Tractive Effort, Horsepower and Speed/Acceleration
Tractive Effort vs. Horsepower
Many people confuse Horsepower (Hp) and Tractive Effort (TE). With this essay I hope to clear that confusion.
I use some terms and formulas in this essay that may not be familiar. I suggest
that you read the Definitions at the end of this document.
Assume a train that weighs 15,000 tons is stopped
on a 1% grade. For every ton of train weight on a 1% grade a force of 20 pounds
is acting to roll the train down the hill. A 15,000 ton train produces a force of
300,000 lbs. (15,000 tons x 20 lbs per ton = 300,000 lbs).
To prevent this train from
rolling back down the hill we must apply an equal force in the opposite direction
to the coupler of the first car. Imagine yourself holding onto the coupler of the
first car and trying to hold the train from rolling back down the hill. It should
be obvious that try as you might you will not be able to hold the train. No matter
how hard you grasp the coupler you cannot hold the train because your shoes will
simply slide across the ties and ballast. The adhesion of your shoes to the ties and ballast is not equal to 300,000 lbs.
Thus not even if you were Superman could you develop 300,000 lbs of pull (traction)
to hold the train.
The level of force required to break the adhesion
of your shoes to the ties and thus slide your shoes is dependent upon two
things
1.
The coefficient of friction
between your shoes and the ties.
2.
The weight on your shoes.
Increase either one of those parameters and your
adhesion to the ties increases enabling you to pull harder before sliding.
For the purpose of this article we will say the factor
of adhesion of a steel wheel on a steel rail is about 30%. (Actually it is more
like 20%-25% but we will get to that later). A factor of adhesion of 30% means that
a wheel will stick to the rail so that a pull equal to 30% of the weight on that
wheel is required to break that adhesion and slide the wheel. Locomotives have a
lot of weight on their wheels. A typical 4 axle GP40 might weigh 280,000 lbs. Four of them weigh 1,120,000 lbs. Thirty
percent of 1.12 million pounds equals 336,000 lbs. In other words, it requires a
force greater than 336,000 lbs to slide four GP40s. Therefore these locomotives,
with their brakes set, will hold the train on the hill since the weight of that
train on the grade is only producing 300,000 lbs of pull.
Tractive Effort is the amount of Pull.
You can have pull with no
HP. Notice that in the example above we are applying a force, a pull, a Tractive
Effort, of 300,000 lbs to that coupler of the first car but we don't need any
diesel engines. All we need is at least 1 million pounds of weight on the wheels
and a factor of adhesion of 30%. We have Tractive Effort (TE) without any Horsepower
(HP).
If I don't need any HP why did I use 4 locomotives?
Why not just use one very heavy loco? Because the maximum weight on any single wheel
that steel rails can withstand is 35,000 lbs. If we put more weight than that on
any wheel it will crush the rail head and create excessive wear or outright failure
of the rail. The two wheels on opposite ends of one axle can each carry 35,000 lbs
of weight. So the maximum weight on an axle assembly is 70,000 lbs. Since we need
a total locomotive weight of at least 1 million pounds that means we need a minimum
of 14.3 axles. Three 4 axle locos would only be 12 axles so we must use four which
gives us 16 axles.
From this discussion we can see that total locomotive
weight determines the maximum amount of tractive effort a locomotive can produce.
The number of wheels or the number of traction motors has nothing to do with it.
We only add additional wheels to spread the required weight out along the rails
to avoid damaging the track. I said that 3 GP40s will not work in this case because
they only have 12 axles and each can be weighted no more than 70,000 lbs for total
loco weight of 840,000 lbs. At 30% adhesion that only produces 252,000 lbs of TE
and we need at least 300,000 lbs. If we take those same 3 GP40s and add two more
axles to each unit but DO NOT increase the weight of each unit, then we still only
have 840,000 lbs of total weight and 30% of that is still only 252,000 lbs. Adding
axles and wheels to locomotives does not increase tractive effort.
HP is the Tractive Effort (pull) times the Speed.
Burn that statement into your brain. It is crucial to understanding this essay.
While our coal train is just sitting on the grade,
held there by the locomotives, there is 300,000 lbs of "pull" on the first
car's drawbar. Because the train is not moving there is no HP required. But try
to move it at 1 mph up that hill and HP is required. The required HP is the TE needed
for the grade (300,000 lbs) times the speed (1 mph or 1.47 ft per second) divided
by the definition of a HP (550 lb-ft per second).
(300,000 lbs) x (1.47 ft per sec) / (550 lb-ft per sec) = 801 HP
The HP required is 801 HP! Yes just 800 hp will move
this coal train up the hill. Amazing isn't it? But will one 800 HP 4 or 6 axle unit
do it? No! Because that one 800 hp unit must have at least 1 million pounds on its
drivers to prevent it from sliding back down the hill. You must have the
weight to get the adhesion required. That means each wheel of an 800 hp 6 axle unit
would have to have 84,000 lbs on it. (That is 168,000 lbs. per axle). Oh my, the
crushed and broken rails that would leave behind! Not to mention the overloading
of bridges and the track structure itself. As I said above, the minimum number of
axles we need to spread out the required weight is 14.3 axles. It doesn't matter
whether we have four 200 HP 4 axle units, or whether we have one 800 HP 4 axle unit
and three 4 axle engineless slugs . It is all the same to the coal train.
