https://www.fronz.org.nz/sites/default/files/technical-papers/B31201_Weighing_of_Locomotives.pdf
pg 5, 1(a) notes one purpose for ensuring "weight is correctly distributed" is to "reduce wheelslip".
greg - Philadelphia & Reading / Reading
gmpullmanThere was a controversial issue between labor, management and the ICC regarding older, non-stoker, locomotives having to be retrofitted with stokers based on their weight on drivers.
"One difference between pessimists and optimists is that while pessimists are more often right, optimists have far more fun."
No major side issue, John, it just seems that traction tires are too artificial for my tastes. I had a steam locomotive some time ago that had traction tires and it was simply a lousy-running locomotive. I realise that such things are much improved, but they simply don't interest me as much as trying to improve the pulling power in other ways. In normal operations, my longest train would likely be 20 cars or less, and many might be much shorter. Most operations, when I get around to that stage of the game, will involve a loco going from town to town, dropping-off or picking-up cars, and perhaps simply re-spotting one or two as needed. Then it's off to the next town.The through trains will run mostly between staging yards, stopping only for water, and they're the ones more likely to be longer and to use more than one locomotive.
As for removing traction tires from your loco, do you have replacement drivers without the grooves? It would be interesting to compare the tractive effort with or without them, but, regardless of the results, I'm unlikely to become a convert.
Wayne
doctorwayne Since I won't use traction tires, I solve the problem of insufficient tractive effort by adding more locomotives....works well, even using DC control. Wayne
Wayne,
At the risk of generating a potentially major side issue, may I ask why you avoid traction tires? Is it simply one of on-going loco maintence that is a pain in the neck, or something else? And as long as I've just threatened this thread, why not juice it a bit more by asking what would happen if I chucked the tires on my MT-4s? I'll be glad to post the matter as a separate subject if it gets in the way of this one.
John
yes. threads like these are interesting and educational.
gregc peahrens: when the force on the light wheel gets to it's tractive effort limit, by it will start to slip. At that point it's tractive effort will go down (below max) as I understand it because it's slipping coefficient of friction is lower(?). greg: i believe once the wheel/axle slips, the force of the piston is redistributed on the remaining wheels. In your 0-4-0 case, it doubles. the force of the piston has to be balanced by frictional forces.
peahrens: when the force on the light wheel gets to it's tractive effort limit, by it will start to slip. At that point it's tractive effort will go down (below max) as I understand it because it's slipping coefficient of friction is lower(?).
greg: i believe once the wheel/axle slips, the force of the piston is redistributed on the remaining wheels. In your 0-4-0 case, it doubles. the force of the piston has to be balanced by frictional forces.
Thanks, Greg,
I get it now. I was very much overlooking the incoming (piston) force aspect as well as the dynamic cycle aspects of the rods to the wheels. I can see how exceeding the first wheel adhesion force starts it spinning and that the other (0-4-0 case) wheel's max adhesion ability cannot handle the whole incoming force, so it all starts spinning.
And that the balancing therefore is important for a steam loco.
I enjoyed the learning process, thanks to you and others.
Paul
Modeling HO with a transition era UP bent
Paul, I think that you might be looking too hard at this tractive effort scenario, although if you're into this sort of analysis, have at it.
If a loco isn't balanced, and the wheels with the lighter portion of the load slip, the weight on those wheels isn't transferred to the non-slipping ones as the weight is still supported by the slipping wheels. And of course, since the wheels are all connected, either by siderods on a steamer or the gears/gearbox of diesels, the other wheels will slip, too - they can't rotate at a speed slower than the slipping ones.
peahrens...Now let's put traction tires on the 3rd (main rod) drivers of our balanced 0-8-0. It is installed (usually in a groove) so that other drivers' rail contact is not affected. Assume the traction tire changes the coefficient of friction to 0.75 (certainly much higher than 0.25). In that case, the pulling power of the axles is 4 + 4 + 12 + 4 = 24 oz, since the max tractive force of the 3rd axle has tripled due to the 0.75 factor.....
