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Which is more energy efficient in the U.S. (in terms of passenger-miles per gallon) a railroad passenger train or an airline passenger airplane? While many people expect the train to be more efficient, the reported statistics, after correcting, indicate that the airplane is perhaps 15% more energy efficient. See Train vs. Plane Statistics.
Trains have the reputation of being energy-efficient due to their low rolling resistance. It turns out that in spite of its low rolling resistance, passenger trains in the US today (including rail transit) are no more more energy efficient than the airplane (and also about the same efficiency as for the automobile --but the comparison to the automobile was the topic of another article by me: Train vs. Auto Energy. See it for the bibliography regarding trains.
Before a detailed comparison of trains vs airplanes let's make a very short comparison which covers 3 type of energy losses and omits many significant details and other factors to be covered later in this article.
The energy-efficient steel wheels of a train, which hold the train off the ground, have a function similar to the wings of an airplane which holds the plane up in the sky. And the airplane expends about 2/3 of its energy used in flight to move its wings thru the air (see Aerodynamic drag report) so its not as efficient as the train for this (although the airplane cruises over ten-thousand times higher off the ground than a train).
One advantage of the airplane is that it loses little energy by braking while trains must apply the brakes to slow down for curves and have such low rolling resistance of the wheels that they need to brake when descending gentle downgrades. An airplane is not confined, like a train is, to the railroad track and can select a curve of almost any radius and fly that curve by banking (tilting) the airplane without any braking. On descents it can glide (coast) down at the required slope so as to recover almost all of the climbing energy expended due to climbing to cruising altitude from its originating airport, (The sum of the fuel used during climb and descent is not much more than if the climb and descent distances were flown at a level cruise.)
Then there is the comparison of the aerodynamic drag (wind force) on the passenger compartments: the fuselage for the airplane excluding the wings and just the whole train for the train. We exclude the wings since the large amount of energy used to move them through the air has already been accounted for as energy used to hold the plane up in the sky. The airplane fuselage is highly streamlined and the airplane files at high altitude where the thin air lowers aerodynamic drag. The airplane's passenger are packed into the fuselage more densely than in a train and there are often less empty seats on a plane. Also, the plane doesn't have wind blowing into its side, since it aims itself to fly directly into the air ahead of it, even if there are winds aloft. Side winds, even on a slightly windy day, significantly increase the aerodynamic drag of a train. But all these aerodynamic advantages of the airplane are more than cancelled out by the much higher speed of the airplane since aerodynamic force is almost directly proportional to the square of the speed (at constant altitude). See v^2
So we have just compared 3 energy types:
Be warned that the above is overly simplistic and there is also the efficiency of the engine, transmission, and aircraft propulsion efficiency to consider. For a more detailed analysis in this article see Numerical Example of the Efficiency Ratio
The table below Historical Energy-Efficiency shows some of the the history of the statistical energy efficiency of trains vs. planes in the U.S.
PM/gal = Passenger-Miles per Gallon (auto gasoline)
BTU/PM = BTU per Passenger-Mile
Mkt. % share = Market share of total intercity passenger-miles by all modes within the U.S. Data source ceased publication after 2007.
Energy in Am. Economy: a book. See energy_in_am
Sources: See Data Sources for the above table
Scheduled Domestic Air | Passenger Rail per Energy in Am. Economy: | David Lawyer | Eric Hirst | David Lawyer Mkt. | diesel trains | all trains | steam | electric PM/gal BTU/PM %share| PM/gal BTU/PM | BTU/PM %share| BTU/PM | BTU/PM 1935: 12.8 9,770 0.1 | 51 2,440* | to-do 8.3 | 23,100*| 5,400* 1940: 17.5 7,150 0.4 | 41 3,080 | to-do 7.5 | 17,900 | to-do 1945: 28.6 4,370 1.2 | 85 1,480 | to-do 27.1 | 8,100 | to-do 1950: 21.8 5,720 1.8 | 42 2,950 | 7,400 6.5 | 15,400 | to-do 1955: 24.2 5,160 3.0 | 41 3,010 | 3,700 4.0 | | to-do per Eric Hirst: * (above) => 1936 | | 1955: 26.0 4,800 3.0 | 41 3,010 | 3,700 4.0 | | to-do 1960: 18.1 6,900 4.1 | 41 3,0l0 | 2,900 2.8 | | to-do 1965: 15.3 8,200 5.9 | 40 3,160* | 2,700 1.9 | | 2,900 1970: 14.9 8,400 9.3 | | 2,900 0.5 (1963) per Transportation Energy Data Book: includes both international | and non-scheduled (charter) | Amtrak: diesel and electric trains. 1970: 13.9 8,980 9.3 | 41 2,900 (this 2,900 per Eric Hirst) 1975: 19.2 6,510 10.1 | 34 3,677 0.5 1980: 26.8 4,660 13.9 | 39 3,176 0.4 1985: 28.5 4,390 17.0 | 45 2,800 0.3 1990: 30.4 4,110 18.7 | 48 2,609 0.3 1995: 35.6 3,510 19.3 | 48 2,590 0.3 2000: 37.9 3,300 24.5 | 38 3,257 0.3 2005: 45.6 2,740 24.7 | 45 2,784 0.3 2010: 52.2 2,437 ? | 55 2,271 ? 2015: 55.9 2,237 ? | 57 2,186 ? ? => Source ceased publication.
Notes: Air: Correction for air freight: Hirst used 1 ton = 5 passengers. Lawyer applied 1 ton = 4 passengers to Transportation Energy Data Book (TEDB) data. See Cargo correction Note the discrepancies for both 1955 Diesel locomotive trains and 1970 due to different authors. No correction was made by anyone for circuity. Pass-Mi/gal (or PM/gal) is shown for the automotive gasoline BTU equivalent and can be though of as per gallon of gasoline. If you want to get PM/gal-of-kerosene (jet fuel) increase PM/gal-of gasoline by 8%. Likewise, for diesel train fuel increase PM/gal-of-gasoline by 10%.
Figures reported by Amtrak but later retracted: 2002: 4,137 BTU/PM, 2003: 4,830 BTU/PM. Per inquiry the author made to TEDB, the reason for this (supposed ?) error is unknown. Amtrak reported high figures for electricity use in these years.
The high efficiency for 1945 was due to planes being full due to wartime conditions. Note the big drop in energy efficiency from 1960 to 1975 due to the introduction of "jet" (really turbofan) aircraft and the recovery starting in 1970 due to larger engine fans (high bypass ratio) and the fuel economies of scale for larger aircraft.
The 0 for the low-order digit of the BTU/PM is due to rounding so as to retain 3 significant figures. Eric Hirst has 00 and only retains 2 significant figures. Due to the uncertainties of allocating fuel to cargo transport, retaining only two or three significant figures is reasonable.
While railroads got their start in the 1830's, airlines didn't appear until 100 years later in the 1930's when railroads had already lost most of their passenger traffic to autos during the 1920's but still retained an 8% share of intercity travel (as compared to only a 0.3 % share today). Airlines were especially competitive with railroads due to the similarities of the service: has schedules, no driving required, need to obtain local transportation at destination such as auto rental or public transportation.
But while the railroad's share of total passenger transportation was low in the 1930's, the airlines share was roughly 100 times lower at only 0.1% (in 1934). See. Passenger-Mile Data. Flying was dangerous, expensive, and slow as compared to today. For example,in 1936 the fastest flights coast-to-coast across the U.S. took 14 hours with 3 intermediate stops for refueling. This was still a lot faster than going by train.
What about fuel economy? Before the introduction of the streamlined Douglas DC-3 airplane in 1936 (and other more efficient airplanes, airlines got only about 13 passenger-miles per gallon (in 1935). This was about as poor as driving alone in a luxury auto but still much better than the steam-powered railroad trains which burned coal or fuel oil and only got the equivalent of 5 passenger-miles per gallon.
While 90% of rail passengers rode on such trains, some went on electric trains that were about twice as efficient as airplanes, and a relatively few went on the new diesel trains which were about 4 times as energy efficient at airplanes. The fact that it then took about twice as much fuel to generate a kilowatt-hour of electricity as it does today mostly explains why diesel trains were about twice as efficient as the electric trains of that era.
After the mid 1930's, airline fuel economy went on a roller coaster ride, rapidly improving until the late 1950's when the inefficient "jet" airplanes replaced the more efficient propeller airplanes. Improvements (before the "jets") included better streamlining, larger planes with less surface area per passenger to cause drag, and flying higher where the air is thinner resulting in less aerodynamic drag. Railroads improved energy efficiency at a faster pace than airlines by simply but slowly replacing the steam locomotives with diesels which were nearly 8 times more efficient. The pace of dieselization of rail passenger travel (which began in the mid 1930's) continued apace after 1949 (when it was half-way completed) becoming fully completed in the mid 1950's. See Car-mi by Diesel (from ICC data)
The result was that by the mid 1950's, passenger rail had surpassed the airlines in fuel efficiency. As airlines introduced inefficient "jets" in the late 1950's trains became more than twice as fuel efficient than flying. But unfortunately, rail traffic was declining and in 1957, airline travel surpassed that of rail so the increase in rail efficiency was becoming of less significance as passengers gave up on rail travel for airplanes. Enticement to the airplane was due in part to air travel being faster and cheaper but the neglect of passenger traffic by railroads as well as "make work" labor union rules played a significant role in rail passenger decline.
