Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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?

share|cite|improve this question
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
Unfortunately i don't think you can get a reliable answer. There are so many variables like performance of engine vs rpm, average speed you want (slower means less air drag), speed limits (might need to waste energy on brakes), drag relation to speed, etc. Any answer would be pretty much a wild guess. – Fermi paradox Apr 16 at 6:27
You should just get an electric car. They will give you the best mileage per gallon and should also be more efficient ;-) – Mark N Apr 17 at 20:09

16 Answers 16

up vote 16 down vote

I believe, the answer is a small but quantifiable, yes, there is a non flat road configuration that would lead to better gas mileage between any two points at the same height. I have numerically solved for such an optimal path. I believe I can give a nice explanation of why that is, but it will take some work, so bear with me. Granted, you can only expect to save about $0.01 on fuel by taking such an optimal path, more or less independent of the distance you need to travel.

tl;dr: There is an optimal velocity to travel at to minimize fuel consumption over a fixed distance. If we next ask what the optimal path is between two fixed points assuming we start and end with zero velocity, the answer is roughly, to speed up to the optimal speed rather quickly, maintain that speed for most of the distance, then slow down at the end. If we allow hills, the best road will have a downward slope to start to help us get up to speed, and an upward slope at the end to help us slow down. This allows for better fuel efficiency.

I found this question very interesting, albeit challenging. In what follows I hope to walk you through my thought process and hopefully introduce some clever techniques and results.

As an overall strategy, I thought the original question was too hard. So, in the grand tradition of science, I started to break down the problem into easier ones. Trying to determine the optimal path was too difficult, as it seemed like I would need to know the optimal velocity profile first. So, then I tried to figure out the optimal velocity profile for a flat path. That itself was difficult, so I instead started with trying to figure out the optimal speed for a car travelling a fixed distance at constant speed. That is where we will begin.

Optimal Fixed Speed for Fixed Distance

We have all heard that there is an optimal speed to travel at to maximize fuel efficiency. But why is that? There ultimately has to be some kind of trade off, wherein we pay a penalty for going too slow, and pay a penalty for going too fast, so the optimum can be somewhere in between. It is easy to see why going to fast is bad, the faster we go, the more we lose out to air resistance. On the other end, if we go very very slowly, this is also inefficient, since it will take a very long time for us to reach our goal, during which we will lose fuel to having the car on and running. As I explored in my answer to How effective is speeding?, a very simple model that does a decent job of fitting measured fuel efficiency curves is to assume that the power drawn by the car takes the form $$ P = A v^3 + P_0 $$ here the $\propto v^3$ term is the cost of air resistance, since $F_{\text{air}} \propto v^2, P \propto F v \propto v^3$, and the $P_0$ denotes our constant power losses due to having the car on. As I showed in that other answer, this does a decent job of modelling the observed fuel efficiencies.

Trying to be a bit more accurate this time, let's include rolling friction and rotational losses in our car, using as our model of the force $$ F = A + Bv + Cv^2 $$ as per this EPA report (p. 8). The benefit then is that we can use their parameters for a 2004 Honda Civic DX (p. 99) $$ A = 105.47 \text{ N} \quad B = 5.4276 \text{ N/mps} \quad C = 0.2670 \text{ N/mps$^2$} \quad m = 1239 \text{ kg} $$ and obtain some level of modest realism.

It remains to determine $P_0$, which I set to $P_0 = 6 \text{ kW}$ in order to obtain a sane value for our optimal speed. Our resulting model gives, for the fuel efficiency of our model car:

Model Car MPG

with its peak at 41 mph in this case. This also qualitatively agrees with observed fuel efficiency curves, i.e. this wikipedia figure or this post from automatic

Now it should be clear that if we are interested in going a decent distance, our optimal trajectory will be to speed up to around 41 mph, then maintain that speed for most of the trip, and slow down at the end. This ensures the optimal tradeoff between the time we need the car to be on and the loses due to the various friction forces our car feels. This is going to give a very different character to my answer than to Edwards, as his strategy was to go as slow as possible to avoid the frictional losses. Unfortunately, taken to the extreme, this would mean we waste a lot of fuel just having the car on.

