The International Simutrans Forum

 

Author Topic: Calibrating the power and speed of steam locomotives  (Read 10692 times)

0 Members and 1 Guest are viewing this topic.

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Calibrating the power and speed of steam locomotives
« on: October 14, 2012, 03:05:38 PM »
Following recent changes to the physics parameters in the code for the next release of Experimental to prevent difficulties caused by low powered, low weight vehicles not being able to go fast enough owing to excessive air resistance, I am in the process of trying to recalibrate the power and maximum speed of all of the steam locomotives. My plan is to calibrate a selection of steam locomotives of different types, sizes and eras precisely based on what real life data that I can find, then interpolate/extrapolate with educated guesses for the remainder.

I am starting with the Liverpool & Manchester Railway "Patentee" locomotive, built between 1833 and 1842. Ahrons at p. 67 refers to a 2-2-2 of the Liverpool and Manchester Railway in 1839, which I am assuming is a locomotive of this class. It is recorded as having run with a maximum speed of 50 miles per hour on a run of 14.5 miles with one stop and a train of 22.3 tons (making about 35t including the locomotive). I have calculated that this requires a power of 38kW using the calibration on the current -devel branch of Experimental.

However, a problem arises, on which I should be grateful for any thoughts as to how to address. As can be seen here, the Liverpool & Manchester Railway had some significant gradients - 1:89 at its steepest. Ahrons does not tell us what 14.5 mile stretch was used for the timings. If a significant downhill gradient was involved, that would skew the maximum speed considerably. It is noteworthy in this connexion that he records a London & South-Western train of 46 tons as reaching 50 miles per hour down a 1:388 gradient.

If we were to assume that the LMR calibration is based on the flat, that would require a fairly significant increase in the speed and performance of early locomotives compared to their current arrangements (60km/h currently compared to the 80km/h necessary to meet this calibration).

I should be very interested in any views on how I might take account, even with some degree of approximation, of the possibility of hills in this calibration, or more generally any thoughts on how better to calibrate steam power and speeds more accurately.
« Last Edit: February 17, 2013, 02:51:31 PM by jamespetts »

Offline o_O

  • *
  • Posts: 12
Re: Calibrating the power and speed of steam locomotives
« Reply #1 on: October 14, 2012, 08:12:49 PM »
If that locomotive achieved 50m/h on a flat stretch then it could have gone faster on a long down gradient and I imagine they would advertise that faster speed.  Wikipedia mentions that a Planet made a 50km run from Liverpool to Manchester in an hour.  It seems unlikely that the broadly similar Patentee would have a 30 kmh higher speed under roughly similar conditions (that would be a ~60% improvement), although a somewhat higher speed then 60 would be plausible. 

In general I don't think you are going to get real accurate steam engine data.  Lots of them had design changes incorporated partway through their production run, or experimental prototype models or special modifications.  Speed data generally fails to mention what exactly the engines were pulling at the time, which makes a huge difference.  Plus, a steam engine could always be run a little faster by overpressuring it at the risk of engine damage or an explosion, so it can be hard to compare record speeds with top speeds during normal operations.  I would just recommend extrapolating roughly accurate properties based on the assumption that later engines tended to be incremental improvements over earlier engines. 

Offline ӔO

  • Devotees (Inactive)
  • *
  • Posts: 2345
  • Hopefully helpful
  • Languages: en, jp
Re: Calibrating the power and speed of steam locomotives
« Reply #2 on: October 14, 2012, 08:30:21 PM »
"top speed" claims (gloating) are usually done on a slight downhill
speed records are supposed to be average speed in both directions and this is more indicative of actual performance. Only downside to speed records, is that they tend to be done with special configurations that you'll never see in regular service.

It doesn't seem entirely impossible that it would hit 50mph, sometime between 1833 and 1842, since the speed records for 1830 and 1848 are 30mph and 60mph, respectively. However, adding fudge factor, I'd say it's actual top speed would have been 40~45mph.

By fudge factor, I mean things like not wanting to pay for increased engine maintenance for operating at full power and having conservative timetables so the train can make up for lost time
« Last Edit: October 14, 2012, 08:37:11 PM by ӔO »

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #3 on: October 14, 2012, 08:33:21 PM »
Thank you for the thoughts, and welcome to the forums!

