Elvas Tower: Train Line Air Pressure. - Elvas Tower

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Train Line Air Pressure. Rate Topic: -----

#1 User is offline   Genma Saotome 

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Posted 01 July 2014 - 05:21 PM

In our .wags there is a parameter named EmergencyBrakeResMaxPressure() and another named BrakeCylinderPressureForMaxBrakeBrakeForce(). Judging by the values I see in use it looks to me that the first one has a value equal to the air pressure on the train line -- essentially how much pressure is dumped when you "big hole" the air line in an emergency braking -- and the second appears to be the equalization pressure (described in the quote below).

It seems to me that the value of the EmergencyBrakeResMaxPressure() of any given car would always be determined by whatever the engineer is doing to pump up the air and little to nothing about the individual car itself. IOW it is a superfluous parameter in .wags but quite necessary to the software... a best captured. Does that sound correct? And the the value of the second parameter should be computed by the software rather than be in a .wag. Does that sound correct?

Or is it just a nice coincidence for positive air pressure brake equipment in North America?

What I'm getting at is this: does OR needs these two parameters in every .wag?



Quote

Air Pressure Variations

The engineer can change the maximum pressure of the brake pipe by adjusting the FEED VALVE at his control stand. I have used 90 psi as the standard pressure to which the brake pipe is initially charged and subsequently recharged. On the railroad I work for, 90 psi is the standard. Some railroads use 80 psi as a standard. Some mountain grade railroads use 100 psi in mountain territory on loaded coal and grain trains.

What is the significance of these different pressures? During normal service braking operations there is none. A 10 psi REDUCTION from a 100 psi brake pipe, a 90 psi brake pipe, or an 80 psi brake pipe all result in 25 psi in the brake cylinder and thus equal braking effort. Remember that 2.5 to 1 ratio between service reservoir volume and brake cylinder volume.

But what happens if you make a 26 psi reduction from a 90 psi brake pipe? 90 minus 26 = 64 psi in the brake pipe. Remember the triple valve moves to the apply position and allows service reservoir air to flow into the brake cylinder until the service reservoir pressure lowers to equal the brake pipe pressure. As the service reservoir pressure flows into the brake cylinder the brake cylinder pressure rises. Because of the 2.5 to 1 ratio of volumes, when enough air has flowed into the brake cylinder to lower the service reservoir 26 psi the BRAKE CYLINDER PRESSURE IS 64 psi !!! (2.5 times 26 = 64). This air came from the service reservoir which is NOW AT 64 psi ALSO. Since the reservoir pressure and the brake cylinder pressure ARE EQUAL no more air will flow into the brake cylinder.

This condition is called a FULL SERVICE brake application because even reducing the brake pipe further, below 64 psi, WILL NOT INCREASE the amount of brake cylinder pressure. Even if you reduce the brake pipe pressure to zero psi the reservoir and brake cylinder pressure will still be 64 psi, the same as it was with only a 26 psi reduction. This full service or "equalization of pressures" occurs at 64 psi for a 90 psi charged system. It occurs at 71 psi for a 100 psi charged system resulting in higher full service brake effort. It occurs at 57 psi for an 80 psi charged system resulting in lower full service braking effort. The corresponding brake pipe REDUCTIONS are 26 psi for the 90 psi brake pipe, 29 psi for the 100 psi brake pipe, and 23 psi for the 80 psi brake pipe. The chart below shows the full service reduction and equalization pressures for various brake pipe settings.
Brake Pipe Setting Full Service Reduction Equalization Pressure
100 psi 29 psi 71 psi
90 psi 26 psi 64 psi
80 psi 23 psi 57 psi
70 psi 20 psi 50 psi

An engineer who makes a reduction greater than these values is just wasting time, no higher braking effort results. This is all academic however since normal train operations seldom require a brake application greater than a 15 psi reduction and any reduction greater than 12 psi is considered heavy braking.