Granted, one could imagine throwing the car in neutral and turning the engine off, but this isn't a safe or realistic strategy. In what follows, I will try to see what kind of answer we get, assuming we don't turn off the car or do any particularly clever hypermiling tricks.

Optimal Control for Fixed Distance and Flat Terrain

Having figured out what the optimal fixed speed for a fixed distance, we would like next to figure out what the optimal velocity profile is for a fixed distance, assuming we start and end at zero velocity. This turns out to be a problem of optimal control.

To formally state the problem, we have the state of our car, $v(x)$ the velocity as a function of distance between two fixed points $x=0, x=X$. We have control variable $u(x)$ which will roughly correspond to how much we are pressing on the accelerator or brake. We are searching for the $u(x)$ that minimizes our fuel consumption, which we will take as:

$$ F = \int dt\, r(v,u) = \int_0^X dx\, \frac{r(v,u)}{v} $$ where $r(u,v)$ is the rate at which we consume fuel for a given $v,u$. Our car is described by the given dynamics. $$ \dot v = a(u,v) = u - \frac{A}{m} (v>0) - \frac{B}{m} v - \frac{C}{m} v^2 $$ here we have the $A,B,C$ terms representing rolling resistance, rotational friction and air resistance, and $u$ is the extra acceleration provided by the car.

What shall we take as our fuel consumption rate? I had a lot of trouble coming up with this, but I think a good definition is $$ r(v,u) = m v u (u>0) + P_0 $$ The intuition is that the fuel consumption should be proportional to the power our car needs to provide, our car provides the acceleration $u$, so power $P=mvu$. But, we only consume fuel if we have positive acceleration. When we brake, unless we have regenerative brakes, that energy is lost and not recovered, so I threshold our fuel consumption function to positive values of $u$. Another important caveat is that cars have a maximum and minimum value of $u$ they can apply. For our model car, I will take $u$ to be bounded in the interval $[-b,a] = [-7.2, 3.2]$ which I obtained from 0-60mph acceleration times ($u \leq a$) and braking distance measurements for honda civics. ($u \geq -b$)

Optimal Control Problem

At this point, we have phrased our problem as a problem in the calculus of variations. Minimize the fuel use over all possible accelerator profiles $u(x)$ subject to the physical constraints and the dynamical constraint.

$$ \min_{u(x) \in [-b, a]} \int dt\, r(v,u) \quad \text{subject to } \dot v = a(v,u), v(0) = 0, v(X) = 0$$ at this point, we could proceed by adding a Lagrange multiplier for our contraints, taking the functional derivative and finding things corresponding to the Euler Lagrange equations for this system. But that was too hard for me. Didn't work out.

Next we could try to find conditions our optimum must satisfy by applying Pontryagin's minimum principle and working things out that way, but again this proved too difficult for me.

So, instead, let's just opt for a numerical solution, which we will find by imploying dynamic programming.

Dynamic Programming

The gist here is that we don't have to solve the whole problem at once, instead let's do something akin to a proof by induction, we will try to break the problem down into its smallest piece, and write the solution to that small problem in terms of the solution to a slightly smaller problem. This will set up a sort of recurrence, which in combination with a solution to the smallest possible problem will allow us to brute force the solution to any problem we want.

We start by replacing our problem with an even larger one. Let's seek optimal paths starting from any intermediate $x$ with any intermediate velocity $v$ and write

$$ F_{x,v} [ u(x) ] = \int_x^X dx\, \frac{ r(u,v) }{v} \quad v(x) = v, v(X) = 0 $$ so that $F_{x,v} [ u(x) ]$ is the fuel consumed for a policy $u(x)$ starting at $x=x$ with $v(x)=v$ and ending at $x=X$ with $v(X)=0$. In terms of this we can define the cost for these optimal partial paths. $$ C(x,v) = \min_{ u(x) \in [-b,a] } F_{x,v} [ u(x) ] $$ So that $C(x,v)$ gives us the fuel used for an optimal path starting at $x$ with velocity $v$ going to $x=X$ with velocity $v=0$. It seems we've just made our lives worse. Instead of determining the cost for a single optimal path ($C(0,0)$ in this case), we've created the problem of solving a whole set of optimal trajectories. The magic happens next. We imagine partitioning $x$ and $v$ into a discrete grid, $x_i$, $v_i = v(x_i)$, and try to write the solution at one of our grid points in terms of the solution at the next grid point.

