Horserpower, Grade, Horsepower, Speed, etc.

I think it's a given that it takes twice the horsepower to double the track speed (not counting rolling resistance, acceleration, curves, etc.). Also, to maintain the same speed when encountering a grade increase I understand that horsepower must be doubled for each one percent increase.

Does anyone know if that would also be true of a car, truck, bike, etc. I know the friction coefficients are different, but it would seem that the principles are the same.

Any thoughts? Thanks.

I don't think that is right. Think about it: it would depend on what the horsepower/ton ratio is to begin with, or, especially at lower speeds, what the tractive effort/ton is.

If a 1500hp. unit can move a train of a given size at 30mph, it may be able to move the same train at 60mph if it isn't being taxed to the limit at 30mph.

A Diesel can develop max. hp. at starting. To move a train, once in motion, will require more power to accelerate.
A steam locomotive develops more hp. as its speed increases. So it can run at speed with any train it is able to start.

Momentum has a great deal to do with train speed. It takes more effort to put a train in motion than it does to accelerate it.

Les

No, no, no! Oh, my.. where to start.

Using my handy U25B Speed-Trailing Ton Calculator (really a slide chart..)

One U25B (2500 HP) hauling 2500 tons will achieve about 48 MPH on level track (grade 0%) while two units (5000 HP) hauling 2500 tons will achieve 63 MPH. So as you can see, horsepower doubled but speed nowhere near doubled. The relationship isn't linear. This sort of relationship does apply to all types of vehicles but there are very many variables.

Diesel locomotives cannot typically develop their maximum horsepower at starting. Diesel locomotives develop a constant horsepower output in a given throttle notch, regardless of the track speed -- at least, this is the goal of the designer of the electric transmission. It's a fact that very large numbers of EMD locomotives were built that deliberately "power limited" at low speeds to roughly 250 HP per axle using a system known generally as "Drag Duty Performance Control" early on, and just "Performance Control" later. This includes all units of models GP-30, GP-35, GP-40 and GP-40-2, as well as SD-45 and SD-45-2. The most extreme example of course is the GP40 series limiting to roughly 2100 HP at around the continuous speed of about 11 MPH, increasing linearly until at about 19 MPH the locomotive is developing its full rated horsepower. This is an artificial limitation placed on the traction equipment to ensure adhesion and compatibility with older locomotives. General Electric offered similar equipment on the U series although as I've documented quite thoroughly before elsewhere it was of two very different styles and different operation. It achieved the same effect, when ordered.

My friend, the statement about momentum and effort and speed is just scientifically incorrect. Momentum is not some magically generated force; it represents energy which had to be developed at some point by locomotives ... even if the locomotives hauled cars to the top of a mountain yesterday and today the cars are going down the other side. Force is just pounds of force - in reference to "effort," this is variable highly. It's true that drawbar pull as a measured parameter reduces as speed increases, but it should be noted that an SD-40 wide open in notch 8 at 25 MPH is developing the same 3000 HP it's developing in notch 8 at 50 MPH. Always remember generally

(HP x 308) ÷ Speed = Tractive Effort

Now to speed and power on grades, looking at it the other way around... Let's take the example of two U25B units hauling 2500 tons. Of course this is 2 HP / ton, as a general thumbrule sometimes quoted as applicable to hotshot freights on the level. I'll just set the U25B calculator for this and note the speeds achieved as grade increases.

0% 63 MPH
0.2% 46 MPH
0.4% 37 MPH
0.6% 31 MPH
0.8% 26 MPH
1.0% 22 MPH
1.25% 18 MPH
1.5% 15.8 MPH

This last number is important, because 15.8 MPH is the continuous rating of these locomotives (the early production U25B, anyway.) These units were rated 50,800 lbs continuous effort at 15.8 MPH.

In reference to doubling the horsepower per one percent grade increase to hold speed constant, we can look at four U25B units instead of two hauling 2500 tons up a one percent grade and see if we can maintain 63 MPH. The answer is "no," because the calculator gives me a speed of about 39 MPH up a one percent grade for four U25B units pulling 2500 tons. Of course, those of us who have wrangled the horsepower - tractive effort formula I gave earlier know this, but I have the feeling that this is becoming an arcane art. I'd like to get it back NON-arcane, at least until folks understand power and speed fully instead of just pounds of force and a quantity of "miles per hour" without realizing what these mean and don't mean. I encourage you to actually plug numbers in to that formula and work it - graph the results. Solve the equation for speed, and select some values and graph those. That is most of what went into the tool that I'm using here to spit out values for the U25.

-Will Davis

Thanks for your responses. I will read them in greater detail.

But first, it takes 20 lbs. of pull per ton per 1% of grade. That means that a 15,000 ton train would need 8000 hp to maintain 10 miles per hour on a 1% grade.

Double the grade (2%) and you're now needing 16,000 hp to maintain that speed.

Again, over-simplified, but...

