I was interested if anyone knows of a way to calculate what the apparant weight of a train is when it is on a particular grade. i.e. If there is a train that weighs 900 tons running level (at grade) what would that same train "feel" like to the engine when it is on a 1.5% grade. Obviously more than 900 tons but how would you figure that. I was hoping to convert this somehow into the model world to figure what a particular engine will be able to pull up a grade and to help figure how much "help" it may need. I know this seems kind of silly for modeling but the prototype does this kind of thing all the time and I thought it would be interesting to play with. PeteC
You might try the train calculator found here: http://www.vcn.com/~alkrug/rrfacts/traincalc.htm Also, while your at that site, take a minute and read through the info. Theres a lot of great info on his site. The answer to your question may be found here: http://www.vcn.com/~alkrug/rrfacts/hp_te.htm
Realize that all of this is theoretical. If you know the weight of the train, you can calculate the pull of the grade as shown above, but the regular rolling resistance of the train on level track, especially in models, is highly variable.
This is a basic problem with trigonometry, combined with a little physics. Assuming frictionless surfaces, the tractive and drawbar forces are dictated by: F = Train Weight X Sin(Rise Angle) The Rise Angle = ArcTan(Rise Rate) For low angles the Sin of an angle very nearly approximates the Tan of that same angle. Example: Assuming a 1" in 36" rise (my current design limit), the tractive effort and drawbar calculations are: Rise Rate = 1/36 = 0.027778 ATan(1/36) = 0.027771 Sin(0.027771) = 0.027767 Assuming 10 coal cars (100 tons each) plus a 200 ton locomotive you get at total train weight of 1200 tons. Total tractive effort is about 33 tons, and total drawbar foce is 28 tons (5 tons to pull the locomotive up the hill). Hope this helps! Mark in Utah
Mark, do you mean that an HO scale "ton" should weigh 22.9885 actual pounds? That would make my HO 1.2 million pound 4-8-8-4 Big Boy actually weigh 13,793.1 pounds! I'm afraid it might fall through one of my bridges! (Just pulling your leg) It has been so long, I really have little memory, but I seem to remember we only counted the weight of the string of cars as the "train" weight. The friction factor was already accounted for on the charts. In addition to the dead weight of all the cars, there is an added draw bar pull required for curves, rain, snow and ice, even sanding required more traction just to ride over the sand. Then we added the engine's weight to see if we had enough tractive effort. I do not know how it is today, but back in steam days, we could add engines to the point so long as we didn't exceed what our couplers could take. You couldn't hold much speed, and sometimes they would get down to just a crawl with throttles wide open, and ever once in awhile you would hear one set of drivers break loose and spin like crazy! It was sometimes required to add helpers on the rear as "pushers" to get up the hill all in one piece. The pushers could drop off when not needed, as the "train" proceeded down grade. Occasionally helpers were also added about mid train too. I do remember seeing a train from across a canyon that had four 2-8-8-2's on the point, two were in the middle, and two more were pushing, with the caboose on behind them to prevent crushing it! It was back in the 40's during the war when they were hauling a whole train loaded with tanks! Dad took movies of it, and wished for a recorder to hear all the whistle "talk" between those engines. I don't know what grade they were pulling but I'll bet you could have easily run along beside them! I suppose today you simply add diesels to achieve the same thing. It would take a club sized layout to setup a train like that, but would really be impressive if you could reproduce the stack pressure they had! It sounded like cannon fire from across the canyon! Those were the days I tell you!
