GELE Surveying (Geomatics) — Horizontal Curves (Simple, Compound, Reverse)Exam Answer Templates

Marks

1

Topic

Simple Curve Elements — Definitions

Difficulty

easy

Template Id

T1

Examiner Tip

VSA questions reward precision. Use the exact engineering terms 'back tangent' and 'forward tangent' rather than vague phrases like 'start' and 'end of the road'.

Model Answer

The PC (Point of Curvature) is the point where the back tangent meets the circular curve — i.e., where the curve begins. The PT (Point of Tangency) is the point where the circular curve meets the forward tangent — i.e., where the curve ends.

Question Type

very_short_answer

Answer Structure

• Line 1: Define PC as the beginning of the curve where the back tangent ends [0.5 mark]
• Line 2: Define PT as the end of the curve where the forward tangent begins [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the definitions of PC and PT (-1 mark)
• Defining PI instead of PC or PT
• Omitting the reference to the tangent lines

Key Phrases To Include

• back tangent
• forward tangent
• beginning of curve
• end of curve
• circular curve

Marks

1

Topic

Degree of Curve

Difficulty

easy

Template Id

T2

Examiner Tip

Memorize 1145.916 as the Philippine/SI arc-definition constant. Board exams frequently use this constant directly in numerical problems.

Model Answer

Using the 20-m arc definition: R = 1145.916 / D, where R is the radius of the circular curve in metres and D is the degree of curve in degrees (the central angle subtending a 20-m arc).

Question Type

very_short_answer

Answer Structure

• Line 1: Write the formula R = 1145.916 / D [0.5 mark]
• Line 2: Define R (in metres) and D (central angle per 20-m arc) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using 5729.58 (the 100-ft/US arc constant) instead of 1145.916 (-1 mark)
• Confusing arc definition with chord definition
• Omitting units for R

Key Phrases To Include

• R = 1145.916 / D
• 20-m arc definition
• central angle
• radius in metres
• degree of curve

Marks

2

Topic

Compound and Reverse Curves

Difficulty

easy

Template Id

T3

Examiner Tip

Examiners reward the practical implication. Mentioning superelevation runoff for reverse curves shows applied knowledge and earns the full second mark.

Model Answer

A compound curve consists of two or more circular arcs of different radii curving in the SAME direction, joined at a common tangent point (PCC — Point of Compound Curvature). A reverse curve (S-curve) consists of two circular arcs curving in OPPOSITE directions, joined at a Point of Reverse Curvature (PRC). On high-speed highways, a tangent segment must be inserted between the two arcs of a reverse curve to allow for superelevation runoff.

Question Type

short_answer

Answer Structure

• Line 1: Define compound curve — same direction, different radii, joined at PCC [1 mark]
• Line 2: Define reverse curve — opposite directions, S-shape, joined at PRC; note the need for a tangent segment on highways [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating both curves turn in the same direction (-1 mark for concept error)
• No mention of superelevation or practical requirement for reverse curves (-0.5 mark)
• Omitting PCC/PRC labels

Key Phrases To Include

• same direction
• opposite directions
• PCC
• PRC
• different radii
• S-curve
• superelevation runoff
• common tangent

Marks

2

Topic

Simple Curve Elements — T and Lc

Difficulty

easy

Template Id

T4

Examiner Tip

Always halve the intersection angle before taking the tangent. Board examiners see 'T = R tan I' (without halving) as the #1 error in curve problems.

Model Answer

Given: R = 400 m, I = 30° Tangent Distance: T = R tan(I/2) = 400 × tan(15°) = 400 × 0.26795 = 107.18 m Length of Curve: Lc = (π R I) / 180° = (π × 400 × 30) / 180 = 209.44 m ∴ T = 107.18 m, Lc = 209.44 m

Question Type

numerical

Answer Structure

• Step 1: List given data: R and I [implicit mark for correct substitution]
• Step 2: Write formula T = R tan(I/2) and substitute to get T = 107.18 m [1 mark]
• Step 3: Write formula Lc = πRI/180 and substitute to get Lc = 209.44 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using I instead of I/2 in the tangent formula: T = R tan(I) — common error (-1 mark)
• Using I in degrees directly without the /180 conversion in Lc formula (-1 mark)
• No units on final answer (-0.5 mark)

Key Phrases To Include

• T = R tan(I/2)
• Lc = πRI/180
• I/2 = 15°
• tan(15°) = 0.26795

Marks

3

Topic

Simple Curve Elements — LC, E, M

Difficulty

medium

Template Id

T5

Examiner Tip

The key distinction: E uses sec (secant stretches outward past the curve to the PI) and M uses (1 − cos) (ordinate measured inward from the long chord). A memory aid: E is External (sec), M is Middle (1 − cos).

