---
title: "Tractive effort up and down a gradient"
subject: "Utilization of Electrical Energy"
chapter: 4
key: "semester_7/utilization_of_electrical_energy:tractive-effort-gradient"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/tractive-effort-gradient
site: BCT Engineering
---

# Tractive effort up and down a gradient

Utilization of Electrical Energy · Chapter 4: Electric Traction

- Asked in: 2082 Baisakh, 4 marks
- Video: [Tractive Effort & its importance |Electric Traction & Control|](https://www.youtube.com/watch?v=VsUj1tr6bZc) — Anish Vora
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/tractive-effort-gradient
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-4

## Question

What is the tractive effort for propulsion of a train up and down a gradient?

## Short answer

Tractive effort is the force at wheel rim to overcome inertia, gradient, and resistance: F_t = 277.8 W_e α ± 98.1 W G + W r (N). On an up gradient the weight component opposes motion, increasing required effort; on a down gradient it assists, reducing effort.

## Answer

**Tractive effort** $F_t$ is the force developed by the locomotive at the rim of the driving wheels. It must supply three components:

#### Components
1. **Force for acceleration** (including rotating parts). With effective mass $W_e = 1.1W$ (10 % allowance for rotational inertia), $W$ in tonnes and $\alpha$ in km/h/s:
$$
F_a = 1000\,W_e \times \frac{1000}{3600}\alpha = 277.8\,W_e\,\alpha\ \ \text{N}
$$
2. **Force to overcome gradient** $G$ (% = rise in metres per 100 m of track):
$$
F_g = 1000\,W\,g\,\sin\theta \approx 1000\,W\times9.81\times\frac{G}{100} = 98.1\,W\,G\ \ \text{N}
$$
3. **Force to overcome train resistance** $r$ (N per tonne: friction, air resistance, track):
$$
F_r = W r\ \ \text{N}
$$

#### Up and down gradient
$$
F_t = 277.8\,W_e\,\alpha \pm 98.1\,W G + W r\ \ \text{N}
$$
- **Up gradient (+)**: the weight component opposes motion, so more tractive effort is needed.
- **Down gradient (−)**: the weight component helps motion, so less effort is needed. If $98.1G > r$, the train accelerates even with power off, and braking is needed to hold speed.

```text
      F_t -->  [train]
            /   | W g
          /     v   (W g sin theta acts down slope)
        /  theta
```

#### Example
A 400 t train ($W_e = 440$ t) accelerating at 1.5 km/h/s with $r = 50$ N/t on a 1 % gradient:
- Up: $F_t = 277.8\times440\times1.5 + 98.1\times400\times1 + 400\times50 = 183348 + 39240 + 20000 = 242588$ N ≈ 242.6 kN
- Down: $F_t = 183348 - 39240 + 20000 = 164108$ N ≈ 164.1 kN

Power at the wheels: $P = F_t v$ (W, with $v$ in m/s).

## Check yourself

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

1. Which three components must the tractive effort supply?
   A. Acceleration, gradient and train resistance
   B. Braking, gradient and voltage drop
   C. Acceleration, adhesion and weight
   D. Gradient, speed and power
   Answer: A. Ft = Fa + Fg + Fr. (Key point: Components of tractive effort)

2. Why is an effective mass We = 1.1W used for acceleration?
   A. To include the locomotive's gradient force
   B. To allow for the 10 percent rotational inertia of wheels and rotating parts
   C. To account for resistance
   D. To convert tonnes to newtons
   Answer: B. Rotating masses need energy to be accelerated, which is covered by a 10 percent increase in mass. (Key point: Effective mass)

3. How does the gradient term enter Ft = 277.8 We α ± 98.1 W G + W r?
   A. Always positive
   B. Always negative
   C. Plus on the down gradient and minus on the up gradient
   D. Plus on the up gradient (opposes motion), minus on the down gradient (helps motion)
   Answer: D. Up the gradient the weight component opposes motion; down it assists. (Key point: Sign of gradient term)

4. When does a train accelerate on a down gradient even with power off?
   A. Whenever G is not zero
   B. When 98.1 G is greater than the train resistance r
   C. Only when We = W
   D. When 98.1 G is less than r
   Answer: B. If the gradient force per tonne exceeds resistance, the net force is forward, so braking is needed to hold speed. (Key point: Down gradient with power off)

## Related in Utilization of Electrical Energy

- [Specific energy of 400 t train on 1% up-gradient](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Atrain-400t-up-gradient-specific-energy) — Utilization of Electrical Energy, chapter 4
- [Quadrilateral speed-time curve on down gradient](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aquadrilateral-curve-down-gradient-energy) — Utilization of Electrical Energy, chapter 4

## Related in other subjects

- [Tractive effort and speed from gear ratios](https://www.bctengineer.com/question.md?key=semester_5%2Fautomotive_technology%3Agear-ratio-tractive-effort-numerical) — Automotive Technology, chapter 5
- [3-bit asynchronous up/down counter](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Aasync-up-down-counter-3bit) — Digital Logic, chapter 8, asked 2×
- [Count up/down and tracking type ADC](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Acount-up-down-tracking-adc) — Advanced Electronics, chapter 3
- [Step-up and step-down single-phase cycloconverter](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-up-step-down-cyclo) — Power Electronics, chapter 6