# How does a car use gas going up and down hills???

November 13, 2014 6:52 PM Subscribe

I think about this at least once a week as I drive myself to work. On my way in, I go up a hill and on my way home, I go down it. Am I using the same amount of gas as I would if my commute were flat both ways?

My commute is just about 4 miles. On the way into work, the first mile is relatively flat, then I hit a stop sign. The next two are a slight incline, and the last mile is a pretty steep uphill. On my way home, I coast down the hill and through the next two miles before hitting the stop sign. I never accelerate during the first three miles down the hill.

My burning question is: Somehow in my brain I'm reasoning that climbing up the hill uses two times as much gas and coasting down the hill uses no gas. Alternatively, I also think that going up the hill and then back down uses the same total amount of gas as it would if both distances were flat. Is this correct? I know absolutely nothing about cars (or physics for that matter)which is embarrassingly obvious. I drive an automatic two door car--its nothing fancy and it's quite old.

I'm not totally sure my question makes sense. Hopefully I can clarify if necessary. I tried to google it but couldn't figure out any good search terms--if anyone has any search results, I'd happily take those too! Yes, this is a ridiculous question but I seriously cannot stop thinking about it.

My commute is just about 4 miles. On the way into work, the first mile is relatively flat, then I hit a stop sign. The next two are a slight incline, and the last mile is a pretty steep uphill. On my way home, I coast down the hill and through the next two miles before hitting the stop sign. I never accelerate during the first three miles down the hill.

My burning question is: Somehow in my brain I'm reasoning that climbing up the hill uses two times as much gas and coasting down the hill uses no gas. Alternatively, I also think that going up the hill and then back down uses the same total amount of gas as it would if both distances were flat. Is this correct? I know absolutely nothing about cars (or physics for that matter)which is embarrassingly obvious. I drive an automatic two door car--its nothing fancy and it's quite old.

I'm not totally sure my question makes sense. Hopefully I can clarify if necessary. I tried to google it but couldn't figure out any good search terms--if anyone has any search results, I'd happily take those too! Yes, this is a ridiculous question but I seriously cannot stop thinking about it.

If there were no air resistance, you would, in fact, "get back" that energy you used going up the hill when you went back down. You're turning chemical energy (gasoline) into mechanical energy (engine) into kinetic energy (vehicle speed) into potential energy (elevation), and then that potential is turned back into kinetic energy on the way down.

However, there is air resistance. If you're keeping an almost steady speed throughout, it's a wash.

Say it takes 10 horsepower to maintain a steady 35 mph on level ground, and an additional 10 hp to climb the hill at 35 (I know you said the last mile is steeper). On the way up the hill, you're using 20 hp. On the way down, the 10 hp gravity is exerting on you is providing enough to overcome drag.

The particular numbers for the vehicle, hill, etc come into play, but overall it's a wash.

I've run this experiment, of a sort, driving across Kansas. Over 420 miles, the elevation gain is about 3100 feet. My past requires that I drive through Kansas without speeding, so I keep it right at the speed limit. I always get about 1 MPG better going down hill compared to uphill.

Now, on steeper hills, more factors come into play - downshifting and keeping the engine at closer to its maximum efficiency, then coasting down, means I actually get a little *better* mileage than steady state.

(n.b. I'm also the guy who experimentally determined that having the tailgate in my pickup down does, indeed, make for worse mileage, with cross, tail, and headwinds factored out, over something like 40,000 miles of St. Louis to Arkansas commuting.)

posted by notsnot at 7:31 PM on November 13, 2014 [1 favorite]

However, there is air resistance. If you're keeping an almost steady speed throughout, it's a wash.

Say it takes 10 horsepower to maintain a steady 35 mph on level ground, and an additional 10 hp to climb the hill at 35 (I know you said the last mile is steeper). On the way up the hill, you're using 20 hp. On the way down, the 10 hp gravity is exerting on you is providing enough to overcome drag.

The particular numbers for the vehicle, hill, etc come into play, but overall it's a wash.

I've run this experiment, of a sort, driving across Kansas. Over 420 miles, the elevation gain is about 3100 feet. My past requires that I drive through Kansas without speeding, so I keep it right at the speed limit. I always get about 1 MPG better going down hill compared to uphill.

