---
title: "Transition curve 6°, Ls=90 m, chainage 13+700"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:transition-t1v-t2v-chainage-13700"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/transition-t1v-t2v-chainage-13700
site: BCT Engineering
---

# Transition curve 6°, Ls=90 m, chainage 13+700

Surveying II · Chapter 6: Curves

- Asked in: 2080 Chaitra, 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-13700
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

Two tangents T₁V and T₂V intersect at chainage of 13+700 km. The bearing of T₂V is 300°30' and T₁V is 80°30'. It is proposed to insert 6° circular curve with transition curve 90 m at each end. Calculate the first two deflection angle for setting out first transition and circular curve only at peg interval of 10 m on transition curve and 30 m at circular curve.

## Short answer

Δ = 40°, R = 286.48 m. Transition shift s = 1.178 m, total tangent Ts = 149.70 m, circular length Lc = 110.00 m. First two transition deflections at TS: 0°02′05″ and 0°08′37″; circular at SC: 0°58′11″ and 3°58′11″.

## 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 $= 13700$ m (13+700 km).

| Point | Chainage (m) |
|---|---|
| TS = V − T_s | 13550.30 |
| SC = TS + 90 | 13640.30 |
| CS = SC + L_c | 13750.30 |
| ST = CS + 90 | 13840.30 |

#### First two deflection angles
Transition curve (theodolite at TS), $\alpha = 573\,l^2/(RL_s)$ minutes, points at multiples of 10 m:

| Point (chainage, m) | l (m) | α (min) | α |
|---|---|---|---|
| 13560.00 | 9.70 | 2.09′ | 0°02′05″ |
| 13570.00 | 19.70 | 8.62′ | 0°08′37″ |

Circular curve (theodolite at SC), $\delta = 1718.87\,c/R$ minutes, points at multiples of 30 m:

| Point (chainage, m) | Chord c (m) | δ | Cumulative |
|---|---|---|---|
| 13650.00 | 9.70 | 0°58′11″ | 0°58′11″ |
| 13680.00 | 30.00 | 3°00′00″ | 3°58′11″ |

**Answer:** transition: 0°02′05″ and 0°08′37″; circular: 0°58′11″ and 3°58′11″.

## Check yourself

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

1. Bearing T1V is 80°30′ and T2V is 300°30′. How is the deflection Δ found?
   A. 300°30′ + 80°30′ = 381°
   B. 300°30′ - 80°30′ = 220°
   C. Reverse 300°30′ to 120°30′ (forward direction) and subtract 80°30′ = 40°
   D. 180° - 220° = -40° ignored
   Answer: C. T2V is observed towards V; the travel direction is 180° opposite, 120°30′, so Δ = 40° to the right. (Key point: Deflection angle between tangents)

2. R for a 6° curve (30 m arc) is
   A. 572.96 m
   B. 143.24 m
   C. 1718.87 m
   D. 286.48 m
   Answer: D. R = 1718.87/6 = 286.48 m. (Key point: Radius of 6° curve)

3. With Ts = 149.70 m and V at 13700, TS is at
   A. 13550.30 m
   B. 13849.70 m
   C. 13700 m
   D. 13640.30 m
   Answer: A. TS = V - Ts = 13550.30 m. (Key point: TS chainage)

4. With TS at 13550.30 and 10 m pegs, the first transition point has l =
   A. 10 m
   B. 9.70 m
   C. 19.70 m
   D. 0.30 m
   Answer: B. The first peg is 13560, which is 9.70 m from TS; its angle is 573 x 9.7^2/(286.48 x 90) = 2.09 min. (Key point: First sub-chord on transition)

5. On the circular curve from SC at 13640.30 with 30 m pegs, the first chord is
   A. 30 m
   B. 19.70 m
   C. 9.70 m to peg 13650
   D. 0.30 m
   Answer: C. Next 30 m multiple is 13650, 9.70 m from SC; δ = 1718.87 x 9.7/286.48 = 58.2 min = 0°58′11″. (Key point: First arc chord)

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