Neutral points

I take it no differential equations are needed.

That' correct.

A simple way to understand this is to consider is a push-pull train with identical engines (same power and weight) front and rear, and an even number (for example six) of identical cars with equal weight and rolling resistance as a function of speed. On this train, the zero coupler-force location would be at the rear coupler of the third car and the front coupler of the fourth car. The front engine would be pulling the front three cars, the rear engine would be pushing the rear three cars, and there would be zero coupler-force at the mid-train location. Essentially, the train would be equivalent to two individual trains with three cars each, one train being pushed and the other train being pulled. In that case, the third and fourth passenger cars would each have a coupler with zero force, equivalent to what occurs when all cars and engines are coupled together as described above. Similar conditions can occur if additional helper engines are added into the train.

MELGAR

@phil gresho posted:

That' correct.

No fun at all then.  But it seems to me, for this to occur at a coupler, there would have to be either a momentary difference in speed of the locomotives or a momentary difference in speed created by the momentum of one of the cars.  If the difference in speed of the engines is constrained to a certain wave pattern so that the push/pull maintains the length limit of the slack on the coupler, then the force on the coupler will always be zero.  If the speed difference oscillates between three slack lengths then I see the zero force switching back and forth between the rear and front couplers of the center car but out of phase taking into account the speed momentum of the car.  The first scenario assumes the speed of the car is unaffected by slowing forces.

The situation is considered as a static problem where the train is in a steady-state condition - constant speed. This allows the steady-state coupler forces to be calculated easily. That is a useful thing to know. It would give the equilibrium coupler force distribution along the train. A coupler might be prone to opening where the coupler force is largest. Dynamic effects are a separate question, but the static forces must first be calculated.

MELGAR

What you are talking about is the point in the train where the couplers go from being "stretched" to being "bunched."

I've seen this point in a train with helpers as it passed by.  Sometimes the couplers knuckles at that point aren't even touching one another. They are just "floating" along.

While theoretically it could occur in the middle of a car somewhere, you can't see that. You can, however, see whether couplers are stretched or bunched.

In the real world, I think it's rare for the zero force location to be exactly at a coupler interface. But, as Rich says, the coupler ahead of the zero force location would be stretched (in tension) and the coupler behind it would be bunched (in compression).

MELGAR

Melgar:  your  case of 6 cars and 2 engines is merely a special case of my general result.  See also my mention of a "rare case".

Rich:  Yes;  that's a better/railroader way to describe my "rare" case.

Phil - I'm not sure I would consider that a "rare" occurrence, just something that happens as a result of physics.  I was thinking of the BNSF Intermodal Consists that run 5...3...2.  Of course, they have sophisticated computer enhanced speed control to keep the proper tension and compression on the intermediate cars.

The rare occurrence would be finding a piece on a railroad with no grade, no curve, and no wind.   I have run drift tests on rail cars to determine the factors for a modified Davis equitation, and finding the right track and getting a windless day is nearly impossible.

In a train with distributed power or just helpers on the rear, the neutral point is constantly changing as the train moves over the road. Consider this...

A train is moving on level track, with a helper set on the rear. The neutral point is somewhere in the middle of the train. The train starts up a grade. Gravity causes the pulling force on the front of the train to drop a bit, but the helpers are still shoving, so the neutral point slowly moves forward from where it was on level track. As the train crests the grade, gravity causes the head end of the train to pull a bit harder. The neutral point now moves slowly towards the rear of the train.

As a train with more than one set of power moves over the road, the neutral point is constantly changing. It is a slow and gentle change as the dynamic forces adjust to the slowly changing profile of the railroad and the changes in pulling force caused by gravity. It is not a dramatic, loud, or forceful change; you would likely not hear the slack adjust, even if standing track side.

A couple of comments:

1. On the railroad, we called the floating point at which slack changed back and forth, from draught to buff, the node.
2. A coal train would be over-powered if it were making 25 MPH on a 2% ascending grade.  15 MPH is more realistic.

This same concept applies to hump yards.  As the cars go over the hill there is sweet spot allowing the pin puller on the ground to uncouple the car (or cars) while in motion as the compressive forces transition to tension when the cut passes over the crest and there is very little pressure on the coupler.  Sometimes on the larger cuts, the cars already are pulling down the hill and it is not practical for the pin puller to walk back in the train and pull the pin, so some hump yards have a pincher retarder on the crest to help grab the cut momentarily and relieve the tension on coupler to allow the pin to be pulled.

