# How to calculate the maximum starting traction force for a train with multiple carriages?

Suppose I have a convoy composed of the following elements:

• A locomotive with 4 driven wheelsets (2 wheels per wheelset and therefore 8 wheels).
• $$n$$ carriages with 2 wheelsets each.

Assuming the weight is uniformly distributed on the wheels, from my understanding the maximum force that can be applied by the locomotive avoiding wheels slipping is given by:

$$F_{max} = m_{locomotive} g \mu$$

Where $$\mu$$ is the adhesion coefficient.

The point is that I'm not sure if I should consider also the passenger carriers and if not, then why. I think I should not consider them since the force that prevents the locomotive's wheels from slipping is just its own weight.

Furthermore, what if the locomotive had only 6 driven wheels out of the 8? Would that change anything?

In that case I think I should multiply the $$F_{max}$$ obtained by 6/8 am I correct?

## 1 Answer

The answer to the first part of question is, you're right, you don't concern yourself with the other carriages friction or weight.

The second part, you have to consider static friction force on the wheels for traction. In steel on steel it is approximately 0.60.

So the approximate traction your locomotive can provide is 60% of its weight tributary to its powered wheels regardless of the number of the wheels.

Say if you reduce the number of the axels to 3 you are puting more weight on each wheel. So the traction remains the same. Unless by changing the geometry of the axels you compromise the even distribution of the weight on the wheels.

Because then the light axel will skid earlier, sinking the power.