One mph is kind of slow. It would take us 24 hours
just to get up Parkman hill. We want to go up the hill about 15 mph. 15 mph is a
good compromise between taking forever and extreme high power costs. To go up this
hill at 15 mph (22 ft per second) requires:
(300,000 lbs) x (22 ft per sec) / (550 lb-ft per sec) = 12,000 HP
Gee now that sounds familiar doesn't it? That is
exactly the HP of four 3,000 HP units! So we can use four 3000 HP GP40s. SD70MACs have 6 axles and weigh 420,000 lbs. Again 70,000 lbs per axle. So
three of them give us 18 axles and 1.26 million pounds on the wheels, more than
enough to produce our required 300,000 lbs of traction. SD70MACs are rated at 4000
HP so three of them are 12,000 HP. Four GP40s or three SD70s, either way you satisfy
both the adhesion needed and the HP needed.
Horsepower Alone
Will two 6000 HP units work?
No. While you have the required 12,000 hp you can only have 70,000 lbs weight per
axle, which times the 12 axles is only 840,000 lbs total. With 30% adhesion and
only 840,000 lbs of weight the two 6000 HP units have only 252,000 lbs of adhesion.
Not enough to hold or move the coal train. The weight of the train on the grade
will slide these two locomotives backwards down the hill. Remember you need at least
300,000 lbs of adhesion (traction).
High HP locomotives on 6 axles create other problems
such as traction motor (TM) overheating. Can we pour 1,000 hp into each TM continuously
at 10-15 mph without frying them? Low gearing helps. But low gearing lowers the
top speed because the TMs will fly apart at high rpms. Here is where AC locos have
an advantage over their DC counterparts. AC TM rotors are much more solid than DC
TM armatures so they can be geared lower and still have a high top end.
Varying Adhesion
All of the above figures
are based on a 30% adhesion factor. IE, the wheels grip the rail with a force equal
to 30% of the loco weight. Locomotives of the GP40/SD40 era and their Dash 2 offspring
are generally considered to have an adhesion factor of about 25% not 30%. Sand will
increase this factor to about 30%. Thus to achieve the 30% adhesion needed for the
examples used in this essay so far, these locos may need to be on sanded rail.
Modern locos such as SD70MACs and C44s claim adhesion factors of 36 to 43%! They do this by using sophisticated anti-wheel slip circuits. These circuits allow the wheels to spin slightly faster than the rail speed warrants. It is called creep. Strangely enough, a creeping wheel has a higher factor of adhesion than a stationary or rolling wheel. Thus in theory two 6,000 HP SD90s weighing 420,000 lbs each and achieving an adhesion factor of 36% will produce a TE of 302,400 lbs and should pull the train up the hill at 15 mph.
Modern locos such as SD70MACs and C44s claim adhesion factors of 36 to 43%! They do this by using sophisticated anti-wheel slip circuits. These circuits allow the wheels to spin slightly faster than the rail speed warrants. It is called creep. Strangely enough, a creeping wheel has a higher factor of adhesion than a stationary or rolling wheel. Thus in theory two 6,000 HP SD90s weighing 420,000 lbs each and achieving an adhesion factor of 36% will produce a TE of 302,400 lbs and should pull the train up the hill at 15 mph.
However, in my experience you cannot count on that
36% adhesion factor in all types of weather and rail conditions. On wet or frosty
rail these units slip and you stall. And when you stall you had better set the train
air brakes in a hurry or the train will slide these units back down the hill. On
the other hand I have had C44s, SD90s and SD70MACs absolutely astound me with what
they are pulling. At times they attain greater than 40% adhesion on dry, sanded,
rail. It is that "at times" that concerns me. You cannot count on them
to do that reliably time after time.
An Actual Experience with High HP Locos and Adhesion
One night I was running a
freight up hill at 7 mph with a Dash 9-44CW on the point. I had previously calculated
that we should have gone up the hill at 11 mph, so why were we only doing 7 mph?
The rail was slightly frosty. I punched up the loco monitor screen on the computer.
It showed that this supposedly 4400 HP unit was only putting out 2930 HP!!! It had
derated to prevent slipping in spite of the sanders being on. So the adhesion factor
of this loco at that time was not the touted 36-43% but instead only 22%. The railroad
had paid for a 4400 HP locomotive with 36% adhesion but was only getting a 2930
HP locomotive with 22% adhesion. The common SD40-2 would have done as good as or
better in this situation than the hi-tech wonder. This was not a one time occurrence.
I have seen similar performances on many occasions,
Horsepower is Speed
Up to now we have assumed
that a locomotive has enough power to slip its wheels. That is true only at low
speeds. Note that HP is TE times speed. If the speed remains the same and the TE
(pull) increases then the HP requirement increases. If the TE remains the same and
the speed increases then the HP requirement increases. If you have a fixed maximum
Hp, such as a loco has, then as speed increases the TE must come down. The
product of the two must remain a constant and is directly related to the HP rating
of the loco. On the 12,000 HP coal train above why can't we go faster than 15 mph
on the 1% grade? Because 15 mph times the required 300,000 lb of drawbar pull divided
by 550 lb-ft per second (the definition of a HP) equals the total locomotive HP.
