[copperpen]

Peter

Using your working example my wag file had a 100t weight. A simple conversion rates that at 5337 newtons. In game it shows 5884 newtons. Why?.

[steamer_ctn]

copperpen, on 08 April 2014 - 06:02 AM, said:

Using your working example my wag file had a 100t weight. A simple conversion rates that at 5337 newtons. In game it shows 5884 newtons. Why?.

Is your wagon a steam locomotive?

One thing I forgot to mention in my info above is that steam locomotive also have resistance due to their mechanics ( valve gear, etc). This is described on pg 43 of the Baldwin publication.

This might account for the difference.

[copperpen]

steamer_ctn, on 08 April 2014 - 12:25 PM, said:

Is your wagon a steam locomotive?

One thing I forgot to mention in my info above is that steam locomotive also have resistance due to their mechanics ( valve gear, etc). This is described on pg 43 of the Baldwin publication.

This might account for the difference.

No. It is my 100 ton test freight car. I am wondering therefore if the extra resistance due to the steam linkages is being carried over to normal passenger and freight cars, and by extension, non steam locomotives.

[mozdonyos]

Hi!
I'm a locomotive driver.
The game features a 2,000-ton freight train real performance values ​​can not move, even though in reality 40 to 50 km / h accelerator.
A few comments viewed it before, I saw someone looking for a table of the actual engine loads. I usually have one, but only the Hungarian locomotives.

[steamer_ctn]

Hi Mervyn,

copperpen, on 08 April 2014 - 01:23 PM, said:

No. It is my 100 ton test freight car. I am wondering therefore if the extra resistance due to the steam linkages is being carried over to normal passenger and freight cars, and by extension, non steam locomotives.

copperpen, on 08 April 2014 - 06:02 AM, said:

Using your working example my wag file had a 100t weight. A simple conversion rates that at 5337 newtons. In game it shows 5884 newtons. Why?.

Thanks for the feedback.

I have had a look at this and believe that 5884N is correct.

The issue seems to be a units of measure issue.

For a wagon with following mass in the WAG file:

Mass ( 100.0t )

The MSTS default value for mass is in tonnes (ie metric).

So the SR calculation will be as follows:

100 tonnes (in wag file) = 110.23 tons us

thus SR = 110.23 * 12 = 1322.76 lbf = 5883.9 N

Aah the joys of mixed units of measure!!!!

Cheers

Peter

[jorgen]

steamer_ctn, on 08 April 2014 - 01:39 AM, said:

Starting Resistance (applies in the range 0-5mph)
The Starting Resistance has been modeled based upon the Baldwin reference from the above post.

Starting Resistance = 20 x wagon weight (US tons) - friction or journal bearings

Starting Resistance = 12 x wagon weight (US tons) - roller bearings
Both of these formulas give units in lbf.
Starting resistance has no relationship to the Davis values or the MSTS values, it is purely calculated based upon the stock weight.
Or will default to the friction or journal bearing if the relevant ORTS parameter is not in the eng file.

Worked example - 50 ton (uk) wagon - friction or journal bearings (default values)

50 ton (uk) = 50.8 tonne = 56 ton (us)

SR = 20 x 56 = 1120 lbf = 4982 N

If OR is told that this is a roller bearing wagon, then the SR should instead be = 2982 N

O
Peter

Hi
Interesting to read what determines the friction of the train.
Baldwin Locomotive Data Book. 1944 (Steam engine) :thumbup3:

You write about a value in eng.file that influence whether that becomes
solid bearings or roller bearings.
What value is that?.

Does that mean that if you have 100 wagons (100 ton each)and turns on a steam engine
so get all the wagons solid berarings.
=20 x wagon weight=2000 ibs 8894 newton*100=totally 889400 N (or 91 ton friction)

Switch to a modern locomotives (with roller bearings value in eng.file) so
get the whole train roller bearings
12 x wagon weight. =1200 ibs 5334 newton*100=totally 5333400 N (or 54 ton friction)

My experience is that all trains get solid bearings.

