Braking distance formula

Does anyone here know a good braking distance/braking curve formula?

What do you mean? And what data do you have? It depends on many things.

Quork wrote:What do you mean? And what data do you have? It depends on many things.

I want to use the train's speed and perhaps a standardized brake power in m/s, gradients, that sort of thing... Distance to go until the next station so that braking can be started at exactly the right spot could be useful too...

Well, you can calculate a quite exact result, but there's no standard formula. Rather an individual multi-step calculation with plain Newton physics...

Even if it's a multi step calculation (that's what's programming all about; making the creator do less work), can you post it?

Well it depends on what type of brakes this is, how they are steered etc. See my posts on the topic of brakes in the "New Year Wish" thread in the Pub Section of the boards to get an idea of what we're talking about.

Let's have an easy example.

X	loco	car
mass	80t	30t
active brake setting	R+E	R+Mg
pneumatic brake tonnage in active brake setting	120t	48t
max. electric brake force	150kN	-
electric brake force for EB	125kN	-
additional brake tonnage for magnetic rail brake	-	12t
length	20m	25m
brake steering	pneumatic only
steering valve	KE-GPR-E-mZ	KE-PR-Mg
brake cylinder pressure rising speed	1240hPa/s	760hPa/s
brake cylinder pressure release speed	310hPa/s	190hPa/s
pressure front speed in brake pipe	190m/s
brake type	discus	block
brake pressure (low)	4200hPa	1700hPa
brake pressure (high)	6200hPa	3800hPa
high br.pr	below threshold speed	above threshold speed
threshold speed (upwards)	160km/h	70km/h
threshold speed (downwards)	130km/h	50km/h

Some basic facts first: While I didn't take exact examples, they're based on real vehicles, the values for mass, brake mass and length are round but realistic. The rest is more or less universally valid and exact. You might wonder now, why maximum brake force isn't fully used at EB; the values are those of the DB-class 101 loco and the reason is, the loco would otherwise overbrake. While it would be more economic (more energy recuperated, less thermical damage to the brake discus) of course to limit the pneumatic brake instead, this would be less safe; the dynamic brake is way more probable to get lost than the pneumatic one, so the latter one is the one used fully. Next aspect; lower brake pressure at higher speeds for the loco. That's the same for highspeed coaches with discus brakes as well. There are basically two reasons:
a) inacceptably high thermical strain on the brake system might lead to great loss of brake force before coming to a stop (hot discs brake worse)
b) block brakes grow weaker with increasing speed while discus brakes have a virtually constant force. The high/low brake pressure of discus braked vehicles prevent inacceptably high forces along the train in mixed trains with both discus and block braked carriages; keep in mind there's already about 150kN along the train due to the loco's electric brake (depending on how strong the pneumatic brakes of the cars are compared to the loco's pneumatic brake; for cargo trains the effect is tendentially (way) stronger, for passenger trains it get's as low as virtually 0)
High/low brake cylinder pressure is only active in brake setting R; in G and P it's always the lower pressure, so 1.7 bar for block braked carriages, and different pressures for locomotives (pressures are class-specific for locos while general for carriages)

What I left out here till now is the plastic block brake; that's a very new thing, not widely used yet, though it is the near future for cargo carriages for ecological, economical and sound emission reasons. "Block brake" means the classical iron block, wherever I mean the plastic thingies I'll state it explicitely.

For people in some countries the terms "brake tonnage" and "brake percentage" might sound totally unfamiliar. With this being a UIC- and OSShD-system since always it's probably rather due to the topic of brakes being widely unpopular to most railway fan scenes around the world than not being used in those countries. Dex can check it up in WP under brzdici procenta and those from the netherlands under rempercentage, the others need to learn Czech, Dutch or German So let's have those here:

