---
title: "Curve resistance and grade compensation"
subject: "Transportation Engineering I"
chapter: 3
key: "semester_6/transportation_engineering_i:curve-resistance-grade-comp"
source: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/curve-resistance-grade-comp
site: BCT Engineering
---

# Curve resistance and grade compensation

Transportation Engineering I · Chapter 3: Geometric Design of Highway

- Asked in: 2081 Chaitra, 4 marks; 2071 Bhadra; 2067 Mangsir (old course), 4 marks (asked 3 times)
- Video: [Geometric Deign of Highway: Curve Resistance](https://www.youtube.com/watch?v=3MIwRv9O4bE) — Sunil Khyaju
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/transportation_engineering_i/curve-resistance-grade-comp
- Whole chapter: https://www.bctengineer.com/md/semester_6/transportation_engineering_i/chapter-3

## Question

What are curve resistance and grade compensation?

## Short answer

Curve resistance is additional tractive resistance on horizontal curves due to turned wheels, tyre slip, lateral friction, and off-tracking. Grade compensation reduces design gradient on curves so total resistance (grade + curve) equals straight resistance. IRC formula: compensation = (30+R)/R % (max 75/R%). Not needed for gradients <4%. Example: R=60m gives 1.25% compensation, reducing 6% gradient to 4.75%.

## Answer

#### Curve resistance
**Curve resistance** is the additional tractive resistance a vehicle has to overcome when it moves along a horizontal curve. When the front wheels are turned through an angle $\alpha$, only the component $T\cos\alpha$ of the tractive force $T$ acts along the direction of motion, so the loss is $T(1-\cos\alpha)$. The sideways slip of the tyres, the lateral friction needed to resist the centrifugal force and the off-tracking of the rear wheels add to it. It increases as the radius becomes smaller, so a gradient and a sharp curve at the same place can overload the vehicle.

#### Grade compensation
**Grade compensation** (grade reduction) is the reduction of the design gradient on a horizontal curve, so that the total resistance (grade + curve) on the curve is not more than the resistance on the straight section.

IRC formula (R in metres):

$$
\text{Grade compensation} = \frac{30 + R}{R}\ \%\ \ \text{(max } \frac{75}{R}\%)
$$

- It is not needed when the gradient is flatter than 4%, and in practice it matters mainly for sharp curves (small $R$).
- Example: $R = 60$ m: $(30+60)/60 = 1.5\%$, maximum $75/60 = 1.25\%$, so adopt 1.25%. If the ruling gradient is 6%, use $6 - 1.25 = 4.75\%$ on the curve.
- Used chiefly on hill roads.

## Check yourself

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

1. Curve resistance is:
   A. Resistance of air
   B. Additional tractive resistance on a horizontal curve
   C. Resistance due to gradient
   D. Rolling resistance
   Answer: B. It arises from tyre slip, lateral friction, steering and off-tracking. (Key point: Curve resistance)

2. Grade compensation is:
   A. Raising the gradient on curves
   B. Doubling the gradient
   C. Reducing the design gradient on a curve so total resistance does not exceed that on a straight
   D. Removing the curve
   Answer: C. It lowers the gradient at curves. (Key point: Grade compensation)

3. IRC grade compensation is (30+R)/R % subject to a maximum of:
   A. 75/R %
   B. 30/R %
   C. R/75 %
   D. 100/R %
   Answer: A. Max is 75/R %. (Key point: Formula)

4. For R = 60 m and ruling gradient 6%, the gradient on the curve is:
   A. 6%
   B. 1.5%
   C. 7.25%
   D. 4.75%
   Answer: D. (30+60)/60 = 1.5 exceeds max 75/60 = 1.25, so 6 - 1.25 = 4.75%. (Key point: Example)

## Related in Transportation Engineering I

- [Reasons for grade compensation](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3Agrade-comp-reasons) — Transportation Engineering I, chapter 3, asked 3×
- [Vertical curve 35 m headlight and momentum grade](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-80c-vc-headlight-35m) — Transportation Engineering I, chapter 3
- [Vertical curve with 300 m grade sections](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-65k-vc-300) — Transportation Engineering I, chapter 3
- [Actual grade on inner and outer lanes](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-64p-grade-inner-outer) — Transportation Engineering I, chapter 3
- [Summit curve limited to 750 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-75bh-vc-summit-750) — Transportation Engineering I, chapter 3, asked 3×
- [Vertical curve -3.5% and +4%, 250 m night](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-78b-vc-night) — Transportation Engineering I, chapter 3, asked 2×

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