200 Mile-Per-Gallon Cars?

Steven Dutch, Natural and Applied Sciences, Universityof Wisconsin - Green Bay
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Please Note:

I occasionally get visits from people with automotive expertise who ask whatabout gas-electric hybrids or who point out that there are far more accurateways to measure engine power needs.

I know all about that stuff. This page is not about those issues. This pageis solely about the widespread myth that there exists some simple gizmo oradditive that can double or triple the mileage of an existing internalcombustion car.

Also I'm perfectly well aware that experimental rigs have gotten hundreds ofmiles a gallon. Thin hard tires, tiny engine, low speeds, and a level track areall you need. It just hasn't been done under highway conditions.

Finally, there are some designs being proposed that supposedly get enormous mileage. There's a concept car that will supposedly get 300 miles to the gallon. It's basically a motor scooter with a fiberglass shell. Granted, it has the cargo capacity of a tricycle and the crash protection of a Styrofoam cup, but it might actually get 300 mpg. And there are a couple of sites with spiffy concept cars and lots on how you can invest in the project (wink, wink, nudge, nudge) but precious little in the way of actual engineering. One touted the revolutionary ideas of reducing mass and air resistance, for those of you driving rectangular boxes made of cast iron.


People often think science deals with ten decimal place accuracy, and there'sa time and place for that: targeting missiles and space probes, buildingsuspension bridges, surveying property and so on. But when we simply want toknow whether something is possible or not, we can often use much simpler andmore approximate calculations and save the refined calculations for when it'sstrictly necessary. That's what we'll do here. Numbers will be rounded forsimplicity. We won't use specialized data, but everyday experiences that anyonefamiliar with cars can relate to, and, if they're skeptical, verify forthemselves.

Also, we're not going to deal with the specifics of what actually happens inan engine, since most of the people who believe the Evil Corporate Conspiracy ishiding a 200-mpg carburetor someplace wouldn't believe the data anyhow. Instead,we'll simply ask what it takes to make a car move down the highway, and how muchof that you can do with a gallon of gasoline.

Some Basic Facts


Scientists use the SI system of units. The main units we'll need are these:

Some Basic Data

The heat of combustion of gasoline-weight hydrocarbons is about 48 million joules/kg.One gram offat contains about 9 kilocalories or 37,700 joules, so a kilogram of fatcontains about 9 million calories or 38 million joules. So the energy content of gasoline issomewhat greater than the food energy of fat, but not a great deal. By the way,energy is energy, oxidation is oxidation, and the energy content of food issometimes determined by simply burning it and measuring the heat output. 

1 gallon = 3.785 liters. Since gasoline has a density of 0.7 gm/cc, a gallonof gas is about 2.65 kilograms, with about 125 million joules of energy.

1 mph = 0.447 meters/second

We will assume our car weighs 1000 kilograms (2200 pounds).

What Can You Do With a Gallon of Gas?

To raise an object, the energy needed is mgh, where h is the mass of theobject, g is the acceleration of gravity (9.8 meters per second squared) and his the height.

To raise a car weighing 1000 kilograms a distance of 100 meters would requireabout a million joules. You could raise a car 12.5 kilometers with the energy ina gallon of gas, or drive up 125 100-meter hills. Driving up Pike's Peak (about a 3000-meter climb) should consume about aquart of gas.

But  these figures assume you can get all the energy out of a gallon ofgas and apply it with perfect efficiency to raising the car. Driving miles ofwinding road up Pike's Peak is a far cry from simply lifting the car vertically.

It takes energy to accelerate. The energy of a moving object, called kineticenergy, is given by K=1/2mv2, where m is the mass of the object and v is thespeed. If a car weighs 1000 kg and accelerates to 60 mph (27 m/sec), the energyrequired is 365,000 joules. About 350 such accelerations would consume a gallonof gas, assuming you could apply all that energy solely to accelerating the car,which you can't.

Just for fun, in outer space, if you could apply all the energy in a gallonof gas to moving a car with perfect efficiency, you could get 125 million joules= 1/2mv2, and for a 1000-kilogram car you'd have v2 = 250,000 and v = 500meters per second or 1120 miles per hour. 

Rolling Resistance

It takes a continuing supply of energy just to keep a car in motion, becauseof air resistance, road friction, internal operating losses, and so on. Toestimate how much, I did a few experiments.

