I’ve often idly wondered, while riding a train, how much extra time it costs to stop at a station as opposed to barrelling through at full speed. Simply looking at the dwell time (the time the train actually stands still at the station) isn’t enough – you also have to factor in acceleration and braking. For the longest time I never carried the thought further than that, but in a recent flash of inspiration I did some research and found a few neat equations.

Let’s dive in… For simplicity we’ll assume a flat, straight track with a single fixed speed limit.

Braking

Source: Railway braking | The Contact Patch

Let’s start with braking as that’s the simplest part. For braking in regular operation (“full service braking”) we can assume constant deceleration, at a rate limited by the passengers’ comfort rather than any technical capabilities of the train. This as opposed to emergency braking which goes a lot harder!

The deceleration rate, let’s call it $b_{max}$, varies between 0.5 and 1.3 m/s² with 0.9 as a typical figure. So we have constant deceleration from a given speed $v_{max}$ to standstill. This takes time:

And distance:

Acceleration: two-stage principle

Sources:

• Physics of train acceleration | Occasionally Coherent
• Tractive effort | Wikipedia

Acceleration is rather more complex as it occurs in two distinct stages: first a constant-force stage limited by the maximum force that can be applied to the wheels without slipping; then a (slower!) constant-power stage limited by the engine’s power and thus how fast it can add kinetic energy to the train.

The maximum force or “tractive effort” $F_{max}$ is a rating of the locomotive/EMU, as is the maximum power $P$. Power is expressed here as “power at rail” i.e. the power that can be effectively delivered to the wheels; to find “power at rail” from the engine rating, an efficiency factor for the drivetrain needs to be applied, typically 82% or 85%. Moreover, the power rating of interest is the peak power the train can deliver for a short time, often 15 or 60 minutes; the power the train can deliver continuously without problems is lower. However this time window is more than enough to complete acceleration to $v_{max}$.

Finding the tipping point

To say anything meaningful about either of the two acceleration stages, we first need to find the tipping-point velocity where the train passes from the first to the second stage (point B in the Wikipedia graph).

From the kinetic energy equation $E = \frac{1}{2} mv^2$and some differentiation rules which I hadn’t seen since university, we find the power limit $P = m \cdot v \cdot a$ where $m$ is the mass of the train, $v$ is velocity and $a$ is acceleration (i.e. the time-derivative of velocity). Combining with Newton’s second law $F = m \cdot a$ we find for the second stage, $F = \frac{P}{v}$. The tipping point is the velocity where this is equal to $F_{max}$ (the force ceiling from the first stage) and thus:

We assume that the maximum velocity is above this tipping point and thus both stages will be relevant.

First stage: constant force

This is pretty much the same as we had in the section on braking, except reversed: constant acceleration from standstill to a known maximum speed. Given that $v_{tip} = \frac{P}{F_{max}}$ we find the time:

And distance:

Second stage: constant power

Sources:

• Poplin, W.M. and Poplin, C.A., “Acceleration Modelling with Constant Power”
• Formula to determine acceleration based on constant energy input | Physics StackExchange

This part is a lot trickier as acceleration (time-derivative of velocity) is non-linearly dependent on velocity itself. However with some mathematical sleight of hand it is possible to find a closed-form expression for the velocity as a function of time:

where we set $t = 0$ as the moment $v_{tip}$ is reached and the second acceleration stage begins.

We find the time spent on the second stage by setting $v = v_{max}$ and solving for $t$, which gives:

Integrating $v(t)$ with respect to time (and this is the part I didn’t fully check by hand as I was getting pretty tired of it at this point!) we find the distance travelled:

Substituting $t_{acc2}$ into the equation for distance, and the expression for $v_{tip}$ into both, we find:

Gathering the equations

We now have equations for the time and distance spent on braking and on both stages of acceleration:

Let’s review our input variables:

• $b_{max}$, maximum braking rate, set at a standard value for passenger comfort
• $v_{max}$, maximum rated speed of the train or the track section, whichever is lower
• $P$, maximum (short-term) power at rail, property of the train
• $F_{max}$, maximum tractive effort from standstill, property of the train
• $m$, mass of the train

Derived variables

Stop penalty

We can now finally derive the “stop penalty” i.e. the additional time spent on an extra stop, compared to passing through at full speed. To find this, we have to:

• sum the braking and acceleration times;
• add a dwell time at the station;
• subtract the time it would have taken to cover the braking + acceleration distances at full speed.

In equation form:

Travel time between stops

We can also reverse the order and derive the fastest possible travel time between two stops that are a distance $d_{tot}$ apart. For this, we:

• subtract the acceleration and braking distances from the total distance between the stations;
• divide the resulting “full-speed distance” by the maximum speed;
• add the acceleration and braking times.

