# Can a car get better mileage driving over hills?

Two towns are at the same elevation and are connected by two roads of the same length. One road is flat, the other road goes up and down some hills. Will an automobile always get the best mileage driving between the two towns on the flat road versus the hilly one, if operated by a perfect driver with knowledge of the route? Is there any hilly road on which a better mileage can be achieved?

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two towns or three towns? Is the flat road doing some curve around the hills, then? –  arivero Nov 21 '13 at 23:58
I am a bit surprised that a brachistochrone with friction and drag is not mentioned even as a comparative example, in any of the answers. –  arivero Nov 22 '13 at 0:19

"Is there any hilly road on which a better mileage can be achieved?"

Let's just use three simple reasonable assumptions: 1) There is rolling friction 2) There is air friction which increases with speed 3) In the graph of power vs gas consumption, there is a peak (a maximum)

note: (power)/(gas/time) has the same units as energy/gas. If there was no air friction and rolling friction was a constant, we'd want to run the engine at this peak until we could just coast the rest. However, because there is air friction, it may actually be better for gas mileage to run below this power output. That is the trade off on the flat road.

An appropriately shaped hill lets us beat this trade off, because we can go into a gear so that we go very slow essentially giving all our engine power into gravitational energy. We just coast to the hill (or after the hill), with the engine off.

Viewed this way, the answer is obvious, since we are essentially using the hill as a gravitational battery. It lets us beat the flat straight away, because we can give that energy back at 100% efficiency compared to the < 100% efficiency of the motor in the "air friction with speed" vs "power output vs gas consumption" tradeoff we are forced into with the flat road.

The best road I believe would then be just enough slope to allow rolling against friction all the way until a steep hill right at the end.

EDIT:

Some of the comments and other answers to this question are quite bizarre. To make it clear, I am not claiming that all hilly roads are better. The question asked "Is there any hilly road on which a better mileage can be achieved?" The answer is YES. Nor am I claiming that this depends only on the road, as it clearly depends on how the driver decides to run the engine for speeds along the route. Again, the question seems clear on this, as it says to consider the car "operated by a perfect driver with knowledge of the route". So I am not sure where the confusion is coming from. So I am extending discussion here, in hopes to clear that confusion up.

There are three places where stored or mechanical energy is lost to heat: the engine performance curve, the rolling friction of the tires on the road, and the air friction. The tire friction is to good approximation a constant, while the air friction increases with speed. Putting this all together:

the total mechanical energy used to get from A to B: $E = mgh|_A^B + \int_A^B (F_{air} + F_{rolling}) ds$

Since A and B are at the same elevation in this problem, the gravitational potential energy terms sum to zero and are the same for both routes. To good approximation the rolling friction is a constant, then if the length of the road is L: $E = L F_{rolling}+ \int_A^B F_{air}(v) \ ds$

So the rolling friction is the same between the two routes. The only thing left then is the engine performance curve (the efficiency at which we can get the mechanical energy from the gas) and the air friction. The answers neglecting the engine performance curves, or air friction, are thus neglecting the real difference between these routes. I hope I made that abundantly clear now.

It is easy to see that if the engine performance curve was flat (a constant), then on a flat road, we'd want to go very slow (the limit v->0 is the best driving strategy for gas mileage in this unrealistic case). However, in realistic cases (and as I took as one of my three assumptions in my answer above) the engine performance curve will have a peak. There is now a trade off between how far off peak performance we run the engine, and how much mechanical energy we waste on air friction. The details of solving this require detailed knowledge of the engine performance curve, but the general result that there is a trade off is clear regardless. The issue is that on a flat road: the engine can only generate mechanical energy in the form of kinetic energy, and kinetic energy in turn causes more energy loss in air friction.

Now consider the case where the road slopes just enough to a hill at the end, that we can just coast down to the hill, and only need to use the engine to get up the hill. (Or, alternatively, as other poster suggested, a hill at the start, and then coast the rest of the way.) When running the engine now, we can generate mechanical energy in the form of kinetic energy and gravitational energy. So a hill allows us to run the engine closer to its peak performance, since we can put the engine output into gravitational energy (which has no loss over the trip) as opposed to just kinetic energy (which we get losses in air friction).