@Number 90 posted:

1. A coal train would be over-powered if it were making 25 MPH on a 2% ascending grade.  15-18 MPH is more realistic.

Tom,

Overpowered would be an understatement! For a tonnage train to make that speed (25mph) up a 2% grade would require knuckle busting power! 2% grade with normal power, my experience is that you would be lucky to be doing 12 mph.

@Big Jim posted:

Overpowered would be an understatement! For a tonnage train to make that speed (25mph) up a 2% grade would require knuckle busting power! 2% grade with normal power, my experience is that you would be lucky to be doing 12 mph.

Agreed! I used to run a line that had a 2-mile long, 2.02% grade right out of the yard. With four GP10s and forty cars (about 5,000 tons), I would be into the traction motor 30-minute short time ratings cresting the grade. That means 9-10 mph.

@Rich Melvin posted:

In a train with distributed power or just helpers on the rear, the neutral point is constantly changing as the train moves over the road. Consider this...

bunch of good stuff snipped-

Question from the peanut gallery, I have no knowledge of real rail operations.

In the situation under discussion, how do you know/adjust the division of effort among the various locomotives?, particularly if they are not identical.

Thanks in advance,

@PLCProf posted:

Question from the peanut gallery, I have no knowledge of real rail operations.

In the situation under discussion, how do you know/adjust the division of effort among the various locomotives?, particularly if they are not identical.

Thanks in advance,

There really is no need to "adjust the division of effort", as each locomotive, steam, diesel or electric, contributes whatever it is capable of, regardless of its size. Thus, when lots of horsepower is required, many locomotives may be required.

@PLCProf posted:

Question from the peanut gallery, I have no knowledge of real rail operations.

In the situation under discussion, how do you know/adjust the division of effort among the various locomotives?, particularly if they are not identical...

There is a common misconception among model train guys that the power and speed of multiple locomotives must somehow be "matched." That's not how it works.

Consider a multiple unit diesel consist consisting of a big, 6-axle, 4,400 HP unit, and a 1,500 HP SW1500. When they get out on the road, are they going to run at the same speed? Of course they are...they are coupled together!

Are they going to pull on the train with the same force? Nope. The big 6-axle has a lot more pulling power than the SW1500, so it is going to pull harder. The SW1500 adds what it can to the overall effort of pulling the train. But they do not have to be "matched" in any way. Each locomotive pulls on the train with whatever pulling force it can generate.

The same holds true with helpers or distributed power. Even if the power setup has different makes, models and horsepower units running together, it doesn't matter. Each locomotive will pull on the train with whatever pulling force it can generate.

Here's another concept that may help to clarify the process. Think of the throttle in a locomotive (any locomotive, steam or diesel) as a POWER control, not a SPEED control. In my example above about climbing the 2% grade right out of the yard, I would have the throttle wide open, yet the train would only be moving at 9-10 mph...sometimes even a bit less! At other times I might be rolling along at 30 mph with the throttle set to less than half. The throttle is directly controlling the power output of the locomotives, but it is the weight of the train and the grade profile that will determine the speed.

PLC,

The "peanut gallery"!  I really haven't heard that great expression for the last quarter of a century.  :-)

I think it came from the original Howdy Doody show, referring to the kids in the live audience stand.  But, since that show ended about a year before I started watching t.v., I am not really sure of that.

Mannyrock

@Rich Melvin posted:

There is a common misconception among model train guys that the power and speed of multiple locomotives must somehow be "matched." That's not how it works.

Consider a multiple unit diesel consist consisting of a big, 6-axle, 4,400 HP unit, and a 1,500 HP SW1500. When they get out on the road, are they going to run at the same speed? Of course they are...they are coupled together!

Are they going to pull on the train with the same force? Nope. The big 6-axle has a lot more pulling power than the SW1500, so it is going to pull harder. The SW1500 adds what it can to the overall effort of pulling the train. But they do not have to be "matched" in any way. Each locomotive pulls on the train with whatever pulling force it can generate.

The same holds true with helpers or distributed power. Even if the power setup has different makes, models and horsepower units running together, it doesn't matter. Each locomotive will pull on the train with whatever pulling force it can generate.

