Steven Dutch, Natural and Applied Sciences, University
of Wisconsin - Green Bay

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

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

Also I'm perfectly well aware that experimental rigs have gotten hundreds of miles a gallon. Thin hard tires, tiny engine, low speeds, and a level track are all 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's a time and place for that: targeting missiles and space probes, building suspension bridges, surveying property and so on. But when we simply want to know whether something is possible or not, we can often use much simpler and more approximate calculations and save the refined calculations for when it's strictly necessary. That's what we'll do here. Numbers will be rounded for simplicity. We won't use specialized data, but everyday experiences that anyone familiar with cars can relate to, and, if they're skeptical, verify for themselves.

Also, we're not going to deal with the specifics of what actually happens in an engine, since most of the people who believe the Evil Corporate Conspiracy is hiding 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 much of that you can do with a gallon of gasoline.

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

- Length: meter (3.28 feet)
- Mass: kilogram (2.2 pounds)
- Volume: liter (0.264 gallons)
- Acceleration: meters per second squared. Earth's gravity = 9.8 m/sec
^{2} - Force: newton (0.225 pounds). Lots of people confuse
*mass*and*weight*. Weight is a force, the pull of earth's gravity on a mass. A newton is the weight - force exerted by - about 100 grams. A newton is about equal to seven fig newtons. You read it here first. - Energy: joules. One large Calorie (energy needed to heat a kilogram of
water one degree C) equals 4190 joules. This is a
*kilocalorie*and is the standard food calorie. - Power: watts. One joule per second. This is the one SI unit in everyday use.

The heat of combustion of gasoline-weight hydrocarbons is about 48 million joules/kg. One gram of fat contains about 9 kilocalories or 37,700 joules, so a kilogram of fat contains about 9 million calories or 38 million joules. So the energy content of gasoline is somewhat 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 is sometimes 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 gallon of 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).

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

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

But these figures assume you can get all the energy out of a gallon of gas and apply it with perfect efficiency to raising the car. Driving miles of winding 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 kinetic
energy, is given by K=1/2mv^{2}, where m is the mass of the object and v is the
speed. If a car weighs 1000 kg and accelerates to 60 mph (27 m/sec), the energy
required is 365,000 joules. About 350 such accelerations would consume a gallon
of 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 gallon
of gas to moving a car with perfect efficiency, you could get 125 million joules
= 1/2mv^{2}, and for a 1000-kilogram car you'd have v^{2} = 250,000 and v = 500
meters per second or 1120 miles per hour.

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

- At highway speeds (60-70) I lose 1/2-1 mph/sec when I take my foot off the gas. Needless to say, this is not an easy measurement to do accurately or safely.
- On an empty, level road, I got up to 30 miles an hour, then decelerated to a stop in neutral. It took 80 seconds. I know the road was level because it had a water-filled ditch alongside. Also, I ran the experiments in both directions.
- Decelerating from 40 to 30 miles an hour in drive required 15 seconds, to 20 miles an hour took 40 seconds.
- Decelerating from 40 to 30 miles an hour in neutral took 25 seconds, to 20 miles an hour took 50 seconds.

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

Velocity Change |
Time |
Deceleration |
Deceleration (m/sec^{2}) |

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 results is pretty good. Note that the deceleration with the car in drive is about 50 per cent greater than in neutral. In neutral, the car is only decelerated by friction with the road and air resistance. In drive, the car is also decelerated by engine compression and internal friction. Also the deceleration at low speeds is 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 level ground at highway speeds, about equal to the force exerted by 20 kilograms or 45 pounds. That's about in line with common experience. Anyone who's pushed a car knows that once the car gets rolling, 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 divided by 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 object versus its weight (which is the force required to lift it). For a 1000-kilogram car, the weight is 1000 times gravity or about 10,000 newtons, but it only takes 200 newtons to push it. So the coefficient of friction is 200/10,000 = .02. This is pretty low. Streamlining (including not carrying a huge car-top rack or driving a boxy SUV) and proper tire inflation can help keep this factor down.