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@Master of Disaster: Can you please be specific in what you believe is a mistake. Currently your comment is not helpful, for I've reread this answer and it sounds correct to me. –  John Mar 22 '11 at 2:06
@Master of Disaster: The answer you provided and the comment here, make it sound like we are disagreeing on basic physics. Can you please update your answer to more clearly state your case, or at least comment here specifically what you feel is incorrect in my answer? Thank you. –  Edward Mar 22 '11 at 3:37
"essentially giving all our engine power into gravitational energy" . At that point you are in the less economic part of the car engine, high rpms slow mileage, not peak performance that you assume. Most of the gas is consumed in heating up the engine so it is not just against gravitational energy. This heat you will not get back downhill, you can only get the gravitational potential to kinetic back. The gas saved from having 0 rpms on the engine versus the economic rpms on the flat will not compensate for this. –  anna v Mar 22 '11 at 8:57
@Edward I drive very hilly roads half the week. I am not talking equal speed here. One incline I have to do 1st gear, the rpms hit the roof, there is one economic rpm range for cars. The engine gets hot.If I go lower rpms the engine will stall. –  anna v Mar 22 '11 at 11:47
@Edward, your claim that the gravitational potential energy sums to zero can be challenged on earth. After ascending a hill, you can not get all that stored up energy back as it will be consumed by increased air friction or brake heat on the way down, thus not being used fully to propel the car. You will be overcoming more gravity on the way up than you can consume on the way down. I would say that calling it a zero sum is valid only in a vacuum. This may be a large enough discrepancy in your logic to sway the conclusion. –  Steve H Mar 22 '11 at 15:21

In a simple minded way I would say that the flat road for the same miles and the same speed would be more economical:

Moving the car uphill against gravity takes extra energy which is not being gained coming downhill because of braking (constant speed) and not turning off the engine for safety.

For each car there is an economic rpm , mine is at 2500 rpm . One can use that on a flat road, but a hill needs lower gears going up , not the economic rpms, and this is not gained back due to braking downhill so I cannot see how a hilly road can be more economical in any case.

edit: Looking at the preferred answer which claims gains by the method of driving the car I searched for a curve of rpms to fuel consumption and power output. Surprisingly there are not many out in the links. Here is the only one I found probably scanned from an engineering textbook. I observe that consumption goes up with rpm. Going uphill needs more rpm. I will also add that most of the inefficiencies are in heating the engine, and the higher rpm the hotter the engine.

Let me simplify the problem: If I pump 100 litres of water up hill and let it run down hill. Will the kinetic energy of the running water be greater than the energy utilized in pumping it up? In the best of conditions, without the inefficiencies of pump, it will break even.

In the car analogue then one is playing with inefficiencies of flat versus hill and maybe a computer program with real car data would give a definitive asnwer.

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"I cannot see how a hilly road can be more economical in any case." - yes, absolutely right. –  delete Mar 21 '11 at 15:41
You are restricting yourself to a constant speed. That defeats the point of the question. So you are oversimplifying the question and then mistakingly concluding that in general "I cannot see how a hilly road can be more economical in any case". As one can see from Master of Disaster's answer, he is restricting himself to an even more limited case (no friction) and trying to conclude something general. These in my opinion are horrific mistakes in logic. –  John Mar 22 '11 at 2:03
@John I cannot think of a more limiting case than energy conservation, which is what I am talking about. I just simplified the description to show up this. All the rest is dressing, like my uncle trying to make a perpetual motion machine with bicycle wheels and gears. Seems car enthusiasts do not believe in energy conservation !! –  anna v Mar 22 '11 at 5:34
@Anna No one is claiming you can make a perpetual motion machine. The issue is that you are making a mistake in logic and essentially arguing against a straw man, and coming to an incorrect conclusion. Energy conservation doesn't let you make the general claim you are making. As just one (of many) simple examples: consider a hill shaped such that the friction is enough that you don't need to use a brake like you assume. The main issue you are missing is the effect of the efficiency curve for an engine. See Edward's and some of the other answers for more information. –  John Mar 22 '11 at 6:09
@John if you do not need to use the brake, you will just break even. On this hill you will be spending more gas going up than on the level, and you will be on the inefficient rpms of your car. You cannot get more energy than you put in. Nobody seems to consider that the hilly road is a smaller distance than on the straight, for the same miles, to increase those miles per gallon for the straight part. –  anna v Mar 22 '11 at 7:16

Unfortunately the answer very much depends on the poorly specified "an automobile" portion of the question.

