Subscriber & Member Login

Login, or register today to interact in our online community, comment on articles, receive our newsletter, manage your account online and more!

Use Of Helpers

3719 views
53 replies
1 rating 2 rating 3 rating 4 rating 5 rating
  • Member since
    September, 2013
  • 1,701 posts
Use Of Helpers
Posted by caldreamer on Friday, April 27, 2018 9:30 AM

I model BNSF in N scale and am looking for opinions as to whether I should run DPU and helpers or just DPU. The mountain (Highiline) subdivision runs eastbound for 421/2 feet of 2.54 percent grade to the summit of the mountain.  Down hill from summit is 481/2 feet of 2.24 percent grade.  I will be running heay trains of greater than 30 cars and over 960 grams (over 34 ounces). I do have an easier grade around the mountain that is double track that begins about half way up the mountain and bypasses the summit.  It is a ,73 percent grade of 371/2 feet. Since I am running a modern railraod and DPU has become the norm.  I am running up long steep grades would helpers be warented or just have 2+2 or 3+2 DPU engines for my trains.  Helpers make for additional interest and running, but are they appropriate for my railroad.

    Thanks In Advance

                 Ira 

  • Member since
    November, 2013
  • 1,179 posts
Posted by VOLKER LANDWEHR on Friday, April 27, 2018 11:15 AM

Do understand correctly, you want to run head end, mid train, and rear units? The last two as helpers or as DPU?

I don't have personal experience using DPUs/helpers. But what I've heard they work best and safest if they are really needed to get the train up the hill.

In model railroading the only difference between manned helpers and DPUs seems to be that helpers are cut in and out somewhere on the road.

On the prototype it is an economical question. If the hill is long and steep enough compared to the run the go with DPU, is it comparatively short they might run helpers. And don't forget dynamic braking downhill.

In times of DPU there are not many helper districts left. CSX's Sand Patch, NS out of Altoona, and MRL's Mullan Pass. As far as I know BNSF and UP us manned helpers on a very limited basis on Cajon Pass and a few other locations. Usually DPUs do the job.
Regards, Volker

Edit: BNSF has power limitations for its trains. Trains are only allowed to have a defined number of "rated powered axles" (RTA): www.smartlocal933.org/BNSF%20SSI%20No.4.pdf
Look at page 70. The RTA for different locomotives can be found on page 7

  • Member since
    September, 2013
  • 1,701 posts
Posted by caldreamer on Friday, April 27, 2018 12:57 PM

So as I understand the chart on page 70, for a manifest train you can have 42 rated axle on the head end and 16 on the rear.  Is  that correct?

  • Member since
    January, 2017
  • 2,428 posts
Posted by NWP SWP on Friday, April 27, 2018 1:02 PM

To model helpers get a friend with a throttle to run the second set of motive power, you have to actually be in sync with each other like they did in the old days. If you have a grade to justify it do it!

Steven

Crooner, Imagineer, High School Graduate, living with Aspergers, President of the Republica Pacifica micronation,  President of the NWP-SWP System.

Hook'em Longhorns! 

  • Member since
    January, 2015
  • From: Southern California
  • 1,340 posts
Posted by Lone Wolf and Santa Fe on Friday, April 27, 2018 1:09 PM

On Cajon Pass the Santa Fe has several locomotives on the head end of each train. For most of the climb that is enough but when the east bound trains reach the steepest part of the hill which is right before the summit they add pushers to the very end of the train to help. The same set of pushers then backs down the pass and pushes the next train. West bound trains don’t require pushers because the grade is not as steep.

Modeling a fictional version of California set in the 1990s Lone Wolf and Santa Fe Railroad
  • Member since
    November, 2013
  • 1,179 posts
Posted by VOLKER LANDWEHR on Friday, April 27, 2018 1:34 PM

caldreamer
So as I understand the chart on page 70, for a manifest train you can have 42 rated axle on the head end and 16 on the rear. Is that correct?

That seems correct as I read it. You have to consider that all six-axle AC locomotives count as 8 RPA (page 8).
Regards, Volker

  • Member since
    September, 2013
  • 1,701 posts
Posted by caldreamer on Friday, April 27, 2018 1:52 PM

Even though the grades are 2.54 and 2,24 from the base of the mountain to the summit, I do have short sections of track as steep as 2.93 percent.  Would I use that when determining which engines based on RTA for a particular train or would I use the 2,54 or 2.24 grades and add an extra engine to the train?