So how far could we move a car with a gallon of gas? Work (energy) equals force times distance, so distance equals work divided by force. 125 million joules 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 keep
a 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 goes
into overcoming the internal resistance of the engine. The slight increase in
deceleration above when the car is in gear is only a tiny piece of the whole
picture. The car is not getting quite enough energy to sustain its speed but it
is getting enough to turn the engine over. The car will decelerate until it's
going slowly enough for the engine idle to keep the car moving at a few miles per
hour.

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

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

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

When it doesn't work (as it usually doesn't), you decelerate very fast. If we estimate - charitably - it takes 15 seconds to stop, you decelerate at 14m/sec/15 seconds or just under one meter per second squared. This is about one tenth g and jibes with the fact that when you try this, you feel a pretty sudden jolt when you pop the car into gear.

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

If you have a fairly small car and engine, you can save huge wear and tear on your 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 will
be earth's gravity times the sine of the slope angle. For a one-degree slope, g sin slope = 0.1745g =
0.17 m/sec^{2}. The force exerted on a 1000 kilogram car would be 1000 times
0.17 or 170 newtons. That's comparable to the effort it takes
to push a car on level ground - in other words, a car in neutral should just
roll on a one degree slope (lots of folks find out the hard way that what looks
like a level spot isn't!), but pushing a car up a one degree slope roughly
doubles the effort. Anyone who's ever pushed a car knows that even a tiny slope
translates to a huge increase in effort. For small angles the extra force is proportional to the
slope, so a five-degree slope would mean 0.85 m/sec^{2} (850 newtons for a 1000
kilogram car), or over four times the force it takes to keep a car rolling in
neutral on level ground.

On a convenient bridge a few days ago I tried this early in the morning when
traffic was minimal. On a three-degree slope (measured on a photo of the
bridge), I quickly decelerated to below 50 mph, even with gravity pulling me
down. A three degree slope would mean a gravitational pull of 0.54 m/sec^{2} (540
newtons for a 1000 kilogram car) or about 2.5 times the force needed to push the
car on level ground. It would take a steeper slope to permit me to coast at
highway speeds.

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

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

When I was in college, I once helped my father reassemble a six-cylinder engine. My task one day while he was at work was to reinstall the pistons. I discovered that, even oiled, there is an astonishing amount of friction between piston 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, I had to resort to a piece of wood and a hammer to pound on the pistons to turn the crankshaft into position. When a piston moves down, it not only has to overcome the friction in its own cylinder, it also has to overcome friction in pushing all the other pistons as well.

Extreme? A two kilogram hammer swung at 10 meters per second packs kinetic
energy 1/2mv^{2} = 1/2 times 2 times 10 squared = 100 joules. The 125 million
joules of energy in a gallon of gas are about equivalent, then, to 1.25 million
hammer blows. A four-cylinder engine at 4000 RPM fires its cylinders on every
other stroke, 8,000
times a minute, or 1.25 million times in 156 minutes. It will probably take you
less time than that to burn a gallon of gas, meaning the energy output is
faster. Even allowing for waste heat, each piston stroke is the equivalent of a
very powerful hammer blow. The things that happen to engines when the process
goes 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 with steel alloys capable of doing this.

If it only takes 200 newtons to make a car roll on level ground, why generate such 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 the problem, using everyday data, is to ask how much energy it takes to crank the engine. A bare-bones auto battery delivers 600 Cold Cranking Amps, which means it can crank out 600 amps for 30 seconds before dropping to half output. At only 6 volts, where you might just possibly get the engine to turn, 600 amps means 3600 watts or 3600 joules per second. Once the engine starts, the gas has to supply this energy. The 125 million joules in a gallon of gas can supply that energy for about 3.7 hours.

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

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

Might there be some other engine design that could cut down drastically on internal 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 it back again when you accelerate. A hybrid design would store that energy in a flywheel or battery. But, on a car with a typical piston engine, you're not going to drop some miracle additive into your tank or stick some gizmo on your fuel injector and get 150 miles per gallon. Maybe 80 is the best you could do - if everything else was perfect. And it isn't.