The 0.25 figure is pretty well universal for non-traction tire equipped locos, and is for the entire loco, not individual wheelsets. Yes, the one with the traction tires will be greater, but it, in effect, will simply be preventing the other wheels from slipping (because of the interconnectivity of all wheelsets), so their contribution, whatever it may be, will nevertheless be a positive input.If you could take the same steam locomotive, and convert it to, f'rinstance, a 4-2-4 , with power pick-up on the lead and trailing trucks, and traction tires on both drivers, with the loco's weight balanced over those drivers' axle, what would be its tractive effort? Perhaps more than the 0.25 figure? Or more even than your hypothetical 0.75? I don't know, but I do know that adding balanced weight to steam locomotives does improve their pulling power, and to roughly that 0.25 figure. Since I won't use traction tires, I solve the problem of insufficient tractive effort by adding more locomotives....works well, even using DC control.
Attuvian By the way, just how would a railroad actually measure the weight and/or tractive equivalent on individual axles? Were instruments developed for direct measurement? Did they lift the engine, place standardized metal slugs atop the axles, lower the engine back down, relift the beast to retrieve the pancaked slugs, and then determine estimates based on how flattened they'd become? Help - my imagination is running wild! John
By the way, just how would a railroad actually measure the weight and/or tractive equivalent on individual axles? Were instruments developed for direct measurement? Did they lift the engine, place standardized metal slugs atop the axles, lower the engine back down, relift the beast to retrieve the pancaked slugs, and then determine estimates based on how flattened they'd become? Help - my imagination is running wild!
I believe it is a very simple process:
Insert one appropriately-sized hydraulic jack in an opening in a rail of a section of track. Roll the locomotive on the track until a wheel is positioned over the jack. Pump the jack until the wheel is lifted. Measure the pressure of the hydraulic fluid and do appropriate calculations. Repeat for all wheels. Turn the engine. Do the other side.
I own a jack like the one above. No, it doesn't have an adequate range for railroads. It's a 20 ton jack with a pressure gauge. And I have used that jack and that gauge to measure the lifting force of the jack.
Quite a surprise: If the typical axle loading of a locomotive is 50,000 pounds, that is 25,000 pound per wheel. Which is 12.5 tons. So my jack would actually do the job. Amazon sells them for about $50 (less gauge).
So it would appear that just about any locomotive shop could come up with these numbers.
Ed
peahrensI'm assuming (disregarding friction) that the force is distributed equally
friction can't be ignored. It the force balancing the force from the piston on the wheel and causing whatever the wheel is attached to to move forward.
peahrensWhen the force on the light wheel gets to it's tractive effort limit, by it will start to slip. At that point it's tractive effort will go down (below max) as I understand it because it's slipping coefficient of friction is lower(?).
i believe once the wheel/axle slips, the force of the piston is redistributed on the remaining wheels. In your 0-4-0 case, it doubles.
the force of the piston has to be balanced by frictional forces. The piston force equals the force on each wheel multipled by the number of wheels.
of course our models are not driven by side rods.
gregcif the piston generates a force F, how much is applied to each wheel?
Greg, I think I may have found my missing ME expert. I'll give some guesses on your quiz questions. Please note that in ChE we got partial credit on test questions.
To your point, I'm assuming (disregarding friction) that the force is distributed equally to each axle as long as each wheel has not gotten to it's max traction point and started to slip.
gregcin the 0-4-0 case, if the force on the lighter wheel exceeds the max friction of the wheel and it starts to slip, what is the force on the remaining wheels?
That helps to think of the dynamics. (I only had to take statics). If I may use my 40 / 60 (front / rear) weight distribution example... When the force on the light wheel gets to it's tractive effort limit, by it will start to slip. At that point it's tractive effort will go down (below max) as I understand it because it's slipping coefficient of friction is lower(?). The unused power (from that piston stroke) will have to find a place to land, so the rear (heavier) axle will attempt to use it. If that force does not exceed the heavier axle's max tractive effort ability, it should will hold traction and roll the loco (and front driver) along. (I wonder if the front driver might moment by moment go in and out of slip in some stages.) But if the loss of front traction leaves power in excess of the rear drivers max tractive effort, that axle slips and the whole she-bang starts spinning. (I guess the excess power all goes into momentum & friction as the drivers accelerate when spooling up to a higher spin speed??)