  • Member since
    November, 2013
  • 1,179 posts
Posted by VOLKER LANDWEHR on Friday, April 27, 2018 1:55 PM

Lone Wolf and Santa Fe
On Cajon Pass the Santa Fe has several locomotives on the head end of each train. For most of the climb that is enough but when the east bound trains reach the steepest part of the hill which is right before the summit they add pushers to the very end of the train to help.

That must be for main 3, the original track. The main 3 has a 3% grade from Cajon to Summit. All other track has 2.2% grade. If help is needed on these tracks it is needed from San Bernadino. Trains had an interesting map of the month: http://trn.trains.com/~/media/files/pdf/map-of-the-month/trnm0811_acajonpass.pdf
Regards, Volker

 

  • Member since
    November, 2013
  • 1,179 posts
Posted by VOLKER LANDWEHR on Friday, April 27, 2018 4:19 PM

caldreamer
Would I use that when determining which engines based on RTA for a particular train or would I use the 2,54 or 2.24 grades and add an extra engine to the train?

Usually the ruling grade on a subdivision determines the power need.

In former years railroads had tonnage ratings for each locomotive on a special route. For example SD45s had a rating of 1,275 tons from Colfax to Sparks across Donner Pass. So a Train of 5,000 tons needed 4 SD45.

Today each railroad uses a slightly different method mostly based on rated power axle (RPA) or equivalent power axle (EPA). For each subdivision/mountain pass is determined how many RPA or EPA a type of train needs for 1,000 tons or how many tons 1 RPA or 1 ETA can move. Sorry I haven't found information about details.

Perhaps you could make up a table determining how many cars a single axle can move on you model railroad and adjust it to the axle limits (or make your own).
Regards, Volker

  • Member since
    September, 2013
  • 1,701 posts
Posted by caldreamer on Friday, April 27, 2018 4:30 PM

Volker:

  What constitutes a ruling grade. the total grade from level to the highest point (on my railroad for example 2.54 percent) or the highest grade, even for a short stretch of track (In my case 2.93 percent)?

  • Member since
    November, 2013
  • 1,179 posts
Posted by VOLKER LANDWEHR on Saturday, April 28, 2018 3:43 AM

caldreamer
What constitutes a ruling grade. the total grade from level to the highest point (on my railroad for example 2.54 percent) or the highest grade, even for a short stretch of track (In my case 2.93 percent)?

The ruling grade eeds the largest tractive or braking effort on a subdivision. It is not necessarily the steepest grade. When the steepest grade is shorter than the trains part of them is on level or less steep grade. A less steep grade longer than the trains then might become the ruling grade.

In your case the 2.93% might not the ruling grade if the trains are longer than the grade. It isn't the lesser grade before the 2.93% either as a train can be partly on both grades at the same time. It would be somewhere in-between.

In case the 2.93% grade is longer than your trains it is the ruling grade.

In helper times railroads would often declare your lesser grade as ruling grade and add helpers for the 2.93%.
Regards, Volker

  • Member since
    January, 2015
  • From: Southern California
  • 1,340 posts
Posted by Lone Wolf and Santa Fe on Saturday, April 28, 2018 12:57 PM

VOLKER LANDWEHR

 

 
Lone Wolf and Santa Fe
On Cajon Pass the Santa Fe has several locomotives on the head end of each train. For most of the climb that is enough but when the east bound trains reach the steepest part of the hill which is right before the summit they add pushers to the very end of the train to help.

 

That must be for main 3, the original track. The main 3 has a 3% grade from Cajon to Summit. All other track has 2.2% grade. If help is needed on these tracks it is needed from San Bernadino. Trains had an interesting map of the month: http://trn.trains.com/~/media/files/pdf/map-of-the-month/trnm0811_acajonpass.pdf
Regards, Volker

 

 

Main 3 according to the map you posted however it used to be the only main. Pushers were added at mile post 62.8 near the Cleghorn exit from I-15.

Modeling a fictional version of California set in the 1990s Lone Wolf and Santa Fe Railroad
  • Member since
    May, 2017
  • 126 posts
Posted by xboxtravis7992 on Saturday, April 28, 2018 2:01 PM

I think it really depends on the era and route. Horseshoe Curve famously still uses helpers. However in comparison, Soldier Summit on the old DRGW is almost entirely DPU units now with no recent manned helper runs to my knowledge.