The timings for the Planet to which you are referring are start to stop timings, including all the intermediate stops and up/down gradients on the whole route, with a train of unknown weight, which is perhaps of limited use. Ahrons' data are a little more useful save for the lack of gradient information (and the timings were done by an independent person, not for promotional purposes of the railway company, so there is not necessarily an inherent bias). The more powerful LSWR locomotive to which he refers could manage 50mph on a much heavier train with a gentle downward gradient; is there any way, I wonder, of controlling mathematically for the effect of that gradient?

As for extrapolating roughly accurate properties based on the assumption that later engines tended to be incremental improvements over earlier engines: the difficulties with that are firstly that one still needs to know what roughly accurate properties are, and secondly, we need, I think, to have data to support the assumption that improvements specifically in power and speed were incremental rather than somewhat haphazard (as there were many other things, mostly relating to maintenance and running cost, that could be improved incrementally without affecting power or speed), and also have some idea as to the rate of improvement of power/speed (and the changes to that rate over time).

Edit: AEO - the problem with average speed is that it cannot be calculated quickly in Simutrans, whereas top speed can be calculated without the train leaving the depot using the top speed indication written by Bernd Gabriel.

Offline ӔO

  • Devotees (Inactive)
  • *
  • Posts: 2345
  • Hopefully helpful
  • Languages: en, jp
Re: Calibrating the power and speed of steam locomotives
« Reply #4 on: October 14, 2012, 08:44:40 PM »
james, you should be able to calculate the gravitational potential energy difference between two heights.
There's also, obviously, some mass loss from the steam and burned coal, but it even without those, you should be able to get a decent estimate.

you can then use that number to make energy/time or energy/distance and subtract it out of the additional energy the train gains when going down hill.

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #5 on: October 14, 2012, 08:48:22 PM »
Hmm - any idea how to do this? My mathematical skills are, I am afraid, somewhat limited...

Offline ӔO

  • Devotees (Inactive)
  • *
  • Posts: 2345
  • Hopefully helpful
  • Languages: en, jp
Re: Calibrating the power and speed of steam locomotives
« Reply #6 on: October 14, 2012, 08:53:52 PM »
well, my physics is not that good, but it's just PE=mgh

PE = potential energy (in Joules)
m = mass
g = gravity
h = height (delta height)

And then to convert Joules to watts, it's just 1W = 1J / 1s

So all you would need is the time it took between A to B and the heights above sea level for each point.

putting that together...
Watts = (mg(hA-hB)) / t

t is in seconds
« Last Edit: October 14, 2012, 09:05:05 PM by ӔO »

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #7 on: October 14, 2012, 09:01:45 PM »
Hmm - I presume that we can take G to equal 1 for these purposes; and M to equal the weight of the train in kilograms (grams? tonnes?). As for the "delta height" - do you mean the height differential between the start and end point...?

Offline ӔO

  • Devotees (Inactive)
  • *
  • Posts: 2345
  • Hopefully helpful
  • Languages: en, jp
Re: Calibrating the power and speed of steam locomotives
« Reply #8 on: October 14, 2012, 09:07:14 PM »
g should be 9.8m/s^2
h should be in meters. Yes, height difference between A and B.
m should be weight in kg.

with PE=mgh, you should end up with a unit of  kg*m^2/s^2, which is equal to Joules

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #9 on: October 14, 2012, 09:11:31 PM »
Right, I see! When I obtain access to the original text again on my return home, and when I have time, I shall look into this. Thank you.

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #10 on: February 16, 2013, 05:18:39 PM »
I have managed to work out, by a process of deduction, that the journey in question to which Ahrons refers must have been between Barton and Sutton Plane (or vice versa). I did this by using the gradient profile below:



and working out what 14.5 miles is as a proportion of 35 miles - the only journey between any pair of stations on the line that exactly fits that proportion is that between Barton and Sutton Plane. That part of the route has relatively gentle gradients, the steepest being 1 in 894. I can now recalculate the power using AEO's formula to take into account this incline.

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #11 on: February 17, 2013, 02:52:48 PM »
I think that I have managed to work this out - I take the weight to be 3,500kg and the speed to be 22.35 metres per second. The delta height per metre is 1/894 of a metre, and, taking your constant for g at 9.8 m/s^2, we get 38.37 joules per metre by multiplying 3,500kg by 9.8 and (1/894), which, multiplied by 22.35 meters per second gets us 857.5 Jules per second - in other words, Watts.