How can we convert these figures to energy? Change in velocity is calledacceleration. It takes energy to accelerate a car, and it takes energy toovercome the deceleration of a car - in other words, to keep the car moving.

Velocity Change Time Deceleration Deceleration (m/sec2)
70-60 (drive) 10-20 sec 1/2 - 1 mph/sec 0.22-0.45 
30-0 (neutral) 80 sec 0.375 mph/sec 0.168
40-30 (neutral) 25 sec 0.4 mph/sec 0.18
40-20 (neutral) 50 sec 0.4 mph/sec 0.18
40-30 (drive) 15 sec 0.67 mph/sec 0.3
40-20 (drive) 40 sec 0.5 mph/sec 0.22

Considering the rough and ready methods used, the consistency of the resultsis pretty good. Note that the deceleration with the car in drive is about 50 percent greater than in neutral. In neutral, the car is only decelerated byfriction with the road and air resistance. In drive, the car is also deceleratedby engine compression and internal friction. Also the deceleration at low speedsis less than at high speeds, in part due to less air resistance.

It takes about 200 newtons of force to keep a car in neutral rolling on levelground at highway speeds, aboutequal to the force exerted by 20 kilograms or 45 pounds. That's about in linewith common experience. Anyone who's pushed a car knows that once the car getsrolling, it doesn't take much to keep it moving, but it's not effortless either.At 40 miles an hour (18 m/sec), the power needed is force times distance dividedby time, or 200 newtons times 18 meters per second, or 3600 watts.

Coefficient of friction is defined as the force it takes to push an objectversus its weight (which is the force required to lift it). For a 1000-kilogramcar, the weight is 1000 times gravity or about 10,000 newtons, but it only takes200 newtons to push it. So the coefficient of friction is 200/10,000 = .02. Thisis pretty low. Streamlining (including not carrying a huge car-top rack ordriving a boxy SUV) andproper tire inflation can help keep this factor down.  

So how far could we move a car with a gallon of gas? Work (energy) equalsforce times distance, so distance equals work divided by force. 125 millionjoules divided by 200 newtons = 625,000 meters, or 625 kilometers, or 400 miles.Wow. Coast to coast on less than ten bucks' worth of gas. You wish.

But this is just rolling friction. The real measure of what it takes to keepa car moving is how much force you'd have to apply with the car in gear.Because when the car is running, a great deal of the energy from your gas goesinto overcoming the internal resistance of the engine. The slight increase indeceleration above when the car is in gear is only a tiny piece of the wholepicture. The car is not getting quite enough energy to sustain its speed but itis getting enough to turn the engine over. The car will decelerate until it'sgoing slowly enough for the engine idle to keep the car moving at a few miles perhour. 

Internal Resistance

So how much energy does it take to turn over an engine? Rather than detailedlab data, let's look for everyday experiences.

Pushing the Car in Gear

Some light cars canbe pushed while in gear but it takes several times as much force as when the caris in neutral (do I need to add that this experiment is only meaningful for astandard transmission? Probably.) That suggests we have to cut our estimates ofmileage based on coasting by a substantial amount. You can get quantitative ifyou want by pushing on a bathroom scale and comparing the results in neutral andin various gears. Maybe I will next time I have access to a standard car and abathroom scale. The times I have done it myself (without a scale) I estimate ittakes 3-5 times as much force.

Push Starting

Anyone who has ever indulged in the thankless task of trying to push start acar with an automatic transmission knows the drill. Get the car going to 30miles an hour (14 meters/second), put it in drive with one hand, turn the keywith the other hand, steer with your third hand. Rinse, repeat, call the towtruck.

When it doesn't work (as it usually doesn't), you decelerate very fast. If weestimate - charitably - it takes 15 seconds to stop, you decelerate at14m/sec/15 seconds or just under one meter per second squared. This is about onetenth g and jibes with the fact that when you try this, you feel a pretty suddenjolt when you pop the car into gear.

Force equals mass times acceleration, so we have 1000 kilograms times onemeter per second squared, or about 1000 newtons, about five times the force ittakes to push the car in neutral.

Engine Braking

If you have a fairly small car and engine, you can save huge wear and tear onyour brakes in the mountains by letting the engine do all the work. 