$$ C(x_i, v_i) = \min \left\{ \int_{x_i}^{X} dx\, \frac{r(u,v)}{v} \right\} = \min \left\{ \int_{x_i}^{x_{i+1}} dx\, \frac{r(u,v)}{v} + \int_{x_{i+1}}^{X} dx\, \frac{r(u,v)}{v} \right\} $$ but that second term is just $C(x_{i+1}, v_{i+1})$, and our integral is a single step so we have: $$ C(x_i, v_i) = \min \left\{ \frac{r(u,v)}{v} \Delta x + C(x_{i+1}, v_{i+1}) \right\} $$

Alas, we've made progress. We have the value of $C(x_i, v_i)$ as a minimum of the fuel we use this step plus the optimal value at the next grid point. Knowing this, and that at the very end of our journey we have $$ C(X, v) = \begin{cases} 0 & v = 0 \\ \infty & v \neq 0 \end{cases} $$ we can proceed to compute $C(x,v)$ for all values of $x$. We just need to define what we mean by $v$ and $u$. For that, we need to ensure that we satisfy our dynamical contraints and do a decent job of approximating the integral by taking $r$ at a single point. So we will take $$ v = \frac 12 \left( v_{i-1} + v_{i} \right) $$ $$ \Delta x = x_{i} - x_{i-1} = \left( v + \frac 12 \dot v \Delta t \right) \Delta t $$ $$ \dot v = \frac{ v_{i} - v_{i-1} }{ \Delta t} = a(v,u) $$ these allow us to treat $\Delta x$ as a constant, and determine $u, r(v,u), a(v,u)$ all in terms of $v_i$, and then take the minimum over all values of $v_i$ on the right. This approach is similar to the one adopted in this paper (doi), which coincidently does a more complicated model with gear ratios and switching and the like and is worth checking out.

Dynamical Program Results for Flat Terrain

We can then code the whole thing up and solve for the optimal profiles for a fixed distance. I'll take as our distance 1 mile. This is what we get for our velocity and accelerator profiles:

Velocity and Accelerator Profiles for Flat Terrain

Notice that we have basically three stages. In the beginning, we accelerate rather rapidly up to nearly optimal speed. Then in the middle portion we travel at nearly optimal speed in order to cover ground, and then at some point, we coast to reduce our speed, and finally brake to end up with zero velocity at the end.

We can also see our fuel use as a function of distance and compute the total fuel use and fuel economy for this trip:

Fuel Use for Flat Terrain

In this case, we use 0.29 Gallons, for an average fuel efficiency of 34.3 mpg.

Notice however that the majority of our fuel use comes from the main stretch at near optimal speed. There isn't anything we are going to be able to due about the portion. If we hope to make any strides towards savings, it is going to have to be in reducing the fuel use in the initial portion of the trajectory. Just that first part where we accelerate up to speed takes up 0.008 gallons of gas or about $0.02 worth at current prices. This is really the only place we should hope to see some kind of benefit, so even if we do manage to do better, it isn't going to be by much.

Intuition for Hill Answer

Alright, having worked out what we should do on flat ground, now we are really in a position to tackle the question at hand. What would be the optimal design for the height of the road in between our two endpoints? Is there any benefit to be gained?

Now that we have set up the framework, it is straight forward to just run the numerical optimization problem again, this time with another dimension to our grid corresponding to road height, which we will do shortly, but let's see if we can extract some intuition based on our answer for flat terrain.

Notice that the optimal control path has basically three stages. Stage I is to accelerate up to nearly optimal speed, in which we incur our greatest fuel hit. Once up to speed in stage II we maintain that speed, which will cost us some small but constant and nearly optimal fuel consumption rate. And finally, as we approach our target, in Stage III we decelerate, first by nearly coasting and finishing up with some brakes.

So, can we do better with hills? Well, since we are dealing with a car, it hurts us to accelerate using the engine, while we don't recover anything for braking, so if we are going to do better we need to do something about that initial acceleration.

But, we could just use a downward slope to achieve that initial acceleration. Let gravity do the work for us and we won't incur any fuel penalty.