If the train went faster the product of the speed times the pull would be higher
and thus the required HP would be higher. But we are limited to 12,000 hp on this
consist so it trudges along at 15 mph. Similarly if we throttle down a notch or
two, reducing the HP, then the speed is going to drop because there is now less
than 12,000 HP available. The drawbar pull account of the grade remains the same
at 300,000 lbs and the lower HP means a lower pull x speed figure so the speed must
drop until the product is proportional to the new lower HP. This why I like to say
HP is speed.
Steeper Grades
We are proceeding at 15 mph
up the 1% grade with our 15,000 ton train powered by four 4 axle GP40s. What happens
when we encounter a steeper grade with this train? The drawbar pull needed to hold
or move a train on a grade is 20 lbs per ton per grade percent. That is where the
300,000 lb figure came from for the 1% grade of the example above. If this train
were to roll onto a 1.5% grade what happens? The drawbar pull needed now is 450,000
lbs.
(15,000 tons) x (20 lbs per ton) x (1.5 grade) = 450,000 lbsWe still have only 12,000 hp available. Since HP
equals pull times speed, if the pull goes up then the speed must come down. The
train speed will drop to 10 mph, the point where the product of the new 450,000
pounds pull and the speed divided by 550 (HP definition) equals the available HP.
But we are in real serious trouble here folks. Our four 4 axle locos can only deliver
336,000 lbs of pull because of their 1.12 million lbs of weight on drivers and the
30% adhesion factor. So our locos are going to slip and stall on the hill. Although
we have enough HP to pull this train up the 1.5% grade at 10 mph we do not have
enough traction to do so. And, again, you had better set the train air brakes as
you stall or the train will drag you back down the hill.
Slugs and SDs
So how do we proceed? Well
we can't increase the weight on drivers by adding weight to our existing four 4
axle locos because they are already weighted to the max for the rail. The only other
way to increase traction, weight on drivers, is to increase the number of drivers.
Add more units. We will need to add two more units (8 more axles) to get the weight
up to at least 1.5 million lbs so the 30% adhesion factor gives us 450,000 lbs of
adhesion. We do not need the added HP of those two units however, they could
be just engineless slugs. Without additional HP the train will go up the hill at 10 mph. At
this point I would like to point out that if we simply add two 4 axle slugs, which
get the power for their traction motors through electrical cables from the original
4 locos that we can do the same thing by switching from GP models to SD models.
From 4 axle units to 6 axle units. An SD40 is simply a GP40 with two more axles (with traction motors) and 50%
more weight. In other words we've added "half a slug" to each of the 4
units. In this manner we once again have the required weight on drivers by using
just four SD40s. These are the same weight and number of drivers as 4 four axle
units plus two 4 axle slugs. If we want to go up the hill at 15 mph instead of 10
mph however, we must add the HP. Slugs or converting to SDs will not do. If the
additional units added are 3000 hp like the rest then we will again go up the hill
at 15 mph. HP is speed.
Helpers
In either case we will not
go up the hill very far, probably not at all! Why not? The figures say we will.
The one word is Kapow! You are going to break in two. We are now trying to
put 450,000 lbs of pull into a coupler rated at 390,000 lbs, it is going to break.
So what can we do? Well we could double the hill. Take half the train up to the
top and leave it there then come back with the engines and get the second half.
When you get both halves to the top, recouple them, make an air test, and proceed.
By taking half the train up the hill at a time the required coupler pull is only
half that of the entire train or only 225,000 lbs. Well under the 390,000 lb strength
of our couplers. This method also requires no more slugs, SDs, or other units. The
original 4 GP40s have enough traction to haul half the train up at a time. Unfortunately
doubling the hill requires a lot of time. The line is blocked while it is being
done and this train and others are delayed for the duration. Alternatively we could put the added two units on
the rear of the train to PUSH. We would need another engineer (a helper engineer)
or a distributed power set-up (radio controlled slaves). The physics are the same,
the coal train and grade could care less where you apply the power just so you have
the right amount to move it. But the couplers do care where you put all that power.
If you try to put it all thru one coupler, the first one, it is going to say "Screw
you, I ain't gonna take this abuse" and it will break to prove its point (you
never knew couplers were so animated did you).
Some things left out
Now for you purists, I know
I have left out a few things. (If you know enough about this to know that I left
things out then you sure don't need to be reading this document).
·
As the grade gets steeper
less and less of the loco weight is felt as pressing directly down on the rails
so effective weight on drivers decreases slightly. (Do the geometry yourself if
you want).
·
I neglected the weight of
the locomotives. They don't go up hill for free.
·
I neglected the efficiency
of the locomotive's mechanical and electrical transmission.
·
I neglected rolling resistance
of the train. At low speeds such as these, on straight rail, rolling resistance
for a loaded coal train is only 10-15% relative to the grade resistance. However
that will increase the total pull and HP required.
·
I neglected acceleration.
The figures given are for steady state running. To accelerate requires more pull
than steady speed.
