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- 3.1 Is Rail Freight Usually More Energy-Efficient than Truck?
- 3.2 Passenger Rail Not So Efficient; Weight Problem

- 6.1 Introduction
- 6.2 Oversimplified Formula (to be modified later on)
- 6.3 Account for Empty Car Return (Backhaul)
- 6.4 Account for Variation of gross EI by Commodity
- 6.5 Circuity and Grades
- 6.6 Grades

- 7.1 Engine-Transmission Efficiencies
- 7.2 Introduction to Resistance Forces
- 7.3 Components of Resistance, Braking

- 8.1 Why it's of primary importance
- 8.2 Values
- 8.3 Examples Hauling Heavy Bulk Commodities (like Coal)

- 9.1 Rolling Resistance
- 9.2 Railroad Resistance References
- 9.3 Train Weights
- 9.4 Truck Weights
- 9.5 Comparing Truck and Train Gross-to-Net Weights

Copyright 2010 by David S. Lawyer. Feel free to make copies but commercial use of it is prohibited. For example, you can't (except to an insignificant degree) combine it with advertising on the Internet. Please let me know of any errors or suggestions for improvement.

This article is still "in progress" but it's complete enough to be useful. At the end of April, 2007 I discovered a serious error in the Rose Study (equation (5) on p. 5-10) which finds energy-intensity by commodity for rail. This error results in Rose reporting a wider variation energy-intensities for rail than was actually the case. So for lightweight goods it made rail look worse than it was and for heavyweight goods it made rail appear better than it actually was. I've discussed this error in his study and corrected equation (5). But correcting the Rose article by recalculation using a corrected formula hasn't been done since it's not so easy to do. One would need to estimate data (Rose omitted some data he used such as assumed train speed), and calculate corrected values of rail energy-intensity for 1972. The results would mainly be of historical interest since the railroad energy situation today has changed due to the higher aerodynamic drag and higher tare weights of container trains (intermodal).

This Rose error also likely invalidates the Vanek Study since Vanek's 1972 results are about the same as Rose's study implying that they may have both used the same erroneous methodology and/or data. But there are also other likely errors in the Vanek study.

In early 2010 I started making comparisons based on vehicle resistance. I've barely started on a comparison of truck with intermodal rail (containers on flat cars, etc.). I've also been improving the clarity and removed some erroneous statements I had previously made regarding the effect of grades.

Perhaps it is but at present no one really knows for sure. Which mode (rail or truck) is more efficient strongly depends on what commodity is being hauled. Some commodities are more energy efficient to haul by rail, while for other commodities it's more efficient to haul them by truck.

Due to the low rolling resistance of the steel wheel on a steel rail, one might expect that hauling freight by railroad would usually be a lot more energy efficient than hauling it by truck. Aerodynamic drag is often less too, since each railroad car often partially shields the car behind it from the headwind. An exception to this is container trains where there are big gaps between the containers. If such gaps are wide enough it may result in high aerodynamic drag roughly as bad as for trucks and it gets much worse if there is a natural side wind.

My article Train vs. Auto Energy contains much information that also is applicable to freight trains. See it for the bibliography regarding train resistance (forces opposing the forward motion of a freight train).

But alas, passengers rail isn't much, if any, more energy efficient than the auto. See Train vs. Auto Energy, and Does Mass Transit Save Energy?

A major reason why a railroad train is usually a lot more efficient for hauling freight than passengers is due to weight and space requirements. Hauling live people takes up a lot of interior space so that people can move about and have space to read, walk, etc. But freight just requires the space to store the freight. For bulk commodities like coal, there is little air space between the coal lumps.

The weight of the passengers in a train is often only a few percent of the total weight of the train so the heavy rail cars and locomotives exact a high toll in energy consumption. In contrast to this, for high density freight, the weight of the freight may constitute roughly half the weight of the train. So the heavy rail cars don't accrue the high weight penalty that they do for passenger rail.

Other reasons why passenger trains use much more energy are:

- Passenger train cars have almost double the specific rolling resistance of freight cars due to the lower weight of the "cargo" (passengers) which result in lower total weight.
- Passenger trains travels at higher speeds than freight and a doubling of speed will quadruple the aerodynamic drag.
- Passenger train cars are heated or cooled (air conditioned) and lighted inside (with electric lights) while freight cars are usually not.

The energy-efficiency of both rail and truck transportation is very much dependent on the commodity hauled. Low density goods, similar to people in trains, are not very energy-efficient to haul if we are measuring such efficiency in say ton-miles per gallon (or tonne-kilometers per liter). There are many sites on the Internet that mislead the viewers by claiming that rail freight is a few times (or even several times) more energy-efficient than truck. While it's true that the estimates show rail to be much more energy efficient overall, such a comparison is misleading since rail and trucks haul quite different good. Rail is mostly hauling the heavy goods that are inherently more efficient to haul, while the trucks are mostly carrying the lighter goods that are inherently less efficient to haul. Thus comparing rail hauling heavy goods to trucks hauling lighter goods is like comparing apples to oranges. A fairer comparison would compare rail and trucks for each commodity.

Many people today see rail cars hauling consumer goods in containers but less seen are the trains hauling of huge amounts of heavy coal and other bulk commodities (such as ores, minerals, sand and gravel, etc.). This is because the trains carrying consumer goods move in urban areas where many people see them, while mineral-hauling trains often operate in more remote areas where fewer people see them. Thus the public may erroneously think that rail hauls mostly consumer goods and that they do this much more efficiently than trucks.

There's the old saw: Which weighs more, a pound of feathers or a pound of lead? Of course they weigh the same but the pound of feathers will occupy much more volume and thus take much more energy to transport regardless of whether it goes by truck or rail. The larger volume of low density goods requires a larger/longer vehicle (or more railroad cars/trucks) to haul the same tonnage of goods. Such a larger vehicle (or more vehicles) will both weigh more and have more air resistance (aerodynamic drag). Thus for very low density goods, hauling them by truck might be more energy efficient than by rail. However, for higher density goods (commodities), rail should be significantly more energy-efficient than trucks.

Thus one should not merely advocate that we should transport as much by rail freight as possible to save energy because in certain situations it would take more energy to transport by rail, especially for low density good where the route by rail is up steep grades and/or significantly longer than the route by truck. Furthermore, unless the rail tracks provide service directly from the shipper to the receiver, trucks will likely be needed to get the goods to the rail station for shipment (except for some cases of ocean transport) and to pick up the goods from the terminating rail station. This also takes energy and results in delay, and delays incur an energy as well as a cost penalty. Thus, access-energy considerations will make it more energy-efficient in some cases to ship via truck.

In transportation, a major potential means of improving energy efficiency is to shift certain freight transportation from truck to rail, especially for higher density goods. But doing this requires better rail service and in many cases an expanded rail network. Both new rail technology and new organizational structures are needed with the later the most difficult to devise and implement. See my articles on rail modernization proposals which are somewhat outdated but still basically valid: Ideas for Freight Railroad Modernization and Railroads, Institutional Issues

We've already discussed the qualitative aspect where very low density commodities are best shipped by truck to save energy, but where is the data that shows this? In the few studies I've seen (except for the erroneous Rose Study) rail is always reported as more energy-efficient than truck but it "ain't necessarily so".