These calculations do not take into account real-world things such as friction. On the flats, that's all you have to work with, but once you start to hit the hills gravity starts to rule. I could work out the calculations for the curves, but it gets EXTREMELY complicated in a hurry. I'm an electrical engineer (doomed to play with electric trains) and I occasionally do cable pulling calculations through conduit. When you hit curves it gets messy fast. I enjoyed that pulling calculator another guy posted, gave some interesting results that really helped to show how badly trains run when they hit the hills going up Ogden Canyon - they CRAWL! Mark in Utah
THE FOLLOWING POSTS AND PHOTOS HAVE BEEN MOVED HERE BECAUSE THIS COMBINES TWO SEPARATE TOPICS INTO ONE COMPLETE. PLEASE ADD REPLIES BELOW AS USUAL. THANKS. sapacif TrainBoard Staff Member # 2572 posted June 27, 2004 09:16 PM -------------------------------------------------------------------------------- Pete: The mass will be the same, but the weight will actually be less than if on the level. I think that the question you are asking is more dynamic, like it takes more power to pull a train up hill than it takes to pull it on the level. That involes momentum which will be somewhat determined by the speed at the entry to the grade. Anyway, maybe some members with more recent physics experience may be able to postulate a better question that can give you the information that you want. BTW, I would say I hope this helps, but after re-reading it, I am sure that it doesn't. Flash Blackman San Antonio, Texas See my layout: http://www.railimages.com/gallery/robertblackman -------------------------------------------------------------------------------- Posts: 418 | From: San Antonio, Texas | Registered: Mar 2002 | : Logged | [ 10. July 2004, 10:55: Message edited by: watash ]
PeteC Member # 2816 posted June 27, 2004 09:49 PM -------------------------------------------------------------------------------- Flash, Yea, I kind of expected this to be a tricky question which is one of the reasons I stated "apparent weight." You are correct of course that the tonnage is the same but it certainly takes more tractive effort up a grade than at grade. Sooo, I guess the more accurate question is... "How do you figure the tractive effort necessary to take a train of given tonnage up a particular grade." After re-reading my original post it does make more sense to look at it this way. Thanks, -------------------------------------------------------------------------------- Posts: 6 | From: NC | Registered: Jul 2002 | : Logged | NYC-BKO TrainBoard Member Member # 4329 posted June 27, 2004 10:59 PM -------------------------------------------------------------------------------- In the real world they can calculate this because there are constants ie: known engine weight and horsepower, known car weights light & loaded, rollablility is pretty much the same, and it is known how much tonnage a given horsepower can handle. In the model world we have too many variables, engine weight and pulling power vary by manufacturer , unless you fine tune every car rollability varies greatly, then there is weight of cars The best way to see how much you can pull with a given engine is to add cars till it stalls on a grade, then reduce till it makes it then calculate the weight of the train, or use a spring scale in ounces ( fish ) rigged up with a coupler and pull it with your engine till the engine spins then you have your pulling power in ozs and you can raise the test track up at one end to simulate a grade and get those values, too. Hope this was simple enough to explain it. Brian On the Road to the Future - in 1/87 -------------------------------------------------------------------------------- Posts: 32 | From: Scottdale, PA | Registered: Jun 2004 | : Logged | Mike Sheridan TrainBoard Member Member # 2032 posted June 28, 2004 12:37 AM -------------------------------------------------------------------------------- Tricky question. There isn't a really simple answer because there are some highly dynamic variables involved. While Googling for more info on train resistance I found an excerpt from "Railway Industry Overview" (on www.arema.org). It had this to say: "Grade Resistance is the force required to overcome gradient and is equal to 20 lb. per ton per percent grade. This force is derived from the resolution of force vectors and is independent of train speed. An up grade produces a resistive force while a down grade produces an accelerating (negative resistive) force. A train moving up a long tangent of 1% grade at 10 mph, a speed that most tonnage trains slow down to at ruling grade locations, will have a train resistance coefficient of 22.4 to 23.5 lb. per ton with the grade resistance accounted for over 85% of the total." So in this