The Weight of the Locomotive
First we'll look at the loco
weight. Four 280,000 lB GP-40-2s weigh 1,120,000 lbs. That 1 million pounds of locos
does not go up the hill for free. It takes just as much HP to move each of those
pounds up the hill as each of the train's pounds. So you should add their weight
to the train when calculating traction, speeds, & HP required. In fact this
is one reason for 4 axle high HP locos. The more the locos weigh the more of their
HP is required to just to move the loco upgrade.
Heavy Haul vs. High Speed.
You may have noticed that
most railroads tend to use 6 axle power on heavy trains such as coal and grain while
they use 4 axle power on their high speed lighter weight intermodal trains.
- If a particular railroad has a good mix of high speed and heavy haul trains its locomotive roster will be a mixed bag of 4 axle and 6 axle power.
- If a railroad, such as BN, has a preponderance of coal and grain trains and/or operates its trains in mountainous territory where grades are steep and speeds are low its roster will be dominated by heavy 6 axle power.
- If a railroad, such as ATSF, has a majority of lighter weight high speed intermodal trains and a lot of relatively flat territory its roster will reflect that with lots of high HP 4 axle power and/or lighter weight 6 axle power.
Speed
|
Tractive
Effort
|
60 mph
|
18,707 lbs
|
40 mph
|
28,060 lbs
|
30 mph
|
37,415 lbs
|
25 mph
|
44,898 lbs
|
15.0 mph
|
75,000 lbs
|
13.4 mph
|
84,000 lbs
|
8.9 mph
|
126,000 lbs
|
Model
|
Weight
|
Max
TE
|
Speed
|
Total
tons on 1% grade
|
Trailing
tons
|
Light GP40
|
250,000
|
75,000
|
15.0
|
3750
|
3625
|
Heavy GP40
|
280,000
|
84,000
|
13.5
|
4200
|
4060
|
Light SD40
|
380,000
|
114,000
|
9.8
|
5700
|
5540
|
Heavy SD40
|
420,000
|
126,000
|
8.9
|
6300
|
6090
|
A first glance at the table
looks as if the heavy SD40 is the best loco. It can pull the most trailing tonnage
up the 1% grade. But we have to ask ourselves "What is the job?” If the job
is to haul as much tonnage up the grade as is possible, then indeed the heavy SD40
is the loco we want. But if the job is to haul as much trailing tonnage up the grade
at 15 mph then the SD40 is not the best choice. The following chart shows
how much tonnage each of these locos can haul up the 1% grade at 15 mph. Since all
the locos are 3,000 HP they all produce the same 75,000 lbs of TE at 15 mph. But
the weight of the loco uses up some of that TE. What is left over can pull the freight
that is paying the bills.
Model
|
Weight
|
TE
|
Speed
|
Total
tons on 1% grade
|
Trailing
tons
|
Light GP40
|
250,000
|
75,000
|
15
|
3750
|
3625
|
Heavy GP40
|
280,000
|
75,000
|
15
|
3750
|
3610
|
Light SD40
|
380,000
|
75,000
|
15
|
3750
|
3590
|
Heavy SD40
|
420,000
|
75,000
|
15
|
3750
|
3540
|
The light weight GP40s can
haul 85 more tons of paying freight per unit up the grade at 15 mph than the heavy
SD40 can.
Those Superpower units
If you haven't been paying
attention you might think that the new 6000 HP single unit locos are destined for
heavy haul service. True they are all heavy 6 axle units. But that is because the
weight is needed to put that 6,000 HP to the rail without slipping. A 6,000 HP unit
that weighs 420,000 lbs and can attain a 43% adhesion factor has an adhesion of
180,600 lbs. The 6,000 HP diesel engine can deliver that 180,600 lbs of Tractive
Effort at a speed of 13 mph. Below that speed you cannot use full throttle on these
locos because they will slip. That was for an astounding adhesion factor of 43%.
What if they cannot maintain that extreme level of adhesion? What if they "only"
get 36%? 36% of 420,000 lbs is 151,200 lbs of TE. The 6000 hp diesel can deliver
that TE at 15 mph so the loco cannot operate below 15 mph in full throttle without
slipping. At an adhesion factor of 30% the lowest full throttle speed is 18 mph.
If the rail is wet or frosty can these modern marvels maintain even a 30% adhesion
factor? My experience with 4400 HP units is a definite no. The C44s often have trouble
maintaining 22% adhesion with bad rail conditions. If a 6,000 HP unit gets down
to 22% adhesion it can only operate at full throttle above 24 mph! Thus if you want
these behemoths to reliably move your trains over the hills in all kinds of weather
you had better dispatch them with trains light enough that they can maintain 24
mph or greater on your steepest hills. That means they are only useful for trains
such as intermodals which get a high HP to tonnage ratio. When it is frosty they
won't work on heavy freights or coal or grain trains which routinely pull up the
hills at 10-12 mph.