One reason for this is as follows: In the published studies energy-intensity has been estimated for certain broad commodity groups (or classes). It always seems to show that rail is more energy-efficient (at least after my correcting for errors; see Calculation of Energy-Intensity by Commodity) The problem with only looking at these broad commodity groups is that there is insufficient disaggregation. That is, a commodity group may include both low and high density goods. We see results for a commodity group as a whole but not for the low or high density goods within that commodity group. Such lack of disaggregation is present in both Rose Study and the Vanek Study. Of course it's easy to criticize for not disaggregating more, but it's not so easy to disaggregate more since the data needed may not be available.

One example of misleading results due to failure to disaggregate is for the commodity class of Electrical Machinery". This class included both heavy electric motors and light consumer electronics. So railroads would likely haul mostly the heavy electrical machinery such as electric motors and transformers which they can do most efficiently. Lightweight types of "electrical machinery" such as consumer electronics, would go mostly by truck, where the truck can haul it at a reasonable energy-efficiency due to the lightness of the truck as compared to the much heavier rail cars. If rail had to haul this lightweight consumer electronics in heavy rail cars, the energy-intensity might be little if any better than hauling it by truck and would be much worse than for the heavy electric motors, etc. that rail actually hauls.

Thus the reported statistics on energy efficiency for "electrical machinery" would strongly favor rail. Someone might erroneously reason that since consumer electronics is "electrical machinery" it would be significantly more energy-efficient to haul it by rail. But unfortunately if rail had to haul the lightweight consumer electronics in heavy rail cars, it might well be less energy efficient than trucks. This type of situation is present for most other commodity classes besides just "electrical machinery".

The rail mode is more likely to attract the heavy goods within a commodity class while the truck mode is more likely to attract the lower density goods within that same commodity class. One reason is that rail tends to attract the goods where there will be significant savings in energy, and that's the heavy, high-density goods. Just look at the reported energy-intensities of such studies and you'll see that for heavy commodity classes like coal, rail is much more energy-efficient than truck. See Estimated Energy Intensities Another reason why rail attracts such heavier goods is that such goods often become obsolete more slowly and thus can tolerate the slower speed of shipment by rail (as compared to trucks).

One commodity class not effected much by this lack-of-disaggregation problem is coal since it's all roughly all the same density. Thus rail is much more energy-efficient than truck for coal shipments and has most of the coal market share.

This method is based on use of statistical data combined with use of some engineering considerations due to lack of detailed statistical data. The method presented is basically that used by Rose. But Rose made a serious error in his equation (5) p. 5-10 resulting in erroneous results. So a correct method will be presented here, with a more detailed explanation than in the Rose report (or study). It's not valid for intermodal container shipments and is presented as a correction to the Rose formula for which the ignoring of containers was not unreasonable for data from the 1970's.

Here's the gist of the method. Definitions: the net weight of a transported commodity is just its weight while the gross weight is the combined weight of the commodity plus the weight of the portion of the train used to haul that commodity. Assume that from statistics we know the overall energy intensity (EI) per gross ton-mile (for example BTU/ton-mi) for the railroads (not differentiated by commodity). This is just the fuel consumed in a year (in heat terms such as BTU) divided by the gross ton-miles of trains (locomotives, cars, and contents). Then to find the energy intensity for commodity i, we just estimate the ratio of gross tons to net tons for trains (or cars) hauling that commodity i and multiply this ratio by the given overall energy intensity (gross) to get the energy intensity (net) for commodity i.

This seems simple enough until we begin to wonder if the EI per gross ton-mi is the same for all commodities. It turns out that it isn't. So then we derive a correction factor to correct the result. If we were trying to find the EI for a certain rail route for a certain commodity, this method might not be the best.

We will now derive an equation for the energy-intensity, EIi (in
BTU/ton-mi) for commodity i similar to the erroneous equation (5)
by Rose. Lower case letters may be considered subscripts. Next, we
must define "net" and "gross" weight. The net weight is just the
weight of the commodity i in tons (including any packaging) in a
railroad car: Wci. The gross weight for a carload of freight is the
total weight (in tons) of the car which is the sum of the commodity
weight Wci and the empty car weight Wei. The commodity weight Wci
is assumed to be a typical carload weight of that commodity and
lower than maximum load the car can carry. Also included in the
gross weight is the amount of locomotive weight needed to haul the
car. This is assumed to be equal to K2 x gross-car-weight where the
constant K2 is about 0.11 (we increase gross car weight by about
11% to account for that part of the locomotive weight necessary to
haul the car). Then the effective gross weight of a carload of the
i th commodity is:

*(Wei + Wci)(1 + K2)*

This gross weight per car may be divided by net weight Wci to
get the gross to net weight ratio for commodity i (G2Ni). Then,
since the energy intensity (EI) per gross ton-mi (EIgtm in
BTU/ton-mi) for rail freight as a whole was known (from collected
statistics) and if it's assumed that it doesn't vary much with
commodity, one could just multiply it by the ratio of gross-to-net
for commodity i (G2Ni) and seemingly obtain the energy-intensity
for commodity i (EIi). The formula for this is:

*EIi = G2Ni x EIgtm* where *G2Ni = (Wei + Wci)(1 +
K2)/Wci*. For example if overall, rails use 400 BTU/ton-mi
gross (EIgtm) and the ratio of gross-to-net (G2Ni) is 2, then the
energy intensity of commodity i is 800 BTU/ton-mi net. Note that
the gross-to-net ratio of car weight (including locomotives) is the
same as the ratio of gross ton-mi to net ton-mi for that
commodity.

The oversimplified equation for EIi (substitute for G2Ni) is
thus:

*EIi = EIgtm[(Wei + Wci)(1 + K2)]/Wci
(1)*

To correct this oversimplified formula requires a few major
modifications which will:

- Account for the energy used for the return of empty cars
- Account for the variation of BTU/ton-mi (gross) with commodity
- Account for the fact that rail routes don't go in straight lines between origins and destinations

It turns out that a high percentage (70% to 100% depending on
commodity) of rail freight car miles are for empty freight cars. So
let K1i be for commodity i, the ratio of empty to loaded car-miles
(perhaps 0.8). Then for a loaded rail car we must add weight Wei x
K1i to account for the return trip of the empty car. And of course
this must the multiplied by (1 + K2) to account for the locomotive
weight needed to haul the empty car. So thus accounting for empty
car mileage the EIi (for commodity i)
becomes:

*EIi = EIgtm[Wei(1 + K1i) + Wci](1 +
K2)/Wci (2)*

This is expressed OK as part of equation (5) of Rose.

The assumption that it takes the same amount of energy to move a gross ton-mile of freight, regardless of commodity is wrong. To see why, let's look at two trains, each one transporting only one commodity and each weighing exactly the same (gross weight) and moving at the same speed side by side in the same direction on the same route. Assume that one commodity is heavy like say coal and the other light, like plastic toys. The coal is much more dense than the plastic toys (with a lot of air spaces) and thus each coal car will carry a much heavier load of coal than a car of the other train has of toys. Because the trains weigh the same it will take many more cars of toys to equal the weight of the coal train. Thus the train hauling toys will be much longer than the coal train and have much greater aerodynamic drag (air resistance).