We'll use a riding lawnmower to prove the hilly case:

• Assume the flat route is flat.
• Assume the "hilly" route is not more than perhaps a dozen degrees going up to the midpoint, then similar angle going down to the endpoint - just barely steep enough that gravity is sufficient to overcome the rolling friction of the lawnmower.
• Assume that at the speed the lawnmower is traveling, about 5-10MPH, there is no significant air friction.

A riding lawnmower's engine and gear train is so inefficient that it will consume the same amount of gas riding on a the specified slight incline as it will riding on a flat surface.

For the first half of both routes, the same amount of fuel is consumed. For the second half, the flat route requires more fuel, but the hill route does not. Therefore with the efficiency of the engine assumed, the hill route could take up to 50% less fuel than the flat route.

If we scale this up to a regular car, then we need only pay attention to the following:

• Does air resistance come into play
• Is the engine efficiency different between the two routes

Internal combustion engines have a lower bound for energy output. You can't get an automobile engine to put out some fraction of this lower bound when it's running - it's spending the same amount of fuel whether you're drawing 100 watts or 500 watts. Once you get into the higher end, the engine consumes fuel at a rate that corresponds to output energy. Fuel consumption only increases from this lower bound.

This is why the riding lawnmower is an easy case - the total range of efficiency of the riding lawnmower is so small, that there's no point in that range where going faster (therefore spending less fuel due to time spent driving) will use less total fuel.

However, some cars will have such low air resistance, and such high efficiency at higher speeds, that the flat route will consume less than the hilly route, because even the slight slope we're positing will make a 2x difference in the energy required to drive the whole route.

Therefore the question is inadequately specified to answer for the general case.

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If the hill permits hypermiling to be done. In the most idea case, at the top of the hill the driver would turn off the ignition and shift into neutral, and descend unpowered.[Don't try this as your power brakes and power steering won't work as expected]. I assume the hill is just the right grade to maintain speed unpowered. The advantage comes from avoiding engine braking (losses within the engine/transmission) during the downhill leg. Of course this also requires the engine to not lose much efficiency during the climb. Hybrid cars work like this, charging the battery resembles the uphill climb, and using the battery in "stealth" mode resembles the downhill portion. So hypermiling is simply using gravity as a low tech (but very efficient) hybrid. If you leave the car in gear on the downhill, and thus suffer from the engine braking during the descent I doubt you get the benefit. So if you are a legal driver, you may only see the milage improvement if you drive a hybrid, which is designed to shut the internal combustion motor off when it doesn't need the power. [Of course the hybrid has the additional advantage, that if the hill is too steep, the excess energy (within limits) can be captured by the battery.

I'm assuming here that speed is constant. It is almost a trivial excercise to show that we can trade off travel time versus fuel consumption. But modern society rarely gives us the option (especially with other cars on the road). If we assume linear engine braking, then the total engine braking loses for a given trip are proportional to the total number of revolutions the engine made. Running at a fixed RPM, but at higher torque, then in neutral with the engine either idling (or turned off) minimizes the total number of engine revolutions for the trip.

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damn, I can't believe you beat me writing this! I saw this after submitting mine... –  Dov Mar 19 '11 at 15:23
I've had similar things happen to me. I think I had one case, where I and another SE responder must have been editing/submitting similar answers nearly simultaneously. I wish stack exchange had a merge mechanism. Trying to improve a given explanation with comments is pretty imperfect. –  Omega Centauri Mar 19 '11 at 19:44
This is the only practical answer. –  Carl Brannen Mar 22 '11 at 0:27

I am not sure whether this is primarily a "car" question or a "physics" question. As a physics question there are a few points to note: the fact that the roads are the same length and that one is hilly means that the flat road is necessarily curved between the two towns. Also the time to travel is not relevant here.

So the flat drive must use petrol/gas for the entire journey between the two towns. However the hilly route only needs to use petrol/gas for the uphill portions: it can glide downhill. So the hilly route will use less fuel if the following equation gets satisfied:

Fuel used uphill < Fuel used on Flat (at any speed)

EDIT: (With more explanation of the physics.)

