Why does Speed affect Mpg?

[Thinktank]

Well, that’s the question- it seems cars use more gas the faster they go.

I don’t mean because you go faster in your car that the fuel is running out faster too.
I’m saying that you use even more gas at an unproportionate rate by traveling at higher speeds on the highway
than if you were to travel at maybe just 10 mph slower.

I’m sorry I have no cite, but it’s in lots of various articles and I’ve noticed it first-hand in my own cars.

Let’s assume however that the vehicle in question has been maintained perfectly to rule out failing/degraded performance due to neglect.

The question is big and it i-s kinda vague; I’m sure there’s more than one answer to this question. If there is, please submit it here.

But all that said, why does this happen?

[Squink]

I blame the atmosphere: Drag equation

[ElvisL1ves]

Fuel burn goes largely to overcoming resistance to motion, which takes the form of both rolling friction and aerodynamic drag. Some components of rolling friction, and all of aerodynamic drag, are proportional to the *square * of speed, not linear. Going at higher speeds reduces mileage because proportionately more fuel is needed to overcome the speed-squared components of drag.

That isn’t true at the lowests speeds, where engine inefficiencies at low power dominate, though. There’s a maximum-mileage speed for any car somewhere in its operating range.

[smiling_bandit]

Efficiency. Pretty much everywhere you look in science, things become less efficient at greater volumes. Larger animals are less energy efficient than small ones, for instance. There’s a number of principles behind it: friction, mass inceasing at a greater rate than surface area, etc.

[FoieGrasIsEvil]

Isn’t engine RPM a factor in all of this? After all, the faster you go, the more RPM’s you’re turning, hence working the engine harder and using more fuel.

I know that normal hwy driving (say, 65mph) has most cars chugging along anywhere from 2500 to 3500 rpm’s in top gear, so it isn’t necessarily applicable to freeway cruising, but it’s certainly applicable during acceleration and when indulging in illegal speeds.

BTW, the drag coefficient is probably the biggest culprit. There’s a reason why vehicles with square-shaped front-ends get worse mileage than an equally equipped and weighted vehicle with a more aerodynamic shape.

Also having all your windows rolled down reduces your mileage for the same reason, especially the faster you go.

[CookingWithGas]

[QUOTE=smiling bandit]
Efficiency. Pretty much everywhere you look in science, things become less efficient at greater volumes. Larger animals are less energy efficient than small ones, for instance. There’s a number of principles behind it: friction, mass inceasing at a greater rate than surface area, etc.
[/QUOTE]
I don’t see how this responds to the OP and I have no idea what your evidence is that larger animals are less energy efficient than small ones. In fact the opposite is true. From The Causes of Aging:

…the greater the area of body surface in relation to body mass, the greater the metabolic rate will have to be in order to replace the lost heat. Consequently, smaller animals have to burn more oxygen and produce more energy in order to maintain their body temperature: or put a different way, a larger body retains heat much more efficiently, and has to generate less heat per unit of body mass than a small body, thus explaining why large animals burn energy more slowly.

A more whimsical example can be found at this site:

@Mighty Dog

For example, studies have shown that a Great Dane weighing 100 pounds needs to consume about 23 calories per pound of body weight every day to maintain condition, whereas a six-pound Pomeranian must take in 47 calories per pound of body weight ever day-more than twice as much!

@FoieGrasIsEvil

Isn’t engine RPM a factor in all of this? After all, the faster you go, the more RPM’s you’re turning, hence working the engine harder and using more fuel.

And the more RPMs you turn, the faster you go, ad infinitum. With a transmission, your engine is turning about the same RPMs at 60 MPH that it is at 15 MPH. But each stroke needs more fuel for a harder push.

[blondebear]

I just heard an interview with a guy from Edmunds on the radio yesterday. He said that speed did was less detrimental to gas mileage than changing speed. In tests driving the interstate in a V8 Tundra using cruise control at 65mph, they got a steady 20mpg. Their mileage dropped to 15mpg once they got into “normal” highway traffic that required constant speed changes required by the flow of traffic.

[flex727]

[QUOTE=Thinktank]
Well, that’s the question- it seems cars use more gas the faster they go.
[/QUOTE]

There’s lots of reasons, as previous posters have stated, but it’s MOSTLY wind resistance with ordinary cars. If you’ve ever ridden a bicycle, or even walked, into a 60 mph wind, you should be able to understand why a car needs more gas to run at that speed.