The railroad I work for uses 12,000 HP on their coal
trains through here and we go up the hills at about 12-13 mph. Note that you can
replace the 12,000 HP of 3 SD70MACs, or the 12,000 HP of 4 SD40-2s, with the 12,000
HP of just two SD90s. You have the same HP so you should go up the hills at the
same 12-13 mph. But it will be awfully iffy. That is because the minimum speed these
6,000 HP units can operate at full throttle is 13 mph even with an adhesion factor
of 43%. If anything causes the train speed to fall below 13 mph even momentarily,
you will never regain the lost speed. The train might be temporarily slowed for
various reasons. Perhaps the SD90s temporarily lost that 43% adhesion factor and
slipped or reduced HP to prevent slipping. Perhaps a wind came up and increased
train resistance. At 12 mph the 6,000 HP locos cannot operate in full throttle even
if they regain that 43% factor of adhesion. They will slip. Operating at reduced
throttle the locos are not producing the 12,000 HP this train needs to travel up
the hill at 13 mph. So the train will never accelerate back up to 13 mph where it
could again operate at full throttle. Four SD40s or 3 SD70MACs would have no difficulty
re-accelerating the train back up to 13 mph.
That is because they are not operating at the limit of their adhesion as the SD90s are. The 4 SD40s have 12,000 HP just like the two SD90s but the SD40s have a total weight of 1,680,000 lbs and even at a 30% factor of adhesion can operate in full throttle down to 9mph! The 3 SD70MACs weigh 1,260,000 lbs and with only a 30% factor adhesion they can operate at full throttle down to 11.9 mph. If they achieve a 36% factor of adhesion they can operate at full throttle down to 9.9 mph. So either the SD40s or the SD70s have enough reserve adhesion they can operate at full throttle after being temporarily slowed. That allows them to accelerate the train back up to the 13 mph.Thus on an equal total HP basis these high HP units are not equal to their lower HP cousins when used in heavy haul service. And heaven help you (more like helpers help you) if the factor of adhesion on these brutes ever falls below 36% because you won't have enough adhesion to pull that 15,000 ton train up that 1% grade, period. You had better hope that it does not rain, frost, or snow.Keep the high HP units in high speed freight service where they do the most good. You are trading 8 axles of weight on two 3000 HP GP40s or 12 axles of weight on two SD40s for the 6 axles of the new units and you have 25-50% less Hp-wasting weight with the two high HP units. Remember that TE decreases as speed increases, so as long as they keep the HP per ton ratio of the trains high enough to maintain high speeds then the TE will be low enough that these high HP single units won't slip. But try to use them in low speed drag service and they will slip as noted in the coal train discussion above. The slower the train goes up a hill the closer these high powered 6,000 HP wonders perform like the good old 3,000 HP SD40.
That is because they are not operating at the limit of their adhesion as the SD90s are. The 4 SD40s have 12,000 HP just like the two SD90s but the SD40s have a total weight of 1,680,000 lbs and even at a 30% factor of adhesion can operate in full throttle down to 9mph! The 3 SD70MACs weigh 1,260,000 lbs and with only a 30% factor adhesion they can operate at full throttle down to 11.9 mph. If they achieve a 36% factor of adhesion they can operate at full throttle down to 9.9 mph. So either the SD40s or the SD70s have enough reserve adhesion they can operate at full throttle after being temporarily slowed. That allows them to accelerate the train back up to the 13 mph.Thus on an equal total HP basis these high HP units are not equal to their lower HP cousins when used in heavy haul service. And heaven help you (more like helpers help you) if the factor of adhesion on these brutes ever falls below 36% because you won't have enough adhesion to pull that 15,000 ton train up that 1% grade, period. You had better hope that it does not rain, frost, or snow.Keep the high HP units in high speed freight service where they do the most good. You are trading 8 axles of weight on two 3000 HP GP40s or 12 axles of weight on two SD40s for the 6 axles of the new units and you have 25-50% less Hp-wasting weight with the two high HP units. Remember that TE decreases as speed increases, so as long as they keep the HP per ton ratio of the trains high enough to maintain high speeds then the TE will be low enough that these high HP single units won't slip. But try to use them in low speed drag service and they will slip as noted in the coal train discussion above. The slower the train goes up a hill the closer these high powered 6,000 HP wonders perform like the good old 3,000 HP SD40.
The Efficiency of the Locomotive
Next we'll look at the efficiency of the locomotives transmission. Their
transmission consists of the generator, traction motors, and gearing. My experience
is that the loco's transmission efficiency normally runs in the 80% range. This
means that if the physics of the train, grade, and speed dictate X HP then you really
need X / .80 HP. If the physics say 12,000 HP then you really need 12,000 /.80 which
is 15,000 HP. Another full unit!
Put another way....The 15,000 ton coal train going up a 1% grade at 10 mph requires 9564 HP. That is 8442 HP for the speed up that grade and 1122 HP for the rolling resistance at that speed. (We'll get to rolling resistance in a minute). But that assumes 100% efficiency. At 80% efficiency this train would need 9564 HP / .80 which is 11,995 HP. SURPRISE! That is three 4,000HP SD70MACs or four 3,000HP SD40-2s to get a 15,000 ton coal train up a 1% grade at 10 mph. Sound familiar?