case the grade has, in a sense, increased the train 'weight' by 85/15 (grade resistance/friction resistance) = 5.7:1 at 10 mph. But if you allow that the train was moving faster on the level, and therefore the friction value was much higher, then the difference would not be so marked since slowing down has made the train 'lighter' from the friction point of view. For example: Working back from their numbers it looks like friction is about 3 lb/ton at 10 mph. So at 20 mph it might be 12 lb/ton and at 40 mph it could be 48 lb/ton (since friction usually increases with the square of speed). If the train had enough power to maintain 40 mph then the train resistance coefficient (for 1% grade) would be 48 + 20 = 68 lb/ton with the grade resistance accounting for only 29% of the total. So at 40 mph the grade has only increased the train 'weight' by 0.4:1. Then you get to factor in curvature ... I think Brian's idea with the fish scale is pretty good actually Mike Website -------------------------------------------------------------------------------- Posts: 321 | From: Cheshire, UK | Registered: Jul 2001 | : Logged | Eagle2 TrainBoard Member Member # 343 posted June 28, 2004 01:53 AM -------------------------------------------------------------------------------- Trying to remember all the way back to when I took physics, I think this is what happens: the friction or drag applied on each car or locomotive is slightly reduced, since not all of the weight is pushing down onto the rails any more (since gravity, the force that helps create weight, is no longer acting perpindicular to rails on a grade). However, at the same time, with an angle off of "true" horizontal, a portion of the mass of each car or locomotive now becomes a weight acting opposite the direction of travel. I hope nobody else gets as lost reading that as I do, but I think it works. Jenks - the REAL Big Blue "Brave Rifles! Veterans! You have been baptized in fire and blood and come out steel!" - GEN Winfield Scott -------------------------------------------------------------------------------- Posts: 538 | From: Fountain, CO, USA | Registered: Mar 2000 | : Logged |
randgust TrainBoard Member Member # 3719 posted June 28, 2004 07:27 PM -------------------------------------------------------------------------------- I'll put up a post of my wacky Dynamometer car here, that's how I measure train resistance in real situations: Albuquerque Third District in N Scale -------------------------------------------------------------------------------- Posts: 71 | From: NW PA | Registered: Dec 2003 | : Logged | [ 10. July 2004, 10:47: Message edited by: watash ]
PeteC Member # 2816 posted June 30, 2004 09:17 PM -------------------------------------------------------------------------------- All of this information has been VERY interesting and much more complex than I originally thought it would be. Thank you to all that replied. I very much like the idea of a scale and I think the spring scale on passenger car trucks is a really wacky and neat idea. I am going to have to find one and set that up for my railroad. Of course now we will have to factor in the weight of the scale when measuring the weight of a train on a grade, but that's still much simpler than all the dynamic physics involved otherwise. PeteC -------------------------------------------------------------------------------- Posts: 6 | From: NC | Registered: Jul 2002 | : Logged | Hooch Runners Member # 4386 posted June 30, 2004 11:43 PM -------------------------------------------------------------------------------- Here is a prototype train forces calculator you might find interesting. http://www.vcn.com/~alkrug/rrfacts/traincalc.htm -------------------------------------------------------------------------------- Posts: 3 | From: Green Bay Wisconsin | Registered: Jun 2004 | : Logged | Eagle2 TrainBoard Member Member # 343 posted July 01, 2004 02:01 AM -------------------------------------------------------------------------------- I for one like the "scale test" car. Jenks - the REAL Big Blue "Brave Rifles! Veterans! You have been baptized in fire and blood and come out steel!" - GEN Winfield Scott -------------------------------------------------------------------------------- Posts: 538 | From: Fountain, CO, USA | Registered: Mar 2000 | : Logged |
You have received many good replys and the fish scale is about as good as you can do for models, As models in general do not scale down from the prototype in weight, horse power, and tractive effort. Now if you really want to get into how the prototype does it you may want to check out A A Krug's Railroad Facts and Figures. It is very comprehensive. You can access his body of work from this site. Warring! Learning this stuff may lead to "A beautyfull Mind" syndrome, For which there is no known cure.