The physics is very similar to a classic demo they show in intro mechanics classes. Consider the following setup:

Conservation of Energy Ramp

We have two ramps. One with a gentle constant slope the whole way and one with a dip in the middle. Written in big red letters across the thing is "CONSERVATION OF ENERGY". The question that is asked of students is: Which ball will reach the end first? Think about it a moment before scrolling down.

Here is the result:

Conservation of Energy Ramp Gif

Notice that the ball that dips down handily beats the first ball. Now, this demo is usually fun because students all say that they balls will take the same amount of time, largely because of the prompting caused by the "CONSERVATION OF ENERGY" written across the apparatus. But, while energy is conserved, the time it takes to travel between the two points is not.

Since some of the fuel loss we incur is some constant power $P_0$ for having our car on in our model, we could benefit from essentially the same design. Use gravity to reduce the time it takes for us to go from A to B, and assist in coming up to nearly optimal speed, and we might be able to save on some fuel.

In particular, as we suggested a moment ago, the real trick would be to try to mitigate the large fuel cost of our initial acceleration in the beginning. If only we could use a slopping road to speed us up, then just maintain constant nearly optimal speed for most of the trajectory and then slope up at the end, wherein again gravity would help us brake without having to apply it ourselves.

In fact, let's look at the integrated acceleration of our optimal path as a function of distance:

Integrated acceleration of Flat Terrain Trajectory

Here I have even divided through by $g$, so that this is a path on which our car would feel the same acceleration profile it takes in the optimal trajectory for the flat terrain case.

Our truly optimal road should look qualitatively like this. We will use an initial downward slope to speed the car up to nearly optimal speed, then we will maintain that speed for most of the trajectory. When we get to the end, we'll use the upward slope to help slow the car down. Notice that our integrated acceleration doesn't quite make it back up to $h=0$. This is due to our frictional losses, and it will mean if we try to use a downward dip, we'll have to throw in some acceleration in the end in order to make it back up.

In our flat trajectory, we didn't need to use extra fuel at the end, since we applied our brakes. Here we'll have to give it some oomph at the end, but it should only take some fraction of the fuel it took in the flat case to get us up to speed. In the language of this hill thing, the downward slope gets us down to speed, and while its true we won't come all the way back up, we will certainly make it some or most of the way back up the final hill.

Let's now actually compute the optimal path for the road, by extending our dynamical program.

Optimal Hill Design

We'll use the same numeric technique we used before, but with a new control variable $h(x)$, the height of our road as a function of distance. We have to modify our dynamical constraint, this time we will have

$$ \dot v = a(v,u,h) = u - \frac{A}{m} (v>0) - \frac{B}{m} v - \frac{C}{m} v^2 - g \frac{dh}{dx} $$ where we have added another term to our acceleration given by the shape of our road. Our fuel use will stay the same, as we only incur a fuel cost for using our accelerator. But now we will need to solve the larger dynamic program:

$$ C(x_i,v_i,h_i) = \min \left\{ \frac{r(v,u,h) }{v} \Delta x + C(x_{i+1}, v_{i+1}, h_{i+1}) \right\} $$ Where we will impose our dynamical constraints in much the same way above, but now our minimization will be a two dimensional minimization over candidate velocities and heights. In either case, the program still runs, just takes some more time and we find for our optimal road:

Optimal Road

Which honestly looks pretty similar to what our intuition suggested. As before, we can look at the optimal $v(x), u(x)$:

Hill Optimal velocity and accelerator

Where our velocity profile looks pretty similar. Our accelerator shows some goofy behavior, but overall has the kinds of trends we expected. We let the car more or less glide down the hill, then maintain a constant nearly optimal speed, and then let the hill slow us down but add some juice at the end.

The real test will be to compute the fuel use in this case:

Hill Fuel Use

and TADA, we use slightly less fuel, 0.027 gallons corresponding to an average efficiency of 37.6 mpg if we allow the road to find its own optimal configuration.

This is 0.002 gallons less than the flat terrain case, or about half a cent's worth of gas at current prices. This saving should also be more or less independent of the distance we are hoping to drive, as in either case we have a long stretch of time where we are travelling at nearly optimal speed, the difference was just in how we try to mitigate the initial speed up from our flat terrain case.