The brake mass is the mass a vehicle can bring to a stop (including itself) from 120km/h in exactly 1000m in the plain and without wind - multiplied by 1.25. Phew... Let's put it this way:
A vehicle of 50 tonnes needs 1000m to brake itself from 120km/h to 0 when being empty. So based on what I said above with X being the brake tonnage of the vehicle it's:
50t = X * 1.25
X = 50t / 1.25
X = 40t
Another vehicle of 50 tonnes needs 1000m to brake itself from 120km/h to 0 - while loaded with another 50t. That's a brake tonnage of 80t. Or the other way round, this vehicle needs only 500m to brake itself from 120km/h to 0 if empty.
The quotient of brake tonnage through mass is called the brake percentage. The first vehicle thus has a brake percentage of 80% and the other one 160%. This means the brake percentage of a vehicle or train tells us about the real brake capacity of itself - way more than a m/s² value, because it includes the effects of brake type, additional brake equipment, activation speed, brake setting (=> cylinder pressure rising speed) etc.
However, the longer a train, the higher the divergence between calculated and real brake percentage would be due to propagation speed. EB accelerators in the carriage can increase propagation speed, EP brake steering virtually annihilates the effect, so the longer a train the more factors add to the problem. To counteract this, there's a whole bunch of correction factors depending on brake settings, brake equipment, length etc. For example, a vehicle in brake setting G (slow action) inside a train with brake setting P or R (fast action) will reach maximum brake pressure long after the rest of the train, so its brake tonnage doesn't have the chance to fully act -> 25% have to be subtracted from the brake tonnage of all vehicles in G. Or in a train with 200m < length < 500m EB accelerators mustn't be counted in if there's two vehicles without EB accelerators directly one after the other (a loco whose brake valve isn't used counts as two vehicles here) inside the train, except if they are at the train's end. That one seems questionnable on the first look, as the EB accelerators before the disturbance would seem to still work; but keep in mind, all the air from the back of the train goes forward to the brake valve, as it can't exit anywhere else when there aren't any active EB accelerators. This means the air from behind goes forward, smoothing the propagation front enough to get it below activation threshold of the EB accelerators. The longer the train is, the smaller a disturbance is enough to produce this effect; thus a whole table of whole and partial correction values exists depending on the number of disturbing vehicles, length in meters (longer pipe -> more air) and length in axles (more axles means more vehicles means more steering valves means more propagation front obstacles).

Okay, that would be clear now, I think... *g* Now let's get back to the physical facts. The exact curves for brake elements depend on several factors, however they're quite close to each other, resulting in the following three formulae being realistic enough:
F_max = maximal brake force of the brake element at 0km/h and maximum brake cylinder pressure for 0 km/h which is the maximal low brake cylinder pressure for block brakes and the maximal high brake cylinder pressure for discus brakes.
F_cur = current brake force of the brake element
p_h,max = maximum high p_c
p_l,max = maximum low p_c
p_c = current brake pressure
v = current speed

So... Now I'm working myself through a number of sources including the Knorr brake encyclopedia, measurement data from the Minden railway technology centre and DB education material to find formulae as generally valid as possible, so I'll post what I have till now here and the rest will follow soon enough.

You should put this into the college of knowledge, it will definately help us developers

When it's done I've been already calculating all night looking for a good formula meeting real measurements best while being simple. The empiric approach (having studied some engineering before going to the railways, I'm rather the type for empirical solutions than calculating everything beginning with the driver's brake valve and ending with mechanical friction inside brake cylinders ) is either

F_cur = a / (b - v) * p_c/p_l,max * F_max

or

F_cur = (a / (b - v) + c) * p_c/p_l,max * F_max

for the block brake. For the first one I used rather schematic model data based on the Knorr brake encyclopedia and DB education material calculating parameters of
a = -2750
b = -27.5
which gives a curve roughly fitting to those simplified curves, but also quite passably to the measured curves you can see in page 16 of this university lecture by DB: http://www.ids.uni-hannover.de/fileadmin/IDS/ids_lehre/SFZ/3._Vorlesung_Bremstechnik_2007.pdf
Mind the curves are influenced by the train needing to start to brake first. Also mind F=m*a means F~a which is why deceleration and force curves are directly comparable.

With this diagram I could establish a greater database to work with than with the simplified curves, thus the try to calculate the additional parameter. Visual comparison of both formulae's curves to the actually real ones will show whether the higher calculation effort is worth it. Visual confirmation is absolutely sufficient in my eyes - just look how big variations are in the diagram with one single test train...

When getting into empirics, it's often not bad to still talk about the things one circumvents this way. So - what does really happen between brake pressure, velocity and eventual brake force/deceleration?

We're specifically talking about the car itself; we do not talk about propagation, brake cylinder fill times and other aspects now. We talk about the braking action with a constant brake pressure only. Everything else will come in later and is subject to other, independant calculations. As I mentioned a few days ago already, this is a multi-step procedure.