Thus makes me aware that the power application to the axles (on steamers) is not at all continuous and smoothly increased, even at steady speed, as each side sees a cycle of increasing power and then power off as the pistons behave. Even in our motorized model steamers, there is a push-nul-pull-null cycle of the rods on the axles. I thought it (the 0-4-0) would be easier to understand than the (mechanical drive train) model diesel, with more conuous power to an axle, but perhaps not.
So, I think I'm well past my ability to figure this all out! Gotta go slice some drives.
if the piston generates a force F, how much is applied to each wheel?
in the 0-4-0 case, if the force on the lighter wheel exceeds the max friction of the wheel and it starts to slip, what is the force on the remaining wheels?
what is the likelyhood of the new force on the remaining wheels to exceed the max friction on the next lightest set of wheels?
"Another thing to keep in mind is that most of our model locomotives, steam or diesel, are driven by one motor through gears or gearboxes, so when one wheelset slips, they all slip.
Wayne"
"So, what happens when you put a traction tire on a steam loco model? Should you then re-arrange the weight distribution to "de-favor" that wheel? It would seem so, if what you're trying to do is get all the wheel's individual T.E.'s to be equal.
Ed"
Skip this if you like, but I'm still mulling the balance / traction issue. Wayne's point of the "unified" multi-driver mechanisms and Ed's traction tire aspects got my Hypothesis Department going again. Consider the following skeptically as my interpretation of this keeps evolving.
So, first, I will back off my prior assertion that a loco's max traction is determined by the lightest weighted driver because, as Wayne reminds us, the drivers are united by the (example: steamer case) side rods.
I'll use the 0-4-0 example. With the main rod connected to the rear driver eccentric and of course the side rods unifying (other than some small tolerances on the prototype) the drivers. If the loco were balanced, it of course will provide 100% of it's maximum traction just before both drivers start to slip (if more power is applied). But let's say the front driver has 40% of the weight and the rear driver 60%.
As the power is increased to just below the slip point of the front drivers, the two axles are providing 40 + 40 = 80% of the loco's potential power. So far, so good. But if power is smoothly increased via the main rod a bit, what happens? Does the front driver start to slip? Previously, I was saying it would because I presumed the power input was shared equally to each axle.
Well, the rear driver is still ready to grab the rail additionally as it has not yet attained it's max tractive effort, is still ready to create more "friction". So as the eccentric turns, with more power on it, the rear driver will not slip, of course, absorbing some of the additional power and creating more traction there. I'm thinking (dangerous) that if the dual drivers, siderods and side rod pins are an ideal system (no slop of tolerances but no friction) that the unified mechanism will behave as a simple load to the incoming power, transferring additional power where there is still "friction" capability. Rather than give 1/2 the additional power to each axle. (Imagine that the front driver had negligible weight atop...it would not start slipping I presume.) IF that is true, the additional power will go to the additional traction capability of the back driver until reaching its max. Beyond which both drivers would slip. And the loco would achieve it's potential traction by 40 + 60 = 100%. So balance would not be a big deal to the extent this hypothesis is correct. And the same outcome would be true if the rear driver were the lighter one, if the drivers act in an ideal unified manner.
Even if that is correct, the reality is that the dual drivers & side rods have some tolerances, a grease layer, etc. involved. And our models can have some pretty sloppy tolerances involved. This might create some "releasing / hitting" action by the side rod on the front driver side rod pin (bolt / sleeve?) that might instigate it's slipping a bit early. So the answer may be somewhat in between, but I'd guessing much closer to that described above (balance importance not that big a factor). Am I sure? Not even close!
Now IF(!!) the above is true, let's consider the traction tire. How about a 0-8-0 with the main rod going to the 3rd driver? Let's have a 4 lbs loco atop 4 axles, equally loaded (1 lb each), with wheels/rail coefficient of friction of 0.25. That means max tractive effort of 16 oz. x 0.25 = 4 oz. per axle, 16 oz. for the balanced loco.
