The key common factor between these physical relationships, is a level
surface, which the LIRR has, by way of its natural geology. In general
practice, railroad grades are kept as low as possible. This sometimes
led to circuitous routes around mountains, or the use of bridges,
tunnels, cuttings, embankments or “switchbacks”. At the other extreme,
some coal railroads in Pennsylvania were powered by gravity.

A look at the specifications and capabilities of some of the very
earliest railway locomotives, and comparisons to modern trucks and
locomotives:

In the formulas used in this piece, the results are expressed in terms
of Power, Speed, Weight and Force. For example, we refer to the formula
HP = PLAN / 33,000, which comes into play a little bit later.

To clarify, Work = (force x distance). However, HP and TP (Tractive Power) are measures of Power = (force x distance) / time.

A less descriptive, but simpler and more versatile (steam, diesel or
electric) method of determining locomotive HP, other than HP =
PLAN/33,000, which only worked for steam locomotives,   is as
per the mathematical relationship between HP, TP and Speed “S” (mph),
described (in the Brotherhood of Locomotive Fireman’s and Enginemen’s
Magazine, Vol XLVI, Jan- June 1909) by the following formula:

HP =(TP x S) / 375

For a brief, but good, explanation of the inter-relationship between
HP, TP, Speed, “Train Resistance”, “Speed Resistance” and “Grade
Resistance”, see the ca 1909 writing cited above, pg 841- 842 here.

It is known*, that the small “Planet” type engines of the period, [at
30 HP power output], produced a Tractive Effort (T.E.) in the range
of  1,450 lbs to 1,550 lbs. on level track. Let’s do the
arithmetic: 1,550 lbs / 8 lbs / Ton = 193.75 Tons total train weight.

Using the equation**

S = (375 x HP) / T.E. (lbs.)

Where:

S = Speed in mph,
HP = Horsepower,
T.E. (lbs) = ( Train Weight (Tons) x 8 lbs/Ton)

And Setting:

HP = 30
T.E. (lbs)  = (193.75 Tons x 8 lbs) /Ton = 1,550 lbs.

S = 7.2 mph

Let’s now speed things up a bit, to 20 mph…and further define Tractive Power (TP):

Tractive Power (TP), also known as Tractive Effort (TE) is only one of
the factors used in calculating the relationships between HP, gross
train weight and speed. Refer back to the formula HP = (TP (lbs) x S) /375.

The required TP in lbs, is the (gross train weight in Tons x Train Resistance in lbs/Ton).

On level track, “Train Resistance“ was cited as 8 lbs/Ton back in 1832,
and was about the same in the ca 1909 writing. The ca 1909 writing
cited basic “Train Resistance”, (or “Friction Resistance” as they
referred to it) as 6 lbs/Ton, but then they added a minimum of an
additional 2lbs/Ton for “Speed Resistance”, bringing us back to a total
“Train Resistance” of 8 lbs/Ton.

By substitution, we get gross train weight

GCW (Tons) = (375 x HP) / (S x 8),
Or,
HP = [GTW (Tons) x 8 lbs/Ton x S] / 375

Tractive Power (TP) of a steam locomotive can also be expressed by the following formula:

Tractive power equals the square of cylin­der diameter, times stroke in
inches, times mean effective pressure per square inch [boiler psi x
0.85], divided by the diameter of the driving wheel in inches. Put in
the shape of a formula this is :

Tractive power in pounds = (d ² x S x P) / D

Where:
d = diameter of cylinder
S = length of stroke in inches,
P = mean effective pressure.
D = diameter of driving wheel.

Source: The Americana Encyclopedia, 1912, Vol 13, pg 51

Now, lets look closer at a key factor in both TP and HP: level track.

As per the the American RR Journal of 1832, once grades are introduced,
the railroad starts to lose its efficiency. This is why railroads had
to be laid on as level right of way as possible. Otherwise, circuitous
routes, tunnels, bridges, cuts, embankments, inclined planes or
switchbacks had to be used. This concept is also restated in the ca
1909 writing.