The brake pressure inside the brake cylinder issues a force onto the piston. I'll add a fact I forgot to mention before, I'll add it to the fact sheet above: We're assuming modern steering valves, KE (Knorr mit Einheitswirkung) concretely, but the same goes for O (Oerlikon), Ch (Charmilles) or other modern steering valves. This means piston travel has no effect on the brake pressure, unlike old single-release brakes like the K (Knorr), W (Westinghouse) or M (Матроссов) steering valves. So, the force onto the piston is lead onto the brake leverage, where the force is changed according on the concrete lever ratio, which depends on the mechanical load change's setting, if there is one (yet another additional brake equipment). Then the force presses the brake blocks onto the wheel surface; the distance between block and wheel in released state has virtually no effect; first, the brake cylinder pressure leads to a proportional force and not a piston travel, and virtually no force is necessary to overcome the distance in comparison to the forces at hand; second, the distance itself is usually negligible, thanks to either regularily calibrated or, "newer" (already decades old), self-calibrating brake leverage. So the p_c ~ F_cyl ~ F_block. Now we leave proportionality as the next step is F_cur = µ_rub,block(v) * F_block which is where we want to get. So the a/(b-v) or a/(b-v) + c in our empirical formulae nothing else than a scalar containing µ_rub,block(v) and momentarily constant factors like the leverage or the quotient of current brake pressure vs. maximal brake pressure. So, now we made this clear, let's get to the results.

The first formula is finished and I went for the three factor version which meets the actual measurements best. The formula thus is:

F_curr = (567.5 / (v + 7.54) + 24.74) * p_c / p_l,max * F_max

This goes for iron brake blocks. I'll add there is yet another type of brake blocks; before I did mention plastic brake blocks (the K brake block) which behave rather similar to disc brakes. Currently, another type of brake blocks, the LL block, is being tested and might get allowed in the very near future. It has the same main aim as the K brake, that is to make goods carriages quieter by 10dB or more; however, the K brake block has a vastly different characteristic than the metal brake block, which means it can't be used in existing carriages. Reworking existing carriages for K blocks is economically imposible while those vehicles still have some decades to live, which would limit the effects of the anti-noise campaign. This is where the LL brake block aims at: It has exactly the same characteristics as the metal block, so it can be (once allowed) used in all vehicles without any construction change to them. Thus of course the above formula is valid for LL blocks as well; however, as I said, that's a thing of the future.

K brake blocks:

F_curr = (9322.97 / (v + 140) + 33.41) * p_c / p _l,max * F_max


Discus brake:

F_curr = (-0.14 * v + 100) * p_c / p _l,max * F_max

I decided to paint some rough, simplified sketches to explain the principle of a modern steering valve (based right off the KE, but afaik true for all modern systems). This way I hope some seemingly illogical connections will become clearer.

This is true for discus and block brakes all the same; the differentiation between the two is via F-pressure, which is

• directly connected to R in brake settings G and P (-> 5bar)
  
• empty in brake setting R for discus brakes
  
• connected to R or emptied by a hysteric centrifugal valve
  • which closes when accelerating above 70km/h (-> 0bar)
    
  • and opens when getting below 50km/h (-> 5bar)

General descriptors:

• A - A-chamber pressure, reference pressure for the whole system
  
• C_v - pilot pressure for the relay valve
  
• C - brake cylinder pressure
  
• F - C_v/C ratio steering pressure (see above)
  
• L - brake pipe pressure (5bar released, 5bar-3.5bar service brake, 3.5bar full brake, 0bar EB [no effect on C])
  
• R - air reservoir pressure (size depends on weight, system etc; however sufficiently dimensioned to keep R-pressure stable)

Own additional descriptors:

• Y - C_v outlet nozzle (hence the 760hPa/s)
  
• Z - C_v inlet nozzle (hence the 190hPa/s)

Let's start off with a general overview:

On the top you can see the brake pipe running along the whole carriage. Just imagine brake pipe couplings at both ends. In the left part of the image you see a series of cylinders forming a vertical column. This is the steering valve in narrow sense. The chamber at the bottom, the A chamber, contains the reference pressure; it is connected to the brake pipe via a one-way valve, it can be filled by the brake pipe but doesn't lose pressure when L < A. Above you see the L chamber, which always contains L pressure. Next is the smaller C_v-chamber containing the pilot pressure. C_v and L push downward, A pushes upward. If the piston rod is pushed upwards, it connects R to C_v (thus increasing C_v), if the piston rod is pushed downwards it connects C_v to the outside (emptying it). This way a balance of forces is achieved on all L changes and also in the event of C_v pressure loss. The C_v chamber is smaller so that a certain sinking of L leads to a greater rise of C_v (F=A*p -> the smaller the face the bigger the pressure for constant force). More specifically, a L loss of 1.5bar (5bar to 3.5bar, full brake) leads to a C_v of 3.8bar, meaning the piston's face in the C_v-chamber is less than half the size of the piston between A and L chambers.
On the right, below the brake pipe, you see the air reservoir (absolutely not to scale), below a horizontal set of cylinders: That's the relay valve. It's most obvious function is to make C out of C_v while taking F under account, but the most important purpose is a different one: It realizes the "E" from "KE" valve. "E" stands for Einheitswirkung, uniform effect. This means the system produces always the same C (3.8bar max) in always the same time (3 to 5 seconds for R and P, 18 to 30 seconds for G [that longer time for G is achieved by rerouting the air through additional nozzles between steering valve and air reservoir / outlet respectively, which aren't depicted in my pics]) no matter how much air is needed to achieve this (discus brakes several liters, block brakes several dozens of liters up to something about 1m³ in goods cars afaik).
After the relay valve there is, finally, the brake cylinder.


