Sherrel, this thread is for you. ;-) The recent discussion of the nature of a HP curve when plotted against speed and tractive effort prompted me to dig-up my Surface Mining textbook. So what does surface mining have to do with locomotive horsepower curves? Well, one section of the text covers the topic of “haulage and transportation”, and in particular it covers the requirements to design a railroad capable of hauling efficiently the mine product to the processing facility. Needless to say when the class covered this topic, I was all ears. The following formulas come from “Surface Mining, edited by Eugene P. Pfleider, and published by The American Institute of Mining, Metallurgical, and Petroleum Engineers, New York, 1968, reprinted 1972, pp531-552. The Davis Formulas are used to calculate train resistance. Locomotive Resistance: Lbs/Ton = (1.3+(29/W))+(0.03V)+(0.0024AV^2/WN) |Journal Res| |Flange | |Wind Resistance| Pgr Car Resistance: Lbs/Ton=(1.3+(29/W))+(0.03V)+(0.00034AV^2/WN) Frt Car Resistance: Lbs/Ton=(1.3+(29/W))+(0.045V)+(0.0005AV^2WN) Where W=Average Tons / Axle N=Number of Axles V= Speed (mph) A=Frontal Area (ft^2) Grade resistance may be calculated: Lbs/Ton=(Wt in lbs * Rise)/Run For example, 2000lb*1ft/100ft So the TE needed to pull 1 ton up a grade of 1% = 20lbs/ton Curve resistance: Lbs/Ton=(0.8lbs/ton)/degree of curvature Most railroads will compensate grades to curvature, where the resistance due to curvature is "subtracted" from the actual grade. So a track that is described as having a grade of 1% “compensated” with 10 degree curves will in actually rise 0.6 feet per 100 feet through those 10 degree curves. In general, these formulas can’t be correlated directly to our modelling efforts, because the masses, velocities, coefficients of friction are so much different. However in the modelling world, compensating our grades for curvature will provide benefit. Acceleration: F=Wa/g Where: W=weight in lbs a= linear acceleration of the weight (mass) ft/sec/sec g=acceleration due to gravity 32.2 ft/sec/sec For example, what is the force required to accelerate 1 Ton at a rate of one mile per hour per sec? F=(2000/32.2)*(5280/3600)=91.1 lbs General Transportation Formulas: Rail HP=(TE in lbs*mph)/375 HP(input to generator)=(TE in lbs*mph)/(375*Efficiency in %) Or TE=(HP*375*Efficiency)/MPH Trailing Tons=(TE/Total resistance in lbs per ton)-Locomotive Wt in Tons Acceleration in mph/sec=Net TE/(100*wt of train in Tons) In summary, these formulas tell us that as velocity increases there is a corresponding increase in train resistance, and on the power side of things, we see a similar relationship, i.e., as velocity increases there is a corresponding decrease in Tractive Effort (TE). Plug the formulas into an EXCEL spreadsheet, plot some graphs, and see how the train dynamics change with speed, curvature, and grade.
Karl - Not exactly sure how I wound up in your gun sight. Don't shoot - I surrender! All I asked about was "rattle-spiking" - and your PM to me answered that very well, plus a few other things that I really had not considered - such as "running rails". BTW - thanks, I learned several things. "More Physics" reminds me of a long time ago story about a salty old captain who was getting his FAA Oral examination for an ATR (airline transport rating) on the Constellation. The FAA examiner had asked him numerous questions about the aircraft concerning weight, horsepower, length, height, wingspan, fuel cap, hyd press, and numerous other differences. (remember, there were seven different models of the Connie + the Starstream - all different) After NOT knowing many of the examiners' questions, the FAA man said, "You really don't know much about this airplane, do you?" Too which the old captain replied, "For Christ's sake, I'm trying to learn how to fly it, NOT build a box for it! Cheers, I am gonna go listen to the rattling spikes, and watch the rails running downhill.
Sherrel, All in fun to be sure. http://www.frisco.org/vb/showthread.php?t=3326 Post number 30 I enjoy the the jargon of railroading and the engineering behind the track and trains. Have a good'n' -keb-
I can tell that you do; You are so far above me that I can barely see your contrail. But don't think for a minute that I am not going to "play" with those equations. I may need to ask you a question or a hundred. Send me, if you would, your private email address. sirfoldalots@gmail.com You may start by explaining to me, if TE is a measurement of starting and/or pulling power, why is a locomotive measured in HP? Does it have something to do with the fact that 746 watts = 1 HP? P.S. All members of the forum are welcome to my direct email as well.
I'd sure like to know how tractive effort relates (If it actually does) to horsepower?? Here and there in the various Frisco books, engine horsepower is mentioned (4500's=5600hp etc.), but mostly tractive effort is shown. Is there any formula Tractive Effort vs. Horsepower? Thanks, Tom
Those formulas made absolutely no since. But my 8th grade brain is in football mode so I wouldn't learn much.
Wow Karl, I can't wait to get out my old Hewlett Packard "digital slide rule" with its reverse Polish notation to have my way with those formulas. There's a latent engineer hiding deep inside....
Karl - I'm still working with my HP-12C. One of the most useful calculators ever made, as long as you can deal with the HP system. And, I still have my slide rules (K&E), but I haven't "slid" either of them in many years! Ken
I cannot find any way to plug those formulas' into my computer? But, I can get you most anywhereon earth.
In the General Formulas section in the first post of this thread there is a formula that relates HP and & TE Fun with Steam Steam Locomotive HP may be calculated with this formula: HP=(P*L*A*N)/33000 Where P=Mean Effective Pressure; try 40-60% of Boiler Pressure L= Stroke (Ft per revolution) A=Area of Cylinder (Inches^2) N=Revolutions per Minute The standard formula used to calculate the TE of a steam locomotive is the Cole Formula. TE=(CD^2*CS*(BP*.85))/DD) CD =Cylinder Dia (inches) CS = Cylinder Stroke (inches) BP = Boiler Pressure (psi) DD= Driver Diameter (inches) Note that the Steam Locomotive TE is independent of the locomotive weight. The formula just gives the TE that the machinery is able to develop. Steam locomotive diagrams include a value called the Factor of Adhesion, which equals Wt on Drivers/TE. Values of about 4 are average, smaller values means that the steam locomotive is prone to slipping, and higher values mean that the locomotive is more sure-footed. It is also interesting to compare efficiencies between different steam locomotives. For example it might be interesting to compare how a 4000 stacks up against a rebuilt 1350 class 2-8-2, or even how a 1306 2-8-0 stacks up against one of its rebuilt sisters. Grate Demand Factor=TE*DD/GA TE=Tractive Effort (lbs) DD=Driver Diameter (inches) GA= Grate Area (Ft^2) The formula predicts the firing rate, and the smaller the ratio the lower the firing rate will be; a good thing. The Boiler Demand Factor=TE*DD/EHS EHS= Evaporative Heating Surface (Ft^2) The boiler demand factor predicts how hard a boiler is working, and the therefore a lower ration is better. It is also useful to compare how efficiently the fuel is converted into HP. This is determined by dividing the HP by the GA. The lower this ratio is the better. The previous formula gives us a relative look at how much heat is going up the stack. It is also useful to determine how efficient our boiler is…..=EHS/HP. A boiler with a high value will have an easier time of producing HP than one with a lower number. Total Machine Efficiency is determined by Engine Wt/HP. A low number is good. Steam fans will note an omission from the Boiler Demand Factor, and that is the Super Heating Surface Area (SHS). When that value is added, a dramatic change in the BDF will be noticed. BDF=(TE*DD)/(EHS+SHS) Again lower is better.