Model Answer

Given: R = 300 m, I = 40° → I/2 = 20° (1) Long Chord: LC = 2R sin(I/2) = 2(300) sin(20°) = 600 × 0.34202 = 205.21 m (2) 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 (3) Middle Ordinate: M = R(1 − cos(I/2)) = 300(1 − cos 20°) = 300(1 − 0.93969) = 300(0.06031) = 18.09 m ∴ LC = 205.21 m, E = 19.25 m, M = 18.09 m

Question Type

numerical

Answer Structure

• Step 1: Note I/2 = 20° [setup — no mark, but shows method]
• Step 2: Write LC = 2R sin(I/2) and compute → 205.21 m [1 mark]
• Step 3: Write E = R(sec(I/2) − 1) and compute, showing sec 20° = 1/cos 20° → 19.25 m [1 mark]
• Step 4: Write M = R(1 − cos(I/2)) and compute → 18.09 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing E and M formulas — using cos for E and sec for M (-2 marks)
• Not showing the sec = 1/cos step: if calculator error occurs, no method mark is earned
• Using full angle I instead of I/2 throughout (-1 to -2 marks)
• Omitting units m from answers (-0.5 mark)

Key Phrases To Include

• LC = 2R sin(I/2)
• E = R(sec(I/2) − 1)
• M = R(1 − cos(I/2))
• sec = 1/cos
• I/2 = 20°

Marks

3

Topic

Stationing

Difficulty

medium

Template Id

T6

Examiner Tip

Sta PT = Sta PC + Lc is the single most tested relationship in stationing problems. Never add T twice or add T to the PI station. The curve length Lc is the actual distance along the arc.

Model Answer

Given: Sta PI = 10 + 120 (= 10 120.00 m), R = 300 m, I = 40° Step 1 — Tangent distance: T = R tan(I/2) = 300 tan(20°) = 300 × 0.36397 = 109.19 m Step 2 — Length of curve: Lc = πRI/180 = π(300)(40)/180 = 209.44 m Step 3 — Station of PC: Sta PC = Sta PI − T = 10 120.00 − 109.19 = 10 010.81 m = 10 + 010.81 Step 4 — Station of PT: Sta PT = Sta PC + Lc = 10 010.81 + 209.44 = 10 220.25 m = 10 + 220.25 ∴ Sta PC = 10 + 010.81, Sta PT = 10 + 220.25

Question Type

numerical

Answer Structure

• Step 1: Compute T = R tan(I/2) = 109.19 m [1 mark]
• Step 2: Compute Lc = πRI/180 = 209.44 m [0.5 mark — may be combined]
• Step 3: Sta PC = Sta PI − T = 10 + 010.81 [1 mark]
• Step 4: Sta PT = Sta PC + Lc = 10 + 220.25 [1 mark] — NOT Sta PI + T or PI + 2T

Scoring Breakdown

Common Mark Deductions

• Computing Sta PT = Sta PI + T (incorrect — this gives a point past the PI on the forward tangent, not the PT) (-1 mark)
• Computing Sta PT = Sta PC + 2T instead of Sta PC + Lc (-1 mark)
• Not writing answer in station format (xx + xxx.xx) (-0.5 mark)
• Arithmetic error in T without showing formula (no method mark recovery)

Key Phrases To Include

• Sta PC = Sta PI − T
• Sta PT = Sta PC + Lc
• T = R tan(I/2)
• Lc = πRI/180
• station format: xx + xxx.xx

Marks

3

Topic

Degree of Curve and Stationing

Difficulty

medium

Template Id

T7

Examiner Tip

The shortcut Lc = (I/D) × 20 m is extremely useful for verification and saves time on board exams when D divides evenly into I. Always show both methods if time permits.

