---
title: "Transition curve 6°, Ls=90 m, chainage (13+7) chains"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:transition-t1v-t2v-chainage-13-7"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-t1v-t2v-chainage-13-7
site: BCT Engineering
---

# Transition curve 6°, Ls=90 m, chainage (13+7) chains

Surveying II · Chapter 6: Curves

- Asked in: 2076 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-t1v-t2v-chainage-13-7
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

Two tangents T₁V and T₂V intersect at chainage of (13+7) Chains. The bearing of forward tangent T₂V is 300°30' and backward tangent T₁V is 80°30'. It is proposed to insert 6° Circular Curve with transition curve 90 m at each end. Make all the calculations necessary for setting out the curve at peg interval of 15 m on the transition curve and 30 m at the circular curve.

## Short answer

Deflection angle Δ = 40° right. R = 1718.87/6 = 286.48 m; transition length Ls = 90 m; shift s = 1.178 m; tangent length Ts = 149.70 m; circular curve length Lc = 110.00 m. Chainages: TS 247.30 m, SC 337.30 m, CS 447.30 m, ST 537.30 m.

## Answer

#### Deflection angle between the tangents
Bearing of T₁V (travelling towards V) $= 80^\circ30'$. Bearing of T₂V is observed from T₂ towards V, $300^\circ30'$, so the forward direction of travel from V to T₂ is $300^\circ30' - 180^\circ = 120^\circ30'$.

$$
\Delta = 120^\circ30' - 80^\circ30' = 40^\circ\ \text{(deflection to the right)}
$$

#### Radius and transition elements
6° curve with a 30 m arc: $R = 1718.87/6 = 286.48$ m. $L_s = 90$ m.

$$
\begin{aligned}
s &= \frac{L_s^2}{24R} = \frac{90^2}{24\times 286.48} = 1.178\ \text{m}\\
\theta_s &= \frac{L_s}{2R} = 9.000^\circ\\
T_s &= (R+s)\tan 20^\circ + \frac{L_s}{2} = 149.70\ \text{m}\\
L_c &= R(\Delta-2\theta_s) = 110.00\ \text{m}
\end{aligned}
$$

Chainage of V $= 397$ m ((13+7) chains read as 13 × 30 + 7 m).

| Point | Chainage (m) |
|---|---|
| TS = V − T_s | 247.30 |
| SC = TS + 90 | 337.30 |
| CS = SC + L_c | 447.30 |
| ST = CS + 90 | 537.30 |

#### Transition curve (theodolite at TS), peg interval 15 m
$\alpha = 573\,l^2/(RL_s)$ minutes:

| Point (chainage, m) | l from TS (m) | α (min) | α (° ′ ″) |
|---|---|---|---|
| 247.30 | 0.00 | 0.00′ | 0°00′00″ |
| 255.00 | 7.70 | 1.32′ | 0°01′19″ |
| 270.00 | 22.70 | 11.45′ | 0°11′27″ |
| 285.00 | 37.70 | 31.58′ | 0°31′35″ |
| 300.00 | 52.70 | 61.72′ | 1°01′43″ |
| 315.00 | 67.70 | 101.85′ | 1°41′51″ |
| 330.00 | 82.70 | 151.99′ | 2°31′59″ |
| 337.30 | 90.00 | 180.01′ | 3°00′01″ |

#### Circular curve (theodolite at SC), peg interval 30 m
$\delta = 1718.87\,c/R$ minutes:

| Point (chainage, m) | Chord c (m) | δ = 1718.87c/R | Cumulative deflection |
|---|---|---|---|
| 337.30 | - | - | 0°00′00″ |
| 360.00 | 22.70 | 2°16′11″ | 2°16′11″ |
| 390.00 | 30.00 | 3°00′00″ | 5°16′11″ |
| 420.00 | 30.00 | 3°00′00″ | 8°16′11″ |
| 447.30 | 27.30 | 2°43′48″ | 11°00′00″ |

Check: last cumulative deflection $= (\Delta - 2\theta_s)/2 = 11.0000^\circ$ = 11°00′00″; at SC the transition angle $\alpha_s = \theta_s/3 = 3.0000^\circ$ = 3°00′00″. The second transition is set out from ST backwards, in the same way.

**Answer:** $\Delta = 40^\circ$, $R = 286.48$ m, $T_s = 149.70$ m, $L_c = 110.00$ m; TS = 247.30, SC = 337.30, CS = 447.30, ST = 537.30 m.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With Ts = 149.70 m and Ls = 90 m, TS and SC are
   A. 247.30 m and 337.30 m
   B. 547.7 m and 637.7 m
   C. 247.30 m and 447.30 m
   D. 397 m and 487 m
   Answer: A. TS = 397 - 149.70 = 247.30; SC = TS + 90 = 337.30. (Key point: TS and SC)

2. At SC the transition deflection angle should equal
   A. θs = 9°
   B. θs/3 = 3°
   C. Δ/2 = 20°
   D. 2θs = 18°
   Answer: B. αs = θs/3 = 9°/3 = 3°. (Key point: Check angle at SC)

3. The last cumulative deflection on the arc at CS is
   A. Δ - 2θs = 22°
   B. θs/3 = 3°
   C. (Δ - 2θs)/2 = 11°
   D. Δ/2 = 20°
   Answer: C. Arc central angle is 40 - 18 = 22°, so deflection is 11°. (Key point: Circular curve end check)

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