Yes you can do slight better, about 1 cent better. The trick is to let the road speed the car up, then just maintain a nearly optimal speed for most of the journey, and let the hill mostly slow you down at the end. This lets you come out slightly ahead.

All of the code used to generate these results is available as an IPython Notebook.

Very fun question.


  • Previous Answer on fuel efficiency at constant speed
  • EPA report on car efficiency pdf
  • Notes on Optimal Control pdf
  • Optimal Control of Automobiles for Fuel Economy doi
  • IPython Notebook of code for this answer ipynb viewer

Appendix A: Hypermiling

As Floris suggests in the comment, we might be interested in how the story changes if we allow our car driver to try to be maximally efficient and turn off their engine when it is not in use. We can solve this case as well. In fact, we can model this scenario by just modifying our fuel consumption function

$$ r(u,v) = \left( m v u + P_0 \right) (u>0) $$

Now, we actually incur no penalty if we are not drawing any acceleration from our engine.

For our flat trajectory, the optimal strategy becomes:

Hypermiling Flat Trajectory

So that our driver just turns on the engine in pulses, and otherwise coasts for most of the journey. This actually cuts our fuel consumption nearly in half:

Hypermiling Fuel Consumption

using only 0.013 gallons, or an average fuel efficiency for this route of 76.8 mpg.

share|cite|improve this answer
Characteristically detailed, as usual - but I think you can do significantly better than your answer by turning the engine off when you don't need power from it. That $P_0 = 6kW$ is bugging me... – Floris Apr 22 at 19:34
I thought about this, but then I also thought that it is probably better if you climb an hill with the engine at its maximum efficiency and then you shut it down completely for the rest of the trip in a slow descend. This avoids a lot of friction from the moving parts of the engine. – DarioP Apr 22 at 19:40
Your hypermiling result for slope sounds wrong. If you climb a single slope at the most efficient rpm, then coast to maintain constant speed, total energy dissipated must be lower than is achieved with the variable speed of the on-off car on the flat. Unless your engine pulsing happens with zero penalty (essentially creating a "class D engine") . If starting the engine has a finite penalty the slope wins because the engine is only used once at the most efficient speed/torque. – Floris Apr 22 at 21:13
The crucial point is to determine at what increase of speed the increased air drag will be greater than the increased kinetic energy. That is the point when the hilly road becomes less convenient: intuitively that point is very high, but, if your calcs should confirm that, than the yes shouldn't be small, I suppose. – user77427 Apr 24 at 5:32
Your answer focuses on optimal speed/ control/ hill design..etc., but the question is not asking for that, and above all, once you found that, both cars on both roads can adopt the same best strategy, so that is quite irrelevant, whatever that is. The question is if/ when one can get better mileage choosing the hilly road, ceteris paribus – user77427 Apr 24 at 6:20

"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.


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).

share|cite|improve this answer
@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
"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
@Edward a hilly road will be longer and require more fuel to climb than is saved in coasting (a car is far from a perpetual motion machine). These factors alone seem to negate your conclusion. – user6972 Nov 21 '13 at 5:21

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.

share|cite|improve this answer
"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
@Anna, I think that the "conservation of energy" argument that is being used here is that friction will prevent the restoration of the initial state of an (unpowered) object in a potential field. This becomes that a car moving uphill/downhill (at a constant speed) must use more energy than a flat (which is how one actually drives). But the answers are being more subtle than this, with some form of non (or low) fuel use downhill in a glide ie the uphill car stops using fuel after half the journey. – Roy Simpson Mar 22 '11 at 13:46

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.

share|cite|improve this answer
I am surprised to see I am the first person to upvote this. I believe you hit the essence of the problem. Eliminating internal friction of the engine is the biggest gain, and a well known trick in the ultra fuel conservation community. – Floris Apr 22 at 20:41

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.

share|cite|improve this answer

A nice answer to this question may be found in the 2009 MIT Course on the Physics of Energy (Lecture 3) by R. Jaffe and W. Taylor. They work out the energy balance of car transport as an example of mechanical energy and its conservation.