Model Answer

Given: D = 4°, I = 50° (arc definition) Step 1 — Radius: R = 1145.916 / D = 1145.916 / 4 = 286.48 m Step 2 — Tangent distance: T = R tan(I/2) = 286.48 × tan(25°) = 286.48 × 0.46631 = 133.57 m Step 3 — Length of curve: Lc = πRI/180 = π(286.48)(50)/180 = 250.00 m [Check: Lc = (I/D) × 20 m = (50/4) × 20 = 250.00 m ✓] ∴ R = 286.48 m, T = 133.57 m, Lc = 250.00 m

Question Type

numerical

Answer Structure

• Step 1: Write R = 1145.916/D and compute R = 286.48 m [1 mark]
• Step 2: Write T = R tan(I/2) and compute T = 133.57 m [1 mark]
• Step 3: Write Lc = πRI/180 and compute Lc = 250.00 m; optionally verify with Lc = (I/D)×20 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using R = 5729.58/D (US 100-ft constant) without conversion (-1 mark)
• Forgetting to halve I when computing T (-1 mark)
• Not stating 'arc definition' — if chord definition is assumed, answer differs

Key Phrases To Include

• R = 1145.916 / D
• arc definition
• 20-m arc
• Lc = (I/D) × 20
• T = R tan(I/2)

Marks

2

Topic

Simple Curve Elements — M

Difficulty

easy

Template Id

T8

Examiner Tip

Examiners reward application context. Mentioning 'stopping sight distance' or 'horizontal sight obstruction' elevates a 1-mark answer into a full 2-mark response.

Model Answer

The middle ordinate M is the perpendicular distance from the midpoint of the long chord to the midpoint of the circular arc. It measures how far the curve departs from the straight chord at its deepest point. Formula: M = R(1 − cos(I/2)) Geometric significance: M represents the maximum offset from the chord to the arc. It is used in sight-distance calculations on horizontal curves — a designer ensures that obstructions (walls, slopes, vegetation) are not within this distance of the chord to maintain adequate stopping sight distance.

Question Type

short_answer

Answer Structure

• Line 1: Define M as the perpendicular distance from chord midpoint to arc midpoint [1 mark]
• Line 2: State formula M = R(1 − cos(I/2)) and explain practical use in sight distance [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing M with E (external distance, which goes beyond the arc to the PI) (-1 mark)
• Giving the formula for E instead of M (-1 mark)
• No mention of the practical/design significance (-0.5 mark)

Key Phrases To Include

• midpoint of the long chord
• midpoint of the arc
• perpendicular distance
• M = R(1 − cos(I/2))
• stopping sight distance
• maximum offset

Marks

5

Topic

Simple Curve Elements — All Five

Difficulty

medium

Template Id

T9

Examiner Tip

On a 5-mark comprehensive element question, write a summary table at the end. This lets the examiner quickly verify all five answers and award all 5 marks without hunting through your working. Neatness and organization translate directly to marks.

Model Answer

Given: R = 250 m, I = 36° → I/2 = 18° (1) Tangent Distance: T = R tan(I/2) = 250 tan(18°) = 250 × 0.32492 = 81.23 m (2) Length of Curve: Lc = πRI/180 = π(250)(36)/180 = 157.08 m (3) Long Chord: LC = 2R sin(I/2) = 2(250) sin(18°) = 500 × 0.30902 = 154.51 m (4) External Distance: E = R(sec(I/2) − 1) = 250(sec 18° − 1) sec 18° = 1/cos 18° = 1/0.95106 = 1.05146 E = 250(1.05146 − 1) = 250(0.05146) = 12.87 m (5) Middle Ordinate: M = R(1 − cos(I/2)) = 250(1 − cos 18°) = 250(1 − 0.95106) = 250(0.04894) = 12.24 m Summary: T = 81.23 m | Lc = 157.08 m | LC = 154.51 m | E = 12.87 m | M = 12.24 m

Question Type

numerical

Answer Structure

• Step 1: State I/2 = 18° and list given data [setup — method mark]
• Step 2: T = R tan(I/2) = 81.23 m [1 mark]
• Step 3: Lc = πRI/180 = 157.08 m [1 mark]
• Step 4: LC = 2R sin(I/2) = 154.51 m [1 mark]
• Step 5: E = R(sec(I/2) − 1) = 12.87 m [1 mark]
• Step 6: M = R(1 − cos(I/2)) = 12.24 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using full I instead of I/2 in all formulas — cascading error (-2 to -3 marks)
• Confusing LC (long chord) with Lc (arc length) — these are different quantities (-1 mark)
• Not showing sec 18° = 1/cos 18° step — if calculator slips, no recovery mark
• Summary table omitted — examiner cannot confirm all five answers (-0.5 mark)
• No units on any answer (-0.5 mark)

Key Phrases To Include

• I/2 = 18°
• T = R tan(I/2)
• Lc = πRI/180
• LC = 2R sin(I/2)
• E = R(sec(I/2) − 1)
• M = R(1 − cos(I/2))
• sec = 1/cos

Marks

2

Topic

Stationing — Conceptual

Difficulty

medium

Template Id

T10

Examiner Tip

This is a conceptual understanding question. Examiners are testing whether you know WHY, not just HOW. Answer by contrasting 'distance along arc' vs 'distance along tangent'.