The physics of this situation involves the use of internal power with which the elementary mechanics examples are misleading. Elementary mechanics examples consider objects moving in frictionless planes. In order to set such an object moving it requires some impulse I for a short time to reach a velocity $v$. Thereafter the object can travel from A to B without further energy. Indeed it can travel to any C (linear with AB) without any further energy. This fact, together with the elementary "conservation of energy" equation in the form

E = T + V

is the basis of both elementary mechanics examples and also fundamental physics examples.

What is different about the "internal power" scenario is that, for reasons to be discussed below, movement from A to B requires energy. Furthermore movement from A to C will require a different amount of energy in general. In elementary physics examples this energy comes from the external potential V, but in "internal power" systems it comes from some compact fuel. For these reasons, although all the basic laws of physics still apply, simplistic applications of "conservation of mechanical energy" do not apply.

The reason why movement from A to B requires energy is that the movement is not over a frictionless plane, but over a friction bearing road surface. The friction contributions are from:

(a) Tyre against road - probably a constant

(b) Internal friction in vehicle - proportional to $v$

(c) Air resistance - proportional to $v^2$

For these reasons a constant use of fuel is required to continue motion; with no additional use of fuel the vehicle will stop (on a flat road).

So the equation has to take account of the amount of fuel used (=amount of energy used), and some approximations need to be made:

So if a journey from A to B uses F units of fuel, a journey of twice the length uses 2F fuel.

The solution proposed here uses 2F fuel for the flat journey and for the uphill journey will use: F + mgH units of fuel. This needs to be less than 2F. However there is still energy transfer in the downhill part resulting from the mgH potential energy transfered to kinetic energy: but this does not use fuel/ internal power in the extreme case.

Recently I have just seen a TV programme showing "gravity racer" cars: no fuel on the downhill and speeds of 40mph+ were reached without much braking - lots of energy transfer of course.

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It is both, and I agree with all of your assertions. Suppose it is possible to coast downhill at a 10% grade without using any fuel. Does it require twice as much fuel to climb uphill at the same grade as it takes to drive flat for the same distance? –  Dan Brumleve Mar 17 '11 at 11:26
One point is that it is not the same distance, but only (perhaps) half the distance. Still I admit that one could develop some more equations to improve the model description. –  Roy Simpson Mar 17 '11 at 11:37
In my previous comment I was supposing that the uphill and downhill parts of the hilly road are of the same length (each part half the length of the flat road): "/\". This is not necessarily the case but I wonder if such a road could satisfy the question positively for some real car? –  Dan Brumleve Mar 17 '11 at 11:56
This is complete nonsense, absolute balderdash. –  delete Mar 21 '11 at 15:40
I have updated the physics description here, to explain that aspect more. –  Roy Simpson Mar 21 '11 at 22:54

It is possible under the following circumstances.
1) The energy spent on itself by the engine (internal friction of engine) is constant (this is true when the engine running at constant speed regardless of the speed of the vehicle (may be by using a infinitely variable gear). Assume the energy spent per second by internal friction is A and the work done (external) is Y. Total fuel consumed is (X + Y).
2) The efficiency of the engine is very low under normal working conditions. That is Y/(X+Y) is very low (say 10%) at 50 km per hour speed.
3) Total frictional force is very high at higher speed. Lower at low speed.
Condition (1) and (3) means that there will be point of maximum efficiency - say at 20 km/hour speed.
Now construct a road with a downward slop such that the vehicle just rolls down without engine (engine off) till a an intermediate point. From that intermediate point the road starts upward slop with maximum steep (depends on the grip of the tyres). That means intermediate point is more close the upward sloping end.
Now vehicle runs till the intermediate point without any fuel cost and the remaining section with marginal increase in fuel cost than a vehicle running at plane road (as most of the energy spent is on internal friction).
In this case the efficiency of the downhill/uphill road is higher than a plane road.
I assume the mathematical calculation behind my arguments are obvious. If not let me know I can elaborate that further.

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absolutely.

If you assume what Edward assumed, I can give you the optimal path. I'm assuming a gas engine, which has an optimal fuel curve and which has a fuel use when its on, even in neutral.