At the same time, the coal train with heavy cars will have heavier axle loads and thus have less rolling resistance per ton. See Economies of Scale of Train Rolling Resistance. This is accounted for by a n term in the "Davis" formula for train resistance where n is the number of axles. See Davis formula and equations (1) and (3) of the Rose report.

So we are now faced with the problem of correcting the erroneous energy-intensity of commodity i (EIi) to account for this phenomena. The equation (5) in the Rose report attempts to include such a correction, but does it incorrectly. What we need is a factor to multiply the simplistic EIi by, so as to account for the variation in EI by commodity. But before showing how to do this we must first explain the Davis formula.

The Davis equation (or formula) finds the resistance force on a
train given its weight and velocity. Outside of the US (Such as
Russia), there are various similar formulas. The Davis formula is
normally just for a single car within a train where the aerodynamic
forces are low due to the train car ahead shielding the car behind
it from the headwind. It's:

*R = W[ 1.3 + 29n/W + bv + (cAv^2)/W]
(3)*

where R is the resistance in pounds. W is the gross car weight in
tons, n=4 is the number of axles, v is the velocity in mi/hr,
b=.045 and c=.0005 are constants, and A=125 ft^2 is the
cross-sectional area of the car. The first 3 terms are the rolling
resistance and the last term is the aerodynamic drag, proportional
to the square of the velocity v. Note in this formula there are no
subscripts. If we divide R by W we obtain the force per unit
weight, r, in pounds per ton. A unit for specific force is parts
per thousand (pounds per thousand-pounds = kilogram-force per
tonne, etc.). This is known as the specific resistance r.

Does this Davis-formula specific resistance r depend on commodity? Of course it does since the loaded car weight W depends on commodity. The two terms of the specific resistance that depend on W are 29n/W and (cAv^2/W). The 29n/W gets lower as W increases and represents an economy of scale in rolling resistance. The (cAv^2)/W term also decreases as W increases and shows that since aerodynamic drag doesn't change as one makes the car heavier (but keeps the exterior dimensions constant), the specific resistance r decreases since more is being transported for the same aerodynamic drag. So we can find the specific resistance r for a certain commodity and compare it to r for the weight W of the average loaded car.

More precisely, we can find the ratio of these two r's where
such ratio for a lightweight commodity might be 1.2. Then multiply
the EIgtm by that ratio to obtain a corrected gross EI for that
commodity (EIgtmi). We might call this 1.2 ratio the "commodity
correction factor" for the overall gross EI. For heavy commodities
it will be less than one and conversely. A way to express the
correction factor in terms of total car resistance Ri for the i th
commodity is

*[Ri/(Wci + Wei)][(Wc + We)/R)
(4)*

where We is the average car weight, Wc the average commodity weight
in a car and R the average resistance. The "average" is taken over
all commodities and it should be weighted over car-miles. The
"average" values may be obtained by dividing system ton-mi by
system car-mi obtained from gathered statistics.

Imagine two trains, one with low density goods and another with high density goods making the same trip on the same route where they both weigh the same. An alternative but equivalent scenario is where there is one heavy train and two lightweight ones, each lightweight one weighing half that of the heavy train. This is more realistic since on single track line, the length of trains is limited to the length of passing sidings. So when we mention a low density train, we may actually mean two low density trains. What is the difference in the train resistance. The lighter trains is longer and thus has both more aerodynamic drag (the cAv^2/W term) and also weighs less per car (the 29n/W term). Both these terms are larger since they are inversely proportional to W which is smaller. Thus the lower density train has higher train resistance even though it weighs exactly the same as the higher density train with fewer but heavier cars.

There are a few major considerations neglected above. One is the empty backhaul. Another is the energy to move the locomotive weight. Still another is that the energy used for moving a train is not proportional to the train resistance since a large amount of energy is expended in braking trains. Trains brake a lot since most lines in the US are single track and trains must frequently brake to a stop to allow other trains (usually moving in the opposite direction) to pass them. A train moving at 50 mi/hr has enough (ref id="ke_" name="kinetic energy to coast"> about 10 miles. Trains must also brake on downgrades and trains must brake even on gently downgrades as low as 0.2 %. This lost braking energy should be roughly proportional to train weight. The needed corrections to the correct correction factor will be made later on, but first let's look at the erroneous correction factor used in the Rose Study.

Rose used the correction factor *(Ri/Wci)(Wc/R)* while
the mine (oversimplifies) was: *[Ri/(Wci + Wei)][(Wc +
We)/R)*

where We is the average empty car weight, Wc the average weight of
the commodity in the car. So Rose is using net commodity weights:
Wci where I am using gross weights: Wci + Wei. Ditto for the
average weights (no subscript i). Ri/Wci (and R/Wc) are the
resistances per net ton of cargo. So it's erroneous to use them
since we want to correct the gross EI, in BTU/ton-mi gross, not the
net EI.

But the Rose correction factor does have a possible use provided one assumes the energy used is directly proportional to car resistance per the Davis formula and that there is no backhaul (both assumptions being invalid). Just take the net EI for railroad freight overall, a figure that is widely available since EI normally means net EI, and then multiply it by Rose's correction factor for commodity i to get the EI for that commodity. But erroneously, Rose multiplies his correction factor by the overall gross EI (see equation (2)) and not by the overall net EI for the railroads. The EIi found from equation (2) has already been adjusted (although not fully) for the commodity i based on the extra tare weight etc. needed to transport commodity i so Rose is sort of double counting since both his correction factor and equation (2) have added on the tare weight to the commodity weight, etc. In other words he is making a correction that has already been made (somewhat like "double counting"). Note that the tare weights in the correction factor occur in the Davis train resistance terms R and Ri and there is no cancellation because Rose divides these by net weights Wc and Wci (not by the gross weights (Wc + We) and (Wci + Wei) as he should have although even this result is oversimplified and need further correction..

Also, Rose failed to account for backhauls and braking energy in his correction factor similar to that explained in Why above correction factor (4) is wrong. Next we'll revise correction factor (4).

To account for empty backhaul and locomotive resistance replace
Wei in equation (4) by Wei(1 + K1i).
Also multiply the gross car weight (including backhaul) by (1 +K2i)
to account for locomotive weight. Replace Ri by Ri + (K1i x R1i) +
(K2i x R2) where R1i is the car resistance on the backhaul and R2
is the locomotive resistance (it may depend on what commodity i is
being hauled. The resulting commodity correction factor is:

*[ {Ri + (K1i x R1i) +(K2i x R2i) } / { Wci + Wei(1 + K1i) } (1
+ K2i) ] [ {Wc + We (1 + K1) } (1 + K2) / { R + (K1 x R1) + (K2 x
R2i) } ] (5)*

Accounting for the fact that much of the energy used for freight trains is used for braking isn't too hard to do but the results are only rough estimates due to lack of accurate data. The first step is to estimate how much energy is used for braking on the railroads. One can do this by converting gross EI to train resistance. In fact, gross EI in BTU/ton-mi is actually a like dimensionless ratio since both the numerator and denominator represent work (or energy). It's something like a coefficient of friction. For example, if it took 2kg-m of work to move 1000kg of mass one meter, then EI = 0.2% and to keep a mass moving on level ground requires a force equal to only 0.2% of its mass. It like it was sliding (on a greased surface) with a coefficient of friction of only 0.002. Of course if the mass units are in kg, then a force of y kg is actually means a force of gy, where g is the acceleration of gravity.

