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Maybe I am missing something but whenever I calculate the maximum speed attainable by a train knowing only its power and mass I always get values that make no sense. If someone could explain my error that would be greatly appreciated. As I understand the relevant formulas are as follows. I am assuming constant mass for simplicity and ignoring the need to accelerate. All I am interested in is the maximum sustainable speed, analogous to terminal velocity.

$$F_\text{tractive effort}=(M_\text{total} \cdot F_g) \cdot C_\text{friction}$$ $$P=F \cdot v$$ $$\therefore P=((M_\text{total} \cdot F_g) \cdot C_F) \cdot v$$ From my reading, a reasonable coefficient of kinetic friction for steel wheels on steel track is $C_\text{friction}=0.5$

So if we try to calculate the speed of the heaviest train ever run we have the following numbers. Sources: https://en.wikipedia.org/wiki/Longest_trains, https://en.wikipedia.org/wiki/GE_AC6000CW $$M_\text{total}=99734\text{t}$$ $$P=8 \cdot 3500\text{kW}=28000\text{kW}$$ $$\therefore v=\frac{28000000}{99743000\cdot 9.81\cdot 0.5}\approx 0.0572\text{m}/\text{s}$$ For reference, according to google a garden snail moves at $0.013\text{m}/\text{s}$

Now clearly my result of point o' 6 meters per second is ridiculous for the speed of even a freight train.

Any insight would be much appreciated.

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2 Answers 2

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The estimation for the coefficient of rolling friction of $C_{\rm friction} = 0.5$ sounds unrealistic. The whole point of a train is that it rolls with very little friction. The above value looks more like the coefficient of sliding friction.

For example if you use a more realistic $C_{\rm friction} = 0.0018$ the resulting rolling resistance force is

$$F_{\rm friction} = C_{\rm friction} M_{\rm total} g = 1760.5\,\mathrm{kN}$$

and the top speed

$$ v_{\rm max} = \frac{P_{\rm total}}{F_{\rm friction}} = 15.9\,\mathrm{m\, s^{-1}}$$

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  • $\begingroup$ It is confusing that a rolling wheel involves static friction, while a locked non-rotating wheel involves kinetic friction, True, nevertheless $\endgroup$
    – DJohnM
    Dec 31, 2021 at 22:31
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Power is force multiplied by the speed of its point of application, so a given power consumption can either be the result of applying a small force at a high speed or an enormous force at a very low speed.

You have come up with a very low speed because you have grossly overestimated the force involved.

The tractive effort formula that you quote describes the potential maximum force that can be applied, as beyond that the tractive wheels will spin. That equation applies to the locomotives pulling the train, not the train as a whole. So you have calculated the tractive effort as being half the weight of an enormous train, which is an enormous force, and hence you end up with a low speed if you assume the application of such a huge force is limited by a relatively low power output. If you could apply such a force in real life, the train would accelerate at 0.5g, which would be something!

Try doing your calculations again with a smaller mass for the locomotives alone. If you assume they are 0.2% of the weight of the train, you will get a speed of around 25m/s I think.

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