Now let's put traction tires on the 3rd (main rod) drivers of our balanced 0-8-0. It is installed (usually in a groove) so that other drivers' rail contact is not affected. Assume the traction tire changes the coefficient of friction to 0.75 (certainly much higher than 0.25). In that case, the pulling power of the axles is 4 + 4 + 12 + 4 = 24 oz, since the max tractive force of the 3rd axle has tripled due to the 0.75 factor.
What if the traction tired 0-8-0 is unbalanced? This is quite different from examples where all wheels have the same coefficient of friction, which above premises that a unified driver set can minimize any effect of imbalance (within the driver wheelbase). Since the traction tire axle has a high coefficient of friction, imbalance can be good or bad. That's because each axle's tractive effort = (coefficient of friction) X (weight applied). If one applied nil weight to the traction tire axle and divided our 4 lb weight between the other 3 axles, the non-traction axles would each provide 21.33 x 0.25 = 5.33 oz tractive effort, adding to our original 16 oz of tractive effort for the loco, because of the same loco weight and the same coefficient of friction for the effective axles. The optimum balance would be maximum (ideally all) weight on the traction tire axle and nil on the other 3 axles. The traction tire axle would generate 64 x 0.75 = 48 oz tractive effort. So we would want max weight on the traction tire axle, ignoring other practicalities that may arise, such as the tire stretching and coming off the rim. (Ask me about my Challenger with too many cars attached up my grade).
So, my (current, at least for several hours) conclusions are:
a) with same type wheels, balance may not matter much if center of gravity is within the drivers wheelbase, to the extent that driver mechanism acts in a unified way
b) caveat to (a) is the extent to which driver mechanism "slop" may involve "hitting" and/or vibration effects that may cause a driver to attain slip force a bit early relative to applied power
c) traction tire locos weight distribution matters because different friction coefficients are involved, so the higher the proportion of weight atop the traction tire axle, the better
Thanks for considering as well as tolerance for my (continuing) guesswork. Now, off to the driving range.
Back to Ed's traction tire question, if the loco is balanced, simply add Bullfrog Snot to the other wheels until attaining the same coefficient of friction as the traction tired wheels. Be careful not to add too many cars and stall / overheat the motor!
gmpullman Of course, the late, great Pennsylvania Railroad had one of the most advanced "Test Plants" http://digital.hagley.org/islandora/object/islandora%3A2365217/datastream/OBJ/view Locomotive_Testing_Plant_(1904_World's_Fair) by Edmund, on Flickr You can see the load-dynamometers below each driver. I'm sure scales were a part of this arrangement as well or at least the weights known prior to testing. Fascinating description of the apparatus here: http://www.catskillarchive.com/rrextra/butest.Html Regards, Ed
http://digital.hagley.org/islandora/object/islandora%3A2365217/datastream/OBJ/view
Locomotive_Testing_Plant_(1904_World's_Fair) by Edmund, on Flickr
You can see the load-dynamometers below each driver. I'm sure scales were a part of this arrangement as well or at least the weights known prior to testing.
Fascinating description of the apparatus here:
http://www.catskillarchive.com/rrextra/butest.Html
Regards, Ed
gmpullmanFascinating description of the apparatus here: http://www.catskillarchive.com/rrextra/butest.Html
that is an interesting read. But it didn't mention measurement of the weight on the driver, although I agree the apparatus may be capable of it.
There are many interesting articals on early railroad technology on the catskillarchive site. I found another confirming that the tractive force is ~25% of the weight on the drivers but also, no mention of how the measurement is made
http://www.catskillarchive.com/rrextra/blclas.Html
gregcso my guess is that such a shop has a track where a short section was a scale. By locating one driver or pilot wheel, they could measure the weight on the wheel/axle.