Rounding this up, this requires deducting only one kilowatt from the calculated power, giving a total power of 37Kw. Thank you very much for your assistance!

Offline ӔO

  • Devotees (Inactive)
  • *
  • Posts: 2345
  • Hopefully helpful
  • Languages: en, jp
Re: Calibrating the power and speed of steam locomotives
« Reply #12 on: February 17, 2013, 07:37:30 PM »
That's an interesting method of calculating power... I'm not entirely sure it's correct. Probably because I gave an incomplete explanation compared to kinetic energy.

PE=mgh only requires a difference in height.

g is always equal to gravity (vertical movement) and not the speed at which something moves laterally (generally speaking, velocity).


Say, if 200kg of mass were to climb 1m above ground, you would get this:

m = 200kg
g = 9.8m/s^2
hA = 0m
hB = 1m

= mg(hA-hB)
= 200 * 9.8 * (0-1)
= -1959 kg*m^2/s^2

Now, because energy is never lost and changes from one form to another, Law of Conservation of Energy, when an object is on the ground (height = 0), it has zero potential energy and all of it is converted into kinetic energy.

Formula for KE: KE = (1/2)mv^2

Since we know the (m) mass to be 200kg and we know that energy conserves itself, or changes forms, KE = PE, we can work out this:

KE = 1959 kg*m^2/s^2
m = 200kg
v = ?

Using the above formula, but rearranged to solve for (v)

v = sqrt (2KE/m)
= sqrt (2*1959/200)
= sqrt (19.59)
= 4.43 m/s


So that would mean, assuming constant velocity, that you lose 4.43m/s of kinetic energy going up a hill and gain 1959 Joules of potential energy.

Conversely, that means you gain 4.43m/s of kinetic energy at the base of the hill by trading it with 1959 Joules of potential energy.


I think that I have managed to work this out - I take the weight to be 3,500kg and the speed to be 22.35 metres per second. The delta height per metre is 1/894 of a metre, and, taking your constant for g at 9.8 m/s^2, we get 38.37 joules per metre by multiplying 3,500kg by 9.8 and (1/894), which, multiplied by 22.35 meters per second gets us 857.5 Jules per second - in other words, Watts.

Rounding this up, this requires deducting only one kilowatt from the calculated power, giving a total power of 37Kw. Thank you very much for your assistance!

Delta height should be total height difference, not a ratio. The gradient ratios are not useful unless we know the horizontal distance to work out total height difference.

using what you have

m = 3500kg
g = 9.8m/s^2
vA = ? (possibly 0?)
vB = 22.35m/s
hA = ? (peak of the hill)
hB = ? (base of the hill)

Hopefully, these would explain it better than I: http://science.howstuffworks.com/engineering/structural/roller-coaster3.htm
http://www.physicsclassroom.com/mmedia/energy/ce.cfm

A key formula, would be this combined one
KE initial + PE initial  = KE final + PE final



This one, which is possibly what you want, is slightly beyond what I have learned or can remember
KE initial + PE initial + W external = KE final + PE final
W = Work done, which would be the locomotive accelerating, or putting additional kinetic energy into the total. I think it should be something like joules * time.

PE final, would always be zero in this case, since we can safely assume the train only reaches top speed at the base of the hill.
« Last Edit: February 17, 2013, 08:10:11 PM by ӔO »

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #13 on: February 17, 2013, 08:08:42 PM »
Hmm - I am very confused. The total height difference would require a distance if we only know the gradient - but there are no figures for distance, and it is not clear what the relevance of this figure would be in any event: what exactly would be measured?

Offline ӔO

  • Devotees (Inactive)
  • *
  • Posts: 2345
  • Hopefully helpful
  • Languages: en, jp
Re: Calibrating the power and speed of steam locomotives
« Reply #14 on: February 17, 2013, 08:14:32 PM »
You would get the total energy difference between the top and base of the hill, which can be subtracted to find the actual power produced by certain locomotives.

Using this: KE initial + PE initial + W external = KE final + PE final

We know this:

KE final = top speed
PE final = 0

What we don't know
KE initial = ? probably 0m/s, assuming a standing start
PE initial = ?