Let me briefly recap their findings in my own words. The two "towns" they consider are Boston and New York, both assumed to be at sea level. Assuming a fuel consumption corresponding to a mileage of 30 miles per gallon, the total energy $E$ used for the 210 miles trip is about 840 MJ. Further assuming an engine efficiency of $\eta = 0.25$ this is reduced to $\eta E$ = 210 MJ which, by energy conservation, must cover the mechanical energy expenditure of the car. The energy balance thus becomes

$$ \eta E = E_\mathrm{kin} + E_\mathrm{pot} + E_{rr} + E_{ar} \; ,$$

where the individual energy contributions and their magnitudes are (see the lecture for the actual calculations):

  • kinetic energy $E_\mathrm{kin} = $ 2 MJ
  • potential energy $E_\mathrm{pot} = $ 27 MJ
  • rolling resistance $E_{rr} = $ 54 MJ
  • air resistance $E_{ar} = $ 133 MJ.

The following remarks are in order: (i) The numbers sum up to 216 MJ which nicely agrees with $\eta E$ above (within the accuracy limits of the basic estimates made). This is a useful reality check. (ii) Kinetic energy is basically negligible. (iii) The bulk of energy loss (to be made up by fuel consumption) is resistance, in particular from air. (iv) Crucially, there is a contribution $E_\mathrm{pot}$ which is due to an imbalance in going over hills: when going downhill one has to brake, so not all the potential energy gained going uphill is recovered as kinetic energy. (Jaffe and Taylor assume a dissipative loss, $E_\mathrm{diss} = E_\mathrm{pot}$, equal to 50% of the uphill potential energy gain, which yields the 27 MJ.)

Now, all other conditions and assumptions remaining unchanged, the only difference between a hilly and a flat road (of equal lengths) is the $E_\mathrm{pot}$ term. Thus, if there were no hills on the road, i.e. for a flat road, the energy consumption would go down by 27 MJ as the up-down imbalance $E_\mathrm{pot} = 0$ in the absence of hills. Thus, in effect, there is a gravitational energy sink rather than a source ("battery"). This could be remedied if the car had perfect regenerative brakes implying vanishing dissipative loss, $E_\mathrm{pot} = E_\mathrm{diss} = 0$, and hence no difference between flat and hilly roads.

A final remark on the largest loss term (air resistance): To arrive at the 133 MJ above, Jaffe and Taylor use the formula

$$ E_{ar} = \frac{1}{2} m_\mathrm{eff} v^2 = \frac{1}{2} c_d AD \rho \, v^2 \; ,$$

where $m_\mathrm{eff}$ is the effective mass of displaced air, $c_d \simeq 1/3$ a typical drag coefficient, $A$ the area swept out by the car, $D$ the distance covered, $\rho$ the density of the air and $v$ the speed of the car. For $E_{ar}$ to be the same on both hilly and flat roads, speed $v$ needs to be kept fixed (which is being assumed). However, the density of air drops at higher altitudes $h$ according to the barometric formula, which states an exponential decrease of $\rho$ with $h$, hence an exponential decrease of air resistance! This seems interesting, but note that the efficiency of car engines drops with altitude as well. (I googled a 3% loss for every 1000 ft, which implies a linear decrease). Modern cars with turbo engines or compressors may not be affected, so one could envisage a theoretical scenario of going from A to B at high altitude to reduce air resistance. In this case, the associated energy loss would get reduced in proportion to $\rho(h) \sim \exp(-h/h_0)$. Thus, the (hypothetical) high altitude road would lead to exponential improvement!

share|cite|improve this answer

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.

share|cite|improve this answer

Answer: YES
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.

share|cite|improve this answer

Yes - but it needs the right sort of hill.

A family member, now sadly passed away, used to conserve fuel during 2nd world war by going into town down a hill with a very long and very gentle incline downwards and turning the engine off - the road is also pretty straight for about 3 or 4 miles. This is hinted at in the answer of Edward.

Sorry, this has even less detail than others, but I hope it is of interest that this principle has been used in practice.

share|cite|improve this answer

I will choose in function of the wind factor.
Wind is a positive source of energy if it blows from the rear and...
The road thru the hills has the potential to minimize the exposure to the wind if it comes from a frontal direction. In the valley the exposure is minimal.
The plain road has the potential to maximize the exposure to the wind if it comes from the rear direction.
Good sailing, enjoy the trip.