Of course, the late, great Pennsylvania Railroad had one of the most advanced "Test Plants"
AttuvianBy the way, just how would a railroad actually measure the weight and/or tractive equivalent on individual axles?
i'll take a guess
based on my visit to Steamtown, engine servicing facilities were a huge investment with enormous capabilities for maintaining locomotive, which I assume include rebuilding locomotives (i.e replace tires, bearings).
but then there were shops where new locomotive designs were built or locomotives modified (e.g. Reading I-10 to T-1). The Alco and Bladwin factories had capabiities typical engine servicing facilities would not need. I'll guess that common servicing facilities didn't have the need to lift boilers or complete engines.
so my guess is that such a shop has a track where a short section was a scale. By locating one driver or pilot wheel, they could measure the weight on the wheel/axle.
There was a controversial issue between labor, management and the ICC regarding older, non-stoker, locomotives having to be retrofitted with stokers based on their weight on drivers.
Locomotives built after 9-1-1937, with a weight on drivers of 150,000 lbs. had to be stoker equipped and switchers with weights of 130,000 lbs. so equipped.
Older locomotives were supposed to have stokers installed in a pre-determined rate, roughly 10% of the locomotives in service per year.
I seem to recall the New York Ontario & Western had some W-2 class 2-8-0s that were "borderline" with the driver weights. In order to skirt the requirement they moved the air pumps and placed large weights on the pilot to transfer some of the weight off the drivers and on to the pony trucks.
I don't know how long this arrangement satisfied the union or the ICC but "Old-Weary" management argued that they could ill-afford the purchase and conversion of stokers on these engines.
Cheers, Ed
AttuvianBy the way, just how would a railroad actually measure the weight and/or tractive equivalent on individual axles?...
In the days of steam, I'm not so sure that they would, but the locomotive builders could certainly determine weights on drivers and by factoring in other info, such as driver size, piston stroke, boiler pressure, etc., could probably get a fairly accurate picture of tractive effort.Many roads also owned dynamometer cars, which could measure various aspects of locomotive performance when coupled between the locomotive and its train.
For my layout, I assign arbitrary weights (expressed as tons) to various car types (and adjusted to accommodate "live" loads), then run test trains over the various grades on the layout, adding or removing cars to determine how many cars (tons) any given locomotive can capably move up a specific grade.
Each locomotive will have a tonnage rating, usually based on its performance on the worst (steepest) grade. When a train's weight (as determined above) is greater than the tonnage rating of the assigned loco, it's time to call for a helper (or helpers). The results seldom look prototypical (two or three locomotives on a 13 car train, f'rinstance), but it does add interest to operations and provides a guideline for ensuring that on-line customers are well-served.
I think this has been a great thread if only because of the thinking that has gone into it. Some of the factors at play here (coefficients of friction, etc.) don't seem to get regular attention on the forum, let alone making connections between their effects on the prototypes and any comparable ones on our models. But I would venture that the consequences for these kinds of issues were more significant for the real railroads (usually in terms of dollars) than for our pikes. Not only that, the prototypes were of a size that better accommodated testing and design modification. In the end we are far more limited in the testing and adjustments we can make to our equipment. I'll have to keep that balance in mind. Nevertheless, the conceptual exercises in our thread challenge and better our understanding of how things work. In that I am thankful for your posts.
So, what happens when you put a traction tire on a steam loco model? Should you then re-arrange the weight distribution to "de-favor" that wheel? It would seem so, if what you're trying to do is get all the wheel's individual T.E.'s to be equal.
I think the effective adhesion needs to be based on the wheels with the lightest load. In the examples I posted earlier, the I5 weight is pretty unbalanced.If you consider that the greatest force that can be applied is limited by the wheel with the lightest loading, the weight on that axle multiplied the number of axless becomes the effective weight of the locomotive that the tractive force is based on.In the I5 case, it works out to be 83% of the total weight on the drivers.
Loco Weight Eff-W Min Ratio Weight-on-wheels B8a 154125 140025 46675 90% 49412 46675 58038 D8a 96425 95150 47575 98% 47575 48850 I5a 140185 117236 29309 83% 41025 37447 32404 29309 I10sa 284190 280760 70190 98% 70910 71510 71580 70190
gregcOn a non-articulated steam engine the force of the engine is coupled to through the side-rods to all the wheels. The engine force is distributed to all the wheels roughly equally. No wheels slip as long as the force on the wheel does not exceed the max friction on the wheel surface. But as soon as one wheel slips, the force is redistributed on the remaining wheels.
