---
title: "Simple circular curve elements and chainages"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-64s-simple-curve"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-64s-simple-curve
site: BCT Engineering
---

# Simple circular curve elements and chainages

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2064 Shrawan (old course), 8 marks
- Video: [How to Calculate Horizontal  or Simple Curve Elements  & Formulas](https://www.youtube.com/watch?v=A1wCuvOkgGE) — Surveying Engineering Design Information
- 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-64s-simple-curve
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

At a certain section of road there is an intersection point (IP) with an angle of 45°30' turning right. The minimum permissible radius is 200 m. The distance between starting point and IP and IP to end point are 500 and 350 m respectively. Calculate the elements of circular curve and chainages of main points of curve assuming the chainage of starting point is to be 20 + 416.60.

## Short answer

T = 83.87 m, Lc = 158.82 m, C = 154.68 m, E = 16.87 m, M = 15.56 m. Chainage of PC = 20832.73 m, PT = 20991.56 m.

## Answer

#### Given
Deflection angle $\Delta=45^\circ30'=45.5^\circ$ (right turn), $R=200$ m. Start chainage = 20+416.60 = 20 416.60 m; start to IP = 500 m; IP to end = 350 m.

#### Elements of the simple circular curve
- Tangent length: $T=R\tan\dfrac{\Delta}{2}=200\tan22.75^\circ=\mathbf{83.87}$ m
- Length of curve: $L_c=\dfrac{\pi R\Delta}{180}=\dfrac{\pi(200)(45.5)}{180}=\mathbf{158.82}$ m
- Long chord: $C=2R\sin\dfrac{\Delta}{2}=2(200)\sin22.75^\circ=\mathbf{154.68}$ m
- External distance: $E=R\left(\sec\dfrac{\Delta}{2}-1\right)=\mathbf{16.872}$ m
- Mid-ordinate: $M=R\left(1-\cos\dfrac{\Delta}{2}\right)=\mathbf{15.560}$ m

#### Chainages
- Chainage of IP $=20\,416.60+500=20916.60$ m (20+916.60)
- Beginning of curve (PC) $=IP-T=20916.60-83.87=\mathbf{20832.73}$ m
- Mid-point of curve $=PC+L_c/2=\mathbf{20912.15}$ m
- End of curve (PT) $=PC+L_c=20832.73+158.82=\mathbf{20991.56}$ m
- End of the road $=PT+(350-T)=20991.56+266.13=\mathbf{21257.69}$ m

The minimum permissible radius is 200 m, and the curve uses exactly this radius.

## Check yourself

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

1. Which formula gives T?
   A. T = R tan(Delta/2) = 83.87 m
   B. T = R tan Delta
   C. T = R sin(Delta/2)
   D. T = R(sec(Delta/2) - 1)
   Answer: A. Tangent length is R tan(Delta/2). (Key point: Tangent length)

2. How is the chainage of the beginning of curve found?
   A. Chainage of IP - T = 20916.60 - 83.87
   B. Chainage of IP + T
   C. Start chainage + T
   D. Chainage of IP - L_c
   Answer: A. PC lies a tangent length before the IP. (Key point: PC chainage)

3. How is the end of curve found?
   A. PC + L_c = 20832.73 + 158.82
   B. IP + L_c
   C. PC + T
   D. IP + 2T
   Answer: A. Curve ends one curve length after PC; L_c = pi R Delta/180. (Key point: PT chainage)

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## Related in other subjects

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