If you incorporate grades (causes power requirement spiking) on your
railroad route, the HP required increases drastically, by a factor of
[20 lbs for each per cent of grade]. As per the ca 1909 writing,
assuming a grade of 3.6%, we then get the following formula:

HP = (70 x 20 x [20 x 3.6]) / 375 = 268.8

Where:

70 = gross train weight in Tons
20 = Speed in mph
[20 x 3.6] = “Grade Resistance” factor

What this means, is if the maximum grade on your railroad is 3.6%, your
engine must be able to produce 268.8 HP, but only while its pulling the
train over the hill at 20 mph. However, the same engine only needs to
produce 30 HP to pull the same 70 Ton train at 20 mph on level track.

Here’s an electrical analogy. My High School music teacher had a
particular stereo amp that could put out a maximum 200 watts / channel
(1 HP = 746 Watts). It had Watt meters on its speaker outputs. At
normal volume, the amp never put out more than 3 to 5 watts / channel
into the big speakers. However, at the loudest crescendos, it sometimes
momentarily spiked up to 100 + watts / channel. They key thing, is the
amp had to have enough reserve power to get over the momentarily
increased peak power demand requirements, which corresponds to the
grades on a railroad.

To improve efficiency in any system, get rid of the “demand peaks”, and
a lower constant power level will be enough to keep things flowing at a
“high” constant rate.

The following calculations were empirically confirmed
by runs made in 1830 and 1831 on the Liverpool and Manchester Railway.
The original (and most primitive) Planet locomotive (9 Tons) drew a
train of 18 “waggons” (four wheel rail cars) weighing some 80 Tons at
14 mph on level track. According to the formula HP = (80 Ton x 8
lbs/Ton x 14 mph) / 375, the Planet engine was exerting 23.8 HP.  The Planet had only two driving wheels (2-2-0) wheel arrangement.

The equally primitive engine Samson (10 Tons), drew a train of 30
“waggons” weighing 164.5 Tons, at a speed of 20 mph on level track.
According to the same formula, the Samson was exerting 70 HP.
The Samson had 4 driving wheels, of  smaller diameter than the
Planet’s, they were of coupled (0-4-0) wheel arrangement, and also had
larger cylinder bore (larger engine displacement) than the Planet.

The Samson consumed its coke fuel at a rate of slightly less than 1/3 pound/mile/Ton.

Note, the Samson was the same basic machine as the John Bull rebuild/replica locomotive currently on display at the Smithsonian.

Further, the American Railroad Journal of Aug 1, 1842, pg 90, states a
train carrying 1,608 barrels of flour, of 200 Tons weight, was drawn
from Albany to Boston.

Let’s now calculate the HP output of the Samson locomotive, at 20 mph and a train weight of 164.5 Tons.

Using the formula***

HP = (PLAN) / 33,000, where:

P = 0.85 x Boiler Pressure in psi.
L = 2 times the stroke length in feet
A = area of piston in inches sq
N = rpm = revolutions / minute

Plugging in the data****;

P = 0.85 x 60 psi = 51 psi
L = (2 x 16″) / 12″ =  2.6 ft
A = 3.14 x [7″ squared] = 154 inch sq
N = 123 rpm – how did we get this number? 1 mi/hr = 5,280 ft / 60 min =
88 ft/min; 20 mph = 20 x 88 ft/min = 1,760 ft/min; the driver diameter
is 4.5 ft; driver circumference = 3.14 x 4.5 = 14.13ft / 1 revolution;
(1,760 ft/min) / 14.31 ft/rev. = 123 rpm.

We get the following results for the Samson engine:

HP = (51 x 2.6 x 154 x 123) / 33,000 = 76 HP. Lets say 70 HP, to agree with formula 1.

Using the same formula, with the Planet engine’s empirical and specification data; (same pressure= 51, same stroke= 2.6, 11″ diameter piston; radius = diameter/2; A= 3.14 x (5.5 squared)= 95 inch sq;
with 5 ft drivers at 14 mph,  we have 3.14 x 5 ft/rev= 15.7 ft/rev
and 14 x 88 ft/min= 1,232 ft/min,  rpm= (1,232ft/min) / (15.7
ft/rev) = 78.5

Therefore, we get 29.96 HP for the Planet type engine. Lets say 30 HP, to agree with the historical information cited above.

As we can see from the formula HP = (TP x S) / 375, circa 1830’s
locomotives could pull a 70 ton train on level track at 20 mph, with
only 29.8 HP.