Why do we not use wag.file
Fcalc 2:0 so is a wagon with solid bearings 50 tons 0 mph
1094.0N/m/s -0.25 5.5mph 19.678N/m/s 1.000
The same carriage with roller bearing
690.3N/m/s -0.10 1.5mph 10.966N/m/s 1.000

Is third value 5.5 = solid bearings or Is that 1.5 Roller bearings.

Bye

[steamer_ctn]

Hi

jorgen, on 11 April 2014 - 12:24 PM, said:

You write about a value in eng.file that influence whether that becomes
solid bearings or roller bearings.
What value is that?.

You add the following statement to the wag file:

ORTSBearingType ( Roller )

See this thread and the document "OR Steam Model_03_02_2014.pdf" in the documentation folder of SVN.

jorgen, on 11 April 2014 - 12:24 PM, said:

Does that mean that if you have 100 wagons (100 ton each)and turns on a steam engine
so get all the wagons solid berarings.

You add the above statement to each of the relevant wag files that you want to have roller bearings. If the above statement is left out, then wagon defaults to a friction bearing.

jorgen, on 11 April 2014 - 12:24 PM, said:

Why do we not use wag.file
Fcalc 2:0 so is a wagon with solid bearings 50 tons 0 mph
1094.0N/m/s -0.25 5.5mph 19.678N/m/s 1.000
The same carriage with roller bearing
690.3N/m/s -0.10 1.5mph 10.966N/m/s 1.000

The values and formulas used by MSTS are an approximation of the Davis formula, so it was desired to use the Davis formula, as these are widely described in railway literature, as opposed to the MSTS which has very little or no info as to how they are calculated in different situations.

Cheers

Peter

[jorgen]

steamer_ctn, on 11 April 2014 - 01:45 PM, said:

Hi
You add the following statement to the wag file:

ORTSBearingType ( Roller )

You add the above statement to each of the relevant wag files that you want to have roller bearings.
If the above statement is left out, then wagon defaults to a friction bearing.
Peter

Hi Thanks for answer Peter.

I'm beginning to finally understand how the whole system fits together.
Each wagon and locomotive that has roller bearing should have this text. ORTSBearingType ( Roller)
If you dont go into each car and locomotive and write this you get solid bearings.

Tested it on a 50 tonne wagon (stack).
Without ( Roller) had stack wagon 5000 newtons (starting resistance)
The same wagon with ( Roller) became the 2881 Newton or 293 kg (starting resistance)
42 percent lower starting resistance .

Now I have to change 90 % of my 797 wag and 390 eng files.
It will probably take a while .... I wish that it had chosen the opposite. that you have typed
solid bearing and no text they had roller bearings .

Here there will be an educational thing when you're done with version 1. Writing about this.
For the first time I drove with this effect , I thought it was something seriously wrong with the train.
It felt glued to the track. I got to sand and running with 80 % of the power of the engine to get the train to move at all (no slope).
I haven't set roller bearings in all wag files for I didn't know it at the time.

---------------------------------------------------------------------------------------------
Another issue on starting resistance roller bearings.

Long trains with low hp/ton
http://youtu.be/n8uL3OkT_vM (three engine 140 loades coal wagon 21000 ton train) (0,62hp/ton)
http://youtu.be/aQimgGr4KKo (two engine 120 loaded grain wagons about 13000-14000 ton) (0,62hp/ton)

These values is very low but the trains is real and running.
in OR I'm not sure that it will work with starting trains and uphill.

Mathematics test on coal train Open rails.
Starting Resistance = 12 x wagon weight ( U.S. ton ) - roller bearings
12 * 143 ton =1716 ibs (778 kg) 7629 N
Totally 7629 N * 140=240240 Ibs 1068060 N Or (109 ton force)
es44ac Tractive Effort (starting) 183,000 lbs. (with lots of sanding)
It takes 1.5 locomotive just to overcome the start no slope.
With uphill will probably locomotives not overcome the friction and the hill.