To roughly estimate this "coefficient of friction" ratio, convert BTU to mechanical work using a 20% average diesel-electric locomotive efficiency as found by the Russians. Then consider tons to be a weight (force) and use conversion factors to cancel out units to get a dimensionless ratio of 0.4%. Since prior to the large number of container trains, aerodynamic drag was fairly low and rolling resistance is only a little over 0.1% it might be reasonable to estimate a resistance of about 0.2% on level ground (included aerodynamic drag and poorly maintained track). This implies that roughly half of the energy expended by trains goes into braking, since then braking would be equivalent to a 0.2% train resistance force in order to get the 0.4% figure. Actually the braking energy will include the energy used idling locomotive engines and it seems reasonable to allocate this on a tonnage basis.

To account for this one can find the resistance for a typical car and then modify the Davis formula by increasing the first term so that it will account for braking energy. Note that this first term represents the energy loss directly proportional to weight and is normally just the fixed component of rolling resistance. So we are just increasing the rolling resistance to account for braking losses.

Braking losses will be proportional to weight for downgrade braking, since it involves converting potential energy to heat in the brakes due to the elevation of the train at the top of a downgrade. But for braking to a stop to allow other trains to pass it will be roughly proportional to the weight times the square of the speed from which the train started braking. It can be shown that prior to such braking, it's optimal to coast for a while to utilize the kinetic energy of the train so that less energy will be wasted in braking since it starts from a slower speed. But for lightweight commodities that often move at higher speeds than bulk commodities, the speeds at the start of braking will likely be higher and this may need to be somehow accounted for.

Thus weight plays a far greater role in efficiency than the Davis equation for level ground implies (as used by Rose). How does this distort Rose's results? Well, the Davis formula is only used in the correction factor and the correction factor takes into account that lighter commodities have more aerodynamic drag and more specific rolling resistance. Without these 2 factors, the correction factor would be 1 (wouldn't be needed at all). So as weight takes a more significant role, the aero drag and the higher specific rolling resistance due to light rail wheel loads become of relative less significance. This tends to drive the correction factor closer to unity.

Railroads and highways sometimes parallel each other but often don't. Usually, the railroad distance between any two locations is longer. Another way of saying this is that rail has greater circuity than the highway routes used by trucks. The actual route distance that one travels by road or rail divided by the straight line distance is known as "circuity". Straight line distance is not actually straight line since the earth is round so it's more correctly called "great-circle" distance.

So is it fair to compare the energy-efficiencies using great-circle distance? Since for a given great-circle distance the rail route is apt to be longer than truck, the rail energy-efficiency suffers a penalty as compared to using actual route distance for the comparison. Well, the higher circuity of rail is a fact of life and since few new rail lines are being built, it's not likely to change. But even if we were to compare rail to road for the case where both the road and railroad were newly built, the circuity for rail would likely be significantly greater.

This is because rail freight cars and locomotives are significantly heavier than trucks and they incur a high energy penalty for crossing over mountains or hills. The rolling resistance of a rail wheel (at best) is only about 1 part in a thousand. Thus, to haul a rail car up one mile in elevation (5280 ft.) requires about the same energy it would take to haul it a thousand miles on the level (at low speed). For a truck to gain a mile in elevation, it uses about the same energy as it would take for it to go about 150 miles on the level at low speed. But highway routes are not designed just to optimize truck shipments but to provide for fast travel by automobiles. Since (overpowered) autos can move uphill almost as fast as they move on the level, highways routes are often selected based more on the shortest distance than on considerations of elevation gain and energy-efficiency.

Thus the longer rail lines do often have an advantage for rail: less energy consumption due to rails routes going around mountains rather than over them. Also in some case the railroad goes through a tunnel which is at a significantly lower elevation that the mountain pass that the truck has to go through. So even with the circuity penalty assessed to rail, this savings in energy may (in some cases) more than compensates for the circuity penalty. This penalty is often a small price to pay for saving the energy to overcome the steep grades which would exist if rail routes followed truck highway routes. In other cases however, both rail and truck have to climb over the same mountain passes so then there isn't much savings for rail due to the circuity. Steep rail grades also incur the penalty of having to use more heavy locomotive units to haul the trains up the grade.

Thus using great-circle distances for comparing rail and truck energy efficiency is generally fair (perhaps even biased a little in favor of rail). The Rose Study included circuity but the Vanek Study didn't. The statistical fuel use figures used reflect the advantages of rail lines having easier grades, so no numerical correction is needed in this regard.

One might wonder why it is that in these calculations there has been no inclusion of grades in the formulas nor any explicit accounting for the energy used for trains to ascend grades A lightweight commodity being hauled up a grade will have a high energy intensity EIi and there is no explicit calculation of this. But it is actually implicitly included.

The major way it's included is that we have started the calculations with the historical EI per gross ton-mile (EIgtm). The total fuel used in obtaining this EIgtm includes fuel used to climb grades. But what about lightweight commodities? Well, they will have a much higher gross-to-net weight which is the major part of (see equation (2)) and EIi (per net ton-mi) is directly proportional to this gross-to-net value.

For finding the correction factor, equation (5), the Davis formula is used, to which a braking energy loss term has been added. This term implicitly includes the energy used in climbing grades because: While some of the grade energy is utilized for descending downgrades and thus is not wasted, the rest is simply dissipated in braking. Thus the cost of climbing grades is fully covered by energy wasted in braking. We haven't double counter either since the effect of the braking energy terms in the correction factor tends to make the correction factor less significant (closer to unity).

Note that the presented formulas (and Rose's formulas) are for finding the average energy intensity (EI) for each commodity. Doing this implicitly assumes that upgrades and stops are the same for all commodities. For example, it's implicitly assumed that certain commodities are not transported more over mountain passes than other commodities.

But suppose we want to find the EI for hauling a low density commodity over a certain route with steep grades. One could do this directly by using the Davis formula, taking into account factors such as empty car mileage, idling, yard switching, excessive consumption due to various defects, etc. Or one could use the method presented here but find a correction factor based on the actual grades. To account for steep grades, one would greatly increase the first term in the Davis formula used to derive the correction factor. This would result in a high correction factor. For example, instead of 1.2 it might turn out to be 5. This would make the EI roughly 4 times larger. Since this factor is multiplied by the high gross-to-net weight of the low density commodity, which might be double that of the typical commodity, then the resulting EI is about (4 x 2) 8 times the usual value. Thus grades significantly increase the energy intensity (EI) if a particular shipment goes up such grades and for a low density commodity, the increase in EI is much larger.

Both rail and trucks use diesel motors, except for electrified railroads which are insignificant for freight in the US. Both diesel trucks and diesel locomotives are of about the same thermal efficiency. While it's widely claimed that diesel engines are about 40% efficient this is only true for a new engine running at the ideal operating point (rpm and torque) and neglects transmission losses. One might think that railroad diesel-electric locomotives would be more energy-efficient in their motor/transmissions than trucks. While the locomotive diesel motor is more efficient that the truck motor since the locomotive motor is larger, the fact that the electric transmission is less efficient than the gear transmission of trucks tends to make the overall efficiency roughly equal.ate for this (provided the truck driver shifts gears properly).