[zut]

[QUOTE=ElvisL1ves]
Fuel burn goes largely to overcoming resistance to motion, which takes the form of both rolling friction and aerodynamic drag. Some components of rolling friction, and all of aerodynamic drag, are proportional to the *square * of speed, not linear. Going at higher speeds reduces mileage because proportionately more fuel is needed to overcome the speed-squared components of drag.
[/QUOTE]
What you’re saying is absolutely true, but be careful. There’s an UNtrue implication here that IF drag were LINEARLY proportional to speed, then mileage would not decrease at higher speeds. This is NOT true. A higher drag force requires more input energy, period, whether the higher force is a quadratic function or linear function of speed. However, since the force is quadratically related to speed, my point is moot, except for the basic understanding of the physics behind the system.

@smiling bandit

Efficiency. Pretty much everywhere you look in science, things become less efficient at greater volumes. Larger animals are less energy efficient than small ones, for instance. There’s a number of principles behind it: friction, mass inceasing at a greater rate than surface area, etc.

This is completely erroneous, as well as having nothing at all to do with the OP.

@FoieGrasIsEvil

Isn’t engine RPM a factor in all of this? After all, the faster you go, the more RPM’s you’re turning, hence working the engine harder and using more fuel.

Yes and no. Or, at least, not because of the reason you’re stating. Broadly, with higher speed, “using more fuel” equates to “going a greater distance,” so when calculating “distance per fuel” (MPG, in other words), the two cancel out. However, at very high speeds, engines tend to be less efficient, in large part due to work required to pump air into the engine and exhaust out.

[Mr.Slant]

Here’s a teaching tool for you:
http://www.cartestsoftware.com/cartest4.5/index.html

Go to the bottom, download the old 4.5 version, which is free.
Once running the program

1. Select EDIT/ANALYZE CAR
2. Type V to move to the section of the same letter.
3. Select VECTOR M12 1996. [1]
4. Type F3 to go into the Choose Analysis menu.
5. Select Power Loss Curves
6. Note the green line, labelled AEROD. Compare it to the other elements of the vehicle’s power loss.

[1] I’m choosing this one because it’s the quickest keystroke set you can get on your way to a very fast car, at 490 horsepower and 3700 lbs. For a light, weak car, I suggest the YUGO GV 1987 model, being the 55 horsepower and 2070 lb behemoth that it is . Finally, for an ‘average’ passenger car, I’d suggest looking at TOYOTA CAMRY DX 1995, with 130 horsepower and 3300 lbs.

[Xema]

The nonlinear increase in drag with speed is easily seen without formulas:

A faster car imparts more energy to each air molecule it encounters, in the same way that a faster golf swing imparts more energy to a golf ball. In addition, the faster car encounters more air molecules per second.

[Xema]

[QUOTE=zut]
There’s an UNtrue implication here that IF drag were LINEARLY proportional to speed, then mileage would not decrease at higher speeds. This is NOT true. A higher drag force requires more input energy, period
[/QUOTE]

I’m a bit puzzled here. If drag were linearly proportional to speed, mileage ought to stay constant - higher speeds would certainly require more energy/time, but would also return more distance/time, resulting in the same energy/distance.

[zut]

[QUOTE=Xema]
I’m a bit puzzled here. If drag were linearly proportional to speed, mileage ought to stay constant - higher speeds would certainly require more energy/time, but would also return more distance/time, resulting in the same energy/distance.
[/QUOTE]

The confusing part, I think, is that velocity is figuring into the calculations in multiple places. Let’s go back to some basic principles for a second:

“Drag” is really “drag force.” Energy is force times distance.

If you have a constant force on the vehicle–one that doesn’t change at all–overcoming that force for a certain distance (a mile, say) would take a constant amount of energy. Doesn’t matter how fast or slow you go, it still takes the same amount of energy: force times distance.

Now, if you double that force, it takes twice the energy to go a mile Again, velocity is irrelevant, because the force is constant.

Now, if the force is linearly proportional to velocity, then the faster you go, the higher the force is. But energy is still force times distance, so the faster you go, the greater the force is, and the more energy you use to go one mile, because energy is still force times distance. If the force is quadratic with distance (as drag force really is), the dependence on velocity is just greater.

(The more confusing part here is that people tend to think in terms of power, which is energy/time, or force*velocity. Thinking this way doesn’t change the physics, but adds another velocity factor, which is confusing.)