Now, on steeper hills, more factors come into play - downshifting and keeping the engine at closer to its maximum efficiency, then coasting down, means I actually get a little *better* mileage than steady state.

(n.b. I'm also the guy who experimentally determined that having the tailgate in my pickup down does, indeed, make for worse mileage, with cross, tail, and headwinds factored out, over something like 40,000 miles of St. Louis to Arkansas commuting.)

posted by notsnot at 7:31 PM on November 13, 2014 [1 favorite]

To answer this question accurately, one would need more info.

Speed: When you're driving fast, wind drag has a bigger effect.

One would need to understand the details of the flat/alternate route as well. Stop signs/lights? At what speed is the flat route driven?

posted by sarah_pdx at 7:54 PM on November 13, 2014

Speed: When you're driving fast, wind drag has a bigger effect.

One would need to understand the details of the flat/alternate route as well. Stop signs/lights? At what speed is the flat route driven?

posted by sarah_pdx at 7:54 PM on November 13, 2014

Response by poster: right!

here is any info i can think of that might be useful

the flat portion is 40mph and basically consists of a large curve that ends at a stop sign; i always drive right at the speed limit

the next two miles are a 30mph gentle winding curves (not so curvy that I need to slow down) going at a slight incline

the last mile is a steep uphill, still 30mph that curves once

there are no other stop lights/signs than the first one and for simplicity's sake, we can take out other factors that might cause slowing down

thanks for all the thoughtful answers. its really fun for me to finally get this figured out!

posted by eggs at 8:02 PM on November 13, 2014

here is any info i can think of that might be useful

the flat portion is 40mph and basically consists of a large curve that ends at a stop sign; i always drive right at the speed limit

the next two miles are a 30mph gentle winding curves (not so curvy that I need to slow down) going at a slight incline

the last mile is a steep uphill, still 30mph that curves once

there are no other stop lights/signs than the first one and for simplicity's sake, we can take out other factors that might cause slowing down

thanks for all the thoughtful answers. its really fun for me to finally get this figured out!

posted by eggs at 8:02 PM on November 13, 2014

If you want to find some people really obsessed with this question, google hypermiling. It's the art of maxing out fuel efficiency by doing things like using trucks to protect them from driving into headwinds, slowing and coasting so that you don't have to brake and then hit the gas to start moving again, etc. Here's an article about a guy who can make a honda accord get 59mpg.

posted by selfmedicating at 8:10 PM on November 13, 2014 [2 favorites]

posted by selfmedicating at 8:10 PM on November 13, 2014 [2 favorites]

Best answer: notsnot, while air resistance does affect fuel efficiency (relative to speed, not gravity), it is not fundamental to this question. With or without air resistance and other friction, you do not "get back" the energy. When you're going uphill, some of the chemical energy ends up as the potential energy of your car at the top of the hill, and some is dissipated as heat. When you roll down the hill the potential energy is converted to kinetic energy until you apply the brakes and dissipate heat. If you got back the energy you'd have more gas in the tank at the bottom of the hill (which is the idea behind regenerative braking in electric vehicles).

If, in fact, going up the hill took twice the fuel of traveling on level ground, and going down took zero fuel, then it would be an equivalent trip, in terms of fuel consumption. But 2X is an arbitrary amount of additional fuel needed to drive up the hill. Depending on the route, the car, environmental factors... you might be able to drive up a hill using less than 2X the fuel, and you might be able to coast the whole way down with the engine off (don't), in which case the hilly commute would be more fuel efficient. For this to work it would have to be a pretty gentle slope, which means coasting back would be... really... slow.

However, if the uphill drive uses more than 2X fuel, then as long as you're not burning negative gas, your commute is less efficient. (Again, EVs with regenerative braking can get negative fuel consumption by charging their battery, so the math is different.)

If you want to know your average fuel consumption across two segments of driving, you need to calculate a harmonic mean, or an average of rates. You can see the complicated, general version on the formula on Wikipedia, but for two equal segments, the formula is: 2 / (1/rate_a + 1/rate_b)

Let's say you normally get 30 mpg. On a flat commute, both segments get the same mpg, so the average is 2 / (1/30 + 1/30) = 2 / (2/30) = 2/2 * 30 = 30. Makes sense.

