Draw Bar Load

[spikejones52002]

There is a train that is 100 ton load on the coupler of the engine.

Now you place three engines on the lead. There is still 100 tons on the last engine. What is the load on the first engine coupler? What is the load on the second engine coupler?  Does the second and third engine put more load on the engines in front of it?

Is there a different load factor when you place the second engine and third engine in the middle and end of the train?

[Mimbrogno]

Your post is a little confusing, but I think I understand what you mean. Just FYI though, a 100ton train is just one loaded hoppercar, so for the purpose of this reply I'll up it to a more realistic 100,000 tons.

 So, you have three engines on the head. The distribution of force between the drawbars is somewhat dependant on the individual pulling abilities of each locomotive, but for the sake of simplicity, we will assume all three are identical. (BLW AS-616's OK for everyone?)

 Say we have the train moving at about 5mph and we are accelorating at full power. the total pulling power of the locomotives in this example would be distributed evenly among them. The coupler between the first and second units would have the full pulling power of just the first engine on it, 1/3 of the total power. The coupler of the second and third unit will have the full pulling power of itself, plus the effort of the first engine added to it, for 2/3rds of the combined power output. The coupler between the third engine and the train will have the full force of all the engines on it. (at 6.0mph on this train, that would translate to 239,250lbs of forward effort.)

This principle also works the same for the cars of the train themselves. Starting from the first car behind the locomotives with the full force applied to it, every car you move back will reduce the force on the next coupler by the amount of the resistance of the cars before it. 

When you put engines in the middle or rear of the train, they relieve the force from the front of the train by pushing the cars ahead of them.

Matthew Imbrogno
-Mechanical Vollenteer, Arizona Railway Museum
www.azrymuseum.org

[KCSfan]

Your post is a little confusing, but I think I understand what you mean. Just FYI though, a 100ton train is just one loaded hoppercar, so for the purpose of this reply I'll up it to a more realistic 100,000 tons.

Wow Matthew that's some train. Me thinks you've got one to many zeros cause a 100,000 ton train made up of 100 ton loaded hoppers would have 1000 cars on it. I'd hate to wait at a grade crossing for that one to pass.

What we are actually talking about in the original question is the drawbar pull at each locomotive. The required drawbar pull is the total force that must be exerted to overcome friction (in the axle bearings and between flanges and the rails) and wind resistance, the horizontal component of the train's weight when ascending a grade, and any force required to accelerate the train. Assuming that all engines exert an equal tractive effort, you are correct in stating that the drawbar pull at the rear of the lead engine of a three engine lash up will be about 1/3 of the total required, that at the rear of the second engine about 2/3 of the total and that at the rear of the trailing engine the full required drawbar pull. I say about 1/3 and about 2/3 because in fact each trailing engines puts an added load on those ahead of it. However this is insignificant in relation to the total.

Ideally when engines are cut in the middle of a train they are positioned so they only pull the cars behind them while the cars ahead of them are pulled by the lead engines.

Mark

[jimrice4449]

Actually, a mid-train helper is both pulling the cars behind it and pushing some of the cars ahead of it.   The point at which there's slack in the couplers of the cars between the head end power and the mid-train helper is pretty much in constant flux.   As the head end starts up a grade, for instance, the point of equilbrium will move towards the front, as the head end tops the grade and accelerates, the pont of slack will run towards the mid-train helper.

[J. Edgar]

   As the head end starts up a grade, for instance, the point of equilbrium will move towards the front, as the head end tops the grade and accelerates, the pont of slack will run towards the mid-train helper.

 

 ....................and thus as the rookie engineer looks back at the conductor on the ground walking back to the broken knuckle....................

[doctorwayne]

There is a train that is 100 ton load on the coupler of the engine.

Now you place three engines on the lead. There is still 100 tons on the last engine. What is the load on the first engine coupler? What is the load on the second engine coupler?  Does the second and third engine put more load on the engines in front of it?

Is there a different load factor when you place the second engine and third engine in the middle and end of the train?

The load on the first engine coupler is zero.  If this wasn't a "trick" question, and you actually meant "first engine's coupler", then see the answers already given.

Wayne

[WSOR 3801]

Actually, a mid-train helper is both pulling the cars behind it and pushing some of the cars ahead of it.   The point at which there's slack in the couplers of the cars between the head end power and the mid-train helper is pretty much in constant flux.   As the head end starts up a grade, for instance, the point of equilbrium will move towards the front, as the head end tops the grade and accelerates, the pont of slack will run towards the mid-train helper.

 

The KCS in the 1970s found that having the mid-train helpers pushing as well as pulling led to some bad days.  They would push cars right off the track.  Helpers would get cut in about 1/2 way back, instead of 2/3 of the way back.  They figured it out.  One time, somebody forgot, and put the helpers in 2/3 back.  Big mess.  Memories came back quickly. 

[doctorwayne]

Here's a link to a related topic: HP vs TE

Wayne