Put another way....The 15,000 ton coal train going up a 1% grade at 10 mph requires 9564 HP. That is 8442 HP for the speed up that grade and 1122 HP for the rolling resistance at that speed. (We'll get to rolling resistance in a minute). But that assumes 100% efficiency. At 80% efficiency this train would need 9564 HP / .80 which is 11,995 HP. SURPRISE! That is three 4,000HP SD70MACs or four 3,000HP SD40-2s to get a 15,000 ton coal train up a 1% grade at 10 mph. Sound familiar?
The Rolling Resistance of the Train
Now we'll look at rolling
resistance. Assume the same train as above, i.e., 15,000 tons plus 840 tons of locos
(4 SD40-2s) rolling at 10 mph on a 1.0% grade. Using the well known Davis formula
we get the following values: (note: the Davis formula )is explained at the end of this article)
The calculated HP required is 9563 HP. Since our locos are only about 80% efficient this means we need a HP rating of 12,000 to actually deliver the required 9563 HP.Put the train on a 2000 ft long 3 degree curve and you get:
Resistance
|
Pull
|
HP
|
Grade
|
316,800 LBs
|
8447 HP
|
Rolling
|
41,880 LBs
|
1116 HP
|
Total
|
358,680 LBs
|
9563 HP
|
The calculated HP required is 9563 HP. Since our locos are only about 80% efficient this means we need a HP rating of 12,000 to actually deliver the required 9563 HP.Put the train on a 2000 ft long 3 degree curve and you get:
Resistance
|
Pull
|
HP
|
Grade
|
316,800 LBs
|
8447 HP
|
Rolling
|
41,880 LBs
|
1116 HP
|
Curve
|
15,840 LBs
|
422 HP
|
Total
|
374,520 LBs
|
9985 HP
|
Using a loco efficiency of 80% the required 9985 HP becomes 12,481 HP. The 12,000 HP of four SD40s is not going to be able to pull this train up the hill and around the curve at 10 mph. The speed will drop until the rolling resistance and grade HP drops enough that the actual HP required equals 80% of 12,000 HP (9600 Hp). That speed is 9.6 mph.
At 9.6 mph we get the following values:
Resistance
|
Pull
|
HP
|
Grade
|
316,800 LBs
|
8109 HP
|
Rolling
|
41,557 LBs
|
1063 HP
|
Curve
|
15,840 LBs
|
405 HP
|
Total
|
374,197 LBs
|
9577 HP
|
Acceleration
Let’s look at accelerating
trains. The force of acceleration is mass times acceleration. Force = mass x acceleration (F = m * a).
A coal train is a very big mass!
So even small acceleration s need a lot of force. That force adds to the drawbar
pull account of the grade alone and it can break the train in two. If you keep the
acceleration low by notching out one notch at a time and allowing speed to increase
slowly you can minimize the force of acceleration. If you are reckless and try to
accelerate quickly you may end up in two pieces.
Let’s say we want to accelerate a 15,000 ton train
at a rate of 1 mph per minute. In other words we want to be going 1 mph faster at
the end of one minute than we are going now. To accelerate at that rate requires
a steady force of 23,450 lbs. Note that it doesn't matter whether we are going uphill
or on the level. We need to supply an additional 23,450 LBs of drawbar pull to accelerate
at 1 mph per minute. Horsepower is pull times speed. Since the force to accelerate
this train at 1mph/min is a constant, the HP required to accelerate the train varies
according to speed. At 10 mph the HP needed is 625 HP, at 40 mph the HP needed is
2500 HP.
An acceleration of 1 mph/minute is slow. It would take a train 60 minutes to go from 0 to 60 mph. But if we want to accelerate at 10 mph per minute it requires 10 times the force and 10 times the HP at each speed. At an acceleration rate of 10 mph per minute the drawbar force needed is 234,520 lbs. At 10 mph that requires 6,250 HP. At 20 mph it requires 12,500 HP. At 40 mph it requires a whopping 25,000 HP. Note that if we have only 12,000 HP then we run out of HP before we reach 20 mph. We can no longer accelerate at 10 mph per minute and will fall back to lower and lower acceleration rate as speed increases.
Keep in mind that these values of drawbar pull and HP are ONLY for acceleration. You still need to supply the normal pull and HP for any grade and rolling resistance. Let’s look at that. A 15,000 ton train on a 1% grade going 8 mph requires 357,104 lbs of pull and 7617 HP. If we have 4 SD40-2s we have 12,000 HP times 80% efficiency = 9600 HP available. 7620 HP is what you get in throttle 7. We have one more throttle notch and 1980 HP (9600-7620) remaining that we can use for acceleration. At the stated 8 mph that equates to 97,345 lbs of additional drawbar pull available. This additional force will accelerate the train at a rate of almost 4 mph per minute. Yeehaw! Put'em in number 8 throttle and we'll be doing 12 mph at the end of the next minute.
Well not quite. A few problems crop up with that assumption. One is that as the train speed increases so does the horsepower required for both the grade and the rolling resistance. So as the speed begins to increase we have less "left over HP" for acceleration. The rate of acceleration will drop; we cannot maintain that 4 mph / minute rate we started with. In fact when we reach 10 mph all of the loco's HP is being used to pull the train and none is left over for acceleration. The speed will level out at 10 mph and stay there. Ain't physics neat?