Then you can figure out W external by rearranging the formula to
W external = KE final + PE final - KE initial - PE initial

looking at the picture, I would assume they measured from Rainhill to, either, Bury lane or Wapping tunnel. Those two look like they would give the most difference in height and most potential for speed.

« Last Edit: February 17, 2013, 08:49:01 PM by ӔO »

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #15 on: February 17, 2013, 08:47:57 PM »
I don't really understand this - why do we need to know about the total height of the hill? Surely it is possible to calculate the power required to haul a train at a given speed even if we assume that the hill is infinitely long? We just need to know how much power is required to reach a certain maximum speed given a certain load and certain gradient - the distance over which the locomotive travels at that speed surely is irrelevant (except to energy consumption)? Also, why is the starting speed relevant?

Also, why do you assume that they measured from Rainhill to Bury Lane or Wapping Tunnel? I calculated the proportionate distances very precisely, and the only thing that worked for 14.5 miles was Barton to Sutton Plane, which was accurate to the nearest pixel.

Offline greenling

  • Lounger
  • *
  • Posts: 1728
  • Simutransarchology it my hobby!
  • Languages: DE,EN
Re: Calibrating the power and speed of steam locomotives
« Reply #16 on: February 17, 2013, 09:08:21 PM »
Hello Jamespetts
By the Calibrating the power and speed of steam locomotives do not forget the gearfactor.
The gearfactor it a part of the power from the steam locomotives.

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #17 on: February 17, 2013, 09:33:21 PM »
I am aware of the gear - this is usually set to zero in Pak128.Britain-Ex for steam locomotives.

Offline ӔO

  • Devotees (Inactive)
  • *
  • Posts: 2345
  • Hopefully helpful
  • Languages: en, jp
Re: Calibrating the power and speed of steam locomotives
« Reply #18 on: February 17, 2013, 09:53:47 PM »
ah, I see.

initial speed is relevant when you want to subtract the initial kinetic energy the train may have at the start of the hill. If it is not, this may skew the numbers.


Given a certain load and gradient, the train will always lose kinetic energy when going up a hill, where as it will always gain kinetic energy when going down a hill.


Either way, I can't calculate watts gained going down hill, because watts is joules/seconds and I can't figure out the time taken for this journey.

example: If I had an extra 45,000,000 Joules gained from losing potential energy over 900 seconds, I would have an additional 50kW of power.

Offline The Hood

  • Moderator
  • *
  • Posts: 2889
  • pak128.Britain developer
Re: Calibrating the power and speed of steam locomotives
« Reply #19 on: February 17, 2013, 10:55:22 PM »
The problem with AEO's analysis is that air and mechanical resistance ARE significant for trains - hence why they have top speeds. This means a simple GPE->Kinetic energy conversion calculation (mgh=1/2 mv2) is not appropriate. That equation essentially states that things will continue to accelerate at the same rate and that the gain in velocity depends simply on the height drop. Clearly that isn't the case with trains - even on a downhill they reach a top speed (terminal velocity in physics speak) and thus the energy conversion becomes GPE->heat (no gain of kinetic energy).

The following site has a useful summary of the physics involved: http://www.twoof.freeserve.co.uk/motion1.htm although it doesn't mention suitable values for the coefficients in the equation for resistive force, only noting they are different for different trains. Also it's rather crucial to know the tractive effort curve. All in all, you've got a difficult job doing anything other than guessing power values for old vehicles - lots of unknowns in the theoretical equations and enough other unknowns in the data you've been given.

I'm not sure how you handle all of this in experimental (other than knowing you include tractive effort on some level) but the relevant section for gradients is that it suffices to divide the weight of the train (mass in kg x 9.81) by the gradient (this assumption is fine for train gradients - technically you need the sine of the angle but that's basically the same as the gradient for small angles). This will give you the extra force you get for accelerating downhill. The power (in Watts) associated with this extra force is equal to the force x velocity (in m/s). Perhaps you could assume a power correction based on this?

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #20 on: February 18, 2013, 12:02:32 AM »
Thank you for that - that is most helpful. However, I am having a little difficulty implementing The Hood's version of this equation. Taking a train with a weight of 3,500Kg (the Patentee example) and a speed of 22.35m/s going down a 1/894 gradient, that would seem to involve multiplying 3,500 by 9.81, giving 34,335 then dividing that by 1/894 (or 0.0011185682), which gives a value of 30,695,490. Multiplying that by 22.35 m/s gives 686,044,201.5W or 686,044.2015Kw, which is obviously wrong by several orders of magnitude. Perhaps I have misunderstood what you mean here by "divide the train (mass in kg x 9.81) by the gradient"?