If you get 20 mpg uphill and 40 mpg downhill, you do not average 30 mpg because you're measuring the average over a set number of miles, not a set number of gallons. You actually get 2 / (1/20 + 1/40) = 2 / (2/40 + 1/40) = 2 / (3/40) = 2/3 * 40 = 26.7 mpg. You'd need to get 60 mpg on the downhill to average out, and more to beat a level 30 mpg commute. If your uphill mpg drops to 15, then no matter how efficient the downhill is (rate_b approaches infinity), you can only approach 30 mpg. At 10 mpg uphill, your average can't exceed 20 mpg.

You can also calculate an average for more than two segments. For your commute, still assuming level mpg is 30. Suppose the on the slight incline you get 25 mpg, the steep climb is 10 mpg, the steep downhill you coast, and the slight downhill is 50 mpg. For an 8 mile round trip, the average is 8 / (1/30 + 2/25 + 1/10 + 0 + 2/50 + 1/30) = 27.9 mpg. It is still possible to make up for the hill, as long as the return trip averages over 75 mpg. However, at 6.5 mpg and below for the steep segment, even if you coasted all the way home you could never beat a level commute.

Anyway, those numbers are made up... YMMV.

posted by domnit at 8:50 PM on November 13, 2014 [6 favorites]

If, in fact, going up the hill took twice the fuel of traveling on level ground, and going down took zero fuel, then it would be an equivalent trip, in terms of fuel consumption. But 2X is an arbitrary amount of additional fuel needed to drive up the hill. Depending on the route, the car, environmental factors... you might be able to drive up a hill using less than 2X the fuel, and you might be able to coast the whole way down with the engine off (don't), in which case the hilly commute would be more fuel efficient. For this to work it would have to be a pretty gentle slope, which means coasting back would be... really... slow.

However, if the uphill drive uses more than 2X fuel, then as long as you're not burning negative gas, your commute is less efficient. (Again, EVs with regenerative braking can get negative fuel consumption by charging their battery, so the math is different.)

If you want to know your average fuel consumption across two segments of driving, you need to calculate a harmonic mean, or an average of rates. You can see the complicated, general version on the formula on Wikipedia, but for two equal segments, the formula is: 2 / (1/rate_a + 1/rate_b)

Let's say you normally get 30 mpg. On a flat commute, both segments get the same mpg, so the average is 2 / (1/30 + 1/30) = 2 / (2/30) = 2/2 * 30 = 30. Makes sense.

If you get 20 mpg uphill and 40 mpg downhill, you do not average 30 mpg because you're measuring the average over a set number of miles, not a set number of gallons. You actually get 2 / (1/20 + 1/40) = 2 / (2/40 + 1/40) = 2 / (3/40) = 2/3 * 40 = 26.7 mpg. You'd need to get 60 mpg on the downhill to average out, and more to beat a level 30 mpg commute. If your uphill mpg drops to 15, then no matter how efficient the downhill is (rate_b approaches infinity), you can only approach 30 mpg. At 10 mpg uphill, your average can't exceed 20 mpg.

You can also calculate an average for more than two segments. For your commute, still assuming level mpg is 30. Suppose the on the slight incline you get 25 mpg, the steep climb is 10 mpg, the steep downhill you coast, and the slight downhill is 50 mpg. For an 8 mile round trip, the average is 8 / (1/30 + 2/25 + 1/10 + 0 + 2/50 + 1/30) = 27.9 mpg. It is still possible to make up for the hill, as long as the return trip averages over 75 mpg. However, at 6.5 mpg and below for the steep segment, even if you coasted all the way home you could never beat a level commute.

Anyway, those numbers are made up... YMMV.

posted by domnit at 8:50 PM on November 13, 2014 [6 favorites]

Ah, I see you say that you coast the whole first 3 miles of the return trip. Then using my made up numbers, you'd get 8 / (1/30 + 2/25 + 1/10 + 0 + 0 + 1/30) = 32.4 mpg, beating the flat trip! That's assuming coasting with the engine off, no power steering, headlights, etc.