Then starting in the 1970's, airlines did something about the poor fuel economy of the "jets". Note that "jet" is quoted because they were (and are today) really turbo-fan aircraft that are similar to turbo-props, except that a concealed fan replaces the propeller. By making these fans much larger and thus making the aircraft more like propeller aircraft, the airlines significantly improved their efficiency. See Turbofan engines more efficient. The result was that by about 2005, flying surpassed trains in energy-efficiency.
So now that planes are just as energy-efficient as trains, aren't we better off since planes travel faster and thus save time? Not necessarily. Before the development of widespread air travel, the slowness of trains deterred people from making long trips. This saved energy since people travelled less. Today with it's high volume of air travel we are using a lot more energy due to the higher volume of air travel even though it's much more energy-efficient than it used to be.
Aeronautical Engineers seldom (if ever) talk with Railroad Engineers (not the train driver sort of engineer) about technical factors affecting energy efficiency. It they did, it would seem that they were speaking different languages since the terminology is almost completely different. What a railroad engineer would call "train resistance" an aeronautical engineer would call "drag". What a railroad engineer would call "specific train resistance" is the inverse of what an aeronautical engineer would call the "lift-to-drag ratio". "Train miles per gallon" becomes "Specific Air Range" for airplanes (measured in nautical miles per pound of fuel). "Freight" in airplanes is called "cargo". But there are many terms for air for which there is no close equivalent for rail and conversely and thus the use of different terminology for different concepts is sometimes inherently necessary.
When a train or plane moves forward, there is a resistance (for trains) or drag (for airplanes) force which resists moving the vehicle forward and is pushing back on the vehicle. It's directed backwards along the path which the vehicle takes. For a train, this path is just the railroad tracks. But for a plane this drag force, caused by the headwind hitting the plane is, directed backwards along the direction of such a headwind (air speed).
For a plane, if the air is still with respect to the earth (no natural wind), then this drag direction is just the flight path of the airplane with respect to the ground. But if there is a natural wind, things are a bit more complex since the velocity of the plane with respect to the ground (ground speed) is the vector sum of the wind speed (with respect to the ground) and the air speed of the plane (with respect to the air it's passing through). What pilots call "speed" (above) is more precisely called "velocity" since it is a vector (has both magnitude and direction). See Airplane in wind.
The amount of force required depends on many things such as the vehicle's weight, speed, degree of streamlining, altitude (air density), etc. It takes still additional force to accelerate the vehicle or to ascend grades (trains) or climb (planes) but some of this energy may be recovered by coasting (trains) or gliding (planes). For the force opposing level travel, drag (for planes) is equivalent to resistance (for trains).
The mechanical energy expended in moving a vehicle is just the resistance force times the distance traveled. Thus if one wanted to compare the mechanical energy used to move two vehicles a fixed distance, and one vehicle had twice the resistance of the other, then it would take twice as much mechanical energy to move the high-resistance vehicle. If the resistance varied over this fixed distance then one would use average resistance (averaged over distance and not time). Thus by comparing the total resistances of two vehicles on a level route one may compare the mechanical energy needed to move the vehicles a given distance.
One may think of vehicle resistance/drag as the mechanical energy required to move the vehicle a unit distance (such as a meter, foot, kilometer, or mile). For example, if the resistance of a train is 100 kilo-newtons then the energy required to move a kilometer is (the force of 100k-newtons times the distance of 1 km) 100 kilo-newtons-kilometers = 100 mega-joules since a joule is just a newton-meter. This is thus an energy intensity of 100-MJ (MegaJoules) per km (kilometer) and this "energy-intensity" is just the inverse of "energy-efficiency".
Trains or planes use more energy per mile when gaining altitude (going uphill) than they do moving on the level. The additional mechanical energy expending in going uphill creates "potential energy" which is equal to the mass times gravity times the distance risen. Since gravity is nearly constant near the earth's surface one may say that potential energy is just weight times height (multiply the weight in pounds by the height risen in feet) resulting in ft-lb (foot-pounds) of energy (or ft-lbf to show that in this case, "lb" is a unit of force "f" and not a mass). The metric system would use different units and it requires multiplying by gravity in converting a mass unit to a force unit, but for airplanes and trains in the United States, the English system of units, distance in feet and mass/weight/force in pounds, is still the standard in transportation.
But what goes up must come down the same amount (if the originating location is at the same elevation as the destination location). Thus the potential energy expended going up returns about the same amount of energy when the vehicle descends down. This stored potential energy is used up some way, such as powering coating/gliding downhill or conversely, much of it may be wasted in braking. In the case of the electric train that uses "regenerative braking", the braking energy is not all wasted. Some of this potential energy is recycled by using the train's electric motors running in generator mode as brakes to generate electricity to use for powering other trains, etc. The energy waste part of braking turns the potential energy into mostly heat that is dispersed in the surrounding air by warming the air.
For grades (trains) or climbs (planes) and the decent of both planes and trains, the forces due to the vehicle weight being pushed up the grade are similar but go by different names. For trains they are called grade resistance and will increase the total train resistance when going uphill and decrease total resistance for downhill. But for climbing airplanes, one doesn't normally call it (uphill) "grade resistance" or "climb drag". Instead it's commonly known as "excess thrust" or "rearward (longitudinal) component of airplane weight". A train descending a grade faces negative grade resistance which helps move the train down the grade. But a descending airplane wouldn't classify this as negative drag. Thus for both a train and plane, the concepts are about the same but the terminology is different.
There are some major differences in performance since, unlike a train which follows a fixed path set by the railroad track location, a plane has options of alternate paths and can select a decent (glide path) that will result in almost no power (thrust) needed for the descent while a train has no control over the downgrade steepness of the fixed rails and is often forced to waste energy by braking. Since a train has such low rolling resistance that it can coast (at a slow speed) down a 0.2 % grade. a lot of train energy is wasted in braking, especially since downgrades are likely to be curved due to mountains and hills where the downgrade speed is limited by the curves. For example, if a train at the slow speed limit downhill will only have a train resistance of 0.4% but the downgrade is 1.0% then only 40% of the trains potential energy will be recovered and the remaining 60% wasted in braking. This calculation makes use of the fact that to obtain the total of rolling and grade resistance, one merely adds the two and if they are expressed in percent terms, one just adds the percentages. In the above case, adding rolling (+0.4%) and grade (-1.0%) results in a -0.6% net resistance so to maintain steady speed requires a +0.6% additional resistance to cancel out the required -0.6% resistance. This resistance must come from the brakes resulting in zero net force applied to the train and thus constant steady speed down the grade.
In addition to the potential energy of a train due to its elevation above sea level, a moving train possesses kinetic energy due to its motion. It's equal to half the mass of the vehicle times the square of the velocity. A vehicle can coast on level ground due to its kinetic energy but due to the resistance force opposing the forward motion of the vehicle, it takes energy to coast and thus slows down the vehicle, using up some of its kinetic energy in the process. If the vehicle were to coast to a standstill, all its initial kinetic energy would be dissipated (for a useful purpose) since it has been used up in overcoming vehicle resistance.
Like potential energy, kinetic energy can either be dissipated by braking (lost as heat) or used up as useful work to overcome vehicle resistance as in the coasting example above (which also ultimately becomes heat). When a vehicle is accelerating, it is accumulating kinetic energy part of which will be utilized later on for overcoming vehicle (or grade) resistance and the remainder wasted in braking. Note that when kinetic energy is utilized for overcoming grade resistance it is actually just being converted to potential energy. The converse of this is a vehicle coasting down a steep grade and accelerating, thereby converting potential energy back into kinetic energy. Note that in these examples of going up or downgrade, not all the potential or kinetic energy gets converted to the complementary form of stored energy since some energy is consumed in overcoming vehicle resistance.
Kinetic energy is created by acceleration (vehicle gains speed) and potential energy is created by climbing grades. Part of the these accumulated energies are recovered in overcoming vehicle resistance (such as aerodynamic drag) but the rest is consumed in braking. Thus the energy required to be produced by the vehicle motors for a certain trip is just the sum of the energy required to overcome "normal vehicle resistance" plus braking losses. Note that "normal vehicle resistance" does not include grade resistance nor the force required for acceleration. The braking losses represent the non-recovered kinetic and potential energies created by the motors on a trip.
This knowledge allows us to estimate the mechanical energy used for a trip based on just the energy used by normal vehicle resistance, and not having to make a detailed analysis of energy used for going up grades and the energy used for acceleration. The catch is that to do this we must know the mechanical energy used in braking. However for airplanes this is trivial since relatively little braking is done. Planes do brake after landing on the runway and "Jet" airplanes actually have air brakes = speed brakes (wikipedia) that can slow down the plane but they are not used much. One possible use for them would be to slow the plane before touchdown. Trains usually have friction brake shoes on each wheel or disk brakes on each axle. They also brake with the locomotive by using the traction electric motors as electric generators. Traction motors are the ones that drive the wheels of either diesel or electric locomotives (or electric train-set where each rail car has its own traction motors-generators..