The second problem is that you just broke the train in two so you are actually stopped. Why? Because the train traveling at 8 mph required 357,104 lbs of drawbar pull to maintain that speed on this grade. When you opened the throttle from notch 7 to notch 8 to accelerate you just put the additional 92,602 lbs of available loco tractive effort into that same drawbar. 357,104 lbs + 92,602 lbs equals a total of 449,706 lbs. Since drawbars are only good for about 390,000 lbs you just pulled one in two. Moral: When you are moving slowly you'd better handle that throttle gently if you want the train to remain in one piece. Acceleration can break trains in two.
Now for the purists, those 4 SD40-2s are not going to develop that 454,449 lbs of TE. That would mean an adhesion factor of 27% and SD40-2s can rarely if ever achieve that. They would most likely slip. But they can develop the 390,000 lb rating of the drawbars either continuously or by slipping and jerking. So either way the train is going to be in two pieces.
Note that the above train theoretically can go up this hill at 10 mph based upon Hp, efficiency, grade, and drawbar strength. Whether it actually can or not is in doubt. If we rolled onto this hill at a speed higher than 10 mph then all would be OK. As the train rolled onto the grade in number 8 throttle it would simply slow down to 10 mph and proceed up the hill. The drawbar force would be that 357,104 lb figure. Well within the rating of the drawbars. But if the train had stopped on this grade or had started from a stop on a lesser grade and was not yet up to 10 mph then we may be in trouble. Under these circumstances we may find ourselves in the situation above where we are only going 8 mph when the entire train is on the hill. We cannot go from notch 7 to notch 8 because the drawbar force will exceed their rating. Thus we cannot get to 10 mph. The only recourse is to slug it out all the way up the hill in the lower throttle position and a lower speed. It is very annoying to know you have the HP to go faster but you can't use it. If you are a real good engineer and really know what you are doing you can get around this obstacle in some cases. How? By applying some independent brakes to the locomotives drivers. Those brakes will absorb some of the extra HP you get when you go from #7 to #8. Therefore that amount of the extra HP and its attendant TE never reaches the train's drawbars. In that manner you can keep the total drawbar force lower than the drawbar rating. As the speed increases you feather off more and more of the independent brake until finally you are at the 10 mph physical limit and the brake is fully released. But make one mistake during that process; fail to coordinate the independent brake just right with the increasing HP as the locos rev up and increase their load....or feather it off too quickly.......and Kapow! You are in two pieces. You have let enough extra TE reach the drawbars that their rating was exceeded. Going 10 mph in #8 vs. 8 mph in #7 saves you 15 minutes on the hour. But if you break it in two attempting to reach 10 mph then you are delayed 2 hours while you chain up a car and set it out and double the hill. If you are not sure of your expertise maybe it is better to just go up the hill at 8 mph in number 7 instead of trying for 10 mph in number 8.
The main point of all this is to hopefully dispel the myth that high HP means lots of pull. It does not. Higher HP means higher pull at higher speeds but the total maximum pull is strictly related to weight on drivers. No HP required. None! Therefore a switch engine which only operates at low speeds does not need, nor can it use, high HP. It needs to be heavy. (but not too heavy that it breaks or turns over light industrial or yard rails). Life is a compromise.
An acceleration of 1 mph/minute is slow. It would take a train 60 minutes to go from 0 to 60 mph. But if we want to accelerate at 10 mph per minute it requires 10 times the force and 10 times the HP at each speed. At an acceleration rate of 10 mph per minute the drawbar force needed is 234,520 lbs. At 10 mph that requires 6,250 HP. At 20 mph it requires 12,500 HP. At 40 mph it requires a whopping 25,000 HP. Note that if we have only 12,000 HP then we run out of HP before we reach 20 mph. We can no longer accelerate at 10 mph per minute and will fall back to lower and lower acceleration rate as speed increases.
Keep in mind that these values of drawbar pull and HP are ONLY for acceleration. You still need to supply the normal pull and HP for any grade and rolling resistance. Let’s look at that. A 15,000 ton train on a 1% grade going 8 mph requires 357,104 lbs of pull and 7617 HP. If we have 4 SD40-2s we have 12,000 HP times 80% efficiency = 9600 HP available. 7620 HP is what you get in throttle 7. We have one more throttle notch and 1980 HP (9600-7620) remaining that we can use for acceleration. At the stated 8 mph that equates to 97,345 lbs of additional drawbar pull available. This additional force will accelerate the train at a rate of almost 4 mph per minute. Yeehaw! Put'em in number 8 throttle and we'll be doing 12 mph at the end of the next minute.
Well not quite. A few problems crop up with that assumption. One is that as the train speed increases so does the horsepower required for both the grade and the rolling resistance. So as the speed begins to increase we have less "left over HP" for acceleration. The rate of acceleration will drop; we cannot maintain that 4 mph / minute rate we started with. In fact when we reach 10 mph all of the loco's HP is being used to pull the train and none is left over for acceleration. The speed will level out at 10 mph and stay there. Ain't physics neat?