Huh... That late already, gotta go, I'll continue in the afternoon. (Remember children, always remember to set an alarm clock if sleeping after lunch. A nine hour post-lunch-nap is not helping your biological clock. Don't try it, unless you work at the railway (or in any other shift-work-job) and don't have a biological clock anyway )

Let's start off with high C pressure (no F pressure)


1. The brakes are released. The brake pipe is filling the R reservoir and the A chamber.


2. L pressure is sinking (driver is braking). With L pressure sinking, the one-way valve towards the A chamber closes, thus preserving full A pressure. The force upward by the A pressure is constant while the force downward sinks together with L pressure, which makes the piston move up. The piston rod opens the connection between R reservoir and C_v chamber, air is filling the C_v system with the speed defined by Z. C_v slowly pushes the steering valve's piston back downward and in the same time moves the relay valve's piston to the right, connecting R reservoir and brake cylinder via a big hole, meaning C follows C_v virtually instanteously.


3. In the meantime, enough air passed through Z to create a C_vsufficient to return the steering valve's piston into middle position. In the relay valve you can see now very well that its right part is without effect when F pressure is at zero. C_v and C push at identical faces, so C = C_v. C is pressing the brake cylinder's piston to the left, the carriage is braking.


4. L pressure is going back up. With L pressure rising, the downward forces in the steering valve grow stronger while the upward one stays constant. This moves the piston rod downward, opening the C_v outlet (via Y). C_v will fall until force equilibrium is reinstated and the piston rod returns into middle position, meaning you can brake and release partially. Exactly the same is happening in the relay valve, only, again, without any nozzle, thus C is, again, synchronous and equal to C_v. With C falling, the carriage is braking less/not anymore (depending on L).


And now to low C pressure (F pressure 5bar)


1. Again, L and A form an equilibrium without any brake pressure. F is pressing the piston inside the relay valve to the left, but that's of no importance with C_v 0.


2a. L sinks, A pushes piston up, pistron rod opens connection R-Z-C_v. C_v fills the relay valve again, but this time the right half is playing a part as well. C_v is pressing against F, however even at maximum pressure it is lower than F (3.8bar vs. 5bar then) so that the right half's piston rod is pressing against the bigger, left one, which is operated by C_v as well. This means, there is less force needed from C, so the relay valve closes already with a C pressure smaller than C_v. Concretely, a C_v pressure of 3.8bar is translated into 1.7bar (block brakes in brake setting P or in brake setting R and low speed [low C]) or 3.0bar (discus brakes in brake setting P). However, the, when comparing to high C pressure mode, additional factor of F doesn't change the reaction time; again, the relay valve translates C_v into the corresponding C pressure almost instantously.


2b. As I just wrote, there is no delay compared to high C pressure mode, but I still depicted the relay valve's action as two steps for better understanding.


3. C_v, A and L create a balance of forces, centering the steering valve again, and the relay valve is in a state of equation as well, the carriage brakes with a now constant C.


4. L rises, moves the steering valve's piston rod down, C_v is released via Y. C and F push the relay valve's piston rod to the left, C is released till either a balance of forces is found again (if brakes are released partially only) or till it's empty.

