Tractive effort up and down a gradient

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.

      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).