The perpetuum mobile notion is dead since a long time ago and a solution on give/take from gravity is useless. The friction is always present but the total path length are equal thus I'will not use this argument.

share|cite|improve this answer

Answer #11: You need 1. downhill slope to get the car to an optimal speed - corresponding the highest gear+optimal rpm. 2. At the optimal performance regime at the point of upslope start, you start the motor and keep adding the energy necessary to get to B with $v=0$, then stop the motor and get to B.

Reasons: I assume there is lower efficiency at lower gears (the acceleration from $v=0$ being the most expensive) and the motor consumes something just to keep itself running.

It is handwaving argument anyway, but I tried...

share|cite|improve this answer

In principle, neglecting rolling friction and air friction, and neglecting the fact that the hilly road is longer, it should make no difference.

On the other hand, if on the hilly road you ever have to either hit the brakes, or use engine braking, because of too much speed, that will cost a lot of energy.

Also on the hilly road, if you have to downshift and run the engine at high RPM so as to get up hills, that will also be costly because of energy lost in the engine.

Bottom line: if the hills are shallow, it won't be much worse than if they are flat. It can never be better.

Added: Fun thing to try - take a vacation to St. John, US Virgin Islands. Rent a small Jeep. Then drive the entire length of the island, about 10 miles. In doing so, you will be in first or second gear almost all the way, and you will get about 10 miles per gallon, or less.

share|cite|improve this answer

The so called Pulse and Glide is a driving technique that is reported to save quite much fuel. It consists of moderate accelerations (that better exploit the petrol engine) interleaved with moderate decelerations (that save fuel). This is even more effective with hybrid vehicles which can completely turn off the petrol engine without issues like the absence of electronic controls (ABS), braking/steering assistance and so on.

If you drive on an ideally-shaped hilly road you inherently apply the Pulse and Glide technique keeping a constant speed, thus avoiding spikes the aerodynamic drag. This improves your fuel efficiency compared to a flat road.

share|cite|improve this answer
I tried to give a simple and very easy to understand answer, to a question that $seems$ simple and very easy to understand! – DarioP Apr 23 at 8:19

...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..?

The question

  1. two roads (between two towns A, B) with a different profile cannot be of same length

enter image description here

  1. The flat road (1) is shorter, therefore the ratio fuel consumption/miles will be always worse, but also the absolute consumption can be greater in some circumstances


Well, there is comment to other answer from OP that state "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." – Vaimsu

I regret I had not noticed this comment, this makes the case for the flat road even more hopeless, especially if there are sharp bends.

  1. even if we overlook that ambiguity in the original question, it is still impossible to give an adequate answer (and alemi's considerable effort is somewhat wasted) since the question is too vague, does not specify any parameter and does not specify if it is purely academical problem or if it refers to real life: for example

....if operated by a perfect driver

a perfect driver knows that it is not lawful and not advisable to run downhill in neutral or, even worse, after switching off the engine, especially if there are sharp bends in the route. Is the driver in our answer allowed to behave in such an irresponsible manner? if he is, to what extent? should we consider the ordinary behaviour/speed at which we travel everyday or is it a sort of Mobil Economy Run,... etc.


Will an automobile always get the best mileage driving between the two towns on the flat road?

... the answer is an unequivocal "no, not always"

The answer

  • If the question is an abstract one and we do not consider cars but a ball or a rollercoaster, the answer is simple: the flatter the road the worst the performance, as you can see in this video.

And here you can learn how with calculus of variations one can find the best-performing curve[s], which are the ones they use in rollercoasters:

enter image description here

on a frictionless mag-lev track a car can go for a quite long way without any fuel consumption at all.

  • If the question is referring to real, responsible driving the answer is still, doubtlessly, : NO, not always:

Supposing the driver is in no hurry and drives normally but with prudence (say at 60 Km/h) on both roads, and the down/up-hill road has a gentle slope, there are no hairpin bends, etc..., then the absolute consumption of fuel on the flat road can be indeed greater, ([Edit: obsolete] and the relative consumption, the ratio fuel/miles definitely a lot worse) When riding downhill, a safe trick which even a perfect driver can use, is to step on the clutch: one does save fuel but, at the same time, facing a curve or any other emergency by releasing it one gets an immediate response.