But as soon as one wheel slips, the force is redistributed on the remaining wheels.
Greg, I agree that your example is an easier way to understand what happens.
Your explanation quickly gets to the key point; i.e., the max traction that the loco will generate (steamer is the simpler example) is just prior to the first wheel starting to slip. That (and all) wheel's max traction (friction force) is directly proportional to the weight on it.
Extending that to the balancing question of a steam model that was raised, the max traction for a loco will be attained when the weight on each axle is the same. That occurs when the center of gravity is centered over the drivers (neglecting spring pressures from lead or training trucks).
If off balance, the max traction, being attained just before the first wheel slips, is reduced to the extent (% wise) that the weight on the lighter wheels (lighter axle) is below the balanced weight desired on that axle. If it is 10% lighter than it would be if all were balanced, then the max traction for the loco would be 90% of desired, as the slipping episode will begin when 90% of max force is applied at the "weaker" wheel(s). And as you note, the force being applied will get them all spinning as the sudden increase in force applied will usually exceed the traction force ("friction") that the other drivers can tolerate.
Thanks for simplifying as well as putting up with my mental exercises. For me, it was something like figuring out a jigsaw puzzle; just for fun and interest.
Another thing to keep in mind is that most of our model locomotives, steam or diesel, are driven by one motor through gears or gearboxes, so when one wheelset slips, they all slip.
not sure i fully understood your explanation. I think you may have been considering the tractive force of a steam engine with wheels slipping.
i think there's an easier way to look at this.
consider a force applied to a wheel. Friction prevents the wheel from spinning and causing it to move what it's attached to forward. As soon as the force exceeds the maximum friction, the wheel slips.
Since a diesel has separate motors for each wheel/axle, if a wheel slips, it doesn't affect any of the other wheels (assuming there's no change in horsepower to each wheel). The locomotive looses traction until electronics can presumably reduce power to the slipping wheel until it regains traction.
On a non-articulated steam engine the force of the engine is coupled to through the side-rods to all the wheels. The engine force is distributed to all the wheels roughly equally. No wheels slip as long as the force on the wheel does not exceed the max friction on the wheel surface.
But as soon as one wheel slips, the force is redistributed on the remaining wheels. There is a sudden increase in force on the remaining wheels making it likely for another wheel to slip. This can quickly cascade resulting in all the wheels on a steam engine to slip until power is reduced.
So a steam engine adhesion is limited by the wheel with the least amount of weight on it.
I feel compelled to add a correction to my earlier submissions about balance & tractive effort that occurred to me while meandering around NH on vacation.
I had myself convinced that, as long as the center of gravity was within the axles, the tractive effort would be the same whether fully centered or off center. That is, the maximum tractive force (just before slippage started) that the various wheels could generate would, for each, be the coefficient of friction times the weight on the wheel. The sum of the tractive force for all the wheels would come out the same even if the (same total) weight was unevenly distributed on the drivers.
But, the key issue is when a first axle's wheels start slipping. That will occur as soon as the applied torque / force to any wheel exceeds the maximum tractive force that wheel can handle (before slipping starts), which is equal to the weight on that wheel times the coefficient of friction. So, only if the drive mechanism could distribute its power to the various axles in proportion to the weight on each axle, the various wheels would start to slip at the same time, none earlier than the others.
But it occurred to me that is not how the mechanism provides the power. The mechanism in our models can be either the main rod & side rods in a steamer or, say, a 6-axle diesel's 3-axle truck gear tower. I believe the reality is that the mechanism will essentially provide equal torque to each axle before slippage occurs (increasing force as power is increased). There may be some inequality due to tolerances and vibration, etc., but let's ignore that and say each axle sees similar torque applied. What then happens is that as soon as the torque applied at the lightest axle exceeds the maximum tractive effort of THAT axle, it loses adhesion and starts to slip. When that happens, those wheels are producing much less than their max tractive effort (regarding their weight atop) because the coefficient of friction is lower at slippage than it was before the slipping starts.
