is used in each of the five curve element formulas. Highlights the critical rule of always using I/2 and the relationship E > M.
Stationing Procedure — Correct vs Incorrect Methods
A flowchart explicitly contrasting the correct stationing method (add arc length Lc to PC) versus the two most common errors (adding T to PI, or using chord LC). Directly addresses the most frequently missed board exam concept.
Distribution of Horizontal Curve Question Types in PRC CE Board Exams (Estimated)
An estimated distribution of horizontal curve question types appearing in PRC CE board exams, based on past exam patterns. Curve elements and stationing together account for approximately 60% of questions — prioritize these two areas in review.
Exam strategy
For PRC board exam problems on horizontal curves, adopt the following 4-step systematic approach: (1) READ AND IDENTIFY: Carefully read the problem, identify curve type (simple, compound, reverse) and all given values (R or D, I or I1+I2, Sta PI or Sta PC). (2) COMPUTE R FIRST: If D is given, immediately compute R = 1145.916/D (arc definition). Identify which definition is used. (3) APPLY THE CORRECT FORMULA SET: For simple curves, apply T=Rtan(I/2), Lc=πRI/180, LC=2Rsin(I/2), E=R(sec(I/2)−1), M=R(1−cos(I/2)) as needed. For compound/reverse curves, handle each sub-curve separately. (4) STATION CORRECTLY: Sta PC = Sta PI − T; Sta PT = Sta PC + Lc (arc, not 2T). For compound curves, add Lc1 then Lc2 sequentially. Time management: Horizontal curve problems typically take 3–5 minutes each. If a problem requires finding more than 3 elements, solve them in the formula-set order (T→Lc→LC→E→M) to avoid backtracking. The most common distractor answer in multiple-choice problems is obtained by incorrectly adding 2T to PI for PT — always verify your PT station is greater than PI station only when Lc > T. Always draw a quick sketch with PC, PI, PT, I, R labeled — this 30-second investment prevents conceptual errors and saves time overall. Expect 3–6 horizontal curve questions per board examination sitting.