Of course you could have manned rescue helper moves (which are kind of common in the modern age were power desks often fail to assign enough DPU's to a heavy train passing through mountain territory). Set up trains to heavy and without enough power to pull up the grade, and once up on the grade and the train stalls out; have the crew call the nearest terminal for a manned helper. Then shove everything else on line into sidings to wait for the light engine set being ran by the second crew to run out from the terminal to the stalled train. Or if there is a local switching crew nearby have them abandon their local train in a siding and relocate their engines to helping the underpowered train. I know its not good practice to intentionally set up underpowered trains, but it does imitate a common ocurance on underpowered trains in the mountainous west in the current day and age.

Lastly clearing the line for the light helper rescue move to reach the struggling train will wreck havoc on operating schedules for the session (especially if passenger trains are involved); and keep the other crews who have to wait for the rescue to pass on the line; extending the operator's time on the layout and providing a 'surprise' extra for another crew to run.

  • Member since
    November, 2013
  • 1,179 posts
Posted by VOLKER LANDWEHR on Saturday, April 28, 2018 2:51 PM

Lone Wolf and Santa Fe
Main 3 according to the map you posted however it used to be the only main. Pushers were added at mile post 62.8 near the Cleghorn exit from I-15.

I'm not familiar with modern operations on Cajon Pass. My last visit there was in 1995. Then all helpers were added near San Bernadino IIRC.

Cajon (MP 62.8) is a logical place for adding helpers on Main 3. From there on it is just single track without sidings to Summit.
Regards, Volker

  • Member since
    September, 2013
  • 1,701 posts
Posted by caldreamer on Saturday, April 28, 2018 8:49 PM

Occasionally having an undrpowered train is an interesting idea.  I will have to keep that in mind. My mainline is fully double tracked.  The Valley subdivision has a short grade  of 1.67 percent east bound. At the hump yard is where I will add the additional power to climb over the mountain.  West bound I do not have a good place to add additional power, so I will have all power on the train at staging.  I have a pretty good tonnage rating system which I will use, since I do not fully underatand how to compute an RPA.  Does anyone have any ideas on how to do this?

 

  • Member since
    November, 2013
  • 1,179 posts
Posted by VOLKER LANDWEHR on Sunday, April 29, 2018 11:27 AM

caldreamer
I have a pretty good tonnage rating system which I will use, since I do not fully underatand how to compute an RPA. Does anyone have any ideas on how to do this?

The tonnage rating system is good enough. SP got along with tonnage per locomotive type for many years.

The prototype calculates the restance of a train depending on grade, curvature, bearing, rolling, air resistance. The Davis formula comes to mind.

Union Pacific has defined 1 "equivalent power axle" (EPA) as 10,000 lbs tractive effort (TE). Knowing the resistance you can devide by 10,000 lbs TE and get the required number of EPAs. The UP System Special Instruction page 32 contains the EPA per locomotive type: https://www.up.com/ert/ssi.pdf

In the 1980s SD45 and SD40 had almost the same tonnage rating across Donner Pass. Now you have different wheel slip software, different adhesion factors between DC and AC locomotives and that shows in the EPA per locomotive.

For model railroading purposes a tonnage rating per locomotive for different grades is good enough, I think.
Regards, Volker

 

  • Member since
    September, 2013
  • 1,701 posts
Posted by caldreamer on Sunday, April 29, 2018 12:30 PM

My  tonnage rating system works out  pretty accurately for model railroad purposes.  You take the tractiv effort of a locomotive, divide it by 160 for N scale to get the Tonnage rating on level track.  Divide that number by the grade and you get a tonnage rating for that grade.  For example. For an SD40-2 the Continuous tractive effort is 82,100 oiunds.  Divide that 160 gives you 513.125 or 513 tons on flat andlevel tracl.  Divide that by the grade, for exampe 2.93 percent and you get 175.127986348 opr 175 tons.  I use grams of weight and use the actual weight of each car, so 175 grams will enable the engine to pull that much weight up the grade.  A 500 gram train would require 4 SD40-2's to negotiaste the grade.

  • Member since
    November, 2013
  • 1,179 posts
Posted by VOLKER LANDWEHR on Sunday, April 29, 2018 1:54 PM

Sounds good to me.