An example would be an 0-4-0, with 3/4 of the weight on one axle and 1/4 on the other. IF the torque could be applied proportional to the weight on each axle, balance would not be critical. The lighter axle would not spin early IF 1/4 of the power could be applied while 3/4 of the power was applied to the heavier axle. But since the power would be roughly evenly applied to each axle, the lighter one would spin much sooner than the heavier one could achieve it's max tractive effort.
That 0-4-0 would then have a max tractive effort roughly equal to 3/4 of that if the weight were evenly distributed. (With the lighter axle having started to spin, that axle provides little tractive effort, while the max the heavier axle can contribute is 3/4 (as it does not carry all the weight of the loco) that of a center balanced loco.)
Playing off that, the maximum tractive effort is achieved with a 50/50 balance. If the balance were 49/51, then the max tractive effort of that loco occurs at the point where the lighter axle is at the verge of slippage, so the two axles then are producing equal 49 + 49 = 98% of a balanced loco. If a 40/60 balanced loco, the two axles before the lighter one slips can provide 40 + 40 = 80% of a balanced loco.
After that, the math gets a bit strange. If a 30/70 balanced loco, the light axle slips quite early but then additional power can be applied to the heavier axle, to achieve 0 + 70 = 70% of a balanced loco. Beyond about a 33/67 balance, the max achieveable (if you don't mind a spinning axle tearing up uur rails) is the % of weight on the heavier axle. If 99% on the heavier axle, the loco can achieve 99% that of a balanced loco. It's back to the max tractive effort of a wheel being the weight atop times the coefficient of friction.
So, the balance does matter. It's the only way to get the maximum tractive effort from all the wheels simultaneously, before any start to slip. As the weight become somewhat imbalanced, the max traction that the loco can create is most likely related to the max tractive effort the lightest axle can apply, times the number of axles.
I was not a mechanical engineer or I probably would have avoided some earlier mistakes. I was a chemical engineer (fluid flow & such), so if you need advice on plumbing just send me a PM.
how valid is a coefficient of friction (adheasion) of 25%? I couldn't find the coefficient of friction between nickle-silver and brass.
i weighed, measured the pull and balance on the few decent locomotives I have. I measured balance by putting the loco on a piece of sectional track and located the center with a round pencil under the track.
the 4-4-0 is light and unbalanced. 0% means it balances under the front driver. room for improvement
I'm not sure about the 2-8-0 because it stalled during the pull test. I think it would do better with a better motor.
Weight Pull Adhesion Balance Engine10.5 2.3 22% 50% B8 (0-6-0) 7.3 0.9 12% 0% D8 (4-4-0)13.0 2.4* 18% 50% I5 (2-8-0)
In response to a PM from Greg, I decided to conduct a couple of simple tests using the Micro-Mark Pull Meter shown in one of Greg's posts on page one of this thread. I didn't have such a device when I modified my locos, and am uncertain if I ever tested all of them when I finally did get the meter.
I have a list of tonnage ratings done in the pre-meter era, but they were based on standard weighted test cars (not NMRA suggested weights) and turned out to be, for the most part, fairly accurate and useable standards for assigning locos to trains, based on car types (open cars with actual loads...usually heavy, and closed cars, like boxcars or reefers, where weights are usually close to NMRA specs) and the number of cars in the train.My guess-timate for the Athearn Mike's pulling power as a percentage of its modified and balanced weight was around 15%.
The locomotive (sans tender) weighs 17.25oz. and a test with the meter showed a drawbar pull of of 2.8oz. - 16.23% of its weight - a tad better than my guess.There is no room left to add hidden weight in the loco. Even if I recast the stock weight all in lead, the increase would be insignificant. About the only way to add weight would be additional air reservoirs atop the boiler - brass tubing, filled with lead. At best, probably less than 2oz., and that would look a bit over-the-top on a loco that already has four reservoirs.
I had a couple of smallish sheets of lead and two zinc weights out of Model Power FAs laying around, so I simply shaped the lead a bit, then draped it over the loco's boiler, and piled the two weights atop that. The whole shebang was pretty-much over the drivers, but I didn't make any further attempt to ensure that the total weight was properly balanced.A test with the meter showed 5.9oz. of drawbar pull, a bit more than double the "stock" version. Total weight (loco and added weights) was 38oz.Calculating the drawbar pull as a percentage of the loco's weight showed it to be 15.52%...a drop of just under 1%, so I'm guessing that my placement of the additional weight was fairly well balanced. Even with that additional weight, the drivers were able to slip.