In equation form:

I assume here that the stations are far enough apart that the train reaches its top speed i.e. the acceleration and braking distances don’t overlap; it’s possible to derive formulas for the case where they do, but I’ve done enough math for this post already!

Examples

Regional train

What if we plug in some numbers? Let’s take a 4-carriage Stadler FLIRT zooming along merrily at 140 km/h (38.9 m/s). With numbers from the table in German Wikipedia:

• $P$ = 2600 kW
• $m$ = 120 000 kg (ignore the mass of the passengers and their belongings for now, let’s say it’s an off-peak hour…)
• $F_{max}$ = 144 kN (derived from max acceleration 1.2 m/s² and mass as above)

In the formulas above, still assuming $b_{max}$ = 0.9 m/s² , this gives:

• $t_b$ = 43.2 s; $d_b$ = 841 m
• $t_{acc1}$ = 15.0 s; $d_{acc1}$ = 136 m
• $t_{acc2}$ = 27.4 s; $d_{acc2}$ = 594 m

Our stop penalty $t_{sp}$, assuming a 30-second dwell time, then becomes 75.2 seconds. The minimum travel time between two stops 5 km apart, would be 173.7 seconds – rather more than the 128.5 seconds you would get from naïvely dividing the distance by the maximum speed, even for this fast-accelerating regional train.

High-speed train

Some very different numbers for an ICE 2, found on this forum, on a 200 km/h track. Note that unlike the regional-train example, this situation will be rare in practice: as a glance at OpenRailwayMap will tell you, any place worth a high-speed train stop will be surrounded by track with significantly lower speed limits (French gares de betteraves notwithstanding).

But let’s plug it in:

• $v_{max}$ = 200 km/h = 55.6 m/s
• $P$ = 4800 kW
• $F_{max}$ = 200 kN
• $m$ = 420 000 kg

From the equations we find:

• $t_b$ = 61.8 s; $d_b$ = 1717 m
• $t_{acc1}$ = 50.4 s; $d_{acc1}$ = 605 m
• $t_{acc2}$ = 110.0 s; $d_{acc2}$ = 3266 m

Note especially how much longer the second acceleration stage takes for this higher speed!

Finding the stop penalty (probably with a slightly longer dwell time, say 2 or 3 minutes) and the minimum travel time (with a longer distance between stations to ensure the top speed is reached) is left as an exercise for the reader.

Classic intercity

One last case: a hulking, bulking, loco-hauled intercity. To be specific, the beloved NS 1700 hauling 10 coaches of ICR. We have a mass of $86 + 10\cdot 40 = 486$ tons; $P$ = 5100 kW and $F_{max}$ = 260 kN. We’ll assume a cruising speed of 140 km/h just as in the regional train example.

Braking distance and time will be identical to the regional train since $v_{max}$ is the same: $t_b$ = 43.2 s; $d_b$ = 841 m. From the formulas for acceleration, we find:

• $t_{acc1}$ = 36.7 s; $d_{acc1}$ = 360 m
• $t_{acc2}$ = 53.8 s; $d_{acc2}$ = 1229 m

with a $v_{tip}$ of 70.6 km/h.

The stop penalty, assuming 90 seconds’ dwell time, will be 161.2 seconds. The “pure” acceleration and braking component, disregarding dwell time, is thus 71.2 seconds compared to 45.2 seconds in the regional-train example. Not too bad for such a juggernaut.

Engineering implications

Researching and writing this post has taught me a lot about how trains work and especially how different trains are engineered to suit specific use cases. Let’s compare the numbers for the 3 cases above in a table:

Train type	Stadler FLIRT, 4 carriages	ICE 2	NS 1700 + 10x ICR
Peak power [kW]	2600	4800	5100
Max tractive effort from standstill [kN]	144	200	260
Empty mass [t]	120	420	486
Power-to-weight [W/kg]	21.7	11.4	10.5
Max acceleration from standstill* [m/s2]	1.20	0.48	0.53
Max passengers	219	381	740**
Weight per passenger [kg]	548	1102	657

• * Found as the max tractive effort from standstill divided by the empty mass.
• ** Assuming 8 second-class and 2 first-class coaches.

What’s interesting to note is that the high-speed train and regular intercity don’t differ that much in how much power or tractive effort they can lay down, per kg of mass. What makes a high-speed train high-speed really is its top speed and not particularly fast acceleration (note also its much higher mass per passenger). The regional train on the other hand has meaningfully higher power-to-weight and traction-to-weight ratios than the other two.

Originally shared on Reddit. With all those awfully formatted equations, this was the post that finally convinced me I needed my own blog for this kind of stuff, free from the limitations of Reddit or LinkedIn…

No AI was used in the making of this post. A lot of Google, Wikipedia and Wolfram|Alpha though 😉 I also reproduced many – though not all – of the derivations on paper to make sure I understood wtf I was doing and to do some sanity checks on the units.