Edit: AEO - I am still confused as to why we need to know the overall distance and/or time of a whole journey for your method. What we are calculating is the power necessary to sustain the train at a constant speed on the gradient in question (the terminal velocity for the train without added power on a gentle gradient would be far less than the speed that it can achieve under power: if not, then we need acceleration information, not top speed information), so why do we need anything other than the rate (i.e., speed)?

Offline ӔO

  • Devotees (Inactive)
  • *
  • Posts: 2345
  • Hopefully helpful
  • Languages: en, jp
Re: Calibrating the power and speed of steam locomotives
« Reply #21 on: February 18, 2013, 12:13:56 AM »
It is because gravitational potential energy can add or subtract a lot more power than you would expect.
Time is necessary to convert joules to watts

If you lose 1200m of height in 600s, versus 900s, the power difference between those two would be quite significant

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #22 on: February 18, 2013, 12:16:06 AM »
But doesn't the speed give us this information? If we are travelling at 10 meters per second on a 1/10 falling gradient, then we know that we lose 1 meter in height every second. Will this not suffice?

Offline TurfIt

  • Dev Team, Coder/patcher
  • Devotee
  • *
  • Posts: 1323
Re: Calibrating the power and speed of steam locomotives
« Reply #23 on: February 18, 2013, 12:34:12 AM »
The result obtained back in post #11 is reasonable. And the same as from The Hood's version (divide by 894, not 1/894).

858W extra required to move up the hill vs run on the flat at that speed.

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #24 on: February 18, 2013, 12:46:02 AM »
Ahh, thank you for the formula correction! Using this method, The Hood's calculations produce nearly identical results to my interpretation of AEO's. That is most helpful - thank you.

Offline greenling

  • Lounger
  • *
  • Posts: 1728
  • Simutransarchology it my hobby!
  • Languages: DE,EN
Re: Calibrating the power and speed of steam locomotives
« Reply #25 on: February 18, 2013, 05:08:31 PM »
Hello Jamespetts
I am aware of the gear - this is usually set to zero in Pak128.Britain-Ex for steam locomotives.
I have remark that the gearfactor not totally can be remove from the engines.
Edit:
James have you loking on the engines out pakgerman exp.?
There without gearfactors to strong that's not break in the topspeed under load.
As exsample rating you to test the DB BR103 she it without gearfactor stronger than the DB Br 151.
« Last Edit: February 18, 2013, 05:27:31 PM by greenling »

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #26 on: February 18, 2013, 07:18:20 PM »
What do you mean that the gear factor cannot be removed? Defining a gear factor in the .dat files is optional.

I do not entirely follow what you mean with respect to the German locomotives, however - can you elaborate?

Offline VS

  • Senior Plumber (Devotee)
  • Devotee
  • *
  • Posts: 4855
  • Vladimír Slávik
    • VS's Simutrans site
  • Languages: CS,EN
Re: Calibrating the power and speed of steam locomotives
« Reply #27 on: February 18, 2013, 07:25:41 PM »
Greenling probably meant that a gear value of 0 is nonsense, default is 1, since it is a multiplier.

Offline jamespetts gb

  • Simutrans-Extended project coordinator
  • Moderator
  • *
  • Posts: 18493
  • Cake baker
    • Bridgewater-Brunel
  • Languages: EN
Re: Calibrating the power and speed of steam locomotives
« Reply #28 on: February 18, 2013, 07:43:40 PM »
Ohh, I see.

Offline greenling

  • Lounger
  • *
  • Posts: 1728
  • Simutransarchology it my hobby!
  • Languages: DE,EN
Re: Calibrating the power and speed of steam locomotives
« Reply #29 on: February 18, 2013, 08:31:04 PM »
What do you mean that the gear factor cannot be removed? Defining a gear factor in the .dat files is optional.
I do not entirely follow what you mean with respect to the German locomotives, however - can you elaborate?
Hello Jamespetts i try so soon to post some photo from that what i have be meant.