I know ,what kind a question am I asking ,what locco ,what manufacter,sound ,dcc ect…long short of it ,I built my helix 261/2"r dwn side 29"r up side ,2 trk , oblong to get 200" per loop for a 4" spread at about 2% grade …Im dcc and some what new @ HO scale, so my one rs-3 atlas w/moter dcc will not handle my 16 car mix of 40’/50’ all the way with out adding a second rs 3 (wheel slip,stalling ) 10/11 CARS YES but no more ,is that about what I can expect ? or am I over loading ? now I plan on always running two powered loccos or concist for all my mainline frieghts and my target is a 9-10’ 16 car trains but just dont want to over load em …my 2nd test train is a ab set of p2k f-7s /dccsound pulling 8 walther streamliners ,made it a bit slow but no wheel slip,would that be about right? I know its a loaded question with many ifs and buts just dont want to over load em,but also dont want 3 or 4 powered loccos pulling a short train. any how any advise would be helpful, Jerry
I have not gotten to the point on my layout where my helix is complete enough to test my engine fleet, but concidering an RS3 is a 4 axle, narrow bodied loco, being what I think is a lighter weight engine, on 26 inch helix at 2 percent, I would expect 10 to 12 is about right, maybe more… if your rolling stock is well maintained.
That combination you have is tight. Nothing less than a 30" at 2p would be my limit.
Your engines will not be overloaded as long as your wheels can slip. Only when you stop the motor/wheels from turning will you risk any damage, most likely in the decoder.
RF&PRR
The way to check for overloading is to measure the current. Easily done on DC, you may want a more specialized meter for DCC. The goal is to keep the current within the limits of the motor. That said, most HO models will slip their wheels (assuming no traction tires) before reaching the motor current limits.
If you can add extra weight, that will improve pulling power. If you could find one, the metal Hobbytown of Boston (later Bear Locomotive Works) Alco road switchers will pull twice as much as your Atlas versions. The primary difference is the weight. The curves in your helix do add extra drag, making the effective grade around 3%, not 2%.
my thoughts, your choices
Fred W
Yes, as mentioned above in an answer, it’s not only the elevation that is getting to your train, but also the curves of the helix.
"Because of the resistance produced by curves, they pose an added difficulty when located on grades. To keep the combined resistance of grade and curve from overwhelming trains, grades are often “compensated” by being reduced on curves so resistance remains constant. A grade so treated would be termed, say, “1.7 percent, compensated.”
from this article, The ABC’s of Railroading
On my layout I have a 3 percent grade on a 22 inch curve. Using one of my Atlas RS-3’s anything over about 9 40/50 foot cars (mixed, with metal wheels) gets to be problematic. Now hook up my Class A steam engine to that same train… no problem, but it is a very much heavier engine. I haven’t tried it but it would probably pull 30 or more cars up the grade without strain.
Jarrell
Jarrell
The very short answer is: you can haul up as many cars as your loco will pull.
Curves add to the effective % of grade, and each loco is different as to what it will pull…on level ground, on an incline, and on a curved incline.
Don’t haul more than it {or them if you consist two or more together} can handle unless you want to wear out your loco.
In “Track Planing for realistic operations” John Armstrong has an interesting table that shows the effect of grade on the number of trains you can pull.
At 2% you will get approximately 50% the number of cars you could over a flat level surface. (This is for extremely free rolling wheels on a straight track) So 10 cars on a level flat load = 5 cars on a 2% incline. You have a double whammy as grades increase. The amount of “traction weight” the wheels put on the rails descreases, and the amount of car weight pulling on the coupler increases.
“Tongue weight” (Amount of weight pulled on coupler from cars) = sine(grade angle) * car weight
“traction weight” in engine = cosine(grade angle) * engine weight * friction ratio at rails
“Friction ratio at rails” = ~.7 for steel on steel and ~.9 with all rubber traction wheels
Ratio of engine tractive effort compared to car weight on grade = (Traction weight)/(tongue weight)
As the incline numbers get bigger, the ratio drops and very quickly. When it gets <1 the wheels slip.
Hi!
My 11x15 two level layout has an honest 2 percent grade that winds from one level to the next, so I have some experience with your area of question.
The obvious answer is “it depends”. I model the Santa Fe (post war) and most of my trains are 2,3, or 4 powered units. With 4 Stewart F units I can pull pretty much what I can fit on the track with ease.
Moving down the line (no pun intended), my two powered P2K E units can pull 9 Walthers full size passenger cars up the grade with ease as well.
Two RSD-15s (BLIs) can pull a 20 car freight with ease, and probably 5-10 more with ease as well.
Steamers (w/o traction tires) pulling alone are good for 12 cars, but double headed I can do 20 with ease.
Small switch engines are good for 5-8, depending.
OK, here are some tips:
Make sure your grade is consistent, and understand that a given loco can pull more up a tangent track than a curved track.
Tune up your rolling stock. Mine have Intermountain wheelsets, and KDs, and are very free rolling.
Make sure your track flows nicely and is “bullet proof” against derailments.
Good Luck!
thanks guys .like I said was planning on always double heading my power but was hopeing I wouldnt need it. Im going to rework the exit after the helix ,this seems to be an area of most grade since its an S curve and useing the woodland scenics 2% foam starter ,I think I got a little sloppy and lost my constant grade ( I know it going to bug me later on) ,so thats on the rework list .also the uphill track in the helix is 29" radius and the track is very constant in there (its all wood const, no foam) so I will have to polish up on foam terrain road bed work [*-)] was never planning trains over 10’ long do to the tite quarters ,but Ilike to keep the options open with sound trackwork…thanks to all, good info …Jerry
There are a number of problems and omissions with this calculation that mean it does not relate well to the real world. Is this something you dreamed up on your own or does it come from another source? It’s definitely not in Armstrong’s Track Planning for Realistic Operation, as you seem to be suggesting.