Please keep in mind that these numbers are totally unreliable, even if they seem reasonable. In a harmonic mean, a small change in the inefficient part of the trip can greatly affect the average, whereas similar changes in already high rates do not affect the average much.

posted by domnit at 9:05 PM on November 13, 2014

Please keep in mind that these numbers are totally unreliable, even if they seem reasonable. In a harmonic mean, a small change in the inefficient part of the trip can greatly affect the average, whereas similar changes in already high rates do not affect the average much.

posted by domnit at 9:05 PM on November 13, 2014

It's a bit simpler in an electric vehicle. My commute is 5 miles downhill to work and 5 miles uphill back home. Going downhill adds a couple miles range (charges the battery). Going 5 miles uphill reduces the range by more than 5 miles. The net effect is that 10 miles round trip costs about 10 miles range.

A gasoline engined vehicle has a lot more losses than an EV so a hilly round trip uses more gas than a flat round trip of the same distance.

posted by monotreme at 11:40 PM on November 13, 2014 [1 favorite]

A gasoline engined vehicle has a lot more losses than an EV so a hilly round trip uses more gas than a flat round trip of the same distance.

posted by monotreme at 11:40 PM on November 13, 2014 [1 favorite]

In theory, an electric or hybrid with regenerative braking will capture the downhill energy and use it to charge the batteries - but not at 100% efficiency. I used to drive a Prius on a commute that involved a 200' change in elevation, and it never got as good mileage as it did when driving on the flat.

posted by mr vino at 5:16 AM on November 14, 2014 [1 favorite]

posted by mr vino at 5:16 AM on November 14, 2014 [1 favorite]

Considering a purely theoretical version of this, imagine a cart of one of two tracks (straight vs. hilly) from a point A to a point B. How much energy do you need to impart to the cart at the offset to get it all the way there?

Well, there are two significant features coming to mind: the friction, and the max. potential energy. With regard to the former, the hilly track is worse simply because it's longer --- remember, the shortest route between two points is a straight line.

However, the maximum potential energy's a bigger issue with this thought experiment. On a flat track, after all, you could probably get there

But this isn't actually a very good approximation of what happens with an actual drive. In the real world, you aren't starting off with a hard shove, so much as trying to maintain a speed throughout, which means over the course of the trip, most of the time your fuel consumption is offsetting only friction/air resistence and (when going uphill) gravity. Except on steep grades, the former's a lot larger, and you

There is, of course, the same issue as I mentioned in the thought experiment: hills are

posted by jackbishop at 7:40 AM on November 14, 2014

Well, there are two significant features coming to mind: the friction, and the max. potential energy. With regard to the former, the hilly track is worse simply because it's longer --- remember, the shortest route between two points is a straight line.

However, the maximum potential energy's a bigger issue with this thought experiment. On a flat track, after all, you could probably get there

*eventually*with a fairly modest shove, which would go slow but as long as it was fast enough to overcome friction you'd be OK, while on a hilly route you'd need to have enough energy to get over the highest hump --- even if that meant you had excess energy at the end.But this isn't actually a very good approximation of what happens with an actual drive. In the real world, you aren't starting off with a hard shove, so much as trying to maintain a speed throughout, which means over the course of the trip, most of the time your fuel consumption is offsetting only friction/air resistence and (when going uphill) gravity. Except on steep grades, the former's a lot larger, and you

*do*recover the energy spent combatting gravity going uphill when you go downhill, and that energy does help with the offsetting-air-resistance issue. Of course, if the downhill is extremely steep or twisty, you end up braking and losing those benefits.There is, of course, the same issue as I mentioned in the thought experiment: hills are

*longer*than the straight route, and thus consume more energy. However, except for a road with extraordinary grades, I'd venture that the fuel-efficiency effect of hills falls below the threshold of actual observability. Particularly when measured against other road factors: the single most notable aspect of fuel efficiency is how often you need to brake, which is a function of the road type and traffic; the second is your cruising speed, which again has more to do with the type of road than its steepness.posted by jackbishop at 7:40 AM on November 14, 2014

I think the short answer to the stated question is No. The longer answer is: depends entirely on the speeds driven in the different scenarios, but almost certainly No.

If you were to drive the exact same speed on the uphill leg as you would in the flat scenario, the answer would be No, you are using more in the hilly scenario. If you drive very fast in the flat scenario and very slowly up the hill you _might_ flip the equation to where you would use less glass in the hilly scenario. It would probably take a huge speed differential to really make a difference since you say the last climb is pretty steep.