[Xema]

Thanks - good explanation.

[Sam_Stone]

Higher engine RPM creates more ‘pumping losses’ - there is a difference in pressure between the ambient air and the air in the engine at less than full throttle settings, and it takes work to move that air into the engine. Also, because the exhaust has some element of air restriction, it takes work to move the exhaust gases out of the vehicle. This is why smaller engines tend to be more fuel efficient - you generally run them at higher throttle settings on the highway, which reduces the pressure differential and the pumping losses of the engine.

However, this isn’t necessarily a factor of speed - it’s a factor of the cruise RPM of the vehicle, which can be different based on different transmission/differential gearings. This is why overdrive makes your car more fuel efficient - it allows you to run at lower RPM at cruise. But once you hit the lowest gearing available, additional speed will increase engine RPM and increase the amount of pumping loss in the engine (as well as internal frictional losses, which go up with RPM).

So… Higher air resistance, higher rolling resistance, and increased pumping losses if the car isn’t geared for optimum RPM at the speed you’re gong.

Quick trivia - the Chevrolet Corvette gets its fuel mileage rating in part because of a ‘skip shift’ feature which causes the car to skip from 1st to 3rd gear to keep RPM low and the throttle more open while accelerating around town. With a big V8 engine, this makes a pretty big difference in fuel economy, and helped GM avoid the ‘gas guzzler’ tax on the Vette.

[HMS_Irruncible]

[QUOTE=Xema]
The nonlinear increase in drag with speed is easily seen without formulas:

A faster car imparts more energy to each air molecule it encounters, in the same way that a faster golf swing imparts more energy to a golf ball.
[/QUOTE]

I’m not sure you’ve eliminated a formula there, isn’t that just a folksy way of saying k=½mv[sup]2[/sup] (thus restating that the drag is proportional to the square of velocity)?

[Xema]

[QUOTE=Cosmic Relief]
I’m not sure you’ve eliminated a formula there, isn’t that just a folksy way of saying k=½mv[sup]2[/sup] ?
[/QUOTE]

It is. I wasn’t really seeking to eliminate a formula - simply trying to show a non-mathematical way of visualising what the formula calculates.

[zut]

[QUOTE=Cosmic Relief]
I’m not sure you’ve eliminated a formula there, isn’t that just a folksy way of saying k=½mv[sup]2[/sup] (thus restating that the drag is proportional to the square of velocity)?
[/QUOTE]

It’s actually a folksy way of saying F[sub]d[/sub] = ½C[sub]d[/sub]A[symbol]r[/symbol]v[sup]2[/sup], where F[sub]d[/sub] is the drag force, C[sub]d[/sub] is the drag coefficient, A is frontal area, [symbol]r[/symbol] is air density, and v is velocity. Your formula (½mv[sup]2[/sup]) is kinetic energy.

[Philster]

For the OP’s benefit:

If you travel at 60 MPH instead of 50 MPH, because of physics, the car does not need just 20% more energy to travel 20% faster. To go 20% faster, the car needs X% more energy.

While you get where you are going faster, you burn way more fuel than you save.

“Why?” is a BRUTAL question when it comes to physics. Why? BECAUSE, that’s how it’s been determined to work. Why does light travel at 186,000 miles per second?

You don’t need to understand all the formulas.

[brad_d]

If we assume that aerodynamic drag is the dominant resistance that a car has to overcome, and that the drag coefficient is nearly constant (both pretty good assumptions for typical cars at the kinds of speeds we’re contemplating), then:

Drag force D ~ V[sup]2[/sup], where V is speed. (I’m taking the other factors in that drag equation that zut posted and noting that they are all constant for a given case. Air density, drag coefficient - see my assumptions - and frontal area will not be changing with speed.)

Power (energy/unit time) is given by force * speed: P = D*V ~ V[sup]2[/sup]*V = V[sup]3[/sup].

The amount of energy (E) actually used on a given journey is power * time. The time it takes to go some distance L is (L/V). So E = P*(L/V) ~ LV[sup]2[/sup]

Total energy used to go a given distance varies as speed squared, then. If you double your speed then sure, you’ll get there in half the time. But you’re working eight times as hard during that shortened interval, so the end result is that you’ll have quadrupled your energy expenditure for the journey.

If power consumption were linear in speed, then it wouldn’t matter how fast you went - you’d use the same amount of energy regardless. But it very much isn’t, as a general rule, when aerodynamic drag is the main opposing force.