We need to know what kind of force a car experiences on hills. This is an exercise in vector mechanics: the extra gravitational force willbe earth's gravity times the sine of the slope angle. For a one-degree slope, g sin slope = 0.1745g =0.17 m/sec2. The force exerted on a 1000 kilogram car would be 1000 times0.17 or 170 newtons. That's comparable to the effort it takesto push a car on level ground - in other words, a car in neutral should justroll on a one degree slope (lots of folks find out the hard way that what lookslike a level spot isn't!), but pushing a car up a one degree slope roughlydoubles the effort. Anyone who's ever pushed a car knows that even a tiny slopetranslates to a huge increase in effort. For small angles the extra force is proportional to theslope, so a five-degree slope would mean 0.85 m/sec2 (850 newtons for a 1000kilogram car), or over four times the force it takes to keep a car rolling inneutral on level ground.

On a convenient bridge a few days ago I tried this early in the morning whentraffic was minimal. On a three-degree slope (measured on a photo of thebridge), I quickly decelerated to below 50 mph, even with gravity pulling medown. A three degree slope would mean a gravitational pull of 0.54 m/sec2 (540newtons for a 1000 kilogram car) or about 2.5 times the force needed to push thecar on level ground. It would take a steeper slope to permit me to coast athighway speeds.

Back when standard cars were the rule, the recipe for parking on a hill wasto turn wheels into the curb, set the parking brake, and put the car in gear. Ifyou forgot, usually leaving the engine in gear would do the trick even on apretty steep slope. (Usually!) This is additional testimony to the power ofengine braking.

One final point. These observations underestimate the force of enginebraking, since if the engine is running at all, it is still providing enoughpower to overcome a good deal of the internal friction of the engine. 

Pounding on the Pistons

When I was in college, I once helped my father reassemble a six-cylinderengine. My task one day while he was at work was to reinstall the pistons. Idiscovered that, even oiled, there is an astonishing amount of friction betweenpiston rings and the cylinder. First one, no sweat. Second, a bit harder. Then,I had to use a wrench to turn the crankshaft to line it up. For the last one, Ihad to resort to a piece of wood and a hammer to pound on the pistons to turnthe crankshaft into position. When a piston moves down, it not only has toovercome the friction in its own cylinder, it also has to overcome friction inpushing all the other pistons as well.

Extreme? A two kilogram hammer swung at 10 meters per second packs kineticenergy 1/2mv2 = 1/2 times 2 times 10 squared = 100 joules. The 125 millionjoules of energy in a gallon of gas are about equivalent, then, to 1.25 millionhammer blows. A four-cylinder engine at 4000 RPM fires its cylinders on every other stroke, 8,000times a minute, or 1.25 million times in 156 minutes. It will probably take youless time than that to burn a gallon of gas, meaning the energy output isfaster. Even allowing for waste heat, each piston stroke is the equivalent of avery powerful hammer blow. The things that happen to engines when the processgoes wrong are testimony to the forces at work.

And the pistons have to put up with this punishment at temperatures up to 1200 C. Let's pause a moment in honor of the metallurgists who came up withsteel alloys capable of doing this.

If it only takes 200 newtons to make a car roll on level ground, why generatesuch extreme forces? First of all, a crankshaft is a lot smaller than a tire.It's like lifting a weight by pushing on the short end of a lever. 


Yet another way to get at theproblem, using everyday data, is to ask how much energy it takes to crank theengine. A bare-bones auto battery delivers 600 Cold Cranking Amps, which meansit can crank out 600 amps for 30 seconds before dropping to half output. At only6 volts, where you might just possibly get the engine to turn, 600 amps means3600 watts or 3600 joules per second. Once the engine starts, the gas has tosupply this energy. The 125 million joules in a gallon of gas can supply that energyfor about 3.7 hours.

Bottom Line So Far

These are all rough and ready estimates of the forces and energies it takesto overcome the internal friction in an auto. They rely on commonplaceobservations rather than detailed lab data, so it's really rather remarkablethat they all converge on similar results, namely, the force required toovercome mechanical resistance in a car is in the ballpark of five times what ittakes simply to push the car in neutral. So our theoretical 400 miles per gallonfor just coasting drops to about 80 miles per gallon once we factor in internalfriction.