Model Answer

Station measurements follow the actual traveled distance along the alignment — along the back tangent to PC, then along the arc from PC to PT. The arc length Lc is the actual distance a traveler covers along the curve. The distance from PI to PT along the forward tangent is also T (by symmetry), but this is measured along the tangent line, NOT along the arc. Since stations are cumulative distances along the centerline alignment (not along tangent projections), adding T to the PI station would give the distance from the start to a point T metres ahead of the PI along the forward tangent — which is a different point from PT. Therefore: Sta PT = Sta PC + Lc is the only correct relationship.

Question Type

short_answer

Answer Structure

• Line 1: Explain that stationing follows actual traveled distance along the alignment (arc, not tangent) [1 mark]
• Line 2: Explain that adding T to Sta PI gives a point on the forward tangent T beyond the PI, NOT the PT on the arc; therefore Sta PT = Sta PC + Lc [1 mark]

Scoring Breakdown

Common Mark Deductions

• Merely stating the formula without explaining the conceptual reason (-1 mark)
• Confusing the issue of tangent length vs arc length

Key Phrases To Include

• actual traveled distance
• along the arc
• centerline alignment
• cumulative stationing
• arc length Lc
• not along the tangent

Marks

2

Topic

Degree of Curve — Chord Definition

Difficulty

hard

Template Id

T11

Examiner Tip

Derivation questions reward the geometric reasoning. Draw the triangle (even a rough sketch), label the sides, and write the trig ratio. This shows mastery beyond formula recall.

Model Answer

Under the chord definition, the degree of curve D is the central angle subtended by a chord of exactly 20 m. Consider an isoceles triangle formed by two radii R and the 20-m chord. Bisecting the central angle D gives a right triangle with: — hypotenuse = R — opposite side = 10 m (half the chord) — angle at center = D/2 Therefore: sin(D/2) = 10 / R ∴ R = 10 / sin(D/2) For small D, sin(D/2) ≈ D/2 (in radians) ≈ πD/360, which approximates to R ≈ 1145.916/D — showing that arc and chord definitions converge for flat (small D) curves.

Question Type

short_answer

Answer Structure

• Line 1: Describe the geometric setup — 20-m chord, central angle D, right triangle by bisection [1 mark]
• Line 2: Apply trigonometry to derive sin(D/2) = 10/R, therefore R = 10/sin(D/2); note convergence with arc definition for small D [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the arc definition formula instead of deriving the chord definition (-1 mark)
• Not showing the geometric/trigonometric derivation — just stating the formula (-0.5 mark)
• Not noting when the two definitions differ (sharp vs flat curves)

Key Phrases To Include

• chord definition
• 20-m chord
• sin(D/2) = 10/R
• R = 10 / sin(D/2)
• bisecting the central angle
• arc and chord definitions converge

Marks

5

Topic

Compound Curves — Stationing

Difficulty

hard

Template Id

T12

Examiner Tip

Compound curve problems are high-value board questions. Always draw a sketch showing the two tangents, PI, PC, PCC, and PT — even a rough one. The sketch helps you set up the triangle correctly and usually earns a method mark.