My original answer considered a flat road shorter than the hilly one, but if the roads have the same length, the second route is almost always more convenient, actually it is really hard to find a case in which it is not. The accepted answer was correct and maybe even too hesitant.

Let's consider a realistic case of a road with a slope of 5%, and the MIT example

enter image description here

No complex calcs are needed, and we do not have to accomodate the roads to our wishes, what matters here is only the difference between the two routes:

both cars can accelerate from 0 to 60 Km/h starting from A and travel any distance to B, then

  • travel the distance BD equal in length and different in profile.
  • rolling friction is practically the same and
  • air drag at 17 m/s is the same: what matters here then is only the difference in air drag when the car (2) accelerates downhill. From the MIT formula (quoted by Tom Heinzl): $$[1/2(0.33)(2.66m^2*330Km)(1.2kg/m^3)*(27.7m/s)^2 = 133MJ]/330,000$$ we know that, for that car (m = 1800 Kg), the energy lost to air resistance is: $400J/m$ at 28 m/s, at 60 it's (400*.6^2) roughly = 114, and at 80 it's (400*.8^2) = 260; the difference is (260-144) = 116 J/m, since the slope is 5% and the speed increase is $\Delta v = 11 m/s$ BD is rougly 240 m,

  • the total loss is **- 28* kJ. On the other hand the increase of energy due to the drop in altitude is $1/2 1800* (28^2-17^2)$

  • the gain from PE is + 455 kJ: a small fraction of that will be spent to compensate the increased air resistance and restore the final speed to 60 Km/h, another tiny fraction might be accounted for an eventual increase of rolling friction, keeping the engine running at min, or other factors (like occasional braking, but also car (1) must occasionally step on the brakes), but the balance is surely positive.

    Since we are told that the road is straigth and the speed of 100 Km/h is quite safe, there is no need to brake. It is not necessary, also, to switch off the engine, to shift into neutral or to disengage the clutch, since an authomatic transmission will adapt itself, but also an old car with overdrive will burn almost no fuel at 100 km/h, and then, lastly, modern cars make use of regenerative braking. In conclusion:

Will an automobile always get the best mileage driving between the two towns on the flat road?

Allowing for some eventual (even huge) imprecisions in my calcs, the conclusion is unequivocal: except for particularly high *speeds, the flat road will get the worst mileage and the car (2) will accumulate hundred of KJ's energy at every drop in altitude. Therefore the correct answer seems : NO, almost never!

*Considering a cruising speed of 28 and a peak of 39m/s at the bottom, the energy lost to air drag doubles,but remains well below the energy gain.

Note :

I did not explicitly mention the ideal path above, since the question did not ask for it, but the profile suggested by alemi doesn't seem the right one. It is plastically shown in the second video at 2:05: it's the curve on the blackboard ,and the one Prof. Pelcovits put his hands on. It is more like a cycloid, a parabola or a catenary, the smoothest the change of direction, the less energy is lost. The engineers that realize roller-coasters choose the best performing curves

share|cite|improve this answer
The other road is flat, but it doesn't have to be straight. So, you can have situation where both roads are of same legth. – Vaimsus Apr 22 at 8:13
Well, there is comment to other answer from OP that state "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." – Vaimsus Apr 22 at 8:55
Also the curve can be continuous over the whole drive in a way that there is no need to slow down or brake. Like the image 2 in your answer, only horizontally. – Vaimsus Apr 22 at 8:58
@Vaimsus, I hadn't noticed that comment, but my answer makes that detail irrelevant. The hilly road gives a better performance anyway, which becomes even better if the distance is the same and the flat road has more bends, even if the driver is not reckless. – user78040 Apr 22 at 9:42
@DanBrumleve, I did not explicitly mention the ideal path in my answer as the question did not ask for it, but the profile suggested by alemi doesn't seem the right one. It is plastically shown in the second video, a 2:09 here, the one on which Prof. Pelcovits put his hand. It is more like a cycloid, a parabola or catenary – user78040 Apr 22 at 12:35


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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

protected by Qmechanic Mar 17 at 9:00

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.