Everyone, driver and environmentalist alike, would love to get 80 miles pergallon. But note that this is way below the 150-200 miles per gallon that energyexperts like Steven Seagal in On Dangerous Ground say we could get if thebad guys weren't concealing secret miracle devices.

Might there be some other engine design that could cut down drastically oninternal waste? Possibly. Hybrid gas-electric and turbine designs come to mind.You throw a lot of hard-won energy away every time you stop, then have to get itback again when you accelerate. A hybrid design would store that energy in aflywheel or battery. But, on a car with a typical piston engine, you're not going to drop somemiracle additive into your tank or stick some gizmo on your fuel injector andget 150 miles per gallon. Maybe 80 is the best you could do - if everything elsewas perfect. And it isn't.

Waste Heat

A kilogram of hydrocarbon consists of about 85 per cent carbon and 15 percent hydrogen by weight, which burns to carbon dioxide and water. 850 grams ofcarbon burns to 3.1 kilograms of carbon dioxide, and 150 grams of hydrogenbecomes 1.35 kilograms of water vapor. Those gases come out at around 500degrees C (put your hand on the exhaust manifold while it's hot forconfirmation!). How much heat is lost that way?

We need a quantity called heat capacity. The heat capacity of carbon dioxidegas is about 850 joules per kilogram per degree. It takes 850 joules to heat akilogram of carbon dioxide one degree C. For water vapor, the figure is 1860joules per kilogram per degree. Our gallon of gas yields 3.1 kilograms of carbondioxide with a heat capacity of 2635 joules/degree and 1.35 kilograms of watervapor with a heat capacity of 2511 joules/degree for a total of 5150joules/degree.

Now the gases come out about 500 C hotter than when they went in, or 500x5150 = 2.6million joules of heat. So our gallon of gas sends at least 2% of its energy outthe exhaust pipe as hot gas.

Now, what about heating other things? If the engine weighs 200 kilograms andgets heated from 20C to 100C while warming up, how much energy is that? It takesabout 440 joules to heat a kilogram of iron one degree C, so heating 200kilograms of iron 80 degrees C takes 440 times 200 times 80 = 7 million joules. Offhand,I'd say we found a significant source of waste. Bear in mind the heat productionis continuous, and requires a fan and water pump to circulate fluid to removethe excess heat. Those things also use energy.

The Second Law of Thermodynamics

The energy of a process goes into two forms, work (symbolized W) and heat(symbolized Q). We write E = W + Q. The energy produced by burning gasoline in acylinder goes into work (pushing the piston) and heat (hot exhaust gases, hotcylinder walls). Entropy, the "disorder" in a process, is defined asdS = dQ/T. Temperature, here, is absolute or Kelvin temperature, where zeromeans no molecular movement at all. That happens at -273 C, called absolutezero. What this formula means:

If everything is as hot as the burning gasoline, burning gas in the cylinderwon't accomplish anything - the gases won't expand. We have to dump the wasteheat somehow. The gases cool as they expand, and the hot gases are ventedoutside. 

According to a fundamental law of physics, the Second Law of Thermodynamics,the entropy of any system always increases. Each piston stroke produces entropy,but the engine is cyclic - it returns to its original state, and that somehowmeans a loss of entropy. That can only happen by dumping the entropy someplacethat can absorb it - outside - into a bigger, cooler system.

So, as the gas burns during ignition, it releases energy, which is initiallyonly heat. Call the amount of heat released q. The heat causes the gases in thecylinder to expand, doing work, which we will call w. Any waste heat we willcall q'. Since energy is conserved, q = w + q'.


Igniting the gas creates entropy. Call the amount created s, and s = q/T,where T is the ignition temperature. In order to get the cylinder ready for itsnext cycle, we will have to dump entropy (waste heat) outside, contributing s' =q'/T' to the environment. We put primes on the quantities after extractinguseful work, and T' means the final temperature after we've extracted all thework we can from the cylinder. Since entropy always increases, s' >= s. Thatmeans q'/T' >= q/Ti. Now, energy has to be conserved, and the only otherplace energy is going is useful work, q-q'=w. In other words, useful work iswhatever doesn't end up as heat. So we have q' = q-w, and we can eliminate q'from the formula to get (q-w)/T' >= q/T . We can rearrange the formula toget w/T' <= q/T'-q/T. We can rearrange still further to get w/q <=(T-T')/T.

