# GATC Braking Model

Ideally for a railway the capacity of the line is often required to be as full as possible. There is a trade of between having lots of trains running on the lines and safety. There are a set of standards on how to design s signalling system, these are the ERTMS set of standards. This page will describe one small aspect of the standard, the of the headway, or specifically the braking algorithm used by the headway. The headway is defined to be the minimum safe distance between two trains. In order to derive an expression for the headway there are several things that  must be known, The distance for the following train to stop, the safty overlap distance (183m) and the delays (things like driver reaction time, time for signal to change, system delay etc). The sum of these distances is called the headway. The big piece of this calculation is the distance to come to a complete standstill. Most headway calculations are taken using a constant decleration rate and the train is modelled as a point particle, so the usual kinematic equations can be used.

With the ERTMS signalling system, the system itself tells the train how to decelerate and provides a deceleration curve to follow. The deceleration curve that the current ERTMS model uses is defined as follows: The usual deceleration is taken to be constant and there is no mass in involved, so the distance it takes to stop from a speed v and acceleration -a is given by:

$x=\frac{v^{2}}{2a}$

The deceleration curve that the train follows down in the ERTMS system is always a distance vT behind this, where T is a given delay. So the distance at speed v that the train on the GATC braking curve will be a distance:

$X=\frac{v^{2}}{2a}+vT$

Which can be rearanged into the following equation:

$v^{2}+2aTv-2aX=0$

From the above, it is possible to to calculate the time taken to decelerate from a speed u to a speed v, to do this differentiate the above equation to obtain:

$v\frac{dv}{dt}+at\frac{dv}{dt}-av=0$

which can be re-arranged as:

$\frac{dv}{dt}=\frac{av}{v+aT}$

Which can be integrated as:

$\int_{u}^{v}1+\frac{aT}{s}ds=at$

which show that the time t to go from a speed u to a speed v are related by:

$at=u-v+aT\ln\left(\frac{u}{v}\right)$

Now the calculation of headway requires the time it takes to decelerate from a given speed to stop. using the above equations it is clear that no sensible anwers are going to be obtained. The way out of this is to generalise the braking model slightly. As we have seen, the current GATC model leads to nonsensical answers when calculating the time to stop. Instead of writing the distance behind the train braking curve as vT, write it as k(T,a)f(v), where k(a,T) is a constant which gets the units correct and f(v) is some function of the velocity. The GATC generalisation becomes then:
$X=k(T,a)f(v)+\frac{v^{2}}{2a}$
The equation then becomes:
$v^{2}+2aK(T,a)f(v)-2aX=0$
Going through the same idea, differentiate the above equation with respect to time to obtain:
$2v\frac{dv}{dt}+2ak(T,a)f'(v)\frac{dv}{dt}-2av=0$
Re-arranging shows that:
$\frac{dv}{dt}=\frac{av}{v+ak(T,a)f'(v)}$
The equation can be integrated as usual to obtain:
$at=\int_{v}^{u}1+\frac{ak(T,a)f'(s)}{s}ds$
If it was desired to example to keep close to the original definition of the braking algorthm, it could be possible to choose f(v)=v^(1+q)\$, where 0<q <<1, say q =0.001 for example. To compute what the constant k(T,a) should be, take a simple form for k as k=a^nT^m, then comparing powers:
$L^{n}T^{-2n}T^{m}L^{1+q}T^{-1-q}=L^{1}$
Which forces m and n to be, n=-q and m=1-q. Computing the integral to calculate the time taken shows that:
$at =u-v+(aT)^{1-q}\frac{u^{q}-v^{q}}{q}$

This provides a well defined answer to calculate the time taken to stop on the braking curve.