So why would mountain grade railroads use 100 psi in the brake pipe? Two reasons.
First, as we just saw, the full service braking effort IS higher if it is needed.
Second, suppose a 10 psi reduction is made from a 100 psi charged system (100 - 10 = 90). This results in 25 psi in the brake cylinders. (10 psi times 2.5). Part way down the mountain the grade lessens and the train speed drops. The engineer releases the brakes and the brake pipe returns to 100 psi. The train immediately begins to accelerate down the grade. He immediately resets the air brakes by making another reduction. But the car reservoirs have only just begun to recharge so they have only 90 psi in them. If he makes a 10 psi reduction of the brake pipe (100 - 10 = 90) he will get no brakes. This is because the brake pipe will be at 90 and the reservoirs are also 90. But if he makes an additional 10 psi reduction, a total of 20, (100 - 20 = 80psi) he will get the same braking effort as the original set, 25 psi in the cylinders. So you can see that the 100 psi charged system AFTER one 10 psi set & release is in the exact state a 90 psi charged system is in when fully charged. This means the 100 psi charged system gives him one additional 10 psi set and release before he begins to run out of air compared to the 90 psi charged system.

So why not use 100 psi everywhere? There are penalties that go along with that extra pressure. One is that any weak hoses or valve gaskets may fail at the higher pressures. Another is if the train should go into emergency for any reason the higher braking effort may be enough to lock up and slide car wheels, especially on empty or lightly loaded cars. This will cause wheel damage at the very least and possibly a derailment from failed wheels later. A third reason is it takes longer to charge a train initially to 100 psi instead of 80 or 90 and the higher pressures cause more leaks in the system.

So why the 80 psi system? Long ago, like in the 1920s, the brake pipe was 70 psi. That was fine for the 40 ton cars of the day. By the 1940s the coal cars had grown to 55 tons and the brake pipe pressure pushed to 80 psi. In the 1950s the cars were 70 tons and in the 1960s had grown to 100 tons. Still 80 psi brake pipe pressure handled the braking chores OK. By the 1970s coal & grain cars had climbed to 135 tons and the 80 psi brake pipe had little margin for error. Especially on unit coal trains of 15,000 gross tons, even on 1.25% grades. In addition the emergency stop distances for heavy trains was growing longer and longer.

During the 1970s our railroad rules dictated an 80 psi brake pipe for all trains EXCEPT loaded unit coal and grain trains which were to use 90 psi. This gave the engineer one extra 10 psi set and release compared to an 80 psi brake pipe and it also shortened emergency stop distances for these heavy trains, but it created other problems. For instance when the trains were unloaded the pressure had to be reduced. If a coal train using a 90 psi pressure gave cars during switching operations to a freight using an 80 psi brake pipe, the "over charge" condition had to be reduced. This didn't always get done properly resulting in stuck brakes on some cars and over heated wheels. As the weight of lumber, tank, and other cars caught up to the coal and grain cars and load/empty sensors were applied the railroad simply mandated a 90 psi brake pipe for all trains. However, railroads that don't operate unit coal trains or don't have steep grades still use 80 psi since it is adequate for their type of operations. Some yard and transfer operations that operate at low speeds still use 70 psi, taking advantage of the shorter charging times.


#2 User is offline   Huecuvoe 

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Posted 02 July 2014 - 07:55 AM

In MSTS at least, the BrakeCylinderPressureForMaxBrakeBrakeForce is used as a scalar along with the MaxBrakeForce parameter to calculate the applied brake force for the wagon or engine.

AppliedBrakeForce = MaxBrakeForce * AppliedBrakeCylinderPressure / BrakeCylinderPressureForMaxBrakeBrakeForce

The BrakeCylinderPressureForMaxBrakeBrakeForce is also the value that the brake cylinder pressure is set to for a wagon or engine when it is uncoupled from the train. Consequently, this value is essential for proper operation of MSTS. I don't know for sure what the EmergencyBrakeResMaxPressure does as I have always simply set this value to the maximum brake line pressure.

I also don't know how OR uses (or doesn't use) these values however, they should probably be retained for compatibility.

#3 User is offline   Genma Saotome 

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Posted 02 July 2014 - 07:30 PM

One of the things I've learned while reading up on brake systems is if the applied brake force exceeds whatever mass * adhesion is, the wheel will slide and in quick order develop a flat spot.

Anyway... in the formula you cite, above, it looks as if maxbrakeforce() could be a constant... perhaps something tied to the physical composition of the brake shoe. Do you know?