Here's another concept that may help to clarify the process. Think of the throttle in a locomotive (any locomotive, steam or diesel) as a POWER control, not a SPEED control. In my example above about climbing the 2% grade right out of the yard, I would have the throttle wide open, yet the train would only be moving at 9-10 mph...sometimes even a bit less! At other times I might be rolling along at 30 mph with the throttle set to less than half. The throttle is directly controlling the power output of the locomotives, but it is the weight of the train and the grade profile that will determine the speed.

Rich, Hot Water, thank you for your concise replies.

What happens when two locomotives are geared with different ratios?

Gunny

@gunny posted:

What happens when two locomotives are geared with different ratios?

Gunny

Nothing. Each of them does it share of the work.

@gunny posted:

What happens when two locomotives are geared with different ratios?

Gunny

Nothing. However the minimum speed, i.e. "short time rating" for the high current through the traction motors, must be monitored for the unit with the highest speed gear ratio (thus the minimum continuous steed will be higher), so as not to overheat/burn up rotating electrical equipment.

#90 and Big Jim:  OK;  change it to 15 mph;  or 10.  As Rich and others have explained,  the location of neutral points is a sort of 'automatic' reaction to the throttle settings.  Also,  these 'details' have no effect on the analysis.

What analysis?

Rich:  I'll send it soon.

HERE'S THE TOTAL PICTURE:

Consider a long train of coal hoppers heading up a 2% grade at 25 mph with no curves and no wind. There are 4 head-end diesels, 3 mid-train helpers, and 2 pushers. All diesels have the same horsepower. It's probably not very well-known, but the fact is that there are 2 'neutral points' in the train, which I define as follows: A neutral point is a location in the train where the [longitudinal] force felt by the cars is exactly zero....On one side of such a point the force is tensile, while on the other side it is compressive. On rare occasions, the neutral point is exactly where 2 couplers are joined. Under these conditions, it is theoretically possible to open the couplers owing to the lack of stress at the joint. There are then 3 independent trains that just happen to be touching each other. The pusher engines are doing just that: pushing, with all cars feeling a compressive stress which is a maximum at the last train car, and decreases to 0 at the first [rear] neutral point. The middle helpers are pulling the rest of their 'trailing' train, with all cars feeling tension, increasing from 0 at the rear to a max at the engine's rear coupler. The engine's front coupler, on the other hand, is pushing, with a maximum compressive force at the 1st car; decreasing to 0 at the other neutral point. Finally we come to the head engines, which are pulling; the tensile force being 0 at the left end of its train, increasing to a max at the engine's rear coupler.
These 2 points can be determined by realizing that the total load for each engine is proportional to the engine's tractive effort, which itself is proportional to its horsepower. OK; the physics is done. The rest is algebra. All that needs to be given are the weights of each engine, as well as that of each car...which we assume is a given constant. As well, of course, as the Hp supplied by each group of engines. TE(tractive effort) = Hp/S, where S is the train’s speed.

If these assertions are sufficiently interesting to this audience, I just might be willing to complete the analysis.

IT IS DONE! See below.

NOMENCLATURE

Tf, Tm, Tr: Tractive Efforts (TE = Hp/speed) of front, middle, and rear engines
Wf, Wm, Wr: Weight of front, middle, and rear engines [For simplicity, the engines are assumed to have 0 length; the train is pulling VERY many cars.]
Lf, Lr: Length of front, rear trains
mg: weight per unit length of each coal car
phi: slope of the terrain, radians
Xf, Xr: These are the locations of the 2 neutral points: distance from rear of rear(trailing) train
Below, please interpret “lifting” to mean “moving up the hill”…..NOT lifting VERTICALLY!

Results of the analysis:

1. Xr = [Tr/sin(phi) - Wr]/mg,

2. Xf = (Lr + Lf) +[Wf - Tf/sin(phi)]/mg

3. Xf = [(Tr + Tm)/sin(phi) - (Wr + Wm)]/mg. Yes; both 2 and 3 apply. Equating 2 and 3 yields

[Tr + Tm + Tf]/sin(phi) = Wr + Wm + Wf + (Lr + Lf)mg, which is the force balance on the entire train.

Bounds on Xf and Xr: (1) 0<=Xr<=Lr and (2) Lr<=Xf<=Lf + Lr; where <= means less than or equal to.

From 1 and (1): 0 <=Tr/sin(phi) - Wr, and from 1 and (2): Tr/sin(phi) - Wr <= mgLr. In words: The pusher engine must be able to lift its own weight, but must NOT be large enuff to push the mid-train helper.