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

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

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

Now, what about heating other things? If the engine weighs 200 kilograms and gets heated from 20C to 100C while warming up, how much energy is that? It takes about 440 joules to heat a kilogram of iron one degree C, so heating 200 kilograms 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 production is continuous, and requires a fan and water pump to circulate fluid to remove the excess heat. Those things also use energy.

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 a cylinder goes into work (pushing the piston) and heat (hot exhaust gases, hot cylinder walls). Entropy, the "disorder" in a process, is defined as dS = dQ/T. Temperature, here, is absolute or Kelvin temperature, where zero means no molecular movement at all. That happens at -273 C, called absolute zero. What this formula means:

- The d means "small change in the quantity." A small change in heat, divided by temperature, equals a small change in entropy.
- Q: Entropy - disorder - is tied to the amount of waste heat a process generates. Burning the gasoline in a cylinder (E) produces great heat. The hot gases expand and push the piston (W). The rest goes to waste as heat (Q).
- 1/T: Also, at very low temperatures where atoms are moving slowly, even a tiny amount of heat introduces a large amount of disorder. At high temperatures, the same change in molecular motion won't have anywhere near as great an effect. Tossing a sock on the floor in the low-entropy living room looks very messy; in a teenager's high-entropy bedroom it would probably go unnoticed.

If everything is as hot as the burning gasoline, burning gas in the cylinder won't accomplish anything - the gases won't expand. We have to dump the waste heat somehow. The gases cool as they expand, and the hot gases are vented outside.

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 somehow means a loss of entropy. That can only happen by dumping the entropy someplace that can absorb it - outside - into a bigger, cooler system.

So, as the gas burns during ignition, it releases energy, which is initially only heat. Call the amount of heat released q. The heat causes the gases in the cylinder to expand, doing work, which we will call w. Any waste heat we will call 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 its next cycle, we will have to dump entropy (waste heat) outside, contributing s' = q'/T' to the environment. We put primes on the quantities after extracting useful work, and T' means the final temperature after we've extracted all the work we can from the cylinder. Since entropy always increases, s' >= s. That means q'/T' >= q/Ti. Now, energy has to be conserved, and the only other place energy is going is useful work, q-q'=w. In other words, useful work is whatever 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 to get 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 ability to lower efficiency. Driving with the parking brake on is a popular method. So are 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 equals part.

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

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 do work, but how are we going to cool them all the way back down to outside temperature? If the cylinder is cold, a lot of heat will just go into heating the cylinder and pistons, and in any case there's no way we could remove heat from the cylinder in the fraction of a second between strokes. If the cylinder is hot, the gases can't get any cooler than the cylinder. Once they are exhausted, they can cool down, but that won't give us any work inside the cylinder. 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 much energy that they get very cool, very fast, all in the same vessel? Merely cooling the gases won't do anything - they have to lose heat by doing work. As the gases get closer to the outside temperature, their ability to expand and do useful work diminishes sharply.

If we could get all the energy out of a gallon of gasoline and apply it to overcoming the rolling resistance of a car, we could get about 400 miles per gallon. But the internal engine resistance multiplies the force required by roughly five, meaning the best we could hope for is about 80 miles a gallon. The engineering constraints of getting useful work out of expanding hot gases cuts that figure by 50 per cent to about 40 miles per gallon. So better fuels and fuel 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, who worried? 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, the
mass of the car. So reduce m. This is in fact how most of the fuel economy of
the past 20 years has been achieved. More plastic, less metal, spare tires the
size of a bagel. If you buy an SUV you have no business carping that the big bad
oil 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 of starvation, but then we have to monitor it extremely carefully. We can also recover and re-burn engine gases that blow by the piston. Gone are the simple carburetors that a shade tree mechanic could fix. Now we need computers, oxygen sensors, and all that other fun stuff that is so expensive and complicated to fix. Also we can improve efficiency by doing away with energy consumers like air conditioning. Try that in 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 as a matter of engineering reality. There may be designs out there that vastly exceed what we currently achieve. Maybe you could even fit them under the hood of an existing car. But you won't get them by slapping a cheap retrofit on an existing piston engine.

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