When engine efficiencies are the same for two vehicles, then one can compare energy efficiencies by just comparing specific vehicle resistances. For example, if a slow moving train has 3 pounds of resistance per gross ton of weight (3 parts per thousand) and a truck has 12 pounds per gross ton, then this train is 4 times (12/3) more energy efficient in hauling a gross ton (on level ground at constant speed). The "gross tons" of a vehicle is just its total weight" the sum of the weight of the empty vehicle (including locomotives or truck tractors) plus the freight (or cargo) it's carrying. Net tons is just the weight of the freight (or cargo) alone.

On level ground the resistance force is the sum of "rolling resistance" and "aerodynamic drag" both these are forces measured in say pounds (English system) or kilograms of force (kgf; metric system). While newtons is the standard for force in the metric system, it's simpler to use kilograms (kg) since specific resistance in parts per thousand is just kg-weight/tonne where a tonne is 1000 kg of mass weight.

It turns out that the specific vehicle resistance is also the mechanical energy-efficiency or the energy required per unit volume of transport (such as ton-miles or tonne-kilometers). For example, 3 parts per thousand is 3 kgf/tonne. To take this tonne one km thus requires 3 kgf-km of energy or 3000 kgf-meters or 3000/g Newton-meters or about 300 joules since g (gravity) is almost 10 m/sec^2. Thus the energy efficiency is 300 joules/tonne-kilometer (gross). So the joules per ton-km is approximately 100 times the resistance in parts per thousand.

Thus since mechanical energy efficiency on level ground at constant speed is proportional to specific vehicle resistance, to compare a truck to a train we just compare the specific resistances such as in parts per thousand (same as kgf/tonne). However, we must bear in mind that this efficiency neglects engine/transmission efficiencies and assumes that the vehicles travel at constant speed on level ground. But it turns out that the energy used for climbing hills and for acceleration is usually of the same order of magnitude as the energy expended in overcoming vehicle resistance. However using resistance to compare efficiencies of trains and truck is still of value as will be seen in the examples to follow.

The vehicle resistance may be subdivided into two major components: rolling resistance and aerodynamic drag. The aerodynamic drag is the force of the wind on the vehicle where much of this wind is headwind due to the motion of the vehicle. Stick your hand out of a window of a moving auto and you'll feel aerodynamic drag. In aeronautics, drag is used as a name for the resistance force on aircraft and components such as the wing. Aerodynamic drag is approximately proportional to the square of the velocity so at slow speeds of say only 10 mi/hr it may be neglected.

Rolling resistance is the force it takes to roll the vehicle forward on level ground, neglecting aerodynamic drag. At slow speed (say 10 mi/hr) where aerodynamic drag is small, most all of the force opposing the forward motion of the vehicle is just rolling resistance. If you've ever pushed an auto by hand, the force which you have to push with to keep it rolling at constant speed on level ground is the rolling resistance. Rolling resistance may be subdivided into various components such as the resistance of the wheels, wheel bearings, shaking the ground, etc.

For trucks, almost all of the rolling resistance is due to the rolling resistance of the rubber tires on the pavement. For trains the rolling resistance of the steel wheel on the rail is so low that the other components of rolling resistance become quite significant. Trains are heavy and shake the ground as they pass by and the ground absorbs this energy. This contributes significantly to rolling resistance as does the friction of the wheel bearings and the absorption of energy by shaking the vehicle. For trains only there's the rubbing of the rail by the wheel flanges which keep train cars on the track. More precisely, the railroad truck on which the wheels are mounted is said to "hunt" as it attempts to steer itself, first to the right and then to the left, but there are devices to minimize this hunting.

There are still two more types of resistance but some (or all) of the energy expended on them may be recovered. These are grade resistance and acceleration resistance. For example, going up a 1% grade results in an additional resistance of 1% (or 10 parts per thousand). Acceleration requires a force equal to the acceleration divided by the mass of the vehicle (Newton?s second law of motion). But since the wheels of the vehicle have rotational momentum, one must use an equivalent mass to account for this, which results in an equivalent mass a few percent higher that the actual mass.

Now about the recovery of these two energies. Acceleration force gives the vehicle kinetic energy (momentum) and climbing up a grade gives it potential energy. Both forms of energy may be recovered by coasting. To recover kinetic energy, just coast to a stop and to recover potential energy due to climbing a grade, just coast downhill. Unfortunately, vehicles frequently need to brake and this will consume kinetic and/or potential energy so that these energies Will be wasted as heat (for example heating the brake shoes, pads, wheels, or banks of resistors on diesel-electric locomotives, or for trucks: heating the engine and exhausting hot air by using the engine as a brake.

Since the net energy used by acceleration and climbing grades is just the Energy dissipated by braking, once way to account for acceleration and grade Resistance is just to ignore them and instead, add braking energy to the energy consumed by rolling resistance and aerodynamic drag, all of which are not recoverable.

The most fundamental and stable component of resistance is specific rolling resistance since there is no way to eliminated it or to significantly reduce it. Aerodynamic drag may be significantly reduced by going slow (or by better streamlining). At slow speeds, like 15 mi/hr, aerodynamic drag is pretty insignificant since it drops by 75% with each halving of velocity (since it's proportional to v^2). Braking energy loss may be eliminated or reduced by moving on routes where there are no stops such as freeways or double-track rail lines, coasting instead of braking, etc. Of course it's often not feasible to significantly reduce aero drag and braking, but it's easier than for rolling resistance. That's why this article will first compare the rolling resistance of trains and trucks and then make allowanced to account for aero drag and braking.

For details see Appendix: Train Resistance. A loaded truck has roughly four times as much specific rolling resistance (rr) as a loaded train car. The tires of a large truck have a rolling resistance (rr) of about 6 (per thousand) as compared to about 1.2 for loaded train cars. The truck's rr is nearly constant regardless of load and speed. But for railroads, rr increases with speed and decreases with increasing load. An empty railroad car may have a rr of 2.0. Thus for a rail car fully loaded with heavy goods such as coal and moving at slow speed such as 20 mi/hr, the truck has about 5 times as much rolling resistance. But for an empty railroad car moving at higher speed, the truck may only have twice the rolling resistance. Thus the comparison depends of load and speed, but to get an average, the rolling resistance of trains should be weighed by gross ton miles which gives much more weight to a fully loaded freight car which may weigh over 3 times as much as an empty one (include the car's share of the locomotive weight). The result is that average rail rolling resistance may be about 1.5 which makes trucks almost 4 times worse than trains in rolling resistance. While rolling resistance increases linearly with velocity and truck resistance doesn't the truck usually has more aerodynamic drag since it doesn't have cars in front of it to shield it from such drag (unless there's a side wind blowing).

One example which supposedly shows the railroad to be infinitely more efficient than the truck is to haul coal or ore downhill from a mine at a slight downgrade of say 0.25%. The train with a resistance of only 0.125% can coast down this grade but a truck with 0.6% resistance will need to apply engine power to go down this gentle grade and must supply a force at the wheels of at least 0.35%(0.6% - 0.25%) of it's weight just to power itself downgrade at a very slow speed. By applying more power to overcome wind resistance, it could also go at a faster speed.