Denver Dude--
Re: "it takes 20 lbs. of pull per ton per 1% of grade." I think that's an oversimplification. Maybe it's about right if you are thinking of using the locomotive's tractive force as a holding brake to prevent gravity from starting to roll a stationary train downhill (something that can be done, I understand, with AC traction motors, but that would very quickly overheat DC motors), but if you are talking about pulling a moving train, you have to specify how fast you want to go uphill.

I think a good starting place would be to look at the chart of what a given locomotive can haul at a given grade at different speeds. A number of such charts have been published (in publicly available books, not just locomotive operating manuals). There ought to be some on the WWWeb: I'll post a link if I find one.

O.k., here is the first graph I could find on the WWWeb for what speed a locomotive can achieve at different power settings:
http://nicwhe8.freehostia.com/d5705/tec ... ption.html" onclick="window.open(this.href);return false;

(It's for a small British diesel-electric from the 1950s (Class 28, nicknamed "Cobo" from its unusual wheel arrangement-- I think there is a diesel character in the "Thomas the Tank Engine" stories of this type), but the overall SHAPE of the diagram would be the same for other locomotives: just relabel the axes to adjust for different power!)

Note that "Notch 10" has about twice the power of "Notch 5". On level track, it looks like this locomotive could pull its test train (420 tons of passenger cars) at about 37mph in Notch 5, and at about 60mph in Notch 10-- so doubling the power IN THIS CASE increases speed by about 60%.

Al Krug seems to have some good information.

http://www.alkrug.vcn.com/rrfacts/hp_te.htm" onclick="window.open(this.href);return false;

I don't think that doubling horsepower automatically doubles speed or enables you to maintain speed when your grade increases by 1%, but you have to double the hp in order to accomplish that. It won't happen if you don't.

Again, I know I have over-simplified it. I'm gonna read the informative posts above.

Denver Dude--
Your Google-fu is stronger than mine! I have read some of Al Krug's railroad-tech posts, and he is good value.

Denver Dude wrote:to maintain the same speed when encountering a grade increase I understand that horsepower must be doubled for each one percent increase.

You can see that makes no sense. If a given train needs 5000 hp to maintain 30 mph on the level, then it needs 10000 hp on 1%, or 20000 hp on 2%, or 40000 hp on 3%, or 80000 hp on 4%, or 160000 on 5%?

If you remember that

Power is Force times Speed

things will make more sense to you.

No one knows exactly how much force is needed to maintain X mph with a given train on the level, but we can make a reasonable guess. Say 100 cars of coal, 14300 tons at 40 mph-- 70000 pounds sounds as good a guess as any.

Like you said, a 1% upgrade adds 20 lb per ton-- so 286000 lb for the coal train. So 40 mph up 1% requires somewhere around five times the force, which means five times the power. 40 mph up 2% requires nine times the power, 40 mph up 3% requires 13 times the power compared to level track. That's at 40 mph, remember.

The same train might only need 30000 lb to make 15 mph on the level-- but the added force needed on the upgrade is the same as ever. 15 mph up 1% needs ten times the power, compared to 15 mph on the level.

timz wrote:

Denver Dude wrote:to maintain the same speed when encountering a grade increase I understand that horsepower must be doubled for each one percent increase.

You can see that makes no sense. If a given train needs 5000 hp to maintain 30 mph on the level, then it needs 10000 hp on 1%, or 20000 hp on 2%, or 40000 hp on 3%, or 80000 hp on 4%, or 160000 on 5%?

If you remember that

Power is Force times Speed

things will make more sense to you.

No one knows exactly how much force is needed to maintain X mph with a given train on the level, but we can make a reasonable guess. Say 100 cars of coal, 14300 tons at 40 mph-- 70000 pounds sounds as good a guess as any.

Like you said, a 1% upgrade adds 20 lb per ton-- so 286000 lb for the coal train. So 40 mph up 1% requires somewhere around five times the force, which means five times the power. 40 mph up 2% requires nine times the power, 40 mph up 3% requires 13 times the power compared to level track. That's at 40 mph, remember.

The same train might only need 30000 lb to make 15 mph on the level-- but the added force needed on the upgrade is the same as ever. 15 mph up 1% needs ten times the power, compared to 15 mph on the level.

Not the way I see it. Yes, that train would need 286000 lbs. of tractive effort to keep from rolling back. Without considering acceleration, curves, engine efficiency, rolling resistance, etc., to maintain 40 MPH with that much weight and a 1% uphill grade would require 30,506 hp. As you said, force times speed:- 286,000 times 40 divided by 375.

Increase the grade by 1%, and the tractive effort needed for that 2% is now 572,000 lbs. just to hold it steady. Get it up to 40 MPH and you'll need 61,013 hp to maintain that speed.

Al Krug's Railroad Forces Calculator based on the Davis Formula is available is available to download here;

http://www.mtnwestrail.com/aak/" onclick="window.open(this.href);return false;