Model Answer

Given: R1 = 200 m, I1 = 30°; R2 = 350 m, I2 = 45° Total intersection angle I = I1 + I2 = 75° Sta PI = 5 + 000 (= 5 000.00 m) Step 1 — Tangent distances for each sub-curve: T1 = R1 tan(I1/2) = 200 tan(15°) = 200 × 0.26795 = 53.59 m T2 = R2 tan(I2/2) = 350 tan(22.5°) = 350 × 0.41421 = 144.97 m Step 2 — Compound curve total tangent lengths: For a compound curve, the total back tangent from PI to PC: total_T_back = T1 + (T2 × sin I1 / sin I) ... [use triangle of tangents method if needed] Note: For the simpler board-exam approach — the PCC is the PT of Curve 1 and the PC of Curve 2. Step 3 — Station of PC (back tangent approach): Using the back tangent length to PC: Back tangent = T1 + T2·sin(I1)/sin(I1 + I2) = 53.59 + 144.97 × sin(30°)/sin(75°) = 53.59 + 144.97 × 0.5/0.9659 = 53.59 + 75.00 = 128.59 m Sta PC = 5 000.00 − 128.59 = 4 871.41 m = 4 + 871.41 Step 4 — Arc length of Curve 1: Lc1 = πR1·I1/180 = π(200)(30)/180 = 104.72 m Step 5 — Station of PCC: Sta PCC = Sta PC + Lc1 = 4 871.41 + 104.72 = 4 976.13 m = 4 + 976.13 Step 6 — Arc length of Curve 2: Lc2 = πR2·I2/180 = π(350)(45)/180 = 274.89 m Step 7 — Station of PT: Sta PT = Sta PCC + Lc2 = 4 976.13 + 274.89 = 5 251.02 m = 5 + 251.02 ∴ Sta PCC = 4 + 976.13, Sta PT = 5 + 251.02

Question Type

numerical

Answer Structure

• Step 1: Identify I1, I2, total I = 75°; compute T1 = 53.59 m and T2 = 144.97 m [1 mark]
• Step 2: Apply compound curve back-tangent formula to find Sta PC = 4 + 871.41 [1 mark]
• Step 3: Compute Lc1 = πR1·I1/180 = 104.72 m [0.5 mark]
• Step 4: Compute Sta PCC = Sta PC + Lc1 = 4 + 976.13 [1 mark]
• Step 5: Compute Lc2 = πR2·I2/180 = 274.89 m [0.5 mark]
• Step 6: Compute Sta PT = Sta PCC + Lc2 = 5 + 251.02 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Adding T1 + T2 directly as back tangent (only valid when tangents are parallel) (-1 mark)
• Forgetting the sin(I)/sin(total I) triangle correction for compound curves (-1 mark)
• Using Sta PT = Sta PC + Lc1 + Lc2 (skipping the PCC station) — not wrong but loses the PCC mark
• Not labeling intermediate answers clearly (PCC vs PT) (-0.5 mark)

Key Phrases To Include

• T1 = R1 tan(I1/2)
• T2 = R2 tan(I2/2)
• Sta PCC = Sta PC + Lc1
• Sta PT = Sta PCC + Lc2
• back tangent formula
• PCC — Point of Compound Curvature

Marks

1

Topic

Simple Curve Elements — E

Difficulty

easy

Template Id

T13

Examiner Tip

Remember: E is External — it goes from the arc outward to the PI. The sec function (>1) ensures E is positive and outward.

Model Answer

The external distance E is the distance from the midpoint of the circular arc (the vertex of the curve) to the PI (Point of Intersection), measured along the bisector of the intersection angle. Formula: E = R(sec(I/2) − 1)

Question Type

very_short_answer

Answer Structure

• Line 1: Define E as the distance from the midpoint of the arc to the PI along the angle bisector [0.5 mark]
• Line 2: State E = R(sec(I/2) − 1) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using cos instead of sec in the formula (-1 mark)
• Defining E as the distance from PC to PT (that is the long chord) (-1 mark)

Key Phrases To Include

• midpoint of the arc
• PI
• angle bisector
• E = R(sec(I/2) − 1)
• sec

Marks

5

Topic

Reverse Curves — Parallel Tangents

Difficulty

hard

Template Id

T14

Examiner Tip

Reverse curve with parallel tangents is a classic board problem. The key insight is: equal radii → equal deflection angles. Once you write p and L in terms of R and I, the Pythagorean identity sin²+cos²=1 always closes the system.

Model Answer

Given: Perpendicular offset between parallel tangents p = 12 m; horizontal distance between tangent points L = 60 m; equal radii R1 = R2 = R. For a reverse curve with parallel tangents and equal radii: Step 1 — Geometry setup: The offset p between parallel tangents is related to the geometry by: p = R1(1 − cos I1) + R2(1 − cos I2) Since R1 = R2 = R and I1 = I2 = I (equal radii → equal deflections for parallel tangents): p = 2R(1 − cos I) ... (Equation 1) Step 2 — Horizontal distance relationship: The horizontal projection of the reverse curve: L = R sin I1 + R sin I2 = 2R sin I ... (Equation 2) Step 3 — From Equation 2: sin I = L / (2R) = 60 / (2R) = 30/R ... (i) Step 4 — From Equation 1: 1 − cos I = p / (2R) = 12 / (2R) = 6/R ... (ii) Step 5 — Use the identity sin²I + cos²I = 1: cos I = 1 − 6/R sin²I + cos²I = 1 (30/R)² + (1 − 6/R)² = 1 900/R² + 1 − 12/R + 36/R² = 1 936/R² − 12/R = 0 R(936/R² − 12/R) = 0 → multiply through by R²: 936 − 12R = 0 12R = 936 R = 78 m ∴ R = 78 m