For an airplane it can select a downgrade (glidepath, approach angle) such that it maintains reasonable speed and lift with the engines idling (power-off approach). Since airplanes brake mainly during landing on the runway (speed brakes first and then friction braking of the rubber tires plus reverse thrust from the engines), For a long trip the energy lost in braking at the destination airport is relatively low and may be neglected in an approximate analysis, such as in this report.
To convert vehicle resistance to what is equivalent to (specific) energy intensity, one may divide the resistance force by the vehicle weight to get a specific resistance (trains) or use the inverse: lift-to-drag ratio (airplanes) since lift is the plane's weight and drag is the plane's resistance. Traditionally, in the US, trains used units of pounds of force per ton. If it was say 20 pounds per ton (20 lb/ton) then it would take a pull of 20 pounds to keep a one ton of train moving forward at steady speed on a level path. Other units of specific resistance are kilogram-force per metric-ton (kgf/tonne) or newtons-force per metric ton. Since a metric ton is just 1000 kg, the unit of kgf/tonne is just parts per thousand (independent of what units are used. For example a resistance of 10 parts per thousand would be the same as 10 pounds force per 1000 pounds weight or 10 kgf/tonne, the same as the last example for a train. For a plane the specific resistance is much higher, about 50 kgf/tonne (a lift to drag ratio of 20). Why a plane with much higher specific resistance than a train can be just as energy-efficient for hauling passengers as a train is the subject of this report.
Both a train and airplane must be elevated off the ground to be able to move forward. The train does this by rolling on wheels while the plane stays off the ground by "floating" in air. In both cases it requires energy to do this. For a train, it takes force to keep the wheels rolling known as "rolling resistance". For an airplane there are various theories of the energy required for lift and why a plane can "float" in the sky. An unfinished analysis of this along with a new idea regarding pressure applied to the earth is found in Appendix-4: Airplane, Energy Needed for Lift
For a train, rolling resistance is the force opposing the forward motion of a train due to the rolling of its wheels. It doesn't include the wind forces. At low speed in still air on level ground, almost all the resistance is rolling resistance. If you've ever pushed an automobile by hand, you should know what rolling resistance is: it's the force you have to push the car with to keep it rolling at steady speed on level ground. Besides rolling resistance, there's another resistance known as aerodynamic drag. This is the wind force against the vehicle. You can feel this force by sticking your open hand out the window of a fast moving car. Aerodynamic drag increases almost proportionally to the square of the velocity so it's not very significant at low speed. It also takes additional force to move a vehicle up a grade or to accelerate its speed but these are neither rolling resistance nor aerodynamic drag.
to-redo Rolling resistance may be given as a percentage of the vehicle weight. For example, if an automobile weighs 3000 lb. and has 1% rolling resistance, then it would take 30 lbs. (1% of 3000) to push it slowly on level ground. But rolling resistance is commonly expressed in units of "per thousand" which is ten times the percentage value. Railroad steel wheels on a steel rail have a low rolling resistance of between 1 to 2. The 1 value is for a fully loaded railroad freight car at low speed, while the 2 value is for an empty freight car at low speed. However, rolling resistance goes up a little as speed increases so passenger train rolling resistance at higher speeds may exceed 2 (parts per thousand)..
For railroads, rolling resistance (in percentage terms) is lower when the vehicle is heavier. With a heavier load, the total rolling resistance goes up, but not as fast as the load increases. This "economy of scale" is due in part to the spreading out of the pressure caused by the heavier wheel load along a longer length of rail. It one doubles the wheel load, the pressure under the rail doesn't double, because the additional force is spread out over a longer length of rail. Of course, there's a trade-off since very heavy loads will cause more damage to the roadbed.
In addition to the wheel energy loss, there are some other losses that contribute to rolling resistance of rail vehicles. These are: friction losses in the wheel bearings, shaking and vibration of both the roadbed and the vehicle (including energy absorbed by the vehicle's shock absorbers), and slight sliding of the wheels on the rail. Such losses are quite significant for rail since pure rail rolling resistance is so low.
For rail, pure rolling resistance (under ideal conditions) is only about a third of the total rolling resistance with a value of about 0.33 for a fully loaded freight car. Significant amounts of rail energy are used in shaking/vibrating the earth (and the rail cars). The wheels not only roll but they also slide a little from side to side, thus using energy. A pair of rail wheels are rigidly mounted on the same axle so the wheels on each side of the rail car spin at the same angular velocity (rpm). This may result in slipping if the wheel diameters are slightly different or when going around curves. It's actually a lot more complicated than this since wheels are made with the tread slightly tilted so that by moving sideways a little the same wheel will in effect become slightly larger/smaller in diameter. While auto tires make contact with the road over their entire width of the tread, rail wheels only make contact over only small part of their tread width (about the size of a dime). Thus they can vary the part of the wheel that they ride on by shifting sideways. This happens automatically and the wheels tend to move such so that the slipping is reduced. But this reduced loss still contributes to the rolling resistance. Rail rolling resistance increases with speed, especially if the track is in poor condition and has dips in it.
The question still remains: Why aren't passenger trains more energy efficient if their rolling resistance is so low? There are a number of reasons, the major one being that trains are usually quite heavy (on a per passenger basis). Previously, the units used were rolling resistance per unit weight. But the unit that counts is the rolling resistance per passenger and this become high due to the high weight of passenger trains.
Just how heavy are passenger trains? There are various types of trains, some pulled by heavy locomotives and some that are driven by electric motors under each car. The ones pulled by locomotives tend to be very heavy and estimates made from US government data for 1963 (the government ceased collecting such data after that date) indicate about 3.7 tons/passenger. Airplanes are roughly one ton/passenger with about 80% of seats occupied (see Airplane weights). Thus rail was (in 1963) about 3.5 times heavier per passenger than airplanes today. The airplane weight includes about 400 lb. of fuel per seat,
Most airplane flights don't make intermediate stops to refuel so they need to carrying a large amount of fuel with them on takeoff. This fuel often weighs several times the weight of the passengers aboard. As a result, he takeoff weight per passenger may be almost double the landing weight per passenger after a long flight that has consumed most of the takeoff fuel before landing. In spite of this extreme burden of carrying a lot of heavy fuel (kerosene type) the airplane is much lighter per passenger than a train.
The Acela electric trainsets introduced by Amtrak in the early 21st century, are 2.1 tons/seat. With the load factor of about 50% this is about 4 times the plane-weight/passenger of airplanes. The heavy weight of trains not only increases rolling resistance, it also increases the energy used for climbing up a grade or accelerating from a stop. If the weight triples, so does such energy use.
As a vehicle goes faster, the wind force against the front of the vehicle increases. You may get a "feel" of this force by holding your open hand out the window of a speeding automobile. Since about 2/3 of the aerodynamic drag of an airplane is a cost of keeping the plane in the air the remaining 1/3 of aerodynamic drag cost is for the purpose of transporting the passengers and crew in an enclosed space: the fuselage.
This aerodynamic drag is almost proportional to the square of the velocity. For a train, it not only acts on the front of a vehicle but also on all other surfaces of the train. It even creates a suction (partial vacuum) on the rear of the train which opposes its forward motion. If there is a natural wind blowing over the land, the aerodynamic drag is likely to increase considerably (even if the wind is blowing from the side). For a train, such a wind constantly injects fresh air between the train cars and energy is consumed by accelerating this air to the speed of the train.
A plane has very high aerodynamic drag due to its high speed, But the drag is much reduced due to extreme streamlining and also due to the low density of air at cruising altitude of 37,000 ft. altitude. About 2/3 of aerodynamic drag is expended in keeping the plane aloft in the sky and is compared to a train.
In order to determine the fuel energy consumed, the mechanical energy used must be divided by thermal efficiency of the engine-system. For a train diesel-electric locomotive, the "engine-system" consists of a diesel engine and an electric transmission. For a plane, it consists of turbofan engine that aided by a large fan which throws a high speed stream of mostly air (along with some exhaust) out the rear of it.
Thus a plane seemingly has no transmission losses, but actually it has something like it, since it transmits force to the atmosphere by pushing air backwards (opposite to the direction of flight) while a train locomotive pushes back directly on the earth via the rails and ties. The air the plane pushes back on is not confined and as a result it flows backwards behind the plane, leaving a backward flowing wake (with respect to the earth) of air behind the plane. This flowing air (with a little exhaust gas mixed in) has "kinetic energy" and represents energy wasted in generating engine thrust. This is something akin to transmission losses for a train. The efficiency of this type of "transmission" is known as 'propulsion efficiency". Manufacturers of turbofan engines for "jet" aircraft don't normally provide data on propulsion efficiency but include such losses in the data on engine efficiency, which for the Pratt & Whitney JT9D-7 turbofan engine used on the Boeing 747 was about 29% efficient. The later Boeing 787 was almost 33% efficient.
For a locomotive, if its transmission is 80% efficient and the engine converts 35% of the fuel energy into mechanical work, the efficiency is 0.8 x 0.35 = 0.28 (i.e. 28%). Since mechanical work is measured at the driving wheels, transmission energy losses must be included. But this 28% is at nominal speed and torque and under actual operation it averages out lower as will be explained.
For electrically powered trains, one should not only consider the efficiency of the electric motors on the train, but the energy loss in transmitting the electric energy from the power plant to the train. Of course, the efficiency of the power plant in converting fossil fuel energy into electricity must be counted too.
