---
title: "Quadrilateral speed-time curve on down gradient"
subject: "Utilization of Electrical Energy"
chapter: 4
key: "semester_7/utilization_of_electrical_energy:quadrilateral-curve-down-gradient-energy"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/quadrilateral-curve-down-gradient-energy
site: BCT Engineering
---

# Quadrilateral speed-time curve on down gradient

Utilization of Electrical Energy · Chapter 4: Electric Traction

- Asked in: 2081 Baisakh, 8 marks
- Video: [ELECTRIC TRACTION - PART - 05 - SPEED TIME CURVE FOR TRAIN MOVEMENT & RAILWAY SERVICES](https://www.youtube.com/watch?v=k7YYHSN0u2k) — zenmurali Sachithanandam
- 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/quadrilateral-curve-down-gradient-energy
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-4

## Question

An electric train has quadrilateral speed-time curve as follows:
i) Uniform acceleration from rest at 2 kmphps for 10 seconds.
ii) Coasting for 50 seconds.
iii) Braking period of 15 seconds.
The train is moving a uniform down gradient of 1% tractive resistance 40 newtons per tonne, rotational inertia effect 10% of dead weight, duration of stop 15 seconds and overall efficiency of transmission gear and motor as 75%. Calculate its schedule speed and specific energy consumption of run.

## Short answer

Compute acceleration, coasting, and braking distances with effective mass 1.1W; divide total distance by run plus stop time. Schedule speed is 17.3 km/h. Specific energy consumption is 13.14 Wh per tonne-km.

## Answer

Assumptions: $W_e = 1.1W$; a 1 tonne train is taken (answers per tonne); power is ON only during acceleration; during coasting the net force is the gradient force minus the resistance.

#### Speeds and distances
Acceleration: $V_1 = 2\times10 = 20$ km/h.

Coasting on a 1 % down gradient: gradient force $= 98.1\times1 = 98.1$ N/t, resistance $= 40$ N/t, so the net force **helps** motion. Coasting acceleration:
$$
\alpha_c = \frac{98.1 - 40}{277.8\times1.1} = 0.1901\ \text{km/h/s}
$$
Speed at end of coasting:
$$
V_2 = 20 + 0.1901\times50 = 29.51\ \text{km/h}
$$
Braking in 15 s: $\beta = 29.51/15 = 1.967$ km/h/s.

| Period | Time (s) | Speeds (km/h) | Distance (m) |
|---|---|---|---|
| Acceleration | 10 | 0 → 20 | $\frac{20}{3.6}\times\frac{10}{2} = 27.78$ |
| Coasting | 50 | 20 → 29.51 | $\frac{20+29.51}{2\times3.6}\times50 = 343.80$ |
| Braking | 15 | 29.51 → 0 | $\frac{29.51}{3.6}\times\frac{15}{2} = 61.47$ |
| Total | 75 | | 433.04 |

#### Schedule speed
$$
\begin{aligned} T_s &= 75 + 15 = 90\ \text{s} \\ V_s &= \frac{433.04}{90}\times3.6 = 17.32\ \text{km/h} \end{aligned}
$$

#### Specific energy consumption
Tractive effort per tonne during acceleration (down gradient, so the gradient force $98.1G$ with $G = 1$ is subtracted):
$$
\begin{aligned} F_t &= 277.8\,(1.1)\,\alpha - 98.1G + r \\ &= 277.8\times1.1\times2 - 98.1\times1 + 40 = 553.06\ \text{N/t} \end{aligned}
$$
Energy at wheels per tonne $= F_t \times d_1 = 553.06\times27.78 = 15362.8$ J/t.

Energy input (efficiency 75 %) $= 15362.8/0.75 = 20483.7$ J/t $= 5.690$ Wh/t.
$$
\text{SEC} = \frac{5.690\ \text{Wh/t}}{0.43304\ \text{km}} = 13.14\ \text{Wh/t-km}
$$

**Answer: Schedule speed = 17.3 km/h; specific energy consumption = 13.14 Wh per tonne-km.**

## Check yourself

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

1. On a 1 percent down gradient with resistance 40 N/t, what happens during coasting?
   A. The train retards
   B. Gradient force 98.1 N/t exceeds resistance, so the net force helps and the train gains speed (about 0.19 km/h/s)
   C. Speed stays at 20 km/h
   D. Power must be supplied
   Answer: B. Net forward force is 98.1 - 40 = 58.1 N/t, giving 58.1/(277.8 x 1.1) = 0.19 km/h/s, so V2 = 29.51 km/h. (Key point: Coasting on a down gradient)

2. The total run is 75 s covering 433.04 m, with a 15 s stop. What is the schedule speed?
   A. 20.8 km/h
   B. 17.3 km/h
   C. 23.1 km/h
   D. 15.0 km/h
   Answer: B. Vs = 433.04/90 x 3.6 = 17.32 km/h; stop time is included in the 90 s. (Key point: Schedule speed)

3. For which period is energy taken from the supply?
   A. Coasting only
   B. Braking
   C. Only the acceleration period, since power is on only then
   D. The entire run
   Answer: C. During coasting and braking no tractive power is drawn, so energy is computed from tractive effort and distance during acceleration. (Key point: Energy period)

4. Which expression gives the tractive effort per tonne during acceleration?
   A. 277.8 x 1.1 x 2 + 98.1 + 40
   B. 277.8 x 2 - 98.1 - 40
   C. 277.8 x 1.1 x 2 - 98.1 + 40 = 553.06 N/t
   D. 98.1 - 40
   Answer: C. Acceleration force with rotational allowance is 611.2 N/t; gradient helps (minus) and resistance adds. With efficiency 75 percent, SEC is 13.14 Wh/t-km. (Key point: Tractive effort on down gradient)

## Related in Utilization of Electrical Energy

- [Durations of acceleration, coasting and braking (2 km run)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aquadrilateral-curve-2km-40kmph-periods) — Utilization of Electrical Energy, chapter 4, asked 2×
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- [Speed-time curves for services; notching, coasting etc.](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aspeed-time-curves-services-periods) — Utilization of Electrical Energy, chapter 4
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