---
title: "Highway circular curve, apex distance 15.876 m"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:highway-curve-apex-distance-tangential-angle"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/highway-curve-apex-distance-tangential-angle
site: BCT Engineering
---

# Highway circular curve, apex distance 15.876 m

Surveying II · Chapter 6: Curves

- Asked in: 2078 Poush
- Video: [Simple Circular Curve| Length Of the Curve| Mid ordinate | Apex Distance | Engineering Surveying](https://www.youtube.com/watch?v=rdiRd7pcsCU) — Nilesh Jha Life
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/highway-curve-apex-distance-tangential-angle
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

In a highway circular curve, the midpoint of curve (mc) passes through an apex distance of 15.876 m from IP having chainage of 0+455.50 km and deflection angle being 52°10'. Calculate the suitable radius of circular curve and design the circular curve by tangential angle method assuming normal chord 30.00 m.

## Short answer

Radius = 140 m, tangent length = 68.53 m, curve length = 127.47 m. Chainage of T1 = 386.97 m, T2 = 514.43 m. Tangential angles for 30 m normal chord and sub-chords are computed, with final cumulative deflection of 26°05′ (Δ/2).

## Answer

Data: apex (external) distance $E = 15.876$ m, deflection angle $\Delta = 52^\circ10'$, chainage of IP = 0+455.50 (455.50 m), normal chord $C = 30$ m.

#### 1. Radius from the apex distance

$$
E = R\left(\sec\frac{\Delta}{2} - 1\right) \quad\Rightarrow\quad R = \frac{E}{\sec 26^\circ05' - 1} = \frac{15.876}{0.11339} = 140.01\ \text{m}
$$

A suitable (rounded) radius of **R = 140 m** is adopted (it gives E = 15.875 m, practically the same as the given 15.876 m).

#### 2. Curve elements

$$
T = R\tan\frac{\Delta}{2} = 140\tan 26^\circ05' = 68.535\ \text{m}
$$

$$
L_c = \frac{\pi R\Delta}{180^\circ} = \frac{\pi\times 140\times 52.1667}{180} = 127.467\ \text{m}
$$

- Chainage of T1 (start) $= 455.50 - 68.535 = 386.965$ m
- Chainage of T2 (end) $= 386.965 + 127.467 = 514.432$ m

#### 3. Tangential (deflection) angles

For a chord $c$ the tangential angle between the tangent and the chord is

$$
\delta = \frac{1718.9\,c}{R}\ \text{minutes}
$$

The first chord is a sub-chord to the first 30 m peg; the last is a sub-chord to T2. Normal chord $= 30$ m, $\delta_{30} = 1718.9\times 30/140 = 6°08'20"$.

| Peg chainage (m) | Chord (m) | Deflection for chord $\delta=1718.9\,c/R$ | Cumulative deflection |
|---|---|---|---|
| 390.00 | 3.03 | 0°37'16" | 0°37'16" |
| 420.00 | 30.00 | 6°08'20" | 6°45'36" |
| 450.00 | 30.00 | 6°08'20" | 12°53'56" |
| 480.00 | 30.00 | 6°08'20" | 19°02'16" |
| 510.00 | 30.00 | 6°08'20" | 25°10'36" |
| 514.43 | 4.43 | 0°54'25" | 26°05'01" |

Check: the final cumulative angle 26°05'01" equals $\Delta/2 = 26^\circ05'$ ✓.

**Answer: R = 140 m (computed 140.01 m), T = 68.53 m, $L_c$ = 127.47 m, T1 at 386.97 m and T2 at 514.43 m; the setting-out table is above.**

## Check yourself

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

1. Which formula gives R from the apex distance E and deflection angle Δ?
   A. E = R(sec(Δ/2) - 1)
   B. E = R tan(Δ/2)
   C. E = R(1 - cos(Δ/2))
   D. E = R sin(Δ/2)
   Answer: A. The external (apex) distance is E = R(sec(Δ/2) - 1), so R = 15.876/0.11339 = 140 m. (Key point: Radius from apex distance)

2. How is the chainage of T1 found?
   A. IP chainage + T
   B. IP chainage - T, where T = R tan(Δ/2) = 68.535 m, giving 386.965
   C. IP chainage - Lc
   D. IP chainage - E
   Answer: B. T1 is a tangent length before the IP: 455.50 - 68.535 = 386.965 m. (Key point: Tangent length and T1)

3. T1 is at 386.965 m and pegs are at 30 m multiples. What is the first chord?
   A. 30.00 m
   B. 4.43 m
   C. 68.54 m
   D. about 3.03 m to the peg at 390
   Answer: D. The first peg is at chainage 390, so the sub-chord is 390 - 386.965 = 3.03 m. (Key point: First sub-chord)

4. What must the final cumulative tangential angle be?
   A. Δ = 52°10'
   B. Δ/2 = 26°05'
   C. Δ/4
   D. δ for the 30 m chord
   Answer: B. The total deflection to T2 is half the deflection angle, 26°05', which the table reaches (26°05'01"). (Key point: Check on deflections)

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