But the above is only part of the story since both the truck and train must climb up the grade empty to get another load and both must expend energy for this climb. To estimate this assume that the gross-to-net weight is 1.5 for both truck and train and that the aerodynamic drag is about equal to the rolling resistance on level ground. Comparing Truck and Train Gross-to-Net Weights Using this data one finds that the train is about 11.5 times more energy efficient than the truck.

Hauling on level ground would be a different story since a truck at slow speed has only about 4 times as much specific rolling resistance at low speed. Aerodynamic drag on the truck could also be a few times higher than the train since each train car shields the car behind it from the headwind. It turns out that the aerodynamic drag a moderately streamlined locomotive is about the same as that of a few cars in the train. However if you look at the front end of trucks and trains you will notice that trucks have done better in streamlining to reduce aerodynamic drag.

So in this case the train would only be a few times more energy-efficient than the truck in specific resistance per gross ton. But the actual efficiency measure is net tons of goods transported. See Comparing Truck and Train Gross-to-Net Weights. For example if for a certain good gross-to-net was 2 for a truck but 3 for a train, for the loaded movement the truck would only be 2 2/3 times worse than the train in rolling resistance per unit of cargo (or energy efficiency due to rolling resistance). But what about empty return? A 2 to 1 gross-to-net ratio for the truck means the the truck tare weight is equal to the good weight, while the 3 to 1 for the train means that the tare weight is double that of the goods. This means that if both the train and the truck return empty, then the train will need to haul back double the weight of the truck, and for empty backhauls the truck is only 3 times worse per gross ton which means it is only 1 1/2 times worse per net ton. So for rolling resistance for a round trip (with empty backhauls) the truck uses 2.4 times as much energy (3/5 x 4)

There are a number of reasons why the train might not be say 2.4 times more energy-efficient, resulting in it being less than twice as energy efficient and in some cases even worse than a truck.

First, trucks operating on freeways don't need to stop to overtake other motor vehicle or to let vehicles going in the opposite direction shoot past them. But single track railroad is like a one lane road with the same lane used for both directions of travel. Trains have to stop each time they meet an opposing train heading the opposite direction.

To-do

A very important phenomena is the economy of scale in rolling resistance for axle loading. As more weight is added to a rail freight car, there is more weight per axle (and per wheel) so the specific resistance drops (in parts per thousand for reasons explained in to-do). As the weight increases, the rolling resistance does increase but if one increases the weight 4 times the rolling resistance might say only increase 2 times, resulting in the specific resistance being cut in half.

Of course the aerodynamic drag of an enclosed freight car is independent of load so there is an economy of scale in aerodynamic drag also. Curves showing this may be found in both Hay (p. 176) and Astakhov (p. 55). To see the economy of scale from these curves, look at the values of resistance at low speed (10 mi/hr and 20 km/hr respectively) where the effect of aerodynamic drag is negligible. For the heaviest loading, Hay shows about 1.6 (per thousand) for 19 tons/axle) while Astakhov shows 1.4 for 20.5 tonnes/axle. For light loads, Hay shows 3.3 for 6.25 tons/axle which Astakhov shows 2.6 for 5.95 tonnes/axle. Note that a tonne (1000 kg) is about 1.1 tons.

Thus per the above the rolling resistance loaded at low speed is about 1.5 and for an empty car that weighs about 25 tonnes, it's about 3. However, these studies were likely for plain bearings so today's roller bearings would make the rolling resistances lower. Astakhov shows the resistance due to both roller and plain bearings in Fig. 4.2 (p. 75) and Fig. 4.4 (p. 78). car the roller bearing decrease rolling resistance (at 20 km/hr) by 0.55 (parts per thousand) for an the empty car and by 0.23 for the loaded car (21 tonnes/axle). So subtracting these values to get the data to apply to modern freight trains with roller bearing the loaded rolling resistance becomes about 1.3 and the empty resistance about 2.5.

The AREMA manual has a newer Davis equation developed by the
Canadian National Railway (CN). It's only different in the
numerical coefficients while the basic formula is the same.
It's

*R = 1.5 + 18N/W + 0.03v + Cav^2/10,000W* where R is the
resistance in lbs/ton, W is the weight of the car in tons, v is the
velocity in mi/hr, a is the frontal area, and C is the streamlining
coefficient per CN. The first 3 terms represent rolling resistance
so for a speed of 15 mi/hr and a maximum load of 36 tons/axle, the
rolling resistance is 2.45 lbs/ton or 1.22 parts per thousand,
close to the 1.3 estimated above. So let's estimate it as 1.25
fully loaded (for heavy loads) and double that (2.5) unloaded.

Using these 2 values one may estimate an average rolling resistance of 1.55. To get an average, one must weight rolling resistance by gross ton miles and a rough estimate is to assume that the average payload is twice the empty weight and that all cars return empty. See Table 5.11 on p. 5-16 of the Rose Study. Thus the gross weight for half the miles (fully loaded) is 3 times the empty weight for the other half of the miles. The result of applying 3 times as much weight to 1.25 as applied to 2.5 is a weighted average of 1.55. Since contrary to the assumption made some backhauls don't go empty, reduce 1.55 to just 1.5.

It's interesting to note that the Rose Study used the same Davis formula for the resistance of plain bearing cars as used by Hay but at the time of his study 1979 most all freight cars had roller bearings (per personal observation of the author).

1. Astakhov, F. N., Soprotivlenie Dvizheniiu Zheleznodorozhnogo Podvizhnogo Sostava (Resistance to Motion of Railroad Rolling Stock) Moskva, Transport (publisher) (in Russian), 1966. Issue 311 of the series: Trudy vsesoiuznogo Nauchno-Issledovatel'skogo Instituta Zheleznodorozhogo Transporta. Chapter 4 (p. 73+) partitions resistance into 6 components with a section on each component: bearing, rolling, sliding, shaking the earth, aerodynamic, vehicle vibration and shock absorbers. The quadratic formulas for rolling resistance used in various countries are compared and plotted (fig. 2.3, p. 35) and includes the "Davis" formula used in the United States.

2. Deev, V. V., Tiaga Poezdov (Train Traction) Moskva, Transport (publisher) (in Russian), 1987. The components of resistance are discussed in section 5.2 (p. 78+). The diagram of forces and pressures acting on a wheel (fig. 5.3 on p. 80) is interesting.

3. Hersey, Mayo D. et. al., Rolling Friction (in 4 parts) by in "Journal of Lubrication Technology" April 1969 pp. 260-275 and Jan. 1970 pp. 83-88. Part II is for cast-iron rail car wheels. Contains no info on rubber tires. See p. 267 for variation of resistance with diameter.

4. Hay, W. W., Transportation Engineering, Wiley, 1961. "Tractive and Road Resistance" pp. 174-178.

5, AREMA Manual for Railway Engineering, 2000 (and other years) (AREMA = American Railway Engineering and Maintenance-of-way Association) Chapter 16: Economics of Railway Engineering and Operations, Part 2: Train Performance 16-2-1, Section 1: Resistance to Movement 16-2-2 and Section 2: Train Performance Calculations 16-2-10.

