--- title: "Curve R=300 m, deflection 36°, with checks" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:curve-r300-deflection-36-check" source: https://www.bctengineer.com/q/semester_4/surveying_ii/curve-r300-deflection-36-check site: BCT Engineering --- # Curve R=300 m, deflection 36°, with checks Surveying II · Chapter 6: Curves - Asked in: 2065 Chaitra (old course), 9 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/curve-r300-deflection-36-check - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question Two tangents intersect at chainages 1190 m, the deflection angle being 36°. Calculate all the necessary data for setting out a curve with a radius of 300 m, by deflection angle method. Take peg interval of 30 m. Also provide check to support the calculation during setting out. ## Short answer Tangent length 97.48 m, curve length 188.50 m. Chainage T1 = 1092.52 m, T2 = 1281.02 m. Standard 30‑m chord deflection is 2°51′53″. Final cumulative deflection equals Δ/2 = 18°00′00″ as angular check; long chord T1T2 = 185.41 m gives a linear check. ## Answer Given: $\Delta = 36^\circ$, $R = 300$ m, chainage of IP $= 1190$ m, peg interval 30 m. #### Curve elements $$ \begin{aligned} T &= R\tan\frac{\Delta}{2} = 300\tan 18^\circ = 97.48\ \text{m}\\ L &= \frac{\pi R\Delta}{180} = 188.50\ \text{m}\\ \text{Long chord} &= 2R\sin\frac{\Delta}{2} = 2\times 300\times\sin 18^\circ = 185.41\ \text{m}\\ \text{Ch. of T}_1 &= 1190 - 97.48 = 1092.52\ \text{m}\\ \text{Ch. of T}_2 &= 1092.52 + 188.50 = 1281.02\ \text{m} \end{aligned} $$ #### Deflection angle table First sub-chord $c_1 = 17.48$ m, last sub-chord $c_n = 21.02$ m, $\delta = 1718.87\,c/R$ minutes. | Point (chainage, m) | Chord c (m) | δ = 1718.87c/R | Cumulative deflection | |---|---|---|---| | 1092.52 | - | - | 0°00′00″ | | 1110.00 | 17.48 | 1°40′08″ | 1°40′08″ | | 1140.00 | 30.00 | 2°51′53″ | 4°32′01″ | | 1170.00 | 30.00 | 2°51′53″ | 7°23′54″ | | 1200.00 | 30.00 | 2°51′53″ | 10°15′47″ | | 1230.00 | 30.00 | 2°51′53″ | 13°07′41″ | | 1260.00 | 30.00 | 2°51′53″ | 15°59′34″ | | 1281.02 | 21.02 | 2°00′26″ | 18°00′00″ | #### Checks during setting out 1. **Angular check:** the final cumulative deflection angle must equal $\Delta/2 = 18^\circ00'00''$. The table gives 18°00′00″. 2. **Chord (linear) check:** the last point, fixed by the last sub-chord from the previous peg, must fall on T₂, which is also located independently by measuring $T = 97.48$ m from the IP along the second tangent. Also, the measured long chord T₁T₂ should agree with $2R\sin(\Delta/2) = 185.41$ m. 3. **Sum of chords:** $c_1 + (n-2)\times 30 + c_n$ must equal $L = 188.50$ m. **Answer:** $T = 97.48$ m, $L = 188.50$ m, long chord $= 185.41$ m, T₁ = 1092.52 m, T₂ = 1281.02 m; deflection angles as tabulated, final = 18°00′00″. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For Δ = 36° and R = 300 m, which formula and value give the tangent length? A. T = 2R sin(Δ/2) = 185.4 m B. T = R tan Δ = 218 m C. T = R sin(Δ/2) = 92.7 m D. T = R tan(Δ/2) = 97.48 m Answer: D. T = 300 tan 18° = 97.48 m. (Key point: Tangent length) 2. With IP at 1190 m, T = 97.48 m and L = 188.50 m, the chainages of T1 and T2 are A. 1092.52 m and 1281.02 m B. 1092.52 m and 1190 m C. 1287.48 m and 1475.98 m D. 1190 m and 1378.5 m Answer: A. T1 = IP - T = 1092.52; T2 = T1 + L = 1281.02 m. (Key point: Chainage of T1 and T2) 3. Why is the first chord 17.48 m and the last 21.02 m? A. They are fixed by the radius B. They are the partial chords needed to reach the first full 30 m peg after T1 and to end at T2 C. They equal T/5 and L/9 D. They are half the normal chord Answer: B. T1 is at 1092.52, so 17.48 m reaches the 1110 peg; after 1260 only 21.02 m remains to T2. (Key point: Sub-chords) 4. What must the final cumulative deflection angle equal? A. Δ/4 = 9° B. 90° - Δ/2 = 72° C. Δ/2 = 18° D. Δ = 36° Answer: C. Tangent-chord angle to T2 is half the central angle, i.e., Δ/2 = 18°. (Key point: Angular check) 5. Which is a valid linear check during setting out? A. The long chord should equal R tan(Δ/2) B. T2 should lie at 2T from T1 C. The sum of deflection angles should equal L D. The long chord measured T1 to T2 should equal 2R sin(Δ/2) Answer: D. The long chord is 2R sin(Δ/2) = 1236.07 m as given; also the last peg must fall on T2 found by measuring T from the IP. 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