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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"Is there any hilly road on which a better mileage can be achieved?" The answer is: YES. 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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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:
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:
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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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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Answer: YES |
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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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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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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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