Climb a hill immediately, at a speed governed by the ratio of gas used to climb it/gas used by idling. There will be some optimal climb rate. Example: if you burn 1 gal/hr, and climbing the hill at 200mph costs you 1 gal in 2 minutes (purely imaginary example) then the right answer is somewhere less than an hour, but more than two minutes, whatever gives you the minimum fuel use. Then, turn off the engine, and crawl downhill toward the goal. The optimal shape is the minimum hill that will overcome rolling friction at a crawl.

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Do the uphill and downhill parts of the road have the same slope, or equivalently (because of the same-elevation constraint), do they have the same length? If not, what determines the ratio? Also, are you saying that the most efficient mileage on the flat road is obtained by driving at the slowest possible speed? (I see that this minimizes air resistance but I just want to confirm). Favorite answer so far but having some trouble working out a specific example. –  Dan Brumleve Mar 20 '11 at 6:02
This answer tries to handle the question with a concrete example, but despite including numbers doesn't actually do so. You essentially just say "There will be some optimal climb rate", and the end with "Then, turn off the engine, and crawl downhill toward the goal." The phrase "at a speed governed by the ratio of gas used to climb it/gas used by idling" is also troubling as it doesn't even give a speed unit wise, nor does it make sense why that ratio is useful (the important issue is the power vs gas efficiency peak ... not the gas idling rate). –  Edward Mar 20 '11 at 12:14
I am curious in all these answers, where is energy conservation? Going uphill, no matter what speed, requires extra energy then on a flat road. This could be gained back going downhill, if one could turn the engine off and ignore speed limits, which cannot be the case. Anyway at most one would just break even as far as energy goes. Am I wrong that mileage is gas consumed, i.e. energy per mile? –  anna v Mar 20 '11 at 14:08
@anna: The gravitational energy is the same at the beginning and end of the trip, since the beginning and end are at the same elevation. So of course the energy put into gravitational energy can be retrieved going down the hill. I don't understand why you are worried about speed limits, for in the ideal case the slope would be small enough that the car is creeping down it. Remember, we're trying to answer if the hilly route can EVER be better, or whether the flat route is always better. It turns out that the hilly route can in at least some cases beat out the flat route. –  John Mar 20 '11 at 16:15

This is a question which might be best for “Click and Clack” on “Car Talk.” The question is a bit unclear on a couple of things. So I will assume that the path length on the straight road and the hilly road is the same. I will also assume that the average velocity of the drive on the two roads is the same. Clearly if the straight road is much longer then the hilly road would be more economical. So suppose you drove the hilly road and you take your foot off the fuel on the descending portions of the drive. The idea is then that the potential energy you “banked” in driving up is converted to kinetic energy, and your gas mileage on the trip down is near “infinity.” If you use the brakes a lot on the way down, which might be advised for safety, then some of that energy is dissipated as heat.

Would the two be equal? I don’t think so. The reason is thermodynamics, for a good chuck of the energy used to climb the hills is lost as heat, so you recover a rather small amount of that energy as kinetic energy on the way down. We might think of the thermodynamics of the hilly drive as thermodynamic losses of the straight drive plus thermodynamic losses which are incurred in raising the potential energy of the car on the various hills. This is even more so if the engine idles on the way down, which is advised with modern cars. An old VW bug might save a bit on saving this energy loss with old fashioned brakes.

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Reference to reality and to thermodynamics is not appreciated in this thread. Good luck! –  Georg Mar 21 '11 at 18:40
So let me be sure what you're assuming. The two roads run between two cities. One road is straight and flat. The other is curved. And yet they are the same length. Hmmmmmmm. Somehow I can't picture this in my mind's eye. –  Carl Brannen Mar 21 '11 at 21:37
One is curved in the vertical direction, the other straight. The distance of the two paths on the tangent plane of the Earth is the same. –  Lawrence B. Crowell Mar 22 '11 at 12:24
Lawrence, the flat road must be curved because it has the same length as the hilly road and connects the same endpoints. Neither road is straight. –  Dan Brumleve Mar 22 '11 at 21:04
We can assume the two roads arc around or something, where one remains flat and the other goes up and down a mountain. I am not sure what the difficulty is with this. One thing seems apparent is that not that many people have driven in mountains. It eats up gas --- big time, even if you try to use physics (gravity etc) to economize. –  Lawrence B. Crowell Mar 23 '11 at 1:44