If you are attempting to derive a calculation from the slope of the lines in the chart in Armstrong’s book, note that it’s a rough empirical display and not a matter of physics and math. Varying motor power, efficiency of drive mechanisms, differing wheel materials, balance of engine weight over the drivers, etc., etc. seem to make a purely mathematical calculation as you describe impossible.
The one thing you said that is true is that when the drawbar pull (not “tongue weight”, there is no tongue) of the train is more than the engine(s) can move, the drive wheels will slip – as long as the engine is not over-weighted.
Byron
Drawbar pull/tongue weight are relatively the same thing calculation wise for this modal. Edit: I am incorrect. I mixed up the proper term. But you are right, there is no tongue (unless it’s old school british railrays)
The 50% on a 2% did indeed come from John Armstrongs chart. And he in fact well researched the topic.
The equations however are mine. And I didn’t pull them out of my tail. But I did signifcantly simplify them as to not confuse the reader too much. For example, I left out a rolling resistance. I also left out acceleration, and inertial factors.
The formula I stated was purely to demonstrate how quickly you lose tractive effort with grade increases. Rolling losses (due to wheel/bearings) are relatively low compared to the weight of the car/engine itself.
The numbers for coefficient of friction are pulled from general tables. I pulled them from my ASTM manual which is used by engineers. However I must admit that I used static friction ratio’s for steel on steel, and steel on rubber. Rolling friction would have been more appropriate, but the ratio’s are still relatively close.
I do make mistakes from time to time, and if you think I made a mistake, I freely invite you or someone to call me on it. (Better to have me slightly embarassed than to have bad information) At the heart of it, it is a pretty basic problem. You’re over thinking the problem for what we are attempting to demonstrate. It truey is really basic high school physics. If you think my general equation is wrong, then I can happily provide the proof with graphics. I also have the credentials to back it up.
Not at all. Tongue weight is the downward force that the tongue of a trailer applies to the hitch of the tow vehicle. There is not really a downward component of drawbar pull, since the couplers can move somewhat freely relative to one another in the vertical dimension – that’s why trains (real and model) become uncoupled when the couplers are set at different heights and/or vertical track grade transitions are too abrupt.
As to the rest of your calculation, false rigor is still, well, false. If we were talking about satellite engineering, of course, your credentials matter, and congratulations on your academic and professional achievements.
Recommending a complex calculation that has little relationship to the real-world problem doesn’t seem to me to help anyone in a hobby situation, no matter where one studied. But that’s just me and I certainly don’t have an engineering degree of which to boast.
I have towed a trailer, though.
Okay I’m about to eat my hat for a second time in two days. My appologies. You are correct. I mislabled the appropriate term. (I am seriously red faced) I forgot the proper term for the horizontal component of the weight on the tongue. My component vectors however are correct.
My calculations are NOT incorrect. Are they 100% accurate? No. I stated this. I simplified them to make them easier to understand.
Would you like me to prove it to you? I really don’t mind.
Please don’t post any more on my account, I won’t be back to this thread.
I never stated your calculations were incorrect. I said only that they don’t accurately describe the model situation. Thus, in my opinion, they are not helpful.
Carry on.
[Edit: I do note that you changed your post after I responded, eliminating your earlier recital of your educational and professional acheivements. Not sure what your purpose was in including them in the first place or removing them later. You also made a number of other changes in response to my reply without acknowledging the edits. Editing one’s post after someone else responds to it certainly makes the responder look foolish. But you’ve succeeded, I won’t bother replying to one of your posts again.]
No I think I should…incase anyone else doubts
Let’s start with the basics…
Can we agree that is the weight being pulled on the engine by the cars is greater than the tractive effort force on the rails of the engine that the engine is going to slip?
Fe = Force of engine
Fc = Counter force cars
When Fe < Fc then we have slip
Now on a level flat surface…assuming no frictional/inertial losses (and they are small compared to actual weight of a loaded car) There is relatively no counter force against the train.
This is why a car can drift well over a mile on it’s own on a flat level surface when no brakes are applied. This is why you have to set brakes on cars in relative flat yards. This is why you can get away with hauling a ton of freight a couple hundred miles on a gallon of fuel.
Now let’s get extreme…Let’s say that train has to pull that car straight off the ground. completely vertical incline. That’s 100% of the car weight fighting the train.
The formula to describe the force of the weight of the car fighting the train is thus:
Weight cars = sine(theta) * car weight;
at 0 degrees it is sine (0) * car weight = 0 * car weight = 0
at vertical wall it is sine(90) * car weight = 1 * car weight = car weight
This is the counter force to the car. This is the very minimum the train has to be able to pull to move without slipping.
Now from basic high school physics:
The force of friction on a non moving object is mu-static * object weight sitting on top of it. This is the force of friction the engine can generate before it wheels begin to slip. According to the ASTM manual, the coefficient of friction (mu-static) of steel in contact with steel is .7. That means for every 100 pounds of downword weight, you can generate 70 pounds of counterforce.
You can read about the forces of friction here…
I removed my education and work experience because I didn’t want to seem like a snoot shoving my credentials in someone’s face. I thought it was in bad form. But I had a conflict in trying to prove I knew what I was talking about versus “you’ll just have to take my word for it”
I just decided it was better to prove it. That way I wouldn’t have to seem arogant about it, and to help others who were trying to understand the problem.
My other edit stating you were correct about tongue weight was so I didn’t leave bad information in the post. I marked it with an edit to make it appear I was not cheating and I admitted so in my post. My attempt was not to make you look bad. I’m the one with egg on my face in that regard.
Again my appologies if I offended.