1: I have myself rolled two freight wagons inside an industry.
The wagons weighed 12 tons each = 24 tons
With Or starting resistance
12 * 12 ton =144 ibs (65 kg) 600 N
Totally 288 ibs 130 kg 1600 N
I don't think I have managed to pull 130 kg sideways
I weighed 80 kg at the time when I was 20.

jtr1962, on 05 March 2014 - 01:29 PM, said:

As a very general rule, for roller-bearing wagons at low speeds the friction force is very roughly 20 pounds per axle plus a pound for every ton of mass. For this wagon then the friction should be in the area of 90-100 pounds, or 40-45 kg. Starting friction would probably be somewhat higher but less than 80 kg sounds about right. It's actually a great demo of how efficient railways are at moving goods and people to show 2 men starting and pushing a wagon which weighs as much as a truck. In fact, it's not outside of the realm of possibility for two men to start a 100-ton locomotive.

Mathematics test start a 100-ton locomotive Open rails.
12 * 100 =1200 ibs (544 kg)
I don't think 2 men will roll open rails locomotive but in reality yes maybe

Have done a downhill test with roller bearing 105 wagon 50 ton each (5500ton).
It is 5,5 per thousand downhill. I release the brake and the train don't to start to roll.
Is that behaviour normal? I asking train engineer here.
I do not know right now in OR what slope it takes to train to roll. (maybe 8)


--------------------------------------------------

The Starting Resistance ( Baldwin reference )
Values ​​come from the locomotive baldwin data from 1944. ( steam engine )

Here are the text

Friction bearing.
This decreases as the weight of the car or locomotive Increases and as the temperature
of the bearings and of the lubricating oil Increases . It has a maximum at starting and ice gene rally assumed to be about 20 pounds per ton, Which figure will Increase in freezing tempera tures and May reach 30 to 35 pounds per ton . After the start has been made, it will Decrease Rapidly to a minimum at about 5 mph and Can Be Considered as constant at higher speeds .

Roller bearing .
The most outstanding feature of the roller bearing concerns its reduction of journal
friction , Particularly static friction at starting as Compared with friction bearings . as a general proposition, it May be Stated That friction is reduced not less than 50% at starting and 10% from 5 to 35 miles per hour with no particular saving above 35 mph

On this basis it has been chosen in OR if i understand that right

Starting Resistance = 20 x wagon weight ( U.S. ton ) - friction or journal bearings
Starting Resistance = 12 x wagon weight ( U.S. ton ) - roller bearings
Both of These formulas give units in lbf.
Starting Resistance ( Applies in the range 0 - 5mph both friction and roller bearings )

Is this really correct after 70 years?. 12 x wagon weight (and range 0-5 mph)
Has it been no development on roller bearings in 70 years?
I'm certainly not an expert but the overall experience here on the forum
can properly speak, how easy rolls a train?.
Train drivers and conductors must surely have an opinion too.

Maybe Starting Resistance = 6 8 10 x wagon weight ( U.S. ton ) - roller bearings

Bye Jorgen

[gpz]

jorgen, on 12 April 2014 - 12:37 PM, said:

Now I have to change 90 % of my 797 wag and 390 eng files.
It will probably take a while .... I wish that it had chosen the opposite. that you have typed
solid bearing and no text they had roller bearings .

All the newer cars have roller bearings, also UIC denies the use of friction bearings. I think the program should default to roller type.

[dcarleton]

gpz, on 12 April 2014 - 12:50 PM, said:

All the newer cars have roller bearings, also UIC denies the use of friction bearings. I think the program should default to roller type.

I agree with Jorgen and Peter, the program should default to Roller Bearings. This has been the standard for decades, and where phased in over a period of decades before that. You have to go way, way back to find a time when everything on the railroad used solid (a.k.a. friction) bearings. The program should not default to that!