I attempted a third test, to see if grossly imbalanced weight would have any unexpected consequences, and added all of the additional weights atop the cab and rear portion of the boiler. The lead truck and first driver set were lifted completely off the track, and the second driver set's wheel treads weren't touching the rails, either. I applied power, and while the motor emitted a barely discernable hum, the loco would not move at all, so no reading on the meter.
For the final test, I used only the two weights from the Model Power diesels, and placed them, one behind the other, atop the boiler/cab area - fairly-well centred.
The loco, plus the weights from the diesels, weighed a total of 25.5oz. and the reading on the meter showed a drawbar pull of 4.7oz., for an 18.43% percentage of power to weight.
The tests were mostly just for infomations sake, with no good way of actually upping the locos' capabilities. Traction tires might do so, but I don't want to go that route. I doublehead locos when needed, and won't hesitate to add a pusher (or pushers) even using DC power. I've found that if a train actually requires multiple locomotives, almost any will run well with any others.
To be honest, Paul, I never really gave too much thought about the two-motor idea. I knew that I was going to add a lot of weight (a friend got all three of them), but if I still had them, I've since thought of ways to add more weight and still have it not noticeable - likely at least 7 or 8oz each.I was aware that most motors in our models are capable of handling more weight than can be fit into them, but wasn't sure about those diesels because I had made mine so much heavier than usual.The two motors are not physically connected, my thought being that if one truck slipped, perhaps the other wouldn't, and would provide enough oomph to bring the slipping one under control. I didn't do any real extensive testing of the U-boats, but did try a train of live-loaded hoppers behind one of them, done on a fairly short (about 15') 2.5% grade, laid out on an ess-bend.The loco couldn't quite make it to the top and only part of the 44 car train was on the actual grade.I added a second U-boat, and re-started the train at the bottom of the hill. With a prototypical slow start, the train easily accelerated up the grade, almost as if there were no trailing cars at all. (I had run out of the Black Beauty blasting medium which I was using for the "live" loads, and had only a couple more empty hoppers available anyway.).I wish now that I had run the same test on the 45' long 2.78% grade to the second level of the layout (the upper level was not in place at that time, though), and with more loaded cars. I have enough Black Beauty now that I could have added another 30 loaded gondolas to the train.
The 44 car loaded train weighed 22lbs, and based on how easily the two moved that train, I'd estimate them to be capable of perhaps 70 loaded cars - equivalent to 35lbs. Who says that we overbuild our benchwork?
I'm interested in getting clear on the strategy of 1 vs 2 motors that Wayne discusses and Steven is considering.
I presume that a main reason for adding a 2nd motor would be if the loco or, more likely, train load on a grade was heavy enough to stall the motor at full voltage, before slipping the drive wheels would occur.
It strikes me that if the motor was strong enough to slip the wheels then the priority would be to add weight to the loco until the max train load, at max grade, would create motor stall. At that point, then a 2nd motor option would be advantageous, plus weight as needed for traction.
And of course a 2nd motor takes space that dense solid weights might occupy with more traction effect than the somewhat lighter motor.
Steven's consideration of a 2 motor F45 made me wonder if max weight addition might be a priority over a 2nd motor (or both). Said another way, if the original loco can slip wheels under max intended load, adding a 2nd motor alone might not add enough incremental weight to achieve the needed additional traction. Just another interesting angle.
On another note, I was pleased that Wayne noted the 0.25 friction coefficient (pull fraction of loco weight) on his model was typical as I was curious how our models' wheel to rail friction coefficient compared to the prototype. Of course the materials differ from prototype to model. I note the 0-6-0, 0-8-0 and GP-7 all are in the 0.23 - 0.25 range for prototypes. So we are doing a good job of modelling prototype coefficient of friction! But what about "gleamed" track vs. not?