Let’s compare the ratios of horsepower to maximum weight, on level
ground, and at 20 mph. Using the same formula above, setting the speed
of the Planet type engine to 20 mph, at 30HP we get a maximum train
weight of 70.3 Tons.

Therefore, for the circa 1830 Planet type railway steam engine, at 20
mph, the horsepower to weight ratio was: 30 HP/70.3 Ton = 0.4267
HP/Ton. For the circa 1831 Samson type locomotive, the horsepower to
weight ratio was 70 HP/164.5 Ton = 0.425 HP/Ton.

Efficiency comparison between Primitive Locomotives and Modern Tractor Trailers (at 20mph Speed)

Compare the above weight ratios to those of a modern highway tractor-
trailer. For example, a typical popular make of truck tractor,
has a “GVW” (gross vehicle weight) of 39 Tons. Its modern diesel engine
produces 440 HP. Its GCW (gross combined weight = tractor + trailer +
load) is 70 Tons. This means the modern highway truck tractor can only
draw less than 2 times (1.79) its own weight.

Compare the tractor- trailer weight ratio numbers to those of the
small, light weight (10 Ton) Samson steam locomotive drawing 16.45
times its own weight, with only 70 HP.  In terms of comparative
“weight only” ratios, the early 19th century railway steam locomotive
was 9.2 times more efficient than today’s highway tractor- trailers
[16.45 Samson locomotive weight ratio / 1.79 tractor- trailer (Mack
“Granite Elite”) weight ratio] !

For the modern tractor- trailer the horsepower to weight ratio is 440 HP/70 Ton = 6.285 HP/Ton.

By dividing 6.285 HP/Ton (tractor- trailer) by 0.4267 HP/Ton (Planet
type steam engine), we find that at 20 mph, in terms of the comparative
horse power to maximum weight ratio, the circa 1830 railway steam
engine is 14.7 times more efficient than a modern tractor- trailer !

To restate our data back into the original circa 1832 terms of
“force per Ton” required to move 1 ton on a railroad, compared  to
the “force per ton” currently required to move 1 ton on a highway, we
perform the following conversions:

Using James Watt’s definition of a horsepower (circa 1783), I derived these conversion factors:

As per Watt, 1 HP = 33,000 (ft x lbsf) / min

Therefore, In the case of the circa 1830 steam locomotives, plugging in data from above, we have:

0.43 Hp / Ton = (33,000 x [0.43 (ft x lbsf)/min]) / (1 Ton/2,000 lbs) =
7.1 lbs / Ton “Starting Resistance”- not accounting for bearing friction

Assuming bearing friction accounts for an additional 11.25 %, we get 7.1 lbs/ Ton x 1.1125 = 7.9 lbs/ Ton, “Starting Resistance” or 1/253 part of the gross combined train weight (locomotive + cars + payload). Note the ca. 1832 empirical values for “Starting Resistance” (train) as cited above, were 8 lbs/ Ton, and 1/248th of the GCW.

For the modern truck tractor (Mack “Granite Elite”), plugging in data from above, we have:

6.285 HP / Ton = (33,000 x [6.285 (ft x lbsf]) / (1 Ton/2,000 lbs) =
103.7 lbs / Ton “Starting Resistance”- not counting bearing friction:

103.7 lb / Ton x 1.1125 = 115.4 lbs / Ton “Starting Resistance”, or 1/17th part of the Gross Combined Weight (GCW)

It appears that at a speed of 20 mph, the efficiency of a standard
highway vehicle has not improved much from the cited circa 1832 value
of 1/12 part of the GCW !!

Therefore, during
the days of primitive steam powered railroads circa 1830’s, moving a
ton by rail at 20 mph was 14.6 times more efficient than moving a ton
at 20 mph by modern truck !!

Efficiency comparison between Primitive Locomotives and Modern Tractor Trailers (at 40mph Speed)

Let’s now consider a speed of 40 mph: Its seems obvious from the formulas, that if you want to double the train speed from 20 mph to 40 mph, you have to double the horsepower, and so forth.

Doubling the speed (and thereby the HP) of the American made “tea
kettle”engines of the 1830’s, from 20 mph to 40 mph, was easy thanks to
Matthias Baldwin and his improved steam fitting joint (ca 1834).