The prototype railroads try to move the maximum freight at a minimum costs. I think that is the reason for going so far into detail.
Regards, Volker

DrW
  • Member since
    January, 2008
  • From: Lubbock, TX
  • 117 posts
Posted by DrW on Sunday, April 29, 2018 2:34 PM

caldreamer

My  tonnage rating system works out  pretty accurately for model railroad purposes.  You take the tractiv effort of a locomotive, divide it by 160 for N scale to get the Tonnage rating on level track.  Divide that number by the grade and you get a tonnage rating for that grade.  For example. For an SD40-2 the Continuous tractive effort is 82,100 oiunds.  Divide that 160 gives you 513.125 or 513 tons on flat andlevel tracl.  Divide that by the grade, for exampe 2.93 percent and you get 175.127986348 opr 175 tons.  I use grams of weight and use the actual weight of each car, so 175 grams will enable the engine to pull that much weight up the grade.  A 500 gram train would require 4 SD40-2's to negotiaste the grade.

 

Your formula gives funny results for grades of 1% or less. For a 1% grade, you would get the same 513 tons you give for the level track. For a 0.1% grade, you would get an astonishing 5130 tons. You might also want to consider that a level track has formally a 0% grade, and the result of a division by zero is undefined.

JW

  • Member since
    September, 2013
  • 1,701 posts
Posted by caldreamer on Sunday, April 29, 2018 4:37 PM

For a level track you divide the tractive effort by 160 which gives you 513.125 which is the tonnage rating on a 0 percent (level) track.

DrW
  • Member since
    January, 2008
  • From: Lubbock, TX
  • 117 posts
Posted by DrW on Sunday, April 29, 2018 4:48 PM

caldreamer

For a level track you divide the tractive effort by 160 which gives you 513.125 which is the tonnage rating on a 0 percent (level) track.

 

Yes, but what about an 0.1% grade?

  • Member since
    November, 2013
  • 1,179 posts
Posted by VOLKER LANDWEHR on Sunday, April 29, 2018 5:26 PM

DrW
Your formula gives funny results for grades of 1% or less. For a 1% grade, you would get the same 513 tons you give for the level track. For a 0.1% grade, you would get an astonishing 5130 tons. You might also want to consider that a level track has formally a 0% grade, and the result of a division by zero is undefined.

It works for his grades and that is good enough I think. If you want a more scientific formular to make it generally usable it can get complicated when using e.g. the prototype formulas.

More usable is perhaps cars per driven axle. The La Mesa Model Railroad Club (2% uncompensated) uses 2 cars per driven axle, another modeler with 4.1% uncompensated + adjusted for curvature) uses 1.5 cars per driven axle.

One can test it or one can just determine the ratio.
It is a model railroad not the prototype.
Regards, Volker

  • Member since
    September, 2013
  • 1,701 posts
Posted by caldreamer on Sunday, April 29, 2018 8:41 PM

In my humble opinion, using 2 cars pe driver axle or 1.5 cars driver axle is not nearly as protypical as my approach.  Different locomotives have different amounts of tractive effort and can therefore pull more or less cars up a given grade.  I use the actual tractive effort of each locomotive type to compute my tonnage rating that locomotive type.

  • Member since
    January, 2017
  • 2,428 posts
Posted by NWP SWP on Sunday, April 29, 2018 10:18 PM

Has anyone tested to see if their locomotive pulls the scale tonnage of the real deal?

 

Steven

Crooner, Imagineer, High School Graduate, living with Aspergers, President of the Republica Pacifica micronation,  President of the NWP-SWP System.

Hook'em Longhorns! 

  • Member since
    November, 2013
  • 1,179 posts
Posted by VOLKER LANDWEHR on Monday, April 30, 2018 4:14 AM

caldreamer
In my humble opinion, using 2 cars pe driver axle or 1.5 cars driver axle is not nearly as protypical as my approach.

The problem with your approach is, it doesn't work for grades smaller 1%. Then the pulling power gets larger than at 0% grade. And pulling power at 1% grade equals that at 0%. DrW already pointed this out. The tractive effort of a locomotive on a grade is not less than on level track, the train's resistance gets larger with the grade. Adjust the resistance for grade (R) and don't change the tractive effort (TE).