So what are typical values for such efficiency? It turns out that for both trains and planes (around the year 2000) this nominal efficiency was roughly 30% under the best conditions with planes a little over 30% and trains a little under. This includes electric trains, which in some cases can also be more efficient overall due to regenerative braking.
But actual conditions for trains are mostly far worse than the best. The actual thermal efficiency is often well below 30% and can even be zero. For example, if a train is going down a gentle grade with the electric motors spinning at cruising rpm, but travelling only slightly faster than the coating speed down the grade the overall efficiency of the motor-generator is quite low. It takes power to keep the motor (diesel and electric) tuning but little net torque is being produced and most of the power from the fuel is being wasted just to keep the motors spinning.
Any motor that is operated at low torque (low power output for the speed it's turning at) will have low efficiency. This is even true for electric motors powering trains at extremely low torque. At very high current (and torque) the efficiency of electric motors drops also since ohmic losses are proportional to the square of the the current. Although the efficiency of the electric motor alone is much higher than that of an internal combustion engine (gasoline or diesel) an electric motor that drops to say 60% efficiency will result in an overall efficiency of about 20% due to the losses in the generation and transmission of electricity.
Thus the overall efficiency of a train motor-transmission (electric) system is perhaps only 2/3 that of an airplane engine. 30% (air) vs. 20% (train). There are also engine idling losses and locomotive engines are sometimes kept idling for hours at destination stations. Plane engines usually idle during the long descents when the plane is gliding towards its destination airport.
This is the energy used to either heat or cool the passenger/crew space plus the energy used to light up such space. For trains it goes by various names: "hotel load", "head-end power". The lighting energy is normally significantly less than the heating/cooling energy.
For passenger trains, a significant amount of energy is used to air condition the cars (including electric heating on diesel trains). When it comes to heating and cooling of passenger/crew space, planes use far less energy than trains (per passenger-mile). Both trains and planes have a lot of air blowing against their outer surfaces and transfer much heat thru them. But planes fly in the cold air at high altitude and get heat as a byproduct of compressing air for the passengers (the compresses air is warm). Trains not only expend more power on heating/cooling, but since a train is in motion for several times longer than a plane trip (due to its slower speed) the amount of energy used (per passenger-mile) is much greater since energy is the product of power times time.
Passenger trains today usually use electric heating which is quite inefficient since there is a high loss of energy in generating the electricity. Airplanes heat the air as a byproduct of compressing it. They don't need special compressors for this since an integral component of the turbofan engine (jet) is the turbine compressor from which a relative small amount of compressed air is diverted for the passengers to breath.
Passenger trains use electric air conditioning. For an airplane, the outside temperature is cold due to very high altitude (just like it's cooler in the mountains at higher elevations). But compressed air for the passengers may be too hot (also considering the natural heat generated by the human body) in which case the cold air outside the airplane is used to cool incoming compressed air by use of a heat exchanger.
Due to its higher speed, an airplane generate several times the passenger-miles per year as a train and also generates several times the revenue. Also, the high airplane speed results in less time needed for the crew per seat-mile and thus a savings in labor costs as compared to rail. Thus there is much more money available to spend on research and manufacturing to create energy efficient airplanes.
If an airline fails to be operated efficiently (including reduction of fuel costs) it goes bankrupt. But passenger rail is subsidized and thus can still continue operating even if it is inefficiently managed and operated.
Railroads (at least in the United States) have a history of bad management and of labor unions which required excessive staffing and excessive pay. This seems to have carried over to Amtrak operations which contract with the railroad companies to haul their trains. See Wyckoff's book
By its very nature, the airplane recovers almost all of this energy, but a train doesn't. So it's only a problem for trains. This additional energy that the heavy train must use to accelerate and climb grades can be partially recovered in two ways.
The first method is by coasting. A train approaching a slower speed zone or a stopping point can coast instead of brake. Technically speaking, it is recovering some of the kinetic energy of the train. Then when the speed is low enough, it can apply the brakes. Such a coasting scheme slows down the average speed of the train and thus is apparently not used much even though it is worthwhile in most cases to at least do a little coasting.
The second method is by use of regenerative braking for the case of an electric railroad. This is where the electric motors on the train work as generators during braking. The electric energy so generated is returned to the overhead wire and used to power other trains (or even returned to the power grid). For low voltage systems, there usually needs to be another train nearby which is under power and can absorb this energy. Due to losses in electric generators, wires, and electric motors, only part of the kinetic energy of a braking train is recovered. Not all electric railroads and equipment can do regenerative braking.
One may show that even if one has regenerative braking available, it is still better in most cases to coast for a ways before applying it. The reason is that for coasting, all of the kinetic energy decrease of the train is recovered (with no losses). For regenerative braking, there is energy lost due to generator losses, ohmic losses of the current in the overhead wire, and losses in the electric motors of the train that eventually receives this regenerated energy. There is also losses in power electronic circuits used to transform voltages, etc. For example, only 60% of the kinetic energy loss of a braking train may find its way into an increase in the kinetic energy of the train that gets the regenerated energy. The other 40% is wasted in heat.
Nevertheless, a combination of coasting and regenerative braking can recover a significant amount of energy and the coasting is almost free, since computers that can control it are quite cheap today. In the old Soviet Union, rail coasting policies were developed and monetary incentives given to locomotive drivers who saved energy. But it didn't work out too well due to congested rail lines and dispatchers giving the green light to friends so that they would get undeserved bonuses. The Soviet railroads reported a 20% savings of electricity due to regeneration on lines with steep grades and high traffic. See Electric Railroads
While a train is confined to the fixed path of the rails, a plane can:
This section will present an approximate example which will show that the efficiencies in terms of passenger-miles per gallon of fuel are about the same for trains and planes, and why it's so. Start with specific resistance (drag) in parts per thousand (ppt). For a plane with a lift to drag ratio of 20, it's 1/20 lbf per lb of plane or 50 ppt. For a fantasy train that had only rolling resistance (and this would be true at a low speed of under 25 mi/hr where aerodynamic drag is negligible) it would be only 2 ppt. Since 50 ppt is 25 times 2 ppt, the plane has 25 times more specific drag (resistance) than the train. That is, for each ton of vehicle mass, there is 25 times more force opposing the plane's forward motion thru the air than for the train. It would thus take 25 times more mechanical work to push it thru the air than the fantasy train. After applying numerous corrections to turn this fantasy train into a real one the train's 25-fold advantage will disappear (no advantage).
Let's begin. A real train goes pretty fast and has aerodynamic drag added to rolling resistance. Looking at Fig. 25 in Hoerner-rail at around 80 mi/hr it looks like aerodynamic drag (plus a slight increase in rolling resistance due to speed) is about 1.5 times the rolling resistance at low speed. Thus we add 3 ppt aero. drag to 2 ppt rolling resistance to get 5 ppt total resistance. But there is also the train kinetic and potential energy dispersed in braking and the train energy used for air conditioning (much higher than for a plane). The typical values are not really known and need more research to estimate, but a guess might put the sum at 3 ppt to be added to the 5 ppt resulting in a new total of 8 ppt. Compare this with the previously mentioned 50 ppt drag for an airplane and we see that the airplane is still over six times more (worse).
But we have only compared the mechanical work to move a ton of vehicle and not the fuel needed to move a passenger one mile. To correct this, one needs to multiply the 8:50 ratio by the product of 3 other efficiency ratios.
Let's estimate these ratios and then find the resulting ratio.
Based on the estimates at the start of this section, the percentage of energy expended by the train (rail) and airplane (air) for various purposes is:
Note that the last two figures for rail, 25% and 12% are guesses based on what the author recalls from a simulation study for NE corridor electric trains and an old article from Trains magazine. The citations were not saved and the author's recollection may be faulty. Furthermore, this data was for only one particular case and not for typical passenger trains. More research is needed.
The question is: why are both trains and planes about equal in energy efficiency (passenger-miles per gallon). Both cruising planes and trains require energy (power) /to stay afloat (off the ground) plus aerodynamic drag of their passenger and crew compartments. Airplanes use much more energy than trains to keep off the ground (keep elevated), but trains are burdened by wasting energy in braking and air-conditioning. Planes, not being confined to a fixed rail trajectory or loading gauge can be designed for optimal streamlining and choose optimal trajectories. But even so they use the same order of magnitude of non-lift-aerodynamic-drag-energy per passenger as a train. But trains, in contrast to planes, use a substantial amount of energy for braking, and air-conditioning.
The key advantage of trains, very low rolling resistance of a steel wheel on a rail, is partially canceled out by the high weight of passenger trains. The higher weight also means more energy used for accelerating and climbing grades although some of this might be recovered by coasting and/or regenerative braking. Aerodynamic drag is low for a train at moderate speed but increases rapidly (almost as the square of the speed). Thus one may say that passenger trains are potentially energy efficient, but in actual practice such trains turn out to be no more energy-efficient than the airplane.
What institution changes are needed to realize the potential of rail's inherent energy-efficiency are not clear. Neither private ownership nor government monopoly has been very efficient in providing economical and self-supporting passenger service in most countries. An exception may have been the Soviet Union (see Rail Transport in the Soviet Union).
Per Transportation Energy Data Book for 2005 Amtrak used 2,709 BTU/pass-mi vs. 3,264 for Certified U.S. air carriers. However, for air, a footnote warns: " These energy intensities may be inflated because all energy use is attributed to passengers --cargo energy use is not taken into account.