5. Hoerner, Sighard F., "Fluid Dynamic Drag", published by the author, 1965. (See Ch. 12 for train data: p. 2-10: B. Drag of Railroad Trains.)

Table 5.11 on p. 5-16 of the Rose Study show both average carload goods weights and mean empty car weights for various commodity classes for 1976. While the weights have gotten heavier since then, the ratios of payload (cargo or goods) to empty weight likely have not increased too much, since cars today tend to be heavier to handle the heavier loads. This ratio is highest for coal: about 3 (a car weighing 30 tons hauls 90 tons of coal). Today they may haul 110 tons. But for electrical machinery (likely mostly consumer electronics) the ratio was the lowest: 34 ton cars hauling 17 tons of goods with a ratio of 0.5. In terms of gross to net ratio we have: 1.33 and 3 respectively for coal and lightweight consumer goods. Since lightweight consumer goods (like consumer electronic) are today mostly shipped via container, the above ratio of 3 is likely wrong: the containers add to the weight but the rail cars that haul containers are lighter (per unit container load hauled). to-do: container research.

The above ratio must be corrected to include the locomotive weight which per Table 5.2 on p. 5-6 of the Rose Study shows the need to increase the weight by about 11.4% to account for locomotive weight. This results in gross to net ratios of about 1.45 and 3.33.

For the Rose Study p. 6-9 estimated the typical empty truck weight was 14.5 tons while the cargo weight per Table 6.5 on p. 6-11 range from about 10 tons to 25 tons. This is a gross to net range of 1.6 to 2.5. What's wrong with this is failure to differentiate empty truck weight by commodity. to-do: Find more recent data and truck weight by commodity.

Per Train Weights and Truck Weights the gross-to-net ranges from: 1.45 to 3.33 for trains and 1.6 to 2.5 for trucks. 1.45 means that for every ton of coal hauled, 1.45 tons of total weight (including the coal) must be hauled. Note that for heavy goods (the low ratios) trains may be relatively lighter than trucks while for light goods (the high ratios) trains are relatively heavier than trucks.

"Energy Intensity and Related Parameters of Selected Transportation Modes: Freight Movements" by A. B. Rose, Oak Ridge national Laboratory, Oak Ridge, TN, June 1979 (a 3/8" thick printed report). See Ch.5 "Rail Freight Movements" and Ch. 6 "Truck Freight Movements". Tables to compare rail vs. truck: Table 5.11 (p. 5-16) "Rail Freight Energy Intensity for Various Commodity Classes, 1976" and Table 6.5 (p. 6-11) "Energy Intensities by Commodity Class" (trucks).

This report presents the energy-intensity (in BTU/ton-mile) for freight transportation in the United States between 1970 and 1977. While these overall estimates can be found elsewhere, this reports disaggregates energy-intensity by commodity for truck (1975) and rail (1976) freight. For example, the energy-intensity (in BTU/ton-mile) to transport the commodity class: "rubber and miscellaneous plastics" is estimated.

It indicates that energy-intensity is highly dependent on commodity. While it contains a disclaimer that the results are not at a sufficient level of disaggregation to compare rail to truck, doing so in violation of this warning indicates that for certain low density commodities, the energy-intensity of truck is roughly the same as rail.

But there's a serious error in the methodology, namely equation (5) on p. 5-10 for the calculation of rail energy-intensity. As a result, the rail energy-intensity figures show a wider variation in energy-intensity between different commodities than they should. See Calculation of Energy-Intensity by Commodity for the correct way to calculate this as well as a discussion of his error.

Inspecting the two cited tables for BTU per
great-circle-ton-mile (energy-intensity for
as-the-crow-flies-mileage) indicates that for many commodities,
rail is a few times more energy efficient than truck. But there are
a few commodity classes where rail and truck are erroneously
reported to be about equal in energy intensity:

`BTU/great-circle-ton-mile Reported Corrected Rail Truck Rail Truck Basic Textiles 3350 3630 to-do Electrical Machinery 4220 4130 to-do Rubber and Misc. Plastic Products 3540 3870 to-do`

But for the following commodity classes, trucks are reported to be higher (worse) in energy intensity than rail:

These figures are estimates for rail in 1976 and for truck in 1975, but there is an error in the rail numbers as mentioned above.`BTU/great-circle-ton-mile Reported Corrected Rail Truck Rail Truck Coal 450 2590 to-do Metallic Ores 510 2630 to-do Non-Metallic Minerals (except fuels) 540 2620 to-do`

In making the above comparisons, I have violated what it says at the bottom of each table "The values in this table are not intended for intermodal comparisons, as they do not include route structures and are not at a sufficient level of disaggregation." But intermodal comparisons can be useful, especially if the effects of route structures and level of disaggregation are discussed.

Let's roughly estimate how much more energy it takes to haul "Rubber and Misc. Plastic Products" (henceforth called "Plastic Goods") as compared to coal. In table 5-11 it shows:

Note that I've appended a "rounded weight (in tons) which I'll use in this example so as to enable one to make the calculations without using a hand calculator (or computer).`Av. Tons Av. Tons of Goods "Rounded" of Empty "Rounded" per Car Wt. Car Wt. Coal 86.2 90 28.1 30 Plastic Goods 18.8 15 33.5 30`

Let's haul one car of coal (90 tons) and also haul 90 tons of low density plastic goods for comparison. Since only 15 tons of plastic goods fit into each rail car, we'll need 6 cars for them. That's 6 times as much aerodynamic drag. But the total weight for the plastic stuff at 45 tons/car gross is 6 x 45 = 270 tons. The carload of 90 tons of coal weighs only 90 + 30 = 120 tons. Grade resistance for climbing hills and energy wasted stooping to let other trains pass is 270/120 times greater for the plastic. But the rolling energy losses for the plastic will be higher than implied by the 270 tons of loaded cars, since the load per axle is lighter than normal. Let's assume that everything considered, the rolling, grade, and stopping energy losses for the plastics are 360/120 times greater than for coal. That's 3 times greater.

Now what about the empty return trip. Per the table, the percent which must be backhauled empty is 70% for plastic goods and 91% for coal. But to simplify, let's say these are both 100%. So the energy of returning 6 empty cars for the plastic case will be 6 times that of returning just one empty coal car.

So how much more energy did the plastic goods take. It's an average of 3 and 6 with the backhaul counting for less (since it uses less energy. So if the energy is proportional to gross ton-miles, for one mile of forward haul the coal used 120 units of energy and the plastic 360, For empty backhaul the coal used 30 units of energy and the plastic 180. Adding we have for the round trip: 150 units of energy for the coal vs. 540 for the plastic. Thus the plastic took about 3 2/3 (say 3.7) times as much energy to transport. Note that this has neglected aerodynamic drag which is 6 times greater for the plastic in both directions. However, it can be estimated that aerodynamic drag is only about 1/4 of the energy used for freight in the 1970's (before the dominance of high-aero-drag container trains). So weighting the two ratios: 6 and 3.7 by 1/4 and 3/4 we obtain a ratio of 4.3.