#4 User is offline   copperpen 

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Posted 02 July 2014 - 11:22 PM

Chap called Bill Prieger who is/was an engineer did a lot of work on MSTS braking and the end result is part of the Route Riter install. It works in the different braking characteristics between cast iron and composite shoes.

#5 User is offline   Huecuvoe 

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Posted 03 July 2014 - 09:55 AM

The applied brake force is physically a function of several parameters including:

- Brake Shoe Composition
- Brake Shoe Size (Area)
- Brake Cylinder Pressure
- Brake Cylinder Size (Area)
- Brake Rigging Mechanical Advantage
- et. al.

It is not a function of the weight of the car. However, the maximum brake force that should be applied is a function of the car weight since (as you pointed out), the wheel will slide if the applied brake force exceeds the wheel adhesion. AAR standards and regulations specify a "Brake Ratio" which is then multiplied by the weight of the car to determine the maximum allowed brake force. Brake Ratios are specified for "Loaded" (Gross Rail Load) and "Empty" (Light Weight) cars as well as Hand Brakes. The applicable Brake Ratios depend on car type, brake shoe composition and date of manufacture so it can get complicated quickly. (Figure 1 is only a portion of the complete AAR specification.)

Attached Image: BrakeRatios.jpg

Since the brake force is a function of the brake cylinder pressure, the Brake Ratios and the derived maximum brake force are always specified for a specific brake cylinder pressure.

Then, in a simple world (for a loaded car):

MaxBrakeForce = AAR specified Brake Ratio * Gross Rail Load
BrakeCylinderPressureForMaxBrakeBrakeForce = AAR specified brake cylinder pressure

However, since MSTS uses these values a scalars, it is sometimes useful to modify them to allow normal braking as well as "kicking" rolling stock.

Several years ago while I was researching some of this information, I had the opportunity to correspond with Bill Prieger who was very helpful. However, I am definitely not an expert, I only know enough to be dangerous.

#6 User is offline   Genma Saotome 

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Posted 03 July 2014 - 11:08 AM

It appears that MSTS/OR should have something to convey whether the car is loaded or not and from that point choose whether a simple solution or more complex one makes sense. For the simple solution something like If CarLoaded = "Yes" Then MaxBrakeForce = Mass * Percentage (either pick one as a literal or do something clever to choose a value from a range) followed by If MaxBreakForce > Mass * Adhesion then MaxBreakForce = Mass * Adhesion. When CarLoaded = "No" the formulas would remain the same but the value of percentage would be rather different.

Given that, the simple solution could drop MaxBreakForce() from the .wag file. OTOH if a more complex solution is preferred then some parameters would need to be added to .wags, stuff like BrakeShoeComposition(), etc., etc.,

Does that sound right?

#7 User is offline   JLChauvin 

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Posted 04 July 2014 - 12:26 AM

View PostGenma Saotome, on 03 July 2014 - 11:08 AM, said:

It appears that MSTS/OR should have something to convey whether the car is loaded or not and from that point choose whether a simple solution or more complex one makes sense. For the simple solution something like If CarLoaded = "Yes" Then MaxBrakeForce = Mass * Percentage (either pick one as a literal or do something clever to choose a value from a range) followed by If MaxBreakForce > Mass * Adhesion then MaxBreakForce = Mass * Adhesion. When CarLoaded = "No" the formulas would remain the same but the value of percentage would be rather different.

Given that, the simple solution could drop MaxBreakForce() from the .wag file. OTOH if a more complex solution is preferred then some parameters would need to be added to .wags, stuff like BrakeShoeComposition(), etc., etc.,

Does that sound right?


There was discussion about that on TrainSim.com forum and 3DTrains.com forum:
http://www.trainsim....keForce-Setting
http://www.3dtrains....rakebrakeforce/
It was for the (wrong) way MSTS is modeling the brake and how to modify that fact, in MSTS.

For a correct simulation, it's a better solution to add some more parameters.