Correcting Amtrak for circuity results in 2,709 x 1.2 = 3251 BTU/pass-mi. The airplane Cargo correction results in 3,264/1.19 = 2,743 BTU/pass-mi. These corrections change the "reported" 20% advantage of the train to about a 18% disadvantage for 2005..
A major problem is how to allocate aircraft fuel between freight and passengers since passenger flights often carry mail and freight (other than the baggage of passengers). The situation is further exacerbated by Transportation Energy Data Book allocating all fuel used in scheduled "freight (cargo) only" airplanes to passenger travel (but the statistics required for a correct allocation are not available). [Hirst] p. 35, considered 400 lbs. of mail/freight equivalent to one passenger. This may be too low so I've used 500 lbs. (or 4 passengers = 1 ton cargo)
Example calculation of BTU/passenger-mile for 2012 using Transportation Energy Data Book, edition 32, Table 9.3 "Summary Statistics for U.S. Domestic and International Certified Route Air Carriers (Combined Totals), 1970-2012": In 2012, 2,307 trillion BTU of jet fuel was used to transport 832.7 billion passenger miles plus 34.9 billion ton-miles of cargo. Step 1: Find the equivalent pass-mi = 832.7 + (4 x 34.9) = 972.3 billion equiv. pass-mi. Step 2: Divide 2,307 trillion BTU of fuel by 972.3 billion equiv-pass-mi giving 2,373 BTU/passenger-mile.
If we neglect freight and erroneously assume that 832.7 pass-mi was transported by this fuel, then energy-intensity is 2,307/.8327 = 2770 (instead of 2,373) BTU/pass-mi, which is too high since we have counted fuel used for transporting freight as having been used to transport passengers. Unfortunately, Transportation Energy Data Book's table 2-14 (in edition 32) "Energy Intensities of Nonhighway Passenger Modes, 1970-2011" makes this error (but has a warning footnote that intensities "may" be inflated due to not accounting for cargo energy use). ..."may"...; what an understatement! This error exists from the beginning of this publication around 1970. The author's complaint to them got no reply but shortly thereafter the above mentioned footnote was added.
Amtrak Revenue Passenger-Miles and Load Factor by the Bureau of Transportation Statistics (US). In 2004-14 it was about 50%.
Load Factor (passenger-miles as a proportion of available seat-miles in percent (%)) All Carriers - All Airports by the Bureau of Transportation Statistics (US). In 2010-4 it was about 82%.
Most airplanes fly more or less on a straight line (great circle) directly to their destinations while passenger trains take a more circuitous route since they must curve around and thru mountains and hills and serve cities enroute which often don't line up in a straight line on the map.
See "Energy Intensity and Related parameters of Selected Transportation Models: Passenger Movements" by A. B. Rose. Oak Ridge National Laboratory, U.S. Dept. of Energy, Jan. 1979. Table B.3 "Mean passenger Rail Circuities by Distance Category" p. B-4 shows a mean circuity ratio of about 1.4. However, city pairs with poor circuities likely have less rail passenger traffic. It turns out that Amtrak had a circuity ratio of only 1.24, counting only riders who didn't switch trains. This report will use 1.2 which is on the low side. Airplane statistics are based on great circle distances so no correction is needed for air flight circuities. Since airplanes actually do have small circuities this is compensated for by understating aircraft energy-intensities due to assumed great-circle flight mileage.
For estimates from 1929 to 1965 see Bus Facts, 1966 (34th. edition) p.6: "Intercity Travel in the United States 1929-1965". Bus Facts was published by NAMBO = National Association of Motor Bus Operators.
From 1939-2007 see: "Transportation in America" (annual) by the Eno Transportation Foundation, Washington DC. (Formerly Transportation Facts and Trends, by the Transportation Association of America) Some of their "statistics" are now found in the "Statistical Abstract of the U.S."). The 18th edition of Trans. in America is available with a "Historical Compendium 1939-1999" which covers years that other editions omit. The table "Domestic Intercity Passenger-Miles by Mode" shows the modal split for intercity travel (but excludes international air travel).
After 2007 see: Bureau of Transportation Statistics (BTS)for rail and air but there appears to be no source for "intercity" highway travel since Transportation in America ceased publication in 2007.
To fairly compare rail and air passenger-miles one should include some international air travel on flights, since portions of these flights were sometimes over U.S. territory. But the figures used here are only for domestic transportation between points within the U.S. In the early days of air travel when there was railroad competition, international travelers from the U.S might first take a train to a seaport to board an ocean ship or to a border point to board a Mexican or Canadian train. This portion of the trip compares to part of a flight on an international air carrier where the flight begins somewhere deep within U.S territory. At the same time, air travel to Hawaii and Alaska shouldn't be counted as there were (and are) no railroads serving those locations. Thus these two errors tend to cancel out each other and no corrections have been made.
See Boeing 787 Aircraft Performance Takeoff weight 476k lb. with 242 passengers (100% load factor) => 1967 lb./pass. Operational weight (with crew and minimal fuel: 239k lb + 47k lb of 242 passenger (read from bar chart) => 286k lb => 286k/242 = 1183 lb./pass (optimistic) landing weight. Since Airplane Load Factors are about 80%, divide the above figures by .8 to get the wts. per pass.: Takeoff: 2460 lb/pass, low fuel landing: 1500 lb/pass. Taking an average of these gives a little under a ton/pass. While most flights will weigh more on landing due to reserve fuel, most will also weight less on takeoff due to flights of less distance than the maximum range, which require less fuel. Thus the typical weight per passenger is about one ton (2000 lb.) per passenger.
Note that for a long trips with the plane full of passengers and fuel, the fuel weight at takeoff may be up to 4 times the weight of the passengers plus luggage.
1. BART Prototype Care Development Program--Volume 1, Program Synopsis. Report No. UMTA-CA-006-0032-73-1 by Rohr Industries, Chula Vista, CA for the US Department of Transportation, Urban Mass Transportation Administration. March 1973. This is the rail transit system of the San Francisco Bay region in California.
p. 72 typical weights 58.5k lb. (B-cars); 60k lb. (A-cars). p. 73: 62k lbs. for tests to simulate zero passenger loads. P. 5: (Car and Train Configuration) 72 seats in either A or B cars. 60k/72 = 833 lbs./seat or 0.417 tons/seat. Crush load (most people standing) is 216. For this case the weight of the passengers is very significant and the vehicle weight becomes 98k lb. (AW-3 on p. 73 assumes 167/lbs per passenger). 98k/216 => 0.227 tons/person.
2. The Acela train-set weighs 1.9 tonnes/seat (equivalent to 2.1 tons/seat). See Acela Express
1. Elektricheskie Zheleznye Dorogi by Plaksa, A. V. et. al. Moskva, Transport 1993. See p. 55 for regenerative braking.
1. Ekonomiia Topliva i Teplo-Tekhnicheskaiia Modernizatsiia Teplovozov (Fuel Economy and the Thermodynamic Improvement of the Diesel Locomotive) by Komich, A. Z. et. al. Moskva, Transport, 1975. See p. 15, fig. 4 for how thermal efficiency depends on power output. Efficiency ranges from 13% at 13% power to 28% at 100% power. Note that the manufacturer claimed 30% maximum, etc. but actual tests showed lower values.
Since a diesel locomotive consumes about 10% of its fuel while idling (see p. 16) and operates at part loads where efficiency is lower than nominal, the average efficiency is only 21-22% (see pp. 6,18).
There exists a 1987 "revision" with a new title: Toplivaia Effektivost' i Vspomogatel'nye Rezhimy Teplovoznykh Diselei (Thermal Efficiency and Non-Standard Modes of Operation of Diesel Locomotive Motors).
2. 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 22% actual efficiency for diesel locomotives as found in reference 1. above. My conclusion: electric and diesel traction have thermal efficiencies of roughly 30% at nominal operating conditions (but significantly lower under actual operating conditions).
1. Soprotivlenie Dvizheniiu Zheleznodorozhnogo Podvizhnogo Sostava (Resistance to Motion of Railroad Rolling Stock) by Astakhov, P. N. 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. Plots rolling resistance vs. speed for various situations showing experiments data points.
2. Tiaga Poezdov (Train Traction) by Deev, V. V. 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. Rolling Friction (in 4 parts) by Hersey, Mayo D. et. al. 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.
1. Hoerner, Sighard F., "Fluid Dynamic Drag", published by the author, 1965. See Ch. 12: "Drag of Land Borne Vehicles": 5B "Drag of Railroad Vehicles" (p.12-10+).
2. Ober, Shatswell, "Air Resistance of the Burlington Zephyr", Diesel Railway Traction, June 14, 1935, pp. 1184-5.
3. Johansen, F. C., "The Air Resistance of Passenger Trains", The Institution of Mechanical Engineers (London), Proceedings, Vol. 134, 1936, pp.91-208. Covers increased drag due to yaw.
4. DeBell, George W., "Effect of Natural Winds on Air Drag", Railway Mechanical Engineer, April, 1936, pp.145-7.
5. Klemin, A. "Aerodynamics of the Railway Train", Railway Mechanical Engineer, Vol. 108 (1934): Aug. p. 282, Sept. p. 312, Oct. p. 357.