Overnight, boiler pressure in Baldwin’s American made locomotives was
doubled, from 60 psi to 120 psi, thereby doubling the engine HP- and
the possible speed. Source: History of the Baldwin Locomotive Works, 1907, pg 20: . Refer to the formula HP = (PLAN) / 33,000.

Efficiency comparison between Primitive Locomotives and Modern Tractor Trailers (acceleration from 0 to 50 mph Speed)

When moving freight, high speed is not the main priority. Fifty mph is
plenty. Some truckers like to use excessive speed, because they want to
do the most runs in the least time possible, to satisfy their own
personal economic reasons- and they waste lots of fuel and peoples
lives doing it.

For empirical data, tests were done on a British railroad (Grand
Junction Rwy) during 1839. Due to their notoriously level track, the
British commonly achieved both high speed, and a high payload, [30 mph
average (includes starting and stopping time at 8 “stoppages”- about
40- 50 mph peak), 82 Ton gross train weight (GCW), over a 190 mile
distance) using a common locomotive of the period, exerting 87 HP
(using the formula HP = [(GCW (Tons) x 8 lbs/Ton) x S] / 375.

This data yields a steam locomotive power to weight ratio of 87 HP / 82 Ton = 1.06 HP / Ton at 50 mph
Source: Railway Machinery, by Daniel Kinnear Clark,
1855, see pg 11, 17, and the table on pg 20, columns 1 and 7:  The
LIRR, and later the high speed (for the period) Hudson River RR, were
built according to this British design paradigm.

A modern truck will use 440 HP to pull a 70 Ton GCW (6.285 HP/Ton) at about the same speed: nearly 6 times more horsepower is required per Ton by truck, than the circa 1830’s locomotive at 50 mph, as per the formula 6.285 HP/Ton (truck) / 1.06 HP/Ton (locomotive).

Since 1 HP equals approximately 2,545 BTU/hour, in terms of thermal
energy required at 50 mph, the modern tractor- trailer requires
1,119,800 BTU/hour to draw a GCW of 70 Tons. This corresponds to 6.285
HP / Ton x (2,545 BTU / hour) = (15,995.325 BTU / hour) / Ton (truck).

The circa 1830’s steam locomotives required only 189,857 BTU’s/hour
(16%), to draw the same GCW at the same speed. This corresponds to 1.06
HP / Ton x (2,545 BTU / hour) = (2,697.7 BTU / hour) / Ton (steam
locomotive).

In summary, if you keep your railroad
track as close to a zero grade as possible, you never need to produce
more than 30 HP to pull your 70 Ton train at up to 20 mph, or 87 HP to
pull an 82 Ton GCW train at up to 50 mph (1.06 HP / Ton). This
corresponds to 74.6 HP required to draw a 70 Ton GCW train at 50 mph,
(1.06 HP/Ton) as per HP = [(GCW (70 Tons)) x 8 lbs/Ton x 50 mph] / 375.

Modern diesel electric railroad locomotives have a much greater energy efficiency advantage over diesel trucks.

According
to the current AREMA Manual, Volume 1, Chapter 2, pg 55- 57, a typical
diesel electric locomotive will produce 3,000 HP, and the “Starting
Resistance” (our “Train Resistance” of ca. 1830’s) for roller bearing
wheels (above 32° F), is cited as 5 lbs/ Ton. Using our simplified HP
equation, we get:

TP = (3,000 HP x 375) / 50 mph = 22,500 lbs

Then,

GCW (Tons) = (375 x HP) / (S (mph) x 5 lbs/Ton),

GCW (Tons) = (375 x 3,000) / (50 x 5) = 4,500

Therefore, the power to weight ratio of a modern railroad locomotive at 50 mph, is 3,000 HP /  4,500 Ton = 0.666 Hp /Ton (diesel locomotive).
Since 1 HP equals approximately 2,545 BTU/hour, in terms of (energy
consumed per hour) per horsepower, for the modern locomotive, we get:

0.666 HP / Ton x 2,545 BTU / hour = (1,694.97 BTU / hour) / Ton (diesel locomotive), as compared to (15,995.325 BTU / hour) / Ton  (diesel truck).