To avoid mathematical problems the equation could look like this:
needed TE = R (0%) + R (g%)

Car per axle is a modification of the prototype approach tons per axle. As the weight difference of model cars of different length and load capacity is much smaller than on the prototype, it is easier to use cars per axle.

@NWP SWP: When model railroading magazines tested locomotives in the past they mostly gave the pulling power in ounces or number of cars on level track and sometimes on a grade. If MR still does this I can't check in the test section of this website as I don't subscribe to MR anymore.
Regards, Volker

 

  • Member since
    September, 2013
  • 1,701 posts
Posted by caldreamer on Monday, April 30, 2018 1:22 PM

Instead of using ounces, I use grams.  Where an SD40-2 is rated at 175 grams on a 2.93 percent grade.

  • Member since
    December, 2001
  • From: AU
  • 534 posts
Posted by xdford on Monday, April 30, 2018 9:30 PM

NWP SWP

Has anyone tested to see if their locomotive pulls the scale tonnage of the real deal?

 

It is nearly impossible to do this as the rolling characteristics are not the same for a model as the prototype.

On the prototype at least here in Australia, the ratio for dead level track to move a freight car was considered to be 8lbs of tractive effort per ton. Where I lived in Adelaide, the ruling grade to Melbourne was 2.5% (we call it 1 in 40) and the required traction was considered to be close to 100lbs TE per ton to get over the hills.

This is with very free rolling prototype freight cars but the smaller scales do not roll proportionately as well as prototype cars. 

I believe that an SD40 (based on the Western Australian L class diesel) puts out around  80,000 lbs starting TE which translates to nearly 10,000 tons on dead level track. A 1% (1 in 100) grade would reduce this by a factor of about 3 so that would make a load of 3300 tons and that does not take into account the friction of curves, which would reduce it even more!

HO freight cars do not roll any where as well as the prototype but think of their weight multiplied by 87 cubed and each one would weigh about 90 tons, accompanied by Delrin trucks which are OK for our purposes but have a higher coefficient of friction compared to wheel bearings.

Regardless of whether you go DPU or helper, I have found (from a DC operator point of view) that taking a leaf out of the KCS operating procedure that I know of (and no doubt there are others), putting more load behind the mid train helper will make the operation smoother.

Cheers from Australia

Trevor

  • Member since
    September, 2013
  • 1,701 posts
Posted by caldreamer on Tuesday, May 01, 2018 8:17 AM

All of my rolling stockhas, or will have metal wheel sets and are tested to be ABSOLUTLY free rolling.  By that I mean the that they will roll on the SLIGHTEST grade of there free will.  Physics is physics no matter the size from 1:1 down to 1:110 z scale.  These laws DO NOT change.  My AC  engones are over rated n my tonnage chart, but since I will be pulling long trains up the grade and will be using from three to four units that really does not matter.  As long I have enough power to get over he mountain.

  • Member since
    November, 2013
  • 1,179 posts
Posted by VOLKER LANDWEHR on Tuesday, May 01, 2018 8:53 AM

caldreamer
Physics is physics no matter the size from 1:1 down to 1:110 z scale. These laws DO NOT change.

You are right, physics are he same and I said before, if it works for you it is OK for me.

But that doesn't mean it is physically correct. First of all the tractive effort (TE) doesn't represent the weight of the train it can pull but it is a train's resistance it can overcome.

Second, TE is dependend on weight on driver (W) and adhesion factor (Fa): TE = W * Fa;
As scale weight is W/160³ for N-scale, physically TE scales down the same way to Scale TE = TE/160³. In reality it is even less than that as adhesion of our locomotives is less than of the prototype. Cause: Different materials, no wheel slip control.

Power assignement doesn't have to be science, only believable and representing prototype practice. In our model world 4-axle and 6-axle locomotive weigh the same and pull the same compared to the real thing.
Regards, Volker

  • Member since
    May, 2010
  • 3,166 posts
Posted by mbinsewi on Tuesday, May 01, 2018 10:10 AM

I guess I've been doing it all wrong.  If one loco doesn't pull the train, I add another, and if needed, I'll add a third.  Usually don't have to.

But if I do, I'll usually add more cars, just to see what I can get away with.

Mike.

Subscriber & Member Login

Login, or register today to interact in our online community, comment on articles, receive our newsletter, manage your account online and more!

Users Online

Search the Community

ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT
Model Railroader Newsletter See all
Sign up for our FREE e-newsletter and get model railroad news in your inbox!