But what does Rose's table say? It says it took 3540/450 or 7.87 times as much energy. This is a lot larger than 4.3. So something is seriously wrong with the BTU/ton-mi estimates shown in the last column of Rose's Table 5.11. It turns out that the error is in the long equation 5 on p. 5-10 (previously discussed) which was used by Rose to calculate the energy-intensities. This error was discovered by the author while using this very rough numerical example presented above.

Neither the railroads (nor any survey of them) reports their energy consumption by commodity. In cases of trains hauling cars with various commodities, one would need to allocate a portion of the fuel used by the locomotives to each different commodity. This could be done on a combined weight and volume basis. Each car contributes to the energy used by the train mainly due to its weight, but also each car has a certain amount of aerodynamic drag which depends on it's shape (including the shape and spacing of cargo lying in the open on flat cars such as container) and by how much the car ahead of it shields it from the headwind.

But the "Carload Waybill Statistics" based on a sample taken in 1977 by the Federal Railroad Administration gives the average empty and loaded car weight for each commodity and the estimated percentage of empty mileage as compared to loaded. For, example, for every 100 miles of hauling coal, 91 miles are run empty to pick up another load of coal so 91% of loaded mileage is run empty. It would be 100% if it were not for the finding some other commodity to put into coal cars on part of the backhaul.

Using this data one can estimate the weight penalty for hauling various commodities. Then using a Davis-like formula for train resistance one may calculate the train resistance per ton of cargo for various commodities.

How relevant is this study today (2008)? If rail and truck energy efficiencies have improved by about the same percent, then the relative status of rail vs. truck remains about the same.

But per the article cited next, this hasn't happened and rail energy efficiency has significantly increased while truck energy efficiency has significantly dropped. If so, then rail would be significantly more energy-efficient than truck in almost all cases. But it's not necessarily so, as will be explained later.

"Freight Energy Use Disaggregated by Commodity" by F. M. Vanek, et. al. In Transportation Research Record with meeting title "Energy, Air Quality and Fuels" 1998, pp. 3-8.

While this study is much more recent than the Rose Study it attempts to extrapolate data estimated in 1972 to 1993. This extrapolation is done by taking 1972 energy-intensive estimates for various commodity classes and multiplying each by: 1.253 (for trucks) and 0.720 (for rail). In other words it assumes energy intensity for trucks has increased by 25.3% while energy intensity for rail has dropped by 28%. And it of course assumes that the percent change in energy intensity is the same for all commodities, the only difference being whether we are talking about rail or trucks. These percentages just don't seem right as will be now explained.

Transportation Energy Data Book (TEDB), table 2.14: Energy Intensity of Freight Modes shows for "Heavy single-unit and combination trucks" a decrease in energy-intensity from 1972 to 1993 (not a 25% increase). The figures are not BTU/ton-mi but BTU/vehicle-mile. Since trucks have gotten larger over that time period, one might expect that the BTU/ton-mi would decrease even more than the figures indicate. The figures shown in TEDB are: 1972: 24,668; 1993: 22,373 (both in BTU/vehicle-mile). This is over a 9% decrease.

This same table 2.14 (in TEDB) shows a big drop in rail freight energy intensity (EI). So could the 28% drop in rail energy intensity mean that EI dropped about that much for all commodities? Likely not. This reported drop in rail energy-intensity may be in large part due to the loss by rail to trucks of commodities with higher rail energy-intensity. Also, rail is transporting more coal which is several times more efficiently hauled by rail than the lower density commodities for which rails lost modal share. In addition, many coal mines are at higher elevations than the power plants that get their coal. Thus coal trains can sometimes coast downhill on much of the trip, thereby using less energy. Neither The Rose Study nor the Vanek Study considered this phenomena. So it could be the case that the rail energy-intensity of transporting various commodities didn't change all that much in the 21 year period, and that most of the overall rail energy-intensity decrease is due to changes in rail's modal shares of various commodities. Longer lengths of hauls and less switching of cars from one train to another (via use of rail yards) also likely accounts for much of the energy-efficiency improvement.

The Vanek Study energy-intensities for 1972 seem to be nearly the same as the Rose Study Thus they likely originate from the same data source. Thus the values estimated by Rose Study could be more accurate than the ones presented by the Vanek Study due to Vanek's likely erroneous extrapolation of 1972 data to 1993. But the serious mistakes in the Rose Study have been previously noted.

A moving train on level ground can coast a long ways if it shuts off its engines. We say that a moving train has kinetic energy equal to 1/2 mv^2 where m is the mass of the train and v is the velocity. Actually, the kinetic energy is a little higher than this since rotating wheels and motors provide additional rotary kinetic energy.

How far can it coast? An example calculation for a train moving at 50 mi/hr would be: 1. Find the height which the kinetic energy could lift up the train. It's equal to v^2/2g where g is the acceleration of gravity. g is 9.8 m/sec^2 or 79036 mi/hr^2. So plugging in 50 mi/hr and g in the formula shows that a train at 50 mi/hr can climb 0.016 miles (about 85 ft.) off the ground (up in the sky) or gain this elevation by rolling uphill with the power off, provided there was no resistance against the train except for gravity.

At a train resistance of 0.1% a train could coast 1000 miles for each mile of elevation. Why? Because 0.1% means that for a unit distance traveled (like one meter) the energy used in 0.1% of the weight of the vehicle while for the same weight falling a unit distance (like one meter) the energy obtained is equal to 100% of the weight of the vehicle or 1000 times as much (since 100%/0.1% is 1000).

So since our train can climb .016 miles from its kinetic energy (or momentum) it could coast for 16 miles if the train resistance were only 0.1% which is overly optimistic since it neglects aerodynamic drag. But if we're lucky we might be able to coast 10 miles (assumes an average resistance of 0.16%). This is also assuming a fully loaded railroad freight cars. Empty cars have both more specific aerodynamic drag and rolling resistance and thus wouldn't coast nearly as far.

At say 60 mil/hr at train could coast further, 14.4 miles if train resistance remained at 0.l6%. But unfortunately, train resistance will increase due to the higher speed, so it might only coast about 13 miles under favorable conditions.

In comparison to pipeline and water transport rail may use more or less energy. It's very much speed dependent. Roughly speaking, If speeds are only a few miles per hour for barge (water) and under 1 mile per hour for pipeline (for a large diameter pipe) then rail freight would usually be less energy efficient. At significantly higher speeds than above, rail freight uses less energy.

See: Energetika Lokomotivov (Locomotive Energy) 2nd ed. by Kazan, MM. Moskva, Transport, 1977. See p. 94+ for a comparison of Diesel vs. Electric efficiency; Nominal values (at full load) diesel 32.6%, electric 31.2% (30.0% after transmission losses from the power plant to the railroad substation). See p. 52). This calculation assumes modern electric generation facilities with 43% efficiency and admits that the such efficiency in the USSR at that time (1977) averaged only 33% and not 43%. However, the efficiencies of various diesel locomotives ranged from 25.7% to 32.5% per Table 17, p. 92. Compare this with the 21.9% actual efficiency for diesel locomotives as found in on p. 93. My conclusion: electric and diesel traction have thermal efficiencies of roughly 30% at nominal operating conditions (but significantly lower under actual operating conditions).

Note that for diesel electric locomotives, the electric transmission efficiency is reported as 80% - 85% per p. 90 and is shown as 82% per table 17.