---
title: "Composite curve R=400 m, Ls=90 m, IP 1+400"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:composite-curve-r400-ls90-ip-1400"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/composite-curve-r400-ls90-ip-1400
site: BCT Engineering
---

# Composite curve R=400 m, Ls=90 m, IP 1+400

Surveying II · Chapter 6: Curves

- Asked in: 2065 Kartik (old course), 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/composite-curve-r400-ls90-ip-1400
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

Two straights AB and BC intersect at the chainage (1+400) kilometer, the deflection angle being 40°00'. It is proposed to layout a circular curve of 400 m radius with a cubic parabola of 90 m length at each end. Peg intervals for circular and transition curve are 20 m and 30 m respectively. Calculate tangential angles for first two points on transition curve, deflection angles for two points on the circular curve and chainage at the beginning and at the end of this composite curve.

## Short answer

Beginning chainage TS = 1209.10 m, end ST = 1578.36 m. Tangential angles on transition: 0°06′57″ (at 1230 m) and 0°41′14″ (at 1260 m). Deflection angles on circular curve: 0°03′51″ and cumulative 1°29′47″ for first two points.

## Answer

Given: $\Delta = 40^\circ$, $R = 400$ m, $L_s = 90$ m, chainage of IP $= 1+400$ km $= 1400$ m, peg interval 30 m on the transitions and 20 m on the circular curve.

#### Curve elements
$$
\begin{aligned}
s &= \frac{L_s^2}{24R} = \frac{90^2}{24\times 400} = 0.844\ \text{m}\\
\theta_s &= \frac{L_s}{2R} = \frac{90}{800}\ \text{rad} = 6.446^\circ\\
T_s &= (R+s)\tan 20^\circ + \frac{L_s}{2} = 190.90\ \text{m}\\
L_c &= R(\Delta - 2\theta_s) = 189.25\ \text{m}
\end{aligned}
$$

#### Chainages
- Beginning of composite curve, TS $= 1400 - 190.90 = 1209.10$ m
- SC $= 1209.10 + 90 = 1299.10$ m
- CS $= 1299.10 + 189.25 = 1488.36$ m
- End of composite curve, ST $= 1488.36 + 90 = 1578.36$ m

#### Tangential angles on the transition (from TS)
$\alpha = 573\,l^2/(R L_s)$ minutes. First two points (the first sub-chord runs to the next 30 m peg):

| Point (chainage, m) | l (m) | α (min) | α |
|---|---|---|---|
| 1230.00 | 20.90 | 6.95′ | 0°06′57″ |
| 1260.00 | 50.90 | 41.23′ | 0°41′14″ |

The full transition table is:

| Point (chainage, m) | l from TS (m) | α (min) | α (° ′ ″) |
|---|---|---|---|
| 1209.10 | 0.00 | 0.00′ | 0°00′00″ |
| 1230.00 | 20.90 | 6.95′ | 0°06′57″ |
| 1260.00 | 50.90 | 41.23′ | 0°41′14″ |
| 1290.00 | 80.90 | 104.16′ | 1°44′10″ |
| 1299.10 | 90.00 | 128.93′ | 2°08′56″ |

#### Deflection angles on the circular curve (from SC)
$\delta = 1718.87\,c/R$ minutes; first two points:

| Point (chainage, m) | Chord c (m) | δ | Cumulative |
|---|---|---|---|
| 1300.00 | 0.90 | 0°03′51″ | 0°03′51″ |
| 1320.00 | 20.00 | 1°25′57″ | 1°29′47″ |

Full table:

| Point (chainage, m) | Chord c (m) | δ = 1718.87c/R | Cumulative deflection |
|---|---|---|---|
| 1299.10 | - | - | 0°00′00″ |
| 1300.00 | 0.90 | 0°03′51″ | 0°03′51″ |
| 1320.00 | 20.00 | 1°25′57″ | 1°29′47″ |
| 1340.00 | 20.00 | 1°25′57″ | 2°55′44″ |
| 1360.00 | 20.00 | 1°25′57″ | 4°21′41″ |
| 1380.00 | 20.00 | 1°25′57″ | 5°47′37″ |
| 1400.00 | 20.00 | 1°25′57″ | 7°13′34″ |
| 1420.00 | 20.00 | 1°25′57″ | 8°39′30″ |
| 1440.00 | 20.00 | 1°25′57″ | 10°05′27″ |
| 1460.00 | 20.00 | 1°25′57″ | 11°31′24″ |
| 1480.00 | 20.00 | 1°25′57″ | 12°57′20″ |
| 1488.36 | 8.36 | 0°35′55″ | 13°33′15″ |

Check: last cumulative angle $= (\Delta - 2\theta_s)/2 = 13.5542^\circ$ = 13°33′15″.

**Answer:** TS = 1209.10 m and ST = 1578.36 m (beginning and end of the composite curve); deflection angles as tabulated.

## Check yourself

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

1. For R = 400 m and Ls = 90 m, what is the spiral angle θs?
   A. Ls/(3R) = 4.30°
   B. Ls/(2R) = 6.446°
   C. Ls/(6R) = 2.15°
   D. Ls/R = 12.89°
   Answer: B. θs = Ls/(2R) = 90/800 rad = 0.1125 rad = 6.446°. (Key point: Spiral angle)

2. Which expression gives the tangent length Ts from IP to TS?
   A. R tan(Δ/2)
   B. (R + s) sec(Δ/2) - R
   C. (R + s) tan(Δ/2) + Ls/2
   D. (R - s) tan(Δ/2) + Ls
   Answer: C. The shifted circle has radius R + s from the tangent, and half of the transition length lies beyond the original tangent point, hence + Ls/2. (Key point: Tangent length from IP)

3. With Ts = 190.90 m, Lc = 189.25 m, Ls = 90 m and IP at 1400 m, what are the chainages of TS and ST?
   A. 1299.10 m and 1488.36 m
   B. 1400 m and 1578.36 m
   C. 1209.10 m and 1590.90 m
   D. 1209.10 m and 1578.36 m
   Answer: D. TS = 1400 - 190.90 = 1209.10; ST = TS + 90 + 189.25 + 90 = 1578.36 m. (Key point: Chainage of TS and ST)

4. The first peg on the transition is at chainage 1230 (l = 20.90 m from TS). Which formula gives its tangential angle?
   A. α = 573 l^2/(R Ls) minutes
   B. α = 573 l/(R Ls) minutes
   C. α = 1718.87 l/R minutes
   D. α = 1718.87 l^2/R minutes
   Answer: A. Along the spiral the angle varies as the square of l: α = 573 l^2/(R Ls) = 573 x 20.9^2/(400 x 90) = 6.95 minutes. (Key point: Deflection angle of first transition peg)

5. The last cumulative deflection angle on the circular curve at CS should equal which value?
   A. θs/3 = 2°09′
   B. (Δ - 2θs)/2 = 13°33′15″
   C. Δ/2 = 20°
   D. Δ - 2θs = 27°06′
   Answer: B. The arc subtends Δ - 2θs at the centre, so the deflection to its end is half of that: 13.554° = 13°33′15″. (Key point: Circular curve check)

## Related in Surveying II

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