Tractive Effort

## Tractive Effort

Tractive Effort is the force needed to simply move the train.

To move one ton on level track (e.g., µ = 0.35), you will need a force of 35 Newton.

For slopes, you will need an additional 100 Newton per percent of steepness.

## Power

Power is the force needed to run a train at a certain speed.

To get the Power required to move a train at a certain speed, multiply the Tractive effort to move the train with the speed in m/s. (P = F * v)

With multiheaded trains, the tractive effort of each locomotive will simply be added together.

As a result of this, the power of each locomotive will also be added together.

## Examples

### Flat Land

Let's assume you have an 800-ton train which needs to reach a speed of 60 km/h.

First, we calculate the tractive effort required to simply move the train.

`800 (tons) * 35 N = 28,000 N`

You will need a locomotive with a tractive effort of at least 28 kN.

Secondly, we calculate power to move that train at that speed

`P = 28 kN * 60 km/h = 28 kN * 16.66 m/s = 466 kW`

You will need a locomotive with a power of min 466 kW.

Taken together, to move an 800 ton with a speed of 60 km/h, on flat land, you need a locomotive with a tractive effort of at least 28 kN and a power of at least 466 kW.

### Incline

Let's assume the same train as in the previous example, but now, it needs to ascend an incline of 3%.

Again, first the tractive effort.

```800 (tons) * 35 N = 28,000 N
800 (tons) * 3 (%) * 100 N = 240,000 N
total = 28,000 N + 240,000 N = 268 kN```

The locomotive needs a tractive effort of at least 268 kN.

Now, for the power needed:

`P = 268 kN * 60 km/h = 268 kN * 16.66 m/s = 4466 kW`

Taken together, to move that same train on an incline of 3%, you need a locomotive with a tractive effort of at least 268 kN and a power of at least 4466 kW.

### A Combination of the two

When encountering a combination of flat land and incline, for example, a longer train moving over a single incline, you will need to calculate how much of the train is actually travelling on the incline.

Assuming the same train which is 5 squares long on the map, encountering an incline of 1 square.

Again, First the tractive effort.

```800 (tons) * 35 N = 28,000 N
800 (tons) * 3 (%) * 100 N = 240,000 N * 2 (incline length) / 10 (number shown in depot) = 48,000 N
total = 28,000 N + 48,000 N = 76 kN```

The locomotive needs a tractive effort of at least 76 kN.

Now, for the power needed:

`P = 76 kN * 60 km/h = 76 kN * 16.66 m/s = 1266 kW`

You will need a locomotive with a power of min 1266 kW.