I was reading a thread in a car forum where someone wrote:

The lower the RPM, the less fuel you use, the less strain you put on everything..

Another person then replied, stating:

Not true. If your pulling high gears and making the engine work more to go how you want your using more gas. As well as putting more strain on the car. By wanting more from the car when it is out of its power your forcing the injectors to open more thus burning more gas. Please don't mislead anyone if your not sure. However, waiting until redline to shift isn't ideal either. Shifting at 3/4 or so of the torque curve is optimal for gas.

Who is correct? Is it more gas efficient to to shift at lower RPMs rather than high RPM? Would something like this vary great from car to car, model to model, or would it be a general rule for all standard engines?

  • 1
    While someone will probably provide a detailed answer, internal combustion engines have a certain range where they are most efficient. This is affected by many factors so it isn't a black and white answer. There is a specific rpm point where any given engine is most efficient. It isn't always at a lower or higher rpm point.
    – Rig
    Commented Oct 20, 2011 at 22:54
  • Thanks Rig. Do you think I should tailor the question specifically to my car or leave it relatively broad as it is and ask another question specific to my car's engine?
    – stoicfury
    Commented Oct 20, 2011 at 22:56
  • Might not hurt but I don't know that there is a specific answer to be had either.
    – Rig
    Commented Oct 20, 2011 at 23:17

9 Answers 9


I think you need to define your terms. Efficiency could mean at least four things in this context:

  1. Least fuel consumed per hour.
  2. Most mileage covered per unit of fuel.
  3. Most kinetic energy created per unit of fuel consumed.
  4. Most acceleration created per second.

These four choices often destructively interfere with each other. Hitting some of the high points of each of the above meanings:

  1. High gears usually mean lower revs meaning lower amounts of fuel injected on average. The squirts of fuel might change in size depending on throttle position but those squirts happen less often.
  2. Mileage per fuel units is a multivariate equation in which aerodynamic drag starts to dominate at higher highway speeds. Even so, if you are driving very slowly, you will also see lower mileage than the optimum as you just aren't covering much linear distance.
  3. Internal combustion engines almost always have a peak energy output point where they are best able to combine fuel and air and transform that into kinetic energy. This is usually the torque peak of the engine. Note: peak torque also means near peak fuel consumption per second. Also remember that an engine's torque peak has nothing to do with the current gear selection.
  4. Peak acceleration is likewise tied to the torque peak but also requires proper gear selection and involves aerodynamic and static drag factors. Best acceleration from a stop will use the lowest gear. At speed, low gears will not be a choice as you'd have to rev the engine past its physical limitations, which is one of the many reasons why your acceleration drops off as speed increases.

Note that all factors are more complicated than can be addressed in a single page. For example, my turbo standard transmission car injects even more fuel than you'd expect at its torque peak. The engine computer is using that extra fuel to cool the higher compression fuel-air mixture in order to avoid premature detonation. This allows the car to run higher boost but isn't the greatest for fuel economy....

  • Note that for sports bikes, particularly with inline 4 engines, low gears may remain an option at all legal speeds. My GSXR750 redlines at 15K, with peak torque ~11k. That's 1st gear at freeway speeds in my location. Any higher gear, just lowers the RPM further below peak torque.
    – Leliel
    Commented Feb 5, 2016 at 2:48
  • @Leliel what that means is that you are driving on the street with track gearing. If the gears were compressed down into the range of street speeds, acceleration would be even higher.
    – Bob Cross
    Commented Feb 5, 2016 at 12:09
  • @Bob_Cross You wouldn't want it to be any higher. Would make the bike nearly uncontrollable.
    – Leliel
    Commented Feb 8, 2016 at 20:55
  • @Leliel that may be but now you're talking about a problem with the vehicle system, not the transmission. The math doesn't change just because the tires don't have enough grip.
    – Bob Cross
    Commented Feb 9, 2016 at 14:35
  • 1
    @adelrahimi like always, it depends but, as a rough approximation, the highest gear that doesn't lug the engine is going to be the most fuel efficient.
    – Bob Cross
    Commented Apr 5, 2016 at 14:44

The specific answer to your question lies in the BSFC (Brake Specific Fuel Consumption) map for your engine, combined with the desired output.

The BSFC map shows how efficiently the engine converts chemical energy to mechanical work at any given condition.