First let's assume a spherical cow. In your actual drive, the first mile is a wash both ways because you are stopped at both ends of that leg. So the factors are all about the hill. Assuming no losses,

Given the real-world numbers in your expanded explanation, it's going 30 MPH up the steep hill that is killing any equality.

posted by achrise at 10:25 AM on November 14, 2014

If you were to drive the exact same speed on the uphill leg as you would in the flat scenario, the answer would be No, you are using more in the hilly scenario. If you drive very fast in the flat scenario and very slowly up the hill you _might_ flip the equation to where you would use less glass in the hilly scenario. It would probably take a huge speed differential to really make a difference since you say the last climb is pretty steep.

First let's assume a spherical cow. In your actual drive, the first mile is a wash both ways because you are stopped at both ends of that leg. So the factors are all about the hill. Assuming no losses,

**notsnot's**analysis about the energy conversion is correct, you're converting gas to height, so the energy spent climbing the hill you're mostly going to "get back" when you descend. Problem is, you have to stop at the end of the descent, and if you are braking at any time (except the end) during the descent you're losing out on recouping the potential energy of climbing the hill. Most likely,**domnit's**analysis considering the real-world averaging math will apply, and you really won't be able to recoup all that potential energy math-wise.Given the real-world numbers in your expanded explanation, it's going 30 MPH up the steep hill that is killing any equality.

posted by achrise at 10:25 AM on November 14, 2014

More approximations!

Your engine injects fuel into your cylinders every other time the crankshaft turns*. Let's assume this quantity is nearly constant. (It isn't, but probably close enough for our purposes.) This means we can use the tachometer to approximate fuel consumption: every time the crankshaft turns, you use X amount of fuel.

Going up the hill, your transmission drops down a gear or two, and you go along at, say, 3000 rpm. Coming back down, you kick into overdrive and cruise at 1000 rpm. So you use 3000X + 1000X = 4000X units of gas.

On the straight highway, you will likely find that you cruise at 1250-1500 rpm. The equivalent time is twice as much, so you use (1500X)*2 = 3000X.

You can fill in actual rpm's if you want to study more accurately. But essentially, your engine is always burning at least some fuel, and it takes much more to accelerate than it does to hold your speed on level ground.

*It uses the other time around to burn the fuel

posted by Huffy Puffy at 10:48 AM on November 14, 2014

Your engine injects fuel into your cylinders every other time the crankshaft turns*. Let's assume this quantity is nearly constant. (It isn't, but probably close enough for our purposes.) This means we can use the tachometer to approximate fuel consumption: every time the crankshaft turns, you use X amount of fuel.

Going up the hill, your transmission drops down a gear or two, and you go along at, say, 3000 rpm. Coming back down, you kick into overdrive and cruise at 1000 rpm. So you use 3000X + 1000X = 4000X units of gas.

On the straight highway, you will likely find that you cruise at 1250-1500 rpm. The equivalent time is twice as much, so you use (1500X)*2 = 3000X.

You can fill in actual rpm's if you want to study more accurately. But essentially, your engine is always burning at least some fuel, and it takes much more to accelerate than it does to hold your speed on level ground.

*It uses the other time around to burn the fuel

posted by Huffy Puffy at 10:48 AM on November 14, 2014

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Accelerating on a flat road means you are primarily using gas to get up to speed, and in a direction perpendicular to gravity. In other words, you are only overcoming your lack of momentum, then using a tiny bit more gas to maintain your speed.

Going uphill means you are using more gas, because in addition to overcoming lack of momentum, you are also going slightly against gravity - so you need to use much more gas to overcome both things in order to get to your desired speed. Additionally, gravity pulling you back down the hill is a constant force, so you need to use more gas to maintain speed than you would on a flat surface.

You probably use very little gas (or possibly none) when coasting downhill, but the uphill portion of your drive has already consumed more gas than the equivalent flat commute, so the "savings" of coasting downhill really just means you aren't making your consumption even higher on the non-flat round trip.

posted by trivia genius at 7:02 PM on November 13, 2014 [1 favorite]