6. Lipetz, A. I. "Simplified Formulas for Calculating the Air Resistance of Trains", Railway mechanical Engineer, April 1935, pp. 129+.
7. Astakhow, P. N. "Soprotivlenie Dvizheniiu Zheleznodorozhnogo Podvizhnogo Sostava", Transport Press, Moscow, 1966 (in Russian). pp. 87-104 (LC call number: TF4M67, Vol. 311)
8. Deev, V. V. "Tiaga Poezdov", Transport Press, Moscow, 1987 (in Russian) pp. 81-2.
Per Energy Conservation - Train Lighting and Air Conditioning (in India) the first paragraph gives an example where about 22% of total energy used by a train is used for these purposes ("hotel load") and most of it is for air conditioning. Similar information is found in "Hotel Load on Indian Railways" by Anula Khare and Saroj Rangnekar in: International Journal of Emerging Technologies (ISSN 0975-8364) 2(1): 53-55 (2011).
In Head-end power | Trains Magazine by Bob Johnson, May 1, 2006, it mentions that when diesel locomotive engines supply both head-end power and traction power from the same engine, the power available for traction is derated by about 10%. This implies that if both the traction and head-end power is being supplied at full load then about only 10% of train energy is used for head-end power. An example of a very cold day in Chicago that needed a lot of head-end power for heating, shows head-end power of 550 kw with a 3000 HP locomotive which means about 25% of the locomotive power is going for head-end power providing the locomotive runs at full power. So in this case the percent of power used for head-end is likely more than just 25% since the locomotive normally doesn't use full power.
CRS Report: 96-22 - Amtrak and Energy Conservation in Intercity Passenger Transportation - NLE. Or alternate site if above link broken
Fuel Efficiency of Travel in 20th Century: Amtrak.
The ICC publication used for car-mi by fuel changed names between 1936 and 1963. In 1936 they were issued by the ICC's "Bureau of Statistics". In 1963 it was "Bureau of Transport Economics and Statistics". In all cases "Switching and Terminal Companies Not Included" appears after the title (in smaller print). Below find the starting name of the publication in 1936 and the ending name in 1963.
1936: "Passenger Train Performance of Class I Steam Railways in
the United States" (Statement M-213)
1963: "Passenger Train Performance of Class I Railroads in the United States" (Statement Q-213)
A secondary source is Bituminous Coal Annual. The division of passenger train car-miles by fuel for the years 1936 thru 1950 may be found in the 1951 issue, p. 115 (table 49). As the use of coal by railroads decreased, later annual issues reduced and then eliminated publication of this data.
Wyckoff, D. Daryl, Railroad Management. Lexington, Massachusetts: D.C. Heath & Co., 1976. See pp.39-46 for a very incomplete and misleading analysis of labor issues. It ignores the "brakeman" issue and falsely implies that the problems were mostly solved. See pp. 96-98 for bias against college education.
Trains (magazine) from about 1966 to 1983 ran a column named "The Professional Iconoclast" by John G, Kneiling who told it like is was regarding abuses by labor in the railroad industry. He offended many and his column was eventually dropped. But it may be the best source of information on rail labor/management problems. He was not only critical of railroad labor but of management as well.
Phillips, Warren F. "Mechanics of Flight", John Wiley & Sons, Hoboken NJ, 2004. For turbofan engines see Ch. 2 "Overview of Propulsion" section 2.1 "Introduction" and section 2.9 "Turbofan
Anderson, John: "Fundamentals of Aerodynamics", 5th edition.
Hoerner, Sighard F., "Fluid Dynamic Drag", published by the author, 1965. For change in the drag coefficient with velocity see Ch. 14: "Complete Aircraft" Fig. 5 "... drag coefficients (on wing area) with full scale conditions", (p. 14-8). Recall that the Reynolds number is directly proportional to velocity.
See Boeing 787 Dreamliner : Analysis - Piano. Detailed technical information for this well known airplane. For example there are a number of color-coded charts that can be used to find "specific range" = the nautical-miles/pound-of-fuel, for various speeds, payloads and altitudes (but it uses Mach # as a proxy for speed).
This report is published, not by Boeing but by consultants in England that are using this as an example of both the data and data reduction programs that one can purchase from them for other aircraft.
This part of the above report is for level flight at 37,000 feet at a speed of 487.5 knots (561 mi/hr, Mach .85) The table: "Zero-Lift Component Breakdown" shows values of (Drag Areas = Cd*S = D/q) for various components of the aircraft (such as the wing) where Cd is the dimensionless drag coefficient and S is the area of the wing (one side --about half the wetted area). q is the dynamic pressure in lbf per square ft.: q = .5rv^2 where v is velocity and r is the mass density of air in lb/ft^3 or in (lbf*sec^2)/ft^4. One gets this later unit by recalling that a pound of force is equivalent to a pound of mass at 32.2 ft/sec^2 (acceleration of gravity). (It would be simpler in the metric system.). See Hoerner-air pp. 1-8. 1-10.
Since q depends only on speed and air density, the drag D (for a given speed and air density) will be directly proportional to Cd*S. Comparing "drag areas" (Cd*S) in this table (labelled Zero-Lift ...) one sees that the wing zero-lift drag is roughly equal to the drag of the fuselage. From the table "Drag Coefficients based on ref. area" one sees that the drag coefficients for the wing are .0127 for zero-lift (parasitic) and .0102 for "Lift induced". (While this report fails to specifically state that these coefficients are for the wing, it's implied since the value of the "ref. area" is the same as the wing area per the "Geometry report" section). All of this shows that there are 3 major drag forces that are very roughly equal: fuselage drag, wing parasitic (zero-lift) drag, and wing induced drag to create lift to hold up the plane in the air. Note that "wing" above means both right and left wings combined and thought of as one wing. This method of allocating energy between energy used for lift and energy used to move passengers/crew/freight inside the fuselage throught the air is only approximate. For a discussion of this approximation see Estimating Lift Energy.
However in estimating the drag needed for lift, one should use the sum of the two wing drags (zero-lift and lift-induced) since if there was no need to provide lift, the wings wouldn't be needed so there would be neither zero-lift nor lift induced drag. The lift induced drag directly provides lift while the zero-lift drag represents the energy needed to move the wings thru the air (without supplying any lift).
In the table "Engine/Airframe Performance" of the "Aerodynamic Drag Report" the specific fuel consumption (sfc) is shown as .5279lb/hr/lbf (pounds of fuel per hour per pound of thrust) at Mach # .85 at 37,000 ft altitude. Compare this to the Boeing 747 (With JT9D-7 Pratt & Whitney engine) with a sfc of .595 lb/hr/lbf at Mach .80 at 35,000 ft (per Pratt & Whitney "Fact Sheet").
Note that sfc is not an energy intensity since the output of an engine should be power and not just force (lbf). To find the energy intensity one must convert force to power output by multiplying it by airplane velocity. The aircraft data doesn't give velocity explicitly but it does show Mach # and altitude from which one can look up velocity on standard graphs showing the relationship between velocity and Mach #.. One must also convert fuel flow in lb/hr of jet fuel (kerosene type) to power (such as watts of fuel energy). The figure one obtains this way is fuel-power-in per unit of mechanical power out (the inverse of thermal efficiency of the turbo-fan engine). It results in a thermal efficiency of roughly 30%.
The lift to drag ratio is shown in a "Drag Tabulation" table and in a "Lift/Drag" = L/D plot which shows the L/D ratio as a function of Mach # and lift coefficient Cl. The L/D ratios shown are typically around 20, but at the specified Mach# and Cl of cruise (at 37,000 ft.) the L/D is 20.8. The D/L is thus 4.8% or 48 parts per thousand specific resistance (in railroad terminology). Since wing drag is roughly 2/3 of the total drag and fuselage drag about 1/3 we have a specific wing drag of 3.2% and a specific fuselage drag of 1.6% (much higher than specific resistances for a railroad train, counting both aerodynamic drag and rolling resistance).
Except for 1935, the market share percentages (percent of intercity passenger-miles by all modes) is from "Transportation in America" (annual) by the Eno Transportation Foundation, Washington DC. (Formerly Transportation Facts and Trends, by the Transportation Association of America) Some of their "statistics" were found in the "Statistical Abstract of the U.S."). The 18th edition of Trans. in America is available with a "Historical Compendium 1939-1999" which covers years that other editions ignore. The table "Domestic Intercity Passenger-Miles by Mode" shows the modal split for intercity travel, but only includes domestic air travel (excludes international). Unfortunately, Transportation in America ceased publication after 2007.
For 1935, the data is from Bus Facts, 34th edition, 1966: Table: "Intercity Travel in the United States" p. 6 (published by NAMBO = National Association of Motor Bus Operators).
Sources of airline energy statistics:
1935-1955: Schurr: Energy in the American Economy, 1850-1975 p.550
1950-1970: Hirst: Energy Intensiveness of Passenger ,,, Modes EI 1950-1970 pp. 9, 21, 34-5
1970-2015: Transportation. Energy Data Book. Annual
For airplane fuel efficiency for 1935-1955 see the book: "Energy in the American Economy, 1850-1975" by Sam H. Schurr and Bruce C. Netschert. The Johns Hopkins Press, Baltimore MD, 1960. See p.550, Table XXVI: "Petroleum Products Consumed Compared to Work Performed by Aircraft, Selected Years 1935-1955".
