---
title: "Transition curve for 310 m radius"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-80c-tc-310"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-80c-tc-310
site: BCT Engineering
---

# Transition curve for 310 m radius

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2080 Chaitra, 8 marks
- Video: [Design SECRET to Perfect Transition Curve (Spiral Curve) - Mastering Road Curves](https://www.youtube.com/watch?v=JVqkgftaHNk) — Civil Guru Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-80c-tc-310
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A national highway of two lane with a curve of 310 m radius is to be set out to connect two straights roads in plain terrain. The maximum speed of moving vehicles on this curve is restricted to 90 Kmph. Transition curve are to be introduced at each end curve. Design the transition curve and determine the chainage of the beginning of transition curve, and end of circular curve.
Given that angle of intersection = 136°
Chainage at the point of intersection: (1+000)
Rate of introduction of super elevation = 1 in 120
Rate of change of centrifugal acceleration (c) = 0.5 m/s³

## Short answer

For a two-lane NH with 310 m radius, 90 km/h design speed, 44° deflection angle (180°-136°), PI at chainage 1000 m, e_max=7%, transition length is 105 m (maximum of 100.8 m from centrifugal acceleration, 64.3 m from superelevation rate 1:120, and 70.5 m from IRC formula). Shift is 1.48 m, tangent length 178.35 m. Beginning of transition curve (TS) at chainage 821.65 m, end of circular curve (CS) at chainage 1059.71 m.

## Answer

**Given:** $R = 310$ m, $V = 90$ km/h, two-lane NH in plain terrain, $c = 0.5$ m/s³, rate of introduction of superelevation 1 in 120, PI at $1+000$ m.

**Reading:** "angle of intersection 136°" is the interior angle between the straights, so the deflection angle $\Delta = 180^\circ - 136^\circ = 44^\circ$.

**Assumptions:** carriageway width $W = 7.0$ m, wheel base $l = 6$ m.

#### Step 1: Superelevation
$$
e = \frac{V^2}{225R} = \frac{90^2}{225\times310} = 0.116 > 0.07
$$
Adopt $e_{max} = 7\%$. Check: $f = \dfrac{90^2}{127\times310} - 0.07 = 0.136 < 0.15$, OK.

#### Step 2: Extra widening
$$
W_e = \frac{nl^2}{2R} + \frac{V}{9.5\sqrt{R}} = \frac{2(6)^2}{2(310)} + \frac{90}{9.5\sqrt{310}} = 0.116 + 0.538 = 0.65\ \text{m}
$$

#### Step 3: Length of transition curve
$v = 90/3.6 = 25$ m/s.
- **Centrifugal acceleration:** $L_s = \dfrac{v^3}{cR} = \dfrac{25^3}{0.5\times310} = 100.8$ m
- **Superelevation rate (1 in 120, rotation about the inner edge):** $L_s = N\,e\,(W+W_e) = 120\times0.07\times7.65 = 64.3$ m
- **IRC empirical:** $L_s = \dfrac{2.7V^2}{R} = \dfrac{2.7\times90^2}{310} = 70.5$ m

Adopt the largest, rounded up: **$L_s = 105$ m**.

#### Step 4: Shift and tangent length
$$
\begin{aligned}
s &= \frac{L_s^2}{24R} = \frac{105^2}{24(310)} = 1.48\ \text{m} \\
T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = (311.48)\tan22^\circ + 52.5 = 125.85 + 52.5 = 178.35\ \text{m}
\end{aligned}
$$

#### Step 5: Length of circular curve
$$
\phi_s = \frac{L_s}{2R} = 9.70^\circ,\qquad
L_c = R\,\Delta_{rad} - L_s = 310\times0.76794 - 105 = 133.06\ \text{m}
$$

#### Step 6: Chainages
| Point | Chainage (m) |
|---|---|
| Beginning of transition curve (TS) $= 1000 - 178.35$ | **821.65** |
| End of first transition (SC) $= 821.65 + 105$ | 926.65 |
| End of circular curve (CS) $= 926.65 + 133.06$ | **1059.71** |
| End of second transition (ST) $= 1059.71 + 105$ | 1164.71 |

**Answer:** $L_s = 105$ m; the curve begins at chainage 821.65 m (0+821.65) and the circular curve ends at 1059.71 m (1+059.71).

*If 136° were the deflection angle itself: TS would be at chainage 176.56 m and the end of the circular curve at 912.39 m.*

## Check yourself

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

1. The angle of intersection of 136 degrees is taken as the interior angle. What is the deflection angle?
   A. 136 degrees
   B. 44 degrees
   C. 68 degrees
   D. 22 degrees
   Answer: B. Deflection angle = 180 - 136 = 44 degrees. (Key point: Deflection angle)

2. e = V^2/(225R) = 0.116 exceeds 0.07. What is done?
   A. Provide 11.6%
   B. Provide e_max = 7% and check that f < 0.15
   C. Increase the radius
   D. Ignore superelevation
   Answer: B. Adopt e = 7% and check f = V^2/127R - 0.07 = 0.136 < 0.15, which is OK. (Key point: Superelevation limit)

3. The criteria gave 100.8 m (centrifugal), 64.3 m (superelevation) and 70.5 m (IRC). The adopted Ls is
   A. 64.3 m
   B. 70.5 m
   C. 100.8 m exactly
   D. 105 m, the largest value rounded
   Answer: D. The largest criterion governs, 100.8 m, rounded to 105 m. (Key point: Length of transition curve)

4. Which expression gives the total tangent length from PI to TS?
   A. Ts = R tan(Delta/2)
   B. Ts = (R + s) tan(Delta/2) + Ls/2
   C. Ts = (R - s) tan(Delta/2) - Ls/2
   D. Ts = Ls + R
   Answer: B. Ts = (R + s) tan(Delta/2) + Ls/2 = 125.85 + 52.5 = 178.35 m, where s = Ls^2/24R is the shift. (Key point: Tangent length with transition)

5. With PI at 1000 m and Ts = 178.35 m, the transition curve begins at chainage
   A. 1178.35 m
   B. 821.65 m
   C. 926.65 m
   D. 1059.72 m
   Answer: B. TS = 1000 - 178.35 = 821.65 m. (Key point: Chainage of TS)

## Related in Transportation Engineering I

- [Transition curve 400 m radius, 85 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-tc-400-85) — Transportation Engineering I, chapter 3, asked 2×
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