How to know see how much train can carry?

[Orxan]

how many wagons I can attach

[Saint Landwalker]

As many as you want or can afford, but the train might not move very fast if you go overboard.

[Orxan]

Originally posted by Landwalker04:

As many as you want or can afford, but the train might not move very fast if you go overboard.

how to know what is the limit until overboard?

[SagaciousZed]

You'll have to do the math between the total weight of the train and the tractive effort on all the locomotives. Also you need to take into the account of the grade of your rail lines. On steep lines you will want a better force to weight ratio.

[Ossiwozzy]

u can get an estimate by using this formula M = ((P / (0.055 * s) + (T / 2))

now this is a rough estimate
M = Pullable weight (dont include the locomotive only the weight of the carts)
P = Power in Kw
0,055 to account for km/h instead of m/s)
s = speed in km/h
T = traction in kN

i say again this is a rough estimate and doesnt account for how hilly your track is.

[Voice From Beyond]

Your trains shouldn't be longer than the stations it will visit. While you can have trains longer than a station, it can cause problems when it reverses and it extends beyond a switch it needs to get on the right track for it's next stop.

[Orxan]

I like this communuty thank you guys

[Galileus]

Originally posted by Ossiwozzy:

u can get an estimate by using this formula M = ((P / (0.055 * s) + (T / 2))

(...)

0,055 to account for km/h instead of m/s)

First off, you open more brackets than you close, and not sure how many are even needed. (0.055 * s) + (T / 2) and 0.055s+T/2 is the same and is easier to read. Not going overboard on brackets would make it easier not to miss some of them. That would then help estimate if you mean (P/(0.055*s)) + T/2 or instead P / (0.055*s+T/2)

Second, when going from km/h to m/s you multiply by 0.28 ( or 0.2(7) to be precise). I'm guessing the value you got is due to including a constant for power needed to move one unit of mass, but it would be nice to break it apart (and maybe include for slope for people to include in their calculations if they want to).

[A.A.Ron]

Originally posted by ShRv Super Jetstream Sam:

Originally posted by Ossiwozzy:

u can get an estimate by using this formula M = ((P / (0.055 * s) + (T / 2))

(...)

0,055 to account for km/h instead of m/s)

First off, you open more brackets than you close, and not sure how many are even needed. (0.055 * s) + (T / 2) and 0.055s+T/2 is the same and is easier to read. Not going overboard on brackets would make it easier not to miss some of them. That would then help estimate if you mean (P/(0.055*s)) + T/2 or instead P / (0.055*s+T/2)

Second, when going from km/h to m/s you multiply by 0.28 ( or 0.2(7) to be precise). I'm guessing the value you got is due to including a constant for power needed to move one unit of mass, but it would be nice to break it apart (and maybe include for slope for people to include in their calculations if they want to).

If you take a look at the units (which are completely missing in the formula) the T/2 seems to be added at the end and the 0.55 appears to be 0.27*2. So there is a factor of 2 in both divisors. And when you check the units these 2s seem to be some sort of acceleration (or the acceleration is 1 m/s² and the 2 is really just a factor).

Oh, and you don't multiply by 0.28/0.27 to get from km/h to m/s, you divide by 3.6

[Galileus]

Originally posted by A.A.Ron:

Oh, and you don't multiply by 0.28/0.27 to get from km/h to m/s, you divide by 3.6

Which is the same operation ^^' At any rate, if I need to know the math and physics behind the formula to fix it, then I don't need the formula in the first place. Because then I can do something like this:

https://www.desmos.com/calculator/yawrtyseop

It's a scrap-build based on openttd calculations, and we dont have the slopes, and it does not include the loco in calculations and yada yada yada. But it has two colorful lines and sliders in it.

Just put in the correct values for your setup (locomotive's tractive effort and power; weight of a single wagon, max speed that the train should reach, how many locomotives you add).

I have absolutely no f* idea if this will work, so testing commences.

[Freedzone]

i've made a simple spreadsheet to help you to calculate all data. https://drive.google.com/open?id=1gl_9e60_e6_ek-4fzDqUaCvXtgr-Dxzw8cTSZTKppGQ (copy this spreadsheet to your drive)
you can calculate needed tractive effort and power for any count of wagons type and entered speed. Also you can calculate maximum possible weight for entered power, tractive effort and speed. Input data only in the STATS column. If you want to enter full weight without wagons separation just use the loco weight cell

[Lagislazuli]

Well, if you want this really more detailled here we go:

P = Fz * v (if we want to know the needed power for certain velocity)
v = P / Fz (if we want to know how fast this train setup can go)

v = velocity in m/s multiply by 3,6 to get km/h, P = locos traction power in W not to be confused with engine/motor power, Fz necesary traction force in N

Fz = m*g*w

m = total mass (loco and waggons) in kg
g = earth gravity = 9.81 m/s²
w = total resistance coefficient

w = wR + wL + wS + wK + wB

wR = rolling resistance, aproximately 0,004 (depends on loco, waggons, axle configuration etc)

wL = areodynamic resistance, depends on a lot of things like train length, vagon and loco type and train speed; might be 0,002 for a freight train at 100km/h up to 0,010 for highspeed or even more with doubledecker etc.

wS = uphill resistance (wS = sin alpha; for gradients between 0 and 6% aproximately the gradient, means for 3% uphill wS=0,03), negative values for downhill

wK = curve resistance, the smaller the curve the higher this value, 0 for straight tracks, 0,002 for 190m radius, interpolate in between that should do it for our needs

wB = speedup resistance, depends on train type, wB=a*mF/g; mF=1,05 (highspeed trains), mF=1,15 (freight trains) or something in between, this is only of importance if we want that the train can still speed up in that situation, for our goal we set it to 0

Consider that this is calculation for real trains and i am not sure how exactly and up to which level this is implemented ingame!

[Docsis]

Originally posted by Ossiwozzy:

u can get an estimate by using this formula M = ((P / (0.055 * s) + (T / 2))

now this is a rough estimate
M = Pullable weight (dont include the locomotive only the weight of the carts)
P = Power in Kw
0,055 to account for km/h instead of m/s)
s = speed in km/h
T = traction in kN

i say again this is a rough estimate and doesnt account for how hilly your track is.

... in layman's terms... just keep adding them until your train stops moving! Then, sell one.

[alphaprior]

I enjoy this game because it's difficult and not dumb like most games today but a formula to count the weights? No way... this is not fun at all! Game should calculate those itself and notify you as you keep adding wagons.

[Freedzone]

Originally posted by alphaprior:

I enjoy this game because it's difficult and not dumb like most games today but a formula to count the weights? No way... this is not fun at all! Game should calculate those itself and notify you as you keep adding wagons.

well, you should count weights only before 1900, cuz later locomotives become so powerful that u will exceed the station lenght limit in most cases, not the weight limit )

[Orxan]

Originally posted by alphaprior:

I enjoy this game because it's difficult and not dumb like most games today but a formula to count the weights? No way... this is not fun at all! Game should calculate those itself and notify you as you keep adding wagons.

agree