This is one for the railroad engineers.?

Question

How many SD40-2's would it take to pull a 100 care loaded T100 hopper coal train up a 2% grade. Would they need mid engine helpers?

Answer 1

ziggie,

I'm not an engineer but having studied railroading for a while I would make an educated guess that at that kind of tonnage on a stiff 2% grade you would at least need two SD40-2s up front followed by two helpers pushing on the rear.  Perhaps even some mid-train helpers would be needed but I couldn't say for sure...

Answer 2

Hmmm...  three ex-SP hoggers in a row... there goes the neighborhood...

But:

The first thing you need to do is calculate total horsepower need, with speed as a factor.

Out target speed is no less than 12 mph.  At 12 or above, the locomotives will not go into their "short time rating."  Simply put, short time rating is a speed at which the traction motors are seeing the maximum amount of amperage they can handle without over heating.

To calculate speed, the formula is HPT x 12 / %G = Speed (horsepower per ton, times 12, divided by the % of grade will equal speed).

Next, figure where the power needs to be placed to stay within the limits of the draft gear.  Standard draft gear is rated at 240,000 lbs.  To figure the pull, the "Rolling Train Resistance" formula is used.  This will give us the figure of the load on the draft gear of a moving train:  This formula is: 20 lbs per ton for each % of grade, + 5, or (20 x T) x %G + 5  =   force.  If the math arrives at a number higher than 240,000, then a helper must be entrained to shove enough tonnage to reduce this number below 240,000.

Finally, to figure the force levels when STARTING a train, the formula is 30 lbs per ton for each % of grade,  or (30 x T) x %G = f.

So, your train is 10,000 tons and needs to ascend the grade 2% at 12 mph.  Your SDs are 3,000 hp each.  So, we use our formula to solve for "x":  X x 12 / 2 = 12 mph.  We find that 2 hpt will allow us to operate at 12 mph.  Since we have 10,000 tons, it will take a minimum of 20,000 hp.  So, a minimum of seven SD40s would be needed to supply 21,000 hp.

Next, use rolling train resistance formula to determine drawbar force levels.  10,000 tons, times 20 = 200,000, times the 2%G +5 = 400,005 lbs.  We are way over 240,000.  We need to split the power up.

Using the formula further, you can calculate how much tonnage the road engine will handle and how many locos should be in the road consist and how many should help.  SP always made it a point to operate at the max, but if a higher target speed is desired, then more power will be needed, distributed accordingly.

There is one other thing to consider.  When placing the helper, it must be entrained such that it shoves 1/3 of the tonnage it will handle and pull 2/3s of the tonnage.  This will keep the helper from shoving the cars ahead of it off the track.  If the tonnage trailing the helper exceeds 240,000 lbs of force, then another helper must be added to reduce these forces as well.  When running with two helpers entrained, the first is referred to as the "swing" helper, with the other being the rear helper.

Often times prospective new hires will ask here, "How much math do I need?"  This is one example.  Most times, most of the math is done for you, with locomotive ratings and drawbar tonnage limits printed in the Timetable for the district you are on.  But, if you ever need to figure it out on your own, this is how its done...  But, you'd probably wind up running 3 by 4, or 3 by 5 (more likely), IF the power-chief can scrape up another loco...  It always a good idea to have one more unit in the helper consist than the road engine has...  Either way, a good helper engineer is worth his weight in gold, because he has the hammer...

Answer 3

On the Southern Pacific Railroad on Beaumont Hill, A SD40 was worth 1635 tons of tractive effort. On a 2% grade the coupler strenght is only about 6400 tons. So 100 coal cars loaded with 100 tons of coal each would have a total train wieght of 13,500 tons. Each car wieghs 35 tons. Simple math dictates you would need 8.25 engines to pull the hill. Helper limitations on the rear of a unit train limit the rear helpers to three units. the three units would have a tractive effort rating of 4905 tons. ( 3units X 1635 = 4905 ). 13,500 - 4905 = 8595. The draw bar is only 6400. 8595 - 4905 = 3690 over drawbar. This train will need mid-train as well as rear helpers. While the math may seem crazy to some, this is what railroaders do EVERYDAY.

Answer 4

We have almost that exact same scenario where I work every day.

As you are probably aware, anything over 1.8% is classed as Mountain Grade, our hill is mostly 2% with a bit of 2.2% and lots of sharp curves.

We run 100 car coal and grain trains that will weigh around 13,500 tons.

Until the newer larger horsepower locomotives were brought in we ran 4 SD-40s on the point, 4 cut in 1/3 of the way back and 4 more 2/3 of the way back.

Another configuration that we have used since then is one set of helpers cut in halfway and another set shoving on the rear but that much horsepower on the rear is very dangerous around curves, you could only do it with very heavy loaded cars.

That combination would get us up the grade at 17 or 18 MPH, you could lose one, maybe even two locos and still make the grade (barely).

If the rear helper lost an engine the added drawbar stress would be very likely to cause you break in two somehwere near the head end, (usually in the frikkin tunnel, Murphy's law multiplies exponentially in tunnels).

So the simple answer is, with 10 SD-40's you could barely make the hill some of the time but to make it reliably everytime you need 12 with mid train helpers.

Edit: had to think about it a bit more, our max grade is 2.2%, for a 2.0 with mostly straight track, you might get by one engine less than what I was saying or about 2.0 HPT. The SD-40 is one h**l of mountain locomotive, I've seen them peg the amp gauge far beyond the short time rating and just hang in there mile after mile, smoke rolling off the traction motors and still going strong. (shhhh, this is a secret so dont tell anyone I did it)

Answer 5

I found Michael S's answer interesting.Having worked for the Southern Pacific for many years before it got sold down he river I can't remember ever getting 8 units on any train(i might have but i don't remember it lol)!As an engineer the number that concerns me is HPT(horsepower per ton).We run our coal trains at about 1.1 hpt.While mid train helpers would be nice I think i could get over the hill without them.So give me 3 on the point and 3 pushing and i think it would make it.The only thing that could be a problem is wheel slip.I might be wrong but hey it wouldn't be the first time lol.

edit..I bow to Rango's expertise in the area of heavy grade.i don't run anything steeper than 1.2 here.I guess we know where i'd be.Sitting on the side of the hill in a puddle of sand(if they had sand) waiting for more power to show up lol