Therefore, in terms of energy demand at starting and acceleration from
0 to 50 mph, as per the formula [(15,995.325 BTU / hour) / Ton (truck)]
/ [(1,694.97 BTU / hour) / Ton (locomotive)], the
modern diesel electric locomotive is 9.4 (say 10X) times more energy
efficient than a diesel truck or bus (upon starting and initial
acceleration) !  See:  The Next Progressive Era: A Blueprint for Broad Prosperity”, by Phillip Longman, pg 151.

Trucks Vs.Trains :  Energy required to keep objects moving at a constant speed

So far, all of our
calculations have been based strictly on “old school” methods of
calculating energy requirements, which do not differentiate between the
energy needed to start a train or a truck from a dead stop, and the
much lower energy input needed to then keep it moving at a constant
speed.

All the foregoing
calculations have essentially been functions of “Starting Resistance”.
Now, let’s be more specific, and take a look at the energy required to
overcome “Rolling Resistance”, and thereby maintain an already moving
vehicle at a constant speed:

Rolling Resistance- the force needed to keep a rolling vehicle moving at a constant speed:
Once any vehicle starts moving, it takes a lot less force (energy) to
keep it moving at the same rate, than what it required to start it
moving in the first place. To keep things simple, by avoiding the use
of trigonometric functions, we will again assume a level surface in all
instances.

At the same speed,
same load (GCW) and on level ground, any steel wheeled railway vehicle
is 24.6 times more energy efficient than any large rubber tire road
vehicle, regardless of the type of power source.

Let’s see why:

We start off with Newton’s famous second law :

F = ma

setting  a = g = 32 ft/second²

We now have

F = W (weight)  = mg

Coulomb’s classic model of friction is given as:

Ff < µFn

Where Ff  is the force exerted by friction (in the case of equality, the maximum possible magnitude of this force), µ is the coefficient of friction, which is an empirical property of the contacting materials, and Fn is the normal force exerted between the surfaces.

Since in our case the track/road is level,

Normal Force = F = W (weight)

And let

µ = Crr

Therefore, the formula for calculating “Rolling Resistance” is given as:

Ff  (lbs) = [W (lbs)] x Crr

Where:
W = weight (lbs) = “GCW” (lbs)
Crr = coefficient of rolling resistance (dimensionless)
Ff  = Rolling Resistance

Let’s now apply the following data: (for sources see ***** end note)

W = 70 tons (140,000 lbs)
Crr (road truck rubber tires on pavement) = 0.01479
Crr (Railway steel wheels on steel rails) =  0.0006

Therefore, in the case of any large rubber tire road vehicle, no matter what the energy source:

Ff (truck) = 140,000 lbs x 0.01479 Crr (truck) = 2,070.6 lbs / 70 Tons = 29.58 lbs/ Ton, is required to keep any large rubber tire road vehicle moving at a constant speed, no matter what the power source.

In the case of any steel wheel railway vehicle, no matter what the power source:

Ff (Railway) = 140,000 lbs x 0.0006 Crr (Railway) = 84 lbs / 70 Tons = 1.2 lbs/Ton, required to keep any steel wheel railway vehicle moving at a constant speed, no matter what the power source.

Next, we need to derive a multiplying factor:
0.01479 Crr (truck) / 0.0006 Crr (Railway) = 24.65

As
we can easily see from the formula HP = (TP x S) / 375, HP is directly
proportional to TP. As we already know, as per the formula TP = [GCW x
Train Resistance], TP is directly proportional to Train Resistance (TR)
when starting/accelerating, and also directly proportional to Rolling
Resistance TP = (Ff = GCW x Crr)
when already moving at operating speed. As we have seen, Energy Demand
(BTU/hour) is also directly proportional to HP (1 HP = 2,545 BTU/hour).
Therefore, if weight (GCW) and speed are held constant, energy demand is directly proportional to Crr.

Therefore,
any steel wheel railway vehicle is 24.65 times more energy efficient
than any large rubber tire road vehicle, no matter what the power
source is, as long as speed and GCW are held constant.

So, why is rail so much more energy efficient than pneumatic
tire road vehicles? Let’s refer back to Engineering Tribology, By John Austin Williams, 2005, pg 409, and the equation for µR:

Referencing
the formula directly above, “the rolling resistance of a pneumatic tire
road wheel is very much greater than that of a steel wheel on a steel
rail, because of the very much lower value of the “contact modulus” E
[elastic stiffness] of a rubber tire on a concrete road, as well as its
much greater value of “fractional hysteretic energy loss” α [the energy loss expressed as a fraction of the total input energy], when compared to those of a steel wheel on a steel rail”.

