Grade Compensation on Railway Curves: Formula, BG/MG/NG Values and Example
Grade compensation is the reduction made in the gradient on a railway curve so that the total resistance on the curve does not exceed the resistance on the straight ruling gradient. On Indian Railways it is taken as 0.04% per degree of curve for Broad Gauge, 0.03% for Metre Gauge and 0.02% for Narrow Gauge. So a 3° BG curve on a 1 in 150 ruling gradient must be eased to about 1 in 183.
Why a curve needs a flatter gradient
A train on a curve meets extra resistance: the wheel flanges rub against the outer rail, the rigid wheelbase does not follow the curve perfectly, and the inner and outer wheels on one axle travel different distances. This curve resistance adds to the gradient resistance. If the curve lies on the ruling gradient, a fully loaded train that can just climb the straight gradient may stall on the curve. Easing the gradient on the curve keeps the total resistance the same as on the straight, so the ruling gradient stays the real limit for the whole section.
Grade compensation values
| Gauge | Track gauge | Compensation per degree of curve |
|---|---|---|
| Broad Gauge (BG) | 1676 mm | 0.04% |
| Metre Gauge (MG) | 1000 mm | 0.03% |
| Narrow Gauge (NG) | 762 mm / 610 mm | 0.02% |
Compensated gradient (%) = ruling gradient (%) − compensation per degree × degree of curve.
Degree of curve from radius
Indian Railways defines the degree of a curve as the angle at the centre subtended by a 30.5 m (100 ft) chord. That gives R = 15.25 / sin(D/2), which for small angles is close to R ≈ 1750 / D (R in metres, D in degrees). Many textbooks use 1750; the exact value for a 30.48 m chord is about 1746.
| Radius R (m) | Degree of curve D ≈ 1750/R | BG compensation (0.04 × D) |
|---|---|---|
| 1750 | 1° | 0.04% |
| 875 | 2° | 0.08% |
| 583 | 3° | 0.12% |
| 437.5 | 4° | 0.16% |
| 350 | 5° | 0.20% |
Exam tip: Convert "1 in N" to a percentage first (100 / N), subtract the compensation, then convert back. Don't subtract 0.04 from the "1 in" number — that is the most common slip in RRB JE and SSC JE answers.
Types of gradient you should know
| Gradient | Meaning |
|---|---|
| Ruling gradient | The steepest gradient to which a section is laid; it decides the maximum load one locomotive can haul. Commonly 1 in 150 to 1 in 200 in plains. |
| Momentum gradient | A gradient steeper than the ruling gradient that a train can climb using the momentum gained on a preceding falling or level stretch. |
| Pusher (helper) gradient | A gradient so steep, as on ghat sections, that an extra banker engine is needed. |
| Gradient in station yards | Kept very flat so standing wagons do not roll and trains can start easily. |
Worked example
A BG section has a ruling gradient of 1 in 150. A curve of radius 583 m falls on this gradient. Find the allowable gradient on the curve.
- Degree of curve: D = 1750 / 583 = 3.0°.
- Grade compensation for BG: 0.04 × 3 = 0.12%.
- Ruling gradient as a percentage: 100 / 150 = 0.667%.
- Allowable gradient on the curve: 0.667 − 0.12 = 0.547%.
- As "1 in N": N = 100 / 0.547 = 183, so the gradient on the curve should be about 1 in 183.
The same curve on Metre Gauge would need only 0.03 × 3 = 0.09% compensation, giving 0.577%, or about 1 in 173.
How it links to cant and speed
Grade compensation deals with the curve's extra resistance; cant (superelevation) deals with the curve's centrifugal force. A sharp curve on a gradient therefore needs both: an eased gradient for haulage and proper cant and transition length for speed and comfort. You can work out cant, cant deficiency and transition length with our free Railway Cant & Transition Calculator on the tools page.
Frequently asked questions
What is grade compensation on curves?
It is the reduction in gradient made on a railway curve to offset the extra curve resistance, so the total resistance does not exceed that of the ruling gradient on straight track.
What is the grade compensation for Broad Gauge?
0.04% per degree of curve on Indian Railways Broad Gauge. Metre Gauge uses 0.03% and Narrow Gauge 0.02%.
How do you find the degree of a railway curve from its radius?
Use D ≈ 1750 / R, with R in metres, based on a 30.5 m chord. A 875 m radius curve is about a 2 degree curve.
Why is grade compensation needed?
Without it, a train loaded to the limit for the ruling gradient could stall on a curve, because curve resistance adds to gradient resistance.
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