mailto:dave@lafn.orgCopyright 2007 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.
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 is actually the case. So for lightweight goods it makes rail look worse than it should be and for heavyweight goods it makes rail appear better than it actually is. I've discussed this error in the article and corrected equation (5). But it still remains to create a spreadsheet, obtain data (Rose omitted some data he used such as assumed train speed), and calculate corrected values of rail energy-intensity for 1972.
This 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. But there are also other likely errors in the Vanek study.
Due mostly to the low rolling resistance of the steel wheel on a steel rail, hauling freight by railroad is usually 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 are container trains where there are big gaps between the containers. If such gaps are wide enough it may result in high aerodynamic drag almost as bad as for trucks.
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:
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 cliaming that rail freight is a few times (or even several times) more energy-efficienct than truck. While it's true that the estimates show rail to be much more energy efficient 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 good to trucks hauling lighter goods is like comparing apples to oranges.
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.).
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 may be more energy efficient. However, for the majority of goods (commodities), rail is 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 much 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 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.
So in transportation, a major potential means of improving energy efficiency is to shift a lot of 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, rail is always 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 the commodity group as a whole but not for the low or high density goods withing the commodity group. Such lack of disaggregation is present in both Rose Study and the Vanek Study. Of course it's easy to criticise for not disagregating more, but it's not so easy to disagregate more since the data needed may not be available.
One example of possibly 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. A lightweight type of "electrical machinery", consumer electronics, would go mostly by truck, where the truck can haul it with a moderately good 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. 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.
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. The 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 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 they 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 reasoning 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.
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 with a more detailed explanation than in the Rose report.
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 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 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. Also included in the gross
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 car weight by about 11% to account for
that part of the locomotive weight necessary to haul the car). Then
the gross weight of a carload of the i th commodity is:
(Wei + Wci)(1 + K2)
This gross weight per car may be divided by Wci to get the gross
to net weight ratio for commodity i (G2Ni). Then, since the energy
intensity per gross ton-mi (EIgtm in BTU/ton-mi) for the rail
freight as a whole is known and if it's assumed that it doesn't
vary much with commodity, one could just multiple it by the ratio
of gross-to-net for commodity i (G2Ni) and seemingly obtain the
energy-intensity for commodity i (EIi). The formula is thus:
EIi = G2Ni x EIgtm. 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 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)]/Wei
(1)
To improve this oversimplified formula requires a few major
modifications:
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 the effective gross to
gross to net formula for commodity i
becomes:
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 the train with the low density plastic will have lighter car weights which implies more cars to obtain the same weight. So the longer train of plastic toys will have 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 n in 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, 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=.00o5 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 have the force per unit weight
in pounds per ton. Another unit is force is parts per thousand
(pounds per thousand-pounds = kilogram-force per tonne, etc.). This
is known as the specific resistance r.
Does this specific resistance r depend on commodity? Of course is does since the loaded car weight W depends on commodity. The two terms of the specific resistance that depend on W are (the terms enclosed in [] brackets) are 29n/W and (cAv^2/W). The 29n/W gets lower at W increases and shows 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, the specific resistance r decreases since more is being transported for the same aerodynamic drag. So we can find the 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 which for
a lightweight commodity might be 1.2 and then multiply the EIgtm by
that ratio to get the 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 for all
commodities and it should be weighted over car-miles. These values
may be obtained by dividing system ton-mi by system car-mi.
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 braking energy should be 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). It's something like equation (4) above but it divides resistance by the weight of the commodity in the car instead of by the gross weight of the car. 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, if 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 net EIi for commodity i per equation (2) not by the overall net EI for the railroads. This EIi 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.
Another error is that Rose failed to account for backhauls and braking energy in his correction factor similar to 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. The result 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 hard to do. 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). First 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% which 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 (before increasing it to account for braking energy) just the fixed component of rolling resistance. So we are increasing the rolling resistance to account for braking losses.
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. So even with the circuity penalty assessed to rail, the savings in energy usually 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 some 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 used circuity but the Vanek Study didn't.
To-do
"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 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 as expressed by 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 these 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, except for the "rounded weight (in tons) which I'll use in this example:
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 counts 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 which was used to calculate the energy-intensities. This error was discovered by this very same rough numerical example.
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.
FLASH: I've just done a "back of the envelope example and found that the results in BTU/ton-mi for low density rail commodities (such are electric machinery) per Rose are far to high, perhaps double what they should be. This is because the long equation (5) (p 5-10 in the Rose study) is dead wrong. To-do: explain the serious error.
How relevant is this study today (2007)? If rail and truck energy efficiencies have remained about the same, then it's valid. And if the efficiencies for both truck and rail have say improved by about the same amount, 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 in energy intensity.
This same table 2.14 (in TEDB) shows a big drop in rail freight energy intensity. So could the 28% drop in rail energy intensity be correct? Perhaps not. This reported drop in rail energy-intensity may be in large part due to the loss by rail to trucks for commodities with higher rail energy-intensity while. At the same time rail is transporting more coal which is several times more efficiently hauled 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 use their coal. Thus coal trains can sometimes coast downhill, thereby using less energy. Neither The Rose Study nor the Vanek Study considered this factor. 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.
The Vanek Study energy-intensities for 1972 seem to be nearly the same as the Rose Study Thus they may 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.
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 locomotives. 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).
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 travelled (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 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 has increased 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.