The less than or equals sign (<=) means there is no limit to our abilityto lower efficiency. Driving with the parking brake on is a popular method. Soare running on less than all cylinders, going a long time between tuneups,increasing friction by neglecting oil changes, and so on. Since we're trying for maximum efficiency, we need only worry about the equalspart.

The quantity w/q is called the efficiency. If all the heat produced byburning the gas goes into useful work, then w = q and efficiency = 1, or 100 percent. This can only happen if T-T' = T, or T' = 0, absolute zero. So, if wecould run engines on Pluto, we might figure a way to approach 100 per centefficiency. In normal operation, we get a lot less. At 20 C, or 293 K, withcylinders burning at 1200 C (1473 K), we'd get (1473-273)/1473 = 0.86, or 86 percent.

Imagine water flowing over a dam into Lake Superior to generate electricity. Lake Superior is 600 feet above sea level. There's a heck of a lot of energy that could be obtained by letting Lake Superior flow to sea level, but that's of no relevance to the dam. That energy is not available. The power the dam can produce is governed solely by how high the dam is. The top of the dam corresponds to the initial temperature of the process, Lake Superior to the final temperature, and sea level to absolute zero. You can't get the total energy available by letting the water fall to sea level, because it can only fall as far as Lake Superior. Likewise, you can't get the total energy out of a heat process because the temperature can only fall as far as the local environment.

In reality we get a lot less. The combustion gases expand and cool as they dowork, but how are we going to cool them all the way back down to outsidetemperature? If the cylinder is cold, a lot of heat will just go into heatingthe cylinder and pistons, and in any case there's no way we could remove heatfrom the cylinder in the fraction of a second between strokes. If the cylinderis hot, the gases can't get any cooler than the cylinder. Once they areexhausted, they can cool down, but that won't give us any work inside thecylinder. What we typically achieve is final temperatures about 500 C (773 K),which gives us efficiency of (1473-773)/1473=0.48, or about 48 per cent.

The big killer is this: how do you get very hot gases to give up so muchenergy that they get very cool, very fast, all in the same vessel? Merelycooling the gases won't do anything - they have to lose heat by doing work. Asthe gases get closer to the outside temperature, their ability to expand and douseful work diminishes sharply.

Bottom Line

If we could get all the energy out of a gallon of gasoline and apply it toovercoming the rolling resistance of a car, we could get about 400 miles pergallon. But the internal engine resistance multiplies the force required byroughly five, meaning the best we could hope for is about 80 miles a gallon. Theengineering constraints of getting useful work out of expanding hot gases cutsthat figure by 50 per cent to about 40 miles per gallon. So better fuels andfuel injection could improve mileage, but not stupendously.

Back in the 1950's, there were cars with mileages in single digits. Heavy,inefficient fuel-air mixtures, overpowered. With gas at 20 cents a gallon, whoworried? A lot of that waste was fairly easy to fix.

What to do next? Well, every single formula for moving the car has involved m, themass of the car. So reduce m. This is in fact how most of the fuel economy ofthe past 20 years has been achieved. More plastic, less metal, spare tires thesize of a bagel. If you buy an SUV you have no business carping that the big badoil companies are conspiring to lower your gas mileage. Actually, SUV's get a bum rap because they actually get better mileage than most cars did a few years ago. (A recent study concluded a Hummer had a smaller environmental footprint over its lifetime than a hybrid, mostly because the Hummer lasts longer and nickel hydride batteries exact a big cost in mining and fabrication impacts.)

We can improve efficiency by cutting the fuel mixture to just this side ofstarvation, but then we have to monitor it extremely carefully. We can alsorecover and re-burn engine gases that blow by the piston. Gone are the simple carburetorsthat a shade tree mechanic could fix. Now we need computers, oxygen sensors, andall that other fun stuff that is so expensive and complicated to fix. Also we can improveefficiency by doing away with energy consumers like air conditioning. Try thatin Houston in the summer.

Obviously a huge amount of the energy in a gallon of gasoline goes to waste.Some of that is unavoidable by the laws of physics, much of it is unavoidable asa matter of engineering reality. There may be designs out there that vastlyexceed what we currently achieve. Maybe you could even fit them under the hoodof an existing car. But you won't get them by slapping a cheapretrofit on an existing piston engine.

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Created 8 December, 2001, Last Update 24 May, 2020

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