I'll add one more thing, which is quite relevant in all this: Filling the air reservoir has "higher priority" (i.e. a bigger inlet) than filling the steering valve's L chamber. This means that if greater pressure loss happens in R, the brakes won't release as long as there isn't enough pressure in R to ensure full braking capacity. I didn't think about this immediately, as the effect is not very strong for any trains I work with; already block brakes with their higher air use are quite exotic to me... However looking through some older charts of cargo trains, the effect can't be overseen. For a goods train of 75 carriages, releasing the brakes after a full braking (the same numbers go for EB; apart from the brake pipe's volume itself, which isn't much, there is not more air loss at EB than at full brakes, as goods carriages don't have neither magnetic brakes nor any other additional brake systems) takes about 120sec. All R chambers have to be filled to 5bar again, which needs a certain amount of pressurized air; however a loco's air pump is limited in output. So for a brake calculation of a goods train you need to know the air amount used per braking. Another effect noticeable is L pressure reaches 5 bar in the first carriage sooner than in the last one, obviously, as pressure falls along the train in every car whose R isn't full yet. For the mentioned 75 carriage goods train, it's about a minute for the first car (and thus also the loco, so the driver doesn't see that the train isn't full yet, one needs to know that, otherwise you'll heat your brakes up nicely ) while, as stated above, 2 minutes for the last one. I'm not quite sure yet how to describe it mathematically with a simple formula.


EDIT: Oh, btw, please, anybody, tell me if what I write is understandable at all^^

Hi Q

Speaking as a complete numpty on these matters - I understood it, well, mostly.

Phil

In the meantime I stumbled upon a brake formula for shoe brakes. It is called the "Minden brake formula" (after the DB research centre in Minden) and reads as follows:

s = 3.5v²/(6.5*phi*(1+lambda/10)+i)
s = 3.5v²/(5.5*phi*(-5+lambda/10)+i)

s is the brake distance in metres
v is the speed in km/h
phi is a speed-dependant coefficient for the brake- and brake shoe-type
lambda is the brake percentage, corrected for train length
i is the grade

This isn't, quite obviously, a physical formula; it is empiric and gives the final result for an EB, already including the brake reaction time etc. Well, it isn one, they are two formulae as you see. The higher one relates to brake settings R and P; the lower one for brake setting G. R and P have a response time of 3 to 5 seconds and a release time of 10 to 20 seconds. G has a response time of 18 to 35 seconds and a release time of 45 to 90 seconds.
Propagation speed depends on the train and the brake stage. For trains with electropneumatic or electric brake steering it obviously is negligible. For the rest, it is
- in EB for modern trains with propagation accelerator ca. 250-280 metres per second
- in EB for old trains without propagation accelerator ca. 90 till 180 metres per second
- in service brake steps ca. 90 till 180 metres per second
That's why the formula requires the brake percentage to be corrected for the train length.

Now depending on your country of origin you maybe ask yourself what the f*** a brake percentage is. That's the brake mass divided by the vehicle/train mass. A 30t carriage with a brake mass of 48t has a brake percentage of 48t/30t = 1.6 = 160 Brh
So what's a brake mass? That's an international unit (again, empiric and not physical) for the brake force of a vehicle. Calculations and definitions are very complicated, but for our goals it will suffice to stick with the classical definition to get a good estimate if you want to calculate the brakes of a vehicle from a country not using this system:
"A vehicle requiring 1000m to brake from120km/h to a full stop has a brake percentage of 80%." This means a deceleration of 0.56m/s² is 80 Brh, or 1m/s² is 143Brh. Mind that this always relates to an EB.

I didn't find a source for how to determine phi, however it is a minor factor and for our purposes it's more than sufficing to assume it to be exactly 1. I did some exemplary calculations and found that with phi=1 the error lies within 5%. Considering how variable the reality is, that's perfectly good enough.

Railways have low slopes, meaning tan(alpha) = alpha is close enough. Grades on the railway are given in permille which thus is the slope in milliradians. I.e. 1° is 18mrad is 18 permille. That's what you enter for i into the formula.

I will do some calculations against real measures to determine how close the formula comes for discus brakes; at the first glance it seems it is more than close enough for that too. If I can confirm this, I'll be using this as the further base for an OpenBVE-usable calculation. That sure is a better and easier approach than the strictly newtonic approach I've been following till now.

Hi,

The information and formulas posted by Quork are excellent.

If anyone wants to know the formula that relates speed, deceleration and braking distance:

v is the speed in km/h
a is the acceleration in m/s^2
s is the braking distance in m

s=(v/3.6)^2/2*a

A train that travels at 120 km/h on a straight level track and has a constant deceleration of 1.2 m/s^2:

s=(120/3.6)^2/2*1.2=463 m

Needs 463 m to stop.

Jorge

That's the problem - it isn't that easy. Constant deceleration isn't exactly something a train has. Brakes take time to apply. And since we don't normally stop with EB, they also need time (even more of that!) to release. That's why simple newtonic basics aren't near to good enough.

Are both braking too fast and braking mathematics a cause for brain damage?