The second problem is that you just broke the train in two so you are actually stopped. Why? Because the train traveling at 8 mph required 357,104 lbs of drawbar pull to maintain that speed on this grade. When you opened the throttle from notch 7 to notch 8 to accelerate you just put the additional 92,602 lbs of available loco tractive effort into that same drawbar. 357,104 lbs + 92,602 lbs equals a total of 449,706 lbs. Since drawbars are only good for about 390,000 lbs you just pulled one in two. Moral: When you are moving slowly you'd better handle that throttle gently if you want the train to remain in one piece. Acceleration can break trains in two.
Now for the purists, those 4 SD40-2s are not going to develop that 454,449 lbs of TE. That would mean an adhesion factor of 27% and SD40-2s can rarely if ever achieve that. They would most likely slip. But they can develop the 390,000 lb rating of the drawbars either continuously or by slipping and jerking. So either way the train is going to be in two pieces.
Note that the above train theoretically can go up this hill at 10 mph based upon Hp, efficiency, grade, and drawbar strength. Whether it actually can or not is in doubt. If we rolled onto this hill at a speed higher than 10 mph then all would be OK. As the train rolled onto the grade in number 8 throttle it would simply slow down to 10 mph and proceed up the hill. The drawbar force would be that 357,104 lb figure. Well within the rating of the drawbars. But if the train had stopped on this grade or had started from a stop on a lesser grade and was not yet up to 10 mph then we may be in trouble. Under these circumstances we may find ourselves in the situation above where we are only going 8 mph when the entire train is on the hill. We cannot go from notch 7 to notch 8 because the drawbar force will exceed their rating. Thus we cannot get to 10 mph. The only recourse is to slug it out all the way up the hill in the lower throttle position and a lower speed. It is very annoying to know you have the HP to go faster but you can't use it. If you are a real good engineer and really know what you are doing you can get around this obstacle in some cases. How? By applying some independent brakes to the locomotives drivers. Those brakes will absorb some of the extra HP you get when you go from #7 to #8. Therefore that amount of the extra HP and its attendant TE never reaches the train's drawbars. In that manner you can keep the total drawbar force lower than the drawbar rating. As the speed increases you feather off more and more of the independent brake until finally you are at the 10 mph physical limit and the brake is fully released. But make one mistake during that process; fail to coordinate the independent brake just right with the increasing HP as the locos rev up and increase their load....or feather it off too quickly.......and Kapow! You are in two pieces. You have let enough extra TE reach the drawbars that their rating was exceeded. Going 10 mph in #8 vs. 8 mph in #7 saves you 15 minutes on the hour. But if you break it in two attempting to reach 10 mph then you are delayed 2 hours while you chain up a car and set it out and double the hill. If you are not sure of your expertise maybe it is better to just go up the hill at 8 mph in number 7 instead of trying for 10 mph in number 8.
The main point of all this is to hopefully dispel the myth that high HP means lots of pull. It does not. Higher HP means higher pull at higher speeds but the total maximum pull is strictly related to weight on drivers. No HP required. None! Therefore a switch engine which only operates at low speeds does not need, nor can it use, high HP. It needs to be heavy. (but not too heavy that it breaks or turns over light industrial or yard rails). Life is a compromise.
The Davis Formula
Substantial research early in the 20th century led to the development of a general formula for train resistance. Developed by W.J. Davis, it is still sometimes referred to as the “Davis” equation.
- Ro = resistance in lbs. per ton
- w = weight per axle (= W/n) (measured in short tons)
- b = an experimental friction coefficient for flanges, shock, etc. (0.045)
- A = cross-sectional area of vehicle (measured in square feet)
- C = drag coefficient based on the shape of the front of the train and other features affecting air turbulence etc. (0.0005)
- V = velocity (feet/second)
- n = number of axles
Definitions
(as used in this document)
·
Axle - Two
wheels and an axle with a traction motor geared to it. All "axles" are
powered, there are no idler axles.
·
Slug - A 4
axle unit that has traction motors but no diesel engine. Its traction motors get
their electrical power from adjacent units. A concrete weight ballasts the slug
to 280,000 lbs.
·
C44- A 6 axle
locomotive of 4,400 HP that weighs 420,000 lbs. Actual model designation is Dash
9-44CW.
·
Grade
Pull or Grade Resistance
- The force required on a grade to prevent a train from rolling back down the hill.
It is expressed as 20 lbs per ton per percent of grade.
·
Adhesion -
The ability of the steel wheels of a locomotive to "stick" to the steel
rails to prevent spinning or sliding of the wheels. The amount of force required
to slide the wheels of a locomotive.
·
Adhesion
Factor - The ratio of the adhesion
to the weight of a locomotive. A good ballpark figure for steel wheels on steel
rails is 30%. IE, it requires a force equal to 30% of the loco's weight to slide
its wheels.
·
Tractive Effort
- The pull developed by a locomotive. The maximum tractive effort value is directly
proportional to the weight on drivers and the adhesion.
·
Horsepower
- Any combination of pull and speed that equals 550 lb-ft per second. Examples:
Pulling with a force of 1 pound for 550 feet and accomplishing that in one second.
Pulling with a force of 550 lbs for 1 foot and accomplishing that in 1 second. Pulling
with a force of 225 lbs for 2 feet and accomplishing it in 1 second. Etc.
If you have an idea for a blog post here for the club, let me know. If I can comment on it, I will or I'll see if someone else can and post it.
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