---
title: "Safe speed from stopping criterion on curve"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:n-78c-speed-stopping"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/n-78c-speed-stopping
site: BCT Engineering
---

# Safe speed from stopping criterion on curve

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2078 Chaitra, 8 marks
- Video: [Stopping Sight Distance Restriction in Horizontal Curve](https://www.youtube.com/watch?v=OAMA8nsyrVM) — Daniel Findley
- 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-78c-speed-stopping
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

A two-lane highway with a ruling gradient of 6% has a compensated gradient of 4.5% at a horizontal curve section of length 60 m. The curve section has a sight obstruction at 2 m from the edge of the carriageway. Determine the possible speed on the curve section based on stopping criterion. Assume coefficient of longitudinal friction of 0.35.

## Short answer

Grade compensation leads to R=50 m. Available SSD=38.3 m. Solving SSD equation for a 4.5% downgrade gives safe speed≈34 km/h.

## Answer

#### Step 1: radius of the curve from grade compensation
IRC grade compensation is $\dfrac{30+R}{R}\%$, subject to a maximum of $\dfrac{75}{R}\%$ ($R$ in m). Here the compensation is $6-4.5=1.5\%$.

- From $\dfrac{30+R}{R}=1.5$: $R=60$ m. But then the maximum allowed is $75/60=1.25\% < 1.5\%$, so a compensation of 1.5% is not possible with $R=60$ m.
- From the limit $\dfrac{75}{R}=1.5$: $R=50$ m. Then $\dfrac{30+50}{50}=1.6\% > 1.5\%$, so the limit governs and the compensation is exactly 1.5%. Consistent.

So $R=50$ m. The curve length is $L=60$ m.

#### Step 2: available sight distance
Assumptions: two-lane carriageway 7.0 m (lane 3.5 m); the driver keeps to the centre of the inner lane; the obstruction is 2 m from the pavement edge. The clearance from the driver's path is $m=2+1.75=3.75$ m, and the radius of the path is $R_i=50-1.75=48.25$ m.

For $S<L$: $m=R_i(1-\cos\theta)$, where $\theta=\dfrac{S}{2R_i}$.

$$
\cos\theta=1-\frac{3.75}{48.25}=0.9223\Rightarrow\theta=22.74^\circ=0.3969\ \text{rad}
$$

$$
S=2R_i\theta=2(48.25)(0.3969)=38.3\ \text{m}
$$

$S<L=60$ m, so the assumption is valid. This is the stopping sight distance available.

#### Step 3: speed from stopping criterion
Take the compensated gradient $n=4.5\%$ as a downgrade (critical case), $f=0.35$, $t=2.5$ s:

$$
S=0.278Vt+\frac{V^2}{254(f-n)}\Rightarrow 38.3=0.695V+\frac{V^2}{254(0.305)}
$$

Solving the quadratic gives $V=33.8$ km/h.

**Answer:** the safe speed on the curve is about **34 km/h** (about 36 km/h if the 4.5% grade is an upgrade). If the limit of $75/R$ is ignored and $R=60$ m is used, the available sight distance is 42.0 m and the speed is about 36 km/h.

## Check yourself

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

1. How is the curve radius found in this problem?
   A. Set ruling gradient minus the IRC compensation (30+R)/R % equal to 4.5% and solve for R
   B. Use R = V^2/(127(e+f)) with the given friction
   C. Use R = L/Δ with the curve length 60 m
   D. Set the compensation equal to 4.5% directly
   Answer: A. The compensated gradient is 6% minus (30+R)/R, so (30+R)/R = 1.5 gives R = 60 m. (Key point: Radius from grade compensation)

2. Why is the clearance used as m = 2 + 1.75 = 3.75 m rather than 2 m?
   A. The obstruction is always 1.75 m taller than the driver's eye
   B. The driver travels along the centre of the inner lane, 1.75 m inside the pavement edge, so the clearance to the obstruction is measured from that path
   C. The 1.75 m is the safety factor added to every clearance
   D. Because the road is 7 m wide, 1.75 m is half the curve length
   Answer: B. The sight line starts at the driver's path (centre of inner lane), so lane half-width is added to the 2 m. (Key point: Sight line path)

3. For S < L, which relation gives the available sight distance on the curve?
   A. S = L, the curve length, whenever the curve is long
   B. S = m x Ri
   C. m = Ri(1 - cosθ) with θ = S/(2Ri), so S = 2Riθ
   D. m = Ri(1 - cosθ) with θ = S/Ri
   Answer: C. The sight line is a chord whose half-angle is S/(2Ri); solving gives S about 42 m, less than L = 60 m. (Key point: Sight distance when S<L)

4. Which case is used with the compensated gradient 4.5% to get the safe speed?
   A. Treat it as an upgrade (f + n), the critical case
   B. Ignore the gradient and use level road
   C. Use the full 6% ruling gradient instead
   D. Treat it as a downgrade (f - n in the braking term), the critical case
   Answer: D. A downgrade lengthens braking distance, so 42 = 0.695V + V^2/(254(0.305)) gives V about 36 km/h. (Key point: Critical gradient case)

## Related in Transportation Engineering I

- [Sight distance with obstruction on curve](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-78b-sd-obstruction) — Transportation Engineering I, chapter 3
- [Setback on four lane carriageway, 400 m radius](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-70bh-setback-four-lane) — Transportation Engineering I, chapter 3
- [Setback on six lane highway, 450 m curve](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-73m-setback-six-lane) — Transportation Engineering I, chapter 3
- [Setback on six lane highway, 550 m radius](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-72a-setback-350) — Transportation Engineering I, chapter 3
- [Vertical curve with horizontal curve, 45 degrees](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-77c-vc-horizontal) — Transportation Engineering I, chapter 3
- [Stopping sight distance at 50 kmph](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-66m-ssd-50) — Transportation Engineering I, chapter 3

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