In using this data, I assumed that one ton of cargo is fuel-equivalent to 4 passengers. All reported jet fuel use by this Table xxvi was disregarded, since per footnote c only an insignificant amount was used by "scheduled air carriers" (most of jet fuel was likely used by the military). International flights (not shown in the table: Historical Energy-Efficiency were significantly less energy-efficient than domestic prior to 1955, but by 1955 there was only a little over 10% difference. For a table which shows the efficiency of international flights prior to 1955 see the webpage Fuel-Efficiency of Travel in the 20th Century
Heat values used represent the higher heat of combustion: automotive gasoline: 125 k BTU/gal, aviation gasoline: 120.2 k BTU/gal and for jet fuel (kerosene type): 135 k BTU/gal. Actual reported gallons consumed was converted to both BTUs and to equivalent automotive gasoline, using these values which are from Transportation Energy Data Book edition 26, 2007, table B.4: "Heat Content for Various Fuels". Note that aviation gasoline was phased out during the period centering around 1960 and replaced by kerosene jet fuel which has a higher heat value.
Energy in the American Economy" has an erroneous Table D-51 (p. 667): "Miles per Gallon in Passenger Service of Domestic Scheduled Air Carriers". It wasn't used due to its failure to account for aircraft fuel used for cargo and mail transport by both cargo and passenger aircraft. The same error was made by TEDB. If one uses Transportation Energy Data Book edition 26, 2007, table 9.2: "Summary Statistics for U.S. Domestic and International Certified Route Air Carriers" and assumes that one ton-mi of cargo is equivalent to 4 pass-mi one obtains the equivalent of almost 38 pass-mi per gallon of automobile gasoline for the year 2000.
For passenger trains in the US from 1936 to 1963 see: USA Railroad Passenger-Miles per Gallon 1936-1963 by David S. Lawyer.
Transportation Energy Data Book (annual) by Stacy Davis, Oak Ridge National Laboratory, U.S. Dept. of Energy. The most significant table is: "Energy Intensities of ... Passenger Modes" in Ch. 2. The latest edition is on the Internet. Ch. 2 is on the web at Transportation Energy Data Book, Ch. 2. Ch. 9 is on the web at Transportation Energy Data Book, Ch. 9 (for table 9.2 on airplane energy).
Energy Intensiveness of Passenger and Freight Transport Modes 1950-1970 by Eric Hirst. Oak Ridge National Laboratory (then part of the U.S. Atomic Energy Commission, now part of the U.S Dept. of Energy), 1973 (ORNL-NSF-EP-44).
Based on the formula used for aerodynamic drag, showing it directly proportional to the velocity squared (v^2), some erroneously think that aerodynamic drag is directly proportional to v^2. But they have failed to note that it's proportional to the product of C and v^2 where C is the drag coefficient which is dependent on both velocity and other factors. As velocity increases, C decreases so that while the v^2 term strongly increases with velocity, this increase is reduced by decreasing C. The change in C with velocity depends on many factors including velocity itself. See Anderson p. ? A rough estimate for aerodynamic drag is that it's directly proportional to v^1.8, based on Hoerner-air .
Note that per Historical Energy-Efficiency Table energy intensity increased after 1955 from about 5k BTU/PM to about 9k BTU/PM in 1970. This was mainly due to the introduction of inefficient "jet" aircraft. But after 1970 the energy intensity decreased to less than it was in 1955, due in large part to the introduction of turbofan engines with high bypass ratios. See Mechanics of Flight
The word "jet" regarding passenger aircraft is actually a misnomer. While some military aircraft use turbojet engines, most passenger commercial aircraft today use turbofan engines. Although people call them "jets", they are similar to propeller craft where the propellers are concealed from view and are called "fans". If you could see thru the sheet metal cylinder (called a nacelle = engine housing) surrounding a "jet" engine you would see a somewhat small diameter engine proper, surrounded in front by a large fan of over twice the diameter. The fan is an integral part of the gas turbine engine, powered by kerosene type jet fuel. The large fan (which surrounds the turbine) sucks air into it from its front and throws this air out the rear of it, pushing the plane forward much like a propeller. Some of the "air" thrown back is exhaust gas from the turbine, but most of it is just plain air. This air bypasses the turbine and is not used for combustion. It just flows backwards around the outside of turbine. The bypass ratio is the mass of the bypass air to the mass of the air used for combustion. So an engine with a high bypass ratio will have a high volume of air go thru the fan and will be more energy-efficient as will be explained later.
Originally, passenger "jet" aircraft were turbojet but by the early 1960s, turbofans with low bypass ratios were being introduced. For example, early Boeing 707's were turbojet but later 707's were turbofan. The introduction of low-bypass ratio turbofans and then increasing the amount of air bypassed (high-bypass ratio) resulted in later models "jet" aircraft being much more energy efficient than earlier models. Some additional improvement was obtained by by increasing compression ratios,
Why are high-bypass-ratio turbofan engines more efficient? We'll take a simplified example where an airplane is flying in level flight at constant speed thru still air (no natural winds). This airplane shoots out a mixture of exhaust gas and air from its turbofan engines, leaving a mixture of air and exhaust moving backwards, which we'll just call exhaust. Now the kinetic energy of this exhaust is just the mass of it times one-half of the square of the velocity (to an observer looking up at the plane from the ground). All this kinetic energy of the exhaust in the sky moving backwards was supplied by the kerosene type jet-fuel. Cut this velocity in half and you reduce this energy loss by a factor of 4 (due to the v-squared law where v is velocity).
Except there's one problem if this velocity is halved. The airplane always sees the velocity of the air entering the engine from the front as equal to the speed of the airplane. The thrust force generated by acceleration of air in the fan (and in the turbine too) is the air flow (in kg/sec) times its change in velocity (as viewed by an observer in the plane) and if you think about it, this change is just equal to the exhaust velocity (with respect to the ground) as viewed by an observer on the ground. So if you halve the velocity of the exhaust you halve the change in velocity (between intake and exhaust) as viewed by an observer on the airplane and thus halve the thrust force. But thrust force must be maintained so as to maintain speed (or climb). Thus if one cuts in half the velocity change, one must at the same time double the exhaust flow to keep thrust constant. This doubling is done by increasing the bypass ratio.
The doubling of exhaust flow doubles the wasted energy in the moving exhaust air. But remember that halving the velocity has decreased the this energy by a factor of 4 so the overall result is cutting in half the wasted energy of the moving exhaust in the sky after the plane passes by.
Note that the fuel consumption has by no means been cut in half. There is still a lot of air being directed downward by the wings so as to push up on the wings to keep the plane from falling out of the sky. And there is air being pushed forward as the front of the aircraft hits the still air. So why did high bypass ratios make such a big difference? It was because the ratios were greatly increased and the velocity was reduced much more than in the example just presented
The above example is overly simplistic as it has neglected the part of the exhaust that represents the kerosene type jet fuel (the carbon and hydrogen components of the exhaust in the compounds of carbon dioxide and water). But carbon and hydrogen constitute only a few percent of the exhaust (for bypass ratios greater than one) and can be neglected in an approximate example.
For an interesting history of the turbofan aircraft engine see: Sharke, Paul "Mechanical Engineering; 100 years of flight" 40 pp. a supplement to Mechanical Engineering Magazine, De. 2003. Was formerly on the Internet at url="http://www.memagazine.org/supparch/flight03/jetsfans/jetsfans.html".
The estimation of the energy needed for lift, as used in this report, is simply the total aerodynamic drag on the wing (times distance travelled of course). But this is just an approximation since there are a couple of other things to consider: fuel tanks in the wing and interference drag at the wing-fuselage junction.
The approximation reasoning implicitly assumed that the since the wings hold up the airplane in the sky (provide lift) that if no lift was needed then the wings wouldn't be needed and then the energy used by the wings (due to aerodynamic drag) would be saved. But it's not quite this simple.
Wings not only hold up the plane in the sky but serve as fuel tanks to hold the aircraft's fuel. No wings, goodby wing fuel tanks. But since fuel tanks are essential they would need to be relocated in the fuselage. Since a plane that didn't need wings (or lift) would only require about 1/3 of its previous power to fly, the new fuel tanks in the fuselage would only need to be about 1/3 the capacity of the tanks in the wings. But if the plane could fly without lift, then it wouldn't take any energy to hold up the weight of the tanks with fuel in them. But it would require somewhat more volume in the fuselage which would mean slightly more aerodynamic drag. This consideration slightly increases the estimated energy used for lift.
But if wings were no longer needed, then there would be no interference drag at the location where the wings attach to the body (fuselage) of the airplane. This interference drag is the extra drag that happens when two components of the airplane (in this case, the wings and fuselage) are placed near each other (or in this case, touching each other). It's covered in Hoerner-air Ch. 8: Interference Drag, Sects. 8-4 thru 8-6 pp. 8-10 thru 8-19. There are two such additional drags near the wing-fuselage junction: one on the wings and another on the fuselage. But the reported wing aerodynamic drag figures used for estimating the energy needed to hold a plane up in the sky, may have included interference drag and thus we may only need to look at the additional drag applied to the fuselage due to the proximity of the winds. Per fig. 36 this interference drag coefficient, based on an area equal to the wing width squared (implied by subscript "c" [c=chord] of the drag coefficient) is about 0.001. It's for the extra interference drag on the fuselage ("body" per fig. 36) and depends on the wind location from the nose of the plane, x/l where l is plane length (See fig. 36).
