--- title: "Simple curve data, roads intersecting at 150°" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:simple-curve-two-roads-150-4deg" source: https://www.bctengineer.com/q/semester_4/surveying_ii/simple-curve-two-roads-150-4deg site: BCT Engineering --- # Simple curve data, roads intersecting at 150° Surveying II · Chapter 6: Curves - Asked in: 2074 Bhadra, 6 marks - Video: [Solved Example on Setting out of Simple Circular Curve by Deflection Angle Method/Setting out Table](https://www.youtube.com/watch?v=huQeH2ynAQo) — CIVIL Solved - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/simple-curve-two-roads-150-4deg - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question Two roads BA and AC intersect at an angle of 150°. They are to be connected by a 4° circular curve. The chainage of point of intersection A is (138+20.3) chains. Compute all data necessary (i.e. deflection angle, tangent length, apex distance, mid ordinate, length of curve, long chord) for laying out the curve if only 30 m chain is used. ## Short answer With a deflection angle Δ = 30° and a 4° curve (30 m chord), radius R = 1718.87/4 = 429.72 m. Tangent length T = 115.14 m, curve length L = 225.00 m, long chord = 222.44 m, apex distance = 15.16 m, and mid-ordinate = 14.64 m. ## Answer Interior angle at A = 150°, so the roads deflect by $\Delta = 180^\circ - 150^\circ = 30^\circ$. A 4° curve with a 30 m chain means the degree of curve is defined by a 30 m arc/chord. #### Radius $$ R = \frac{1718.87}{D} = \frac{1718.87}{4} = 429.72\ \text{m} $$ #### Curve elements $$ \begin{aligned} T &= R\tan\frac{\Delta}{2} = 429.72\tan 15^\circ = 115.14\ \text{m}\\ L &= \frac{\pi R\Delta}{180} = \frac{\pi\times 429.72\times 30}{180} = 225.00\ \text{m}\\ \text{Long chord} &= 2R\sin\frac{\Delta}{2} = 2\times 429.72\sin 15^\circ = 222.44\ \text{m}\\ \text{Apex distance} &= R\left(\sec\frac{\Delta}{2} - 1\right) = 15.16\ \text{m}\\ \text{Mid-ordinate} &= R\left(1-\cos\frac{\Delta}{2}\right) = 14.64\ \text{m} \end{aligned} $$ #### Chainages Chainage of A $= 138\times 30 + 20.3 = 4160.30$ m. - Chainage of T₁ (start) $= 4160.30 - 115.14 = 4045.16$ m - Chainage of T₂ (end) $= 4045.16 + 225.00 = 4270.16$ m #### Deflection angles (30 m chain, Rankine) The first sub-chord is the distance from T₁ to the next whole 30 m station, then full chords, and a last sub-chord to T₂. | Point (chainage, m) | Chord c (m) | δ = 1718.87c/R | Cumulative deflection | |---|---|---|---| | 4045.16 | - | - | 0°00′00″ | | 4050.00 | 4.84 | 0°19′22″ | 0°19′22″ | | 4080.00 | 30.00 | 2°00′00″ | 2°19′22″ | | 4110.00 | 30.00 | 2°00′00″ | 4°19′22″ | | 4140.00 | 30.00 | 2°00′00″ | 6°19′22″ | | 4170.00 | 30.00 | 2°00′00″ | 8°19′22″ | | 4200.00 | 30.00 | 2°00′00″ | 10°19′22″ | | 4230.00 | 30.00 | 2°00′00″ | 12°19′22″ | | 4260.00 | 30.00 | 2°00′00″ | 14°19′22″ | | 4270.16 | 10.16 | 0°40′38″ | 15°00′00″ | The last cumulative deflection equals $\Delta/2 = 15^\circ$ (check). **Answer:** $\Delta = 30^\circ$, $R = 429.72$ m, $T = 115.14$ m, apex distance $= 15.16$ m, mid-ordinate $= 14.64$ m, $L = 225.00$ m, long chord $= 222.44$ m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Roads intersect at 150 deg. The deflection angle is A. 30 deg (180 - 150) B. 150 deg C. 75 deg D. 15 deg Answer: A. Deflection angle = 180 deg minus interior angle. (Key point: Deflection angle) 2. With a 4 deg curve on a 30 m chain, the radius is A. R = 1718.87 x 4 = 6875 m B. R = 1718.87/D = 429.72 m C. R = 4/1718.87 D. R = 30/4 = 7.5 m Answer: B. The degree of curve gives R = 1718.87/D for 30 m (chord/arc approximated). (Key point: Radius) 3. Which pair of formulas is correct? A. Mid-ordinate = R(sec(D/2) - 1); apex distance = R(1 - cos(D/2)) B. Both equal R tan(D/2) C. Mid-ordinate = R(1 - cos(D/2)); apex distance = R(sec(D/2) - 1) D. Both equal 2R sin(D/2) Answer: C. Apex distance is larger (15.16 m) than the mid-ordinate (14.64 m). (Key point: Mid-ordinate vs apex distance) 4. How is chainage of A (138+20.3 chains) converted? A. 138 x 20 + 20.3 B. 138 + 20.3 = 158.3 m C. 138 x 100 + 20.3 D. 138 x 30 + 20.3 = 4160.30 m Answer: D. One chain here is 30 m. 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