Question Type

numerical

Answer Structure

• Step 1: Set up p = 2R(1 − cos I) for parallel tangents with equal radii [1 mark]
• Step 2: Set up L = 2R sin I [1 mark]
• Step 3: Express sin I = 30/R and (1 − cos I) = 6/R [1 mark]
• Step 4: Apply sin²I + cos²I = 1 to form a single equation in R [1 mark]
• Step 5: Solve to get R = 78 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not recognizing that equal radii with parallel tangents imply equal deflection angles (-1 mark)
• Setting up p = L (confusing offset with horizontal distance) (-2 marks)
• Not using the trig identity to eliminate I — leaving two unknowns unsolved (-2 marks)
• Arithmetic error in expanding (1 − 6/R)² without showing the binomial expansion

Key Phrases To Include

• parallel tangents
• equal radii
• p = 2R(1 − cos I)
• L = 2R sin I
• sin²I + cos²I = 1
• Pythagorean identity

Marks

5

Topic

Simple Curve — Complete Solution

Difficulty

medium

Template Id

T15

Examiner Tip

This is a comprehensive 5-part problem typical of the PRC board exam. Manage time by doing parts (a)–(c) first (pure substitution), then (d) and (e). Use the Lc = (I/D)×20 shortcut to verify (c) instantly. A wrong R in part (a) cascades into all other parts — double-check the constant 1145.916.

Model Answer

Given: D = 5°, I = 60°, Sta PI = 3 + 450.00 (= 3 450.00 m) (a) Radius: R = 1145.916 / D = 1145.916 / 5 = 229.18 m (b) Tangent Distance: T = R tan(I/2) = 229.18 × tan(30°) = 229.18 × 0.57735 = 132.34 m (c) Arc Length: Lc = πRI/180 = π(229.18)(60)/180 = 240.00 m [Verify: Lc = (I/D) × 20 = (60/5) × 20 = 12 × 20 = 240.00 m ✓] (d) External Distance: E = R(sec(I/2) − 1) = 229.18(sec 30° − 1) sec 30° = 1/cos 30° = 1/0.86603 = 1.15470 E = 229.18(1.15470 − 1) = 229.18(0.15470) = 35.45 m (e) Station of PC: Sta PC = Sta PI − T = 3 450.00 − 132.34 = 3 317.66 m = 3 + 317.66 Station of PT: Sta PT = Sta PC + Lc = 3 317.66 + 240.00 = 3 557.66 m = 3 + 557.66 Summary: R = 229.18 m | T = 132.34 m | Lc = 240.00 m | E = 35.45 m Sta PC = 3 + 317.66 | Sta PT = 3 + 557.66

Question Type

numerical

Answer Structure

• Step 1: R = 1145.916/D = 229.18 m [1 mark]
• Step 2: T = R tan(I/2) = 132.34 m [1 mark]
• Step 3: Lc = πRI/180 = 240.00 m with verification [0.5 mark]
• Step 4: E = R(sec(I/2) − 1) = 35.45 m [1 mark]
• Step 5: Sta PC = Sta PI − T = 3 + 317.66 and Sta PT = Sta PC + Lc = 3 + 557.66 [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Forgetting to divide I by 2 in T and E formulas (-2 marks, cascading)
• Not using the arc definition constant 1145.916 (-1 mark)
• Computing Sta PT = Sta PI + T instead of Sta PC + Lc (-1 mark)
• Not writing answer in station format (-0.5 mark)
• Not showing sec = 1/cos step for E (-0.5 mark)

Key Phrases To Include

• R = 1145.916/D
• arc definition
• T = R tan(I/2)
• Lc = πRI/180
• Lc = (I/D) × 20
• E = R(sec(I/2) − 1)
• sec 30° = 1/cos 30°
• Sta PC = Sta PI − T
• Sta PT = Sta PC + Lc
• station format 3 + xxx.xx