How much does this elimination of interference drag decrease the drag of the fuselage, thereby increasing the % of lift attributed to the wings? Inspecting drag coefficients for fuselages in fig. 22, p.6-16 of Hoerner-air it looks like almost 0.003 which is three times higher than the 0.001 for the interference drag. But we must multiply such coefficients by the respective areas they apply to in order to get numbers proportional to aerodynamic drag. For the 0.003 the area is the entire "wetted" surface of the fuselage. For 0.001 its just the square of c (chord = wing width). If the later area is 1/3 of the wetted area then the drag of the fuselage will decease by 10%. This is likely larger than the increase in fuselage drag due to the fuel transfer from the wings to the fuselage. So the reduction of interference drag more than compensates for the increase in fuselage aero. drag.
There is a significant amount of aerodynamic drag due to the airplane's turbo-fan ("jet") engines but much of it is the internal friction of air and combustion products flowing through the engine and is accounted for by the engine efficiency. But there is also the aerodynamic drag of the engine housings (nacelles) which mount on the wings. If a plane didn't need lift (or wings) the engines (and nacelles) would need to be mounted on the fuselage, thereby increasing it's drag. How is this accounted for?
Well, the engines and their nacelles were neglected in the simple analysis of just looking only at wing drag and fuselage drag. The airplane "data sheet" also shows a numerical value for nacelle drag but it wasn't explicitly taken into account. However, there is an implicitly accounting created by this neglecting of it. That accounting is: if the total drag of the wing is twice that of the fuselage and it thus takes twice as much energy to hold up the plane and it does to move the passengers thru the air then 2/3 of the engine's energy cost of the engines is allocated to holding up the plane and 1/3 of it to move passenger thru the air. In that case, there would be no change in energy allocation if the engine nacelles were included in the accounting. It would remain 2/3, 1/3.
So moving the engines to the fuselage doesn't have any effect on this implicit allocation. Note that such engines would only need to be of 1/3 the power (and weight) of the original engines since without wings, only 1/3 of the normal power is required for the airplane to cruise through the sky..
This section was started before numerical estimates were found on the Internet for the Boeing 787's wing drag from which numerical values for the energy used to create lift can be estimated. Thus this section is no longer essential to this report and considering that simplistic theories are being challenged, and that this section is incomplete, it is therefore consigned to the appendix: Appendix: Airplane, Energy Needed for Lift. It discusses the supporting of the airplane's weight while in flight by pressure on the ground.
Both a train and airplane must be elevated off the ground to be able to move forward. The train does this by rolling on wheels while the plane stays off the ground by "floating" in air. In both cases it requires energy to do this. For a train, it takes force to keep the wheels rolling known as "rolling resistance". For an airplane, the wings push and pull air downward, thereby putting an upward force on the wings to hold the plane up in the sky. But pushing air down puts a backwards force on the wing known as "drag due to lift" and it takes energy to overcome this drag. But there is also the overhead energy cost of moving the wings through the air with no lift, known as "zero-lift-drag" or "parasitic drag". Trains have less rolling resistance than planes have drag-due-to-lift (including the overhead energy cost "zero-lift-drag" of moving the wing through the air). The details of all this (for both trains and airplanes) is far from simple an we'll start by describing the drag-due-to-lift of airplanes.
To help understand drag due to lift (as well as the zero-lift-drag on the wing) we'll explain how an airplane is held up in the sky by its wing(s). There are various ways of explaining this and the emphasis here is that the airplane is held up in the sky by applying a slight pressure to the ground underneath the airplane. see: Lift (Wikipedia) last paragraph of Momentum section. When a plane is say 7 miles up in the sky, this pressure is applied over an area of many square miles, and is so slight that it is hardly noticeable, yet the total force due to this pressure over a huge area on the earth, must support the entire weight of the airplane so it is not insignificant when considered as a whole. While there is little written about this, it is analogous to the case of the train where the earth supports the weight of the train (and the train expends significant energy by shaking the earth as it passes by). Airplanes overhead also slightly shake the earth but use most of their "lift" energy for compressing the air underneath the plane and throwing air downward, which results in a pressure at ground level sufficient to support the plane.
As a plane flies through the air, the wing (= wing pair) both scoops up air and pushes it downward and pulls air downward due to a partial vacuum that forms on top of the wings. The result is a large mass of air (called a downdraft) flowing downward just after a plane passes by. Pushing or pulling on something results in a reaction force in the opposite direction (per Newton's 3rd law) so throwing air down by the wing results in pushing up on the wing by the air along with pulling up the plane by the suction on top of the wings. Try pushing on a wall with your hands and you'll notice that there's a reaction force of the wall pushing you backwards, away from the wall.
But there are major problems when it comes to calculating the force that it take to throw the air down. One might think the Newtons 2nd law (force=mass times acceleration) would give the force (provided one could figure out what the masses were and how much the acceleration was for the various regions of the wing). But the actual situation is even more complex, since air is being thrown down, not into free space, but into existing air which resists being pushed down since it's under atmospheric pressure before the plane gets near and also has inertia which resists movement.
It's something like a fan blowing air forward. Not only does the fan accelerate the air as it goes through the fan but pressure develops ahead of the fan that puts back pressure on the fan blades. On the intake side of the fan, suction develops and this requires more energy to push air thru the fan. So the fan must not only overcome the pressure increases between the fan input and output but must also impart kinetic energy to the air per Newtons 2nd law. It's similar for a wing which is somewhat analogous to a fan blade
Just how much pressure is put on the ground by an airplane flying high overhead? Well, passenger aircraft may weigh about a half-million pounds, flies 7 miles high creating a large patch on the ground below it where pressure is applied. There seems to be no investigation as to the shape, size, and pressure distribution of such patches perhaps since the low pressures are in the noise level of varying air pressure due to natural winds, etc. But the higher pressures under the wing spread out below the airplane and not only are transmitted downward but also sideways, thus resulting in wide pressure patch on the ground. The patch obviously has fuzzy boundaries since the pressure is highest in the center and tapers off as one goes out from the center of the patch. Since the plane is 7 miles high, the patch might be considered to be roughly about this radius or diameter (a big difference) consisting of billions of square inches. Division of plane weight by area gives a pressure of under 1/1000 of a PSI which is less than 1/10,000 of an atmosphere.
There is downwash of air flowing downward, as well as increased pressure and this phenomena continues all the way down to the ground, although velocity may get transformed into pressure as it descends. When the ground (about 37,000 feet below) finally stops any remaining downward velocity, all that remains is pressure on the earth which supports the airplane weight. The downward air flow gets weaker as the air descends and the area over which it's descending gets larger so that long before the air reaches the ground, the downward flow of air has almost petered out although there must still be enough pressure put on the surface of the earth to hold up the plane. The picture here is of various volumes of air descending (and undisturbed air being slowly accelerated downward) at various speeds as well as pressure increases and velocity being slowed and transformed into pressure. This model doesn't seem to be found in classical mathematical aerodynamic "theory".
Another way to explain why a plane "floats" in the air is that there is low pressure above the surface of the wing and higher pressure below it, resulting in a net force pushing the wing up (so as to hold the plane up in the sky). It is sometimes claimed that the low pressure above the wing is due to the increased velocity of flow there as explained by the Bernoulli effect. See Bernoulli's principle But this explanation doesn't explain the actual skewed distribution of partial vacuum on top of the wing, which is much higher nearer to the leading edge of the wing. What really happens it that the air flows around the top of the wing in a curve so that the centrifugal force of this steam of air tries to push it away from the curved top of the wing, resulting in a partial vacuum at the wing surface. The wing-top is curved more near it's leading edge which explains the higher lift there. But air that is directly in front of the rounded leading edge is deflected by the high pressure in front of leading edge so that it flows around the wing rather that collides with it and the curvature here is such as to create centrifugal pressure pushing this edge both backwards (drag) and downward (anti-lift). See Low pressure above and Lift from flow turning (NASA) Note that it's double counting if one claims that a plane is held up in the sky due both to throwing air downward and by pressure differences on the surfaces of the wing. These are just two different ways of looking at the same phenomena.
The "pushing air down" approach (aka "momentum" method) makes it easy to write some equations estimating the energy it takes to "float" an airplane, but the actual phenomena is complicated by the questions: just how much air is pushed down and how fast does it descend? The complexity of this question is illustrated by case where one assumes that the tilted wing just slices off a thin layer of air ahead of it and directs it downward (something like a fan blade slicing into a chunk of air and pushing it out of a fan). The air flows downward, but below the wing is still more air which the first layer of downward flowing layer of air pushes downward. And below that is still more air which gets pushed down, etc.
Above the wing, air gets pulled downward and this phenomena extends upward far above the wing resulting in a complex picture similar to that below the wing. There are also things going on that don't help keep the plane afloat and only contribute to parasitic drag, such as when still air ahead of the plane first hits the leading edge of the wing and is deflected upward instead of downward.
It takes energy to throw air downward to support an airplane but how much energy? However some claim the the volume flow of air diverted downward at an angle (not straight down) is directly proportional to the square of the wingspan. See Deriving the Power for Flight Equations