Now lets calculate the relative amount of energy required by any steel
wheel rail vehicle Vs. any large rubber tire road vehicle, as long as
speed and GCW are held constant:

[(1.2 lbs/Ton (rail) / 29.58 lbs/Ton (truck)] x 100 =  4.06%

The steel wheel rail
vehicle will require less than 5% (1/20) the energy required by a
rubber tire road vehicle, to do the same amount of work.
However, this calculation does not include bearing friction, grade and curve friction, or aerodynamic friction.

If we deduct 1/5,
or  20% of our multiplying factor of 24.55 to account for the
other forms of mechanical friction, rail still enjoys precisely the
same 20X energy advantage over highway vehicles that it had back in
1832 !

CONCLUSION:

At
the same constant speed, on level ground, drawing the same load, any
steel wheeled railway vehicle already in motion, will use only 5%
(1/20) of the energy consumed by any large pneumatic tire road vehicle
already in motion. Upon starting and initial acceleration, any steel
wheeled railway vehicle will only use 10% (1/10) of the energy demanded
by any large pneumatic tire road vehicle. Further, only in the case of
railroads, Train Resistance, or Rolling Resistance, is inversely
proportional to GCW (train weight). This means, the heavier the train,
the more energy efficient it becomes.

As
a nod to the Electric Automobile industry, its noteworthy that
theoretically, the energy efficiency (range) of any electric automobile
on any paved asphalt or concrete road, can be increased up to 2X
(doubled), through improved tire design (i.e., by using “special
pneumatic” or “special solid” rubber tires, or a hybrid of the two).

For example, a modern light weight automobile with small “footprint”
pneumatic tires, has a rolling resistance of about 20 lbsf/ Ton on
pavement. Back in 1909, they had the standard rolling resistance of an
electric car down to 15 lbsf/ Ton on asphalt.  With our modern
materials, it could come down still lower.

This
information has been publicly
available for over a hundred years (since at least 1909).
However,  historical events in the early 20th century led to an
abundant, plentiful and seemingly inexhaustable oil supply.  The
British, in dire need of an oil supply to fuel their Navy, discovered
oil in the Persian Gulf, and by 1911, (then) cheap oil was being pumped
out of Iran. After World War I, the Standard Oil Company of California
followed suit, and began pumping vast quanities of oil out of the
politically unstable countries all around the Persian Gulf.
Accordingly, anything electrically powered or “energy efficient” was
immediately relegated to the scrap heap of history- along with all the
scientific know- how and technology that went with it.

See Standard Handbook for Electrical Engineers, by Frank F. Fowl, 1916, pg 1,461  (you have to download the entire PDF to view this page).  Also see: Electric Traction, by E.H. Armstrong, 1909, pg 807- 808: And Alexander Churchward’s original 1909 paper on the Energy Consumption of Commercial Vehicles
(rubber tires:  pneumatic, solid and vehicle resistance),
presented before the SAE.  See Norton’s (of B.F. Goodrich Tire
Co.) circa 1916 paper on Tires for Electric Vehicles,
presented before the Electric Light Institute, on pg 96- 113.  If
you can find it, also see the paper Electric Vehicle Tires, presented
before the Electric Vehicle Association of America, by F. E. Whitney,
Oct. 27, 1913

* Steam Passenger Locomotives, by Brian Hollingsworth, 1982, pg 20- 22
** Source: Locomotive Fireman and Engineers Magazine, 1909, pg 841.
*** Source: Railway and Locomotive Engineering, Dec. 1907, pg 548.
**** Source: English Mechanic and World of Science, Oct. 18, 1889, pg 158
***** Note: Truck tire Crr is the
average of data from SAE Technical Paper 880584, 1988, pg 4, Table 6.
The average Railway Crr is from these sources: Engineering Tribology By John Austin Williams, 2005, pg 409- 410, and Bicycling Science By David Gordon Wilson, 2004, pgs 217 & 218, and Tractive
Resistance of Rolling- Stock, by J.L. Koffman, British Railways Board,
Railway Gazette International, Vol. 120, Nov. 1964, pg 899- 902. .