Again; from 2 and (2), Lr <= (Lr + Lf) + (Wf - Tf/sin(phi))/mg <= Lr + Lf; i.e.,

Tf/sin(phi) <= Wf +mgLf and Tf/sin(phi) >= Wf. In words: The lead engine must be able to lift its own weight but NOT more than its weight + the entire front train.

Finally, from (2) and 3, Lr <= [(Tr + Tm)/sin(phi) - (Wr + Wm)]/mg <= (Lr + Lf); i.e.,

(Tr + Tm)/sin(phi} <= (Wr + Wm) + (Lr + Lf)mg and (Tr + Tm)/sin(phi) >= (Wr + Wm) + mgLr. In words: The pusher + the mid-train helpers must be able to lift both of their own weights + the weight of the rear train but NOT more than their weights + the weight of the whole train. [Else the front engine is not needed.]

THAT’S ALL, FOLKS!

P.S. I leave it to the reader to obtain various special or limiting cases; e.g., set Tm and Wm = 0, to obtain the case of just 2 engines: lead + pusher.

For what it's worth, LOADED coal trains do NOT ascend 2% grades at 25 MPH! One would be luck to make 10 to 15 MPH.

OK,Hot Water:  Change 25 t0 15;  then re-read it!

Just curious, what is the point of all of this speculation?   Too much time on our hands?

@phil gresho posted:

OK,Hot Water:  Change 25 t0 15;  then re-read it!

I couldn't read it the first time. I'm not a math guy, so your analysis is all just gibberish to me.

When’s the test? I gotta study my a** off!

Just my opinion but, this appears to be a case of a "toy train" person with a PHD (or maybe more than one PHD) with absolutely no knowledge or experience in real railroading.

@Rich Melvin posted:

I couldn't read it the first time. I'm not a math guy, so your analysis is all just gibberish to me.

I can read it, but I'm not willing to spend the time to slog through all of it for no apparent purpose!

Phil, you're way over my head on the math.  I tip my hat.  

There have been mathematical calculations done by equipment manufacturers and railroads, with the intention of calculating the tonnage, train length, horsepower per ton, and distribution of power, that will provide reliable performance.  There is a point at which failures begin to increase significantly, and that is the target of the railroad.

In the process of calculating the predictable failure rate, the node(s) have been found to float with even minute changes in gradient and/or curvature, rail lubrication, ambient and roller bearing temperatures, wind, and elevation above sea level.  The important thing, really, is not precisely where the nodes are, but generally where they are, because, behind the nodes the couplers are in draught, and there is a finite strength to drawbars.  There is not any such thing as a continuous railroad grade that does not have at least small variations.  Railroad track charts show these, and they are so closely spaced that a coal train could never entirely be on the same ascending grade.

So, it is not just theory that, within a train using mid-train or rear end helpers, there is one pair of couplers so slightly compressed that the uncoupling lever could be operated.  It does exist.  But, in real life, within a short time after this uncoupling lever was operated, the node would float toward the head end of the train, causing the train to become uncoupled, or it would float toward the rear and the increased buff force between the couplers would cause the coupler pin to reengage and the train would remain coupled.  

However it is only theory that one pair of couplers could be so precisely aligned that they were perfectly between buff and draught state for more than an instant.  It could happen, but the odds are astronomical that it would not.

@Number 90 posted:

However it is only theory that one pair of couplers could be so precisely aligned that they were perfectly between buff and draught state for more than an instant.  It could happen, but the odds are astronomical that it would not.

And what practical application does all this math gymnastics accomplish?  Surely, nobody in their right mind is going to try to uncouple them!

To all:  Yes,  I'm a retired PhD analytical engineer with too much times on my hands.  Questions like this occur to me in bed before falling asleep.  My nick name from wifey is THE QUESTION MAN.

Tom made a very good post.

After the "gibberish",  I stated  the results "in words".....for the masses.  [And for myself!]

Believe it or not, I DO have some real  RR experience....as a brakeman on the Niles Canyon Railway .

Maybe I should prepare a numerical example.....

@gunrunnerjohn posted:

. . . Surely, nobody in their right mind is going to try to uncouple them!

If only everybody were in their right mind.  (Sigh . . .)  

But despite all the crazies, road ragers, and general meatheads out there, malicious uncoupling of trains is very rare.  

This entire thread started with a theoretical question, just like many great inventions and processes started.  If we do not ask, "Why?" and "What if . . . ,"  we cannot make progress.

Thank you, Tom.