#8 User is offline   Huecuvoe 

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Posted 04 July 2014 - 07:49 AM

View PostGenma Saotome, on 03 July 2014 - 11:08 AM, said:

It appears that MSTS/OR should have something to convey whether the car is loaded or not and from that point choose whether a simple solution or more complex one makes sense. For the simple solution something like If CarLoaded = "Yes" Then MaxBrakeForce = Mass * Percentage (either pick one as a literal or do something clever to choose a value from a range) followed by If MaxBreakForce > Mass * Adhesion then MaxBreakForce = Mass * Adhesion. When CarLoaded = "No" the formulas would remain the same but the value of percentage would be rather different.

Given that, the simple solution could drop MaxBreakForce() from the .wag file. OTOH if a more complex solution is preferred then some parameters would need to be added to .wags, stuff like BrakeShoeComposition(), etc., etc.,

Does that sound right?

The simplest and most widely used solution is to use separate loaded and empty WAG files (I append an "x" to my empty WAG names). By retaining the MaxBrakeForce and BrakeCylinderPressureForMaxBrakeBrakeForce parameters in the WAG file then MSTS or OR doesn't need to know anything about the loaded or empty condition of the car or the composition of the Brake Shoes or the type of car or the AAR Brake Ratio look up table.

#9 User is offline   Genma Saotome 

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Posted 04 July 2014 - 10:44 AM

View PostHuecuvoe, on 04 July 2014 - 07:49 AM, said:

The simplest and most widely used solution is to use separate loaded and empty WAG files (I append an "x" to my empty WAG names). By retaining the MaxBrakeForce and BrakeCylinderPressureForMaxBrakeBrakeForce parameters in the WAG file then MSTS or OR doesn't need to know anything about the loaded or empty condition of the car or the composition of the Brake Shoes or the type of car or the AAR Brake Ratio look up table.


A different opinion: Move the lading specific data to the consist file, stuff like LadingWeight() and if desirable, LadingName(), ConsigneeName(), etc., etc. The leaves all of the .wags as empty cars, meaning for every car's total shape (i.e., its own .s and Freight Animation meshes) there is one and only one .wag file -- what the car is as modeled. All variants, be they by weight or texture occur by some other means. It is, in essence, a question about what the .wag file is supposed to do: Describe what cannot be described in the .s file and no more or describe all possible variations as they might appear in the game. It is a bit like distinguishing how we use an .sd file in comparison to how we use a .world file... the first occurs only once for each .s whereas the in the second we can have as many references to the .s as anybody wants.

Anyway, WRT to what OR is doing, AFAIK the code is this:
[color="#000000"]
            float f = MaxBrakeForceN * CylPressurePSI / MaxCylPressurePSI;
            if (f < MaxHandbrakeForceN * HandbrakePercent / 100)
                f = MaxHandbrakeForceN * HandbrakePercent / 100;
            Car.BrakeForceN = f;[/color]



So as that stands it seems to be pretty important to know what value to plug into both MaxBrakeForce() and MaxHandbrakeForce(). I've no idea what that might be... and I've spent a fair bit of time over the last few days googling those phrases looking for an answer. AFAIK, nobody else knows what those values should be.

What is not present in the code, again AFAIK, is a check to see if the computed value of "f" exceeds the adhesion of the wheel on the rail; It it does the wheel would slide and that's not what the railroads want to see. I'm guessing that additional check would not be terribly difficult to code.

But that still leaves (1) not knowing the right values to plug into those two parameters and (2) where to deal with data whose values will vary with the value of Mass().

#10 User is offline   Genma Saotome 

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Posted 04 July 2014 - 02:39 PM

View PostJLChauvin, on 04 July 2014 - 12:26 AM, said:

There was discussion about that on TrainSim.com forum and 3DTrains.com forum:
http://www.trainsim....keForce-Setting
http://www.3dtrains....rakebrakeforce/
It was for the (wrong) way MSTS is modeling the brake and how to modify that fact, in MSTS.

For a correct simulation, it's a better solution to add some more parameters.


I'm familiar w/ the one over at TS.com... I participated in it but the one at 3dtrains was new to me. I read it this morning and noted one of the posts you made had a table for maxhandbrakeforce(). I took down the numbers and w/ a few minutes in excel concluded your numbers were almost equal to Mass() * .425; I found that very interesting as some years ago Leland suggested using mass() * .4 and I've been using that ever since.

It is not at all clear to me that that formula should apply to a value of mass() that is less than 50t, which was the low limit in your table. Do you think it should?

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