Here is an example:


  • 1
    Is there a database or a place to find such graphs?
    – FarO
    Commented Jul 6, 2017 at 14:06


  • You want the largest possible throttle opening (on naturally aspirated cars anyways) to reduce the vacuum and hence pumping losses, however, you don't want to force the computer out of closed-loop mode... So, roughly 50-75% throttle depending on the vehicle (assuming manual transmission as an automatic will probably grab a less efficient gear in those circumstances).

  • You want the RPMs as low as possible as it truly is RPMs that really burn fuel (frictional losses with higher RPMs). However, if the RPMs are too low, then you don't get the benefit of timing advance...

So, "it depends" on naturally aspirated manuals. They give you the most control of how to optimize your fuel usage, but you need to know your specific car. :-) The easiest car to optimize for is the turbo automatic... Drive as slow as possible and hope for the best since you don't really have much control beyond that... :-)


Basically, yes lower rpms == lower consumption but there is a caveat Looking at what your throttle is doing, if you are at low rpm in a high gear but your foot is on the throttle heavy then there is more air coming in and at low rpm valves are open longer so the computer responds with the right amount of fuel for the air you are sucking up with the wide open throttle. You want to get your rpms low but not at the cost of having to open up the throttle. So there is a limit. So pick the highest gearing that a light throttle can handle.

  • 2
    Incorrect. Just because the throttle is wide open doesn't mean that there is actually more air coming in. Only as much air comes in as the engine can pump out. Only as much fuel is injected as is necessary for the air coming in. Commented Oct 24, 2011 at 13:35

The answer is complicated. Lets say we have two conditions - low revs and high injector times, and twice the revs and half the injector times. In both cases the fuel burned will be the same. This is normal - we need work done and this comes from the fuel. We have to account something else - when engine turns there is friction - the more revolutions will produce more wasted energy. So we want low revs/high. But thats one side of the story - when low rev/ high torque situation is present, there are more vibration which is again wasted energy and not comfortable at all. Some of the peripherals as water and oil pump made to work at high rpm may not be efficient in low rpm , and not enough oil may be present, in this high torque situation leading to even more friction. The alternator has its own efficiency dependent on rpm. Balance is everything - do not force the engine with 1000 rpm , but do not go past half the engine speed if economy is the desired.


Generally speaking, all automotive internal combustion engines are at their most efficient at around 0.3 of peak power, which works out, generally, to about 2000 rpm in the highest gear. The designers tend to set them up that way because these operating points produce a car that feels drivable. All cars are different. Driving behavior plays a huge role in efficieny. You can generally get vastly greater increases in efficiency by learning how to drive efficiently than by detailedly analysing engine efficiency. If you keep your speeds as low as reasonably safe, empty useless weight from the car, keep your tires inflated, and stay off the brake, you're doing about as well as you can. Based on several decades of study and research into fuel economy.


More intuitive, less scientific reply. Fuel economy of my motorcycle (250cc, electronic ignition and FI, thus comparable in behaviour to EFI automobiles), recently dropped off fairly considerably (below 80 with lowest of 76 MPG per fuel-up). Disregarding possible negative influence of winter additives (is there any truth to that?), one thing in common between those less-efficient fuel ups was me lugging it, i.e. shifting gears early, and slowly accelerating.

However, even though I took it easy, the amount of work it had to perform turned out to be higher than normal. Last two fuel-ups I went back to manufacturer-recommended shifting intervals, and my fuel economy went back over 82MPG (85 latest), and with much sprightlier acceleration.


In the case of heavy trucks keeping the RPM between 10-15 regardless of gear shifts is the best for fuel efficiency.

  • Do you have any references to back up your statement? As it reads, it's just your opinion. You may want to flesh your answer out some with links, charts, and information to corroborate your claims. Please take a look at the Help Section for some specifics on how to write good answers. Commented Dec 8, 2015 at 0:52

Obviously there are two major things that make a car energy to use. (others effects are not much)

  1. Accelerate the Car.
  2. maintain the velocity.

1- for the first how much you increase the speed then break, you will waste more energy.

Driving in a fix Speed is More Economy

the energy required for a car to increase its speed can be calculated with this formula:

Energy [Joules] = 0.5 * mass[kg] * ( velocity[meter/second] )^2

for example:

two drivers drive a route with average 108 km/h

driver A accelerate a car to 108 km/h you will need 450 Kj.

0.5 * 1000 * (30)^2 = 450 Kj

driver B would accelerate the car 3 times then breaks. would required 3 times energy for acceleration.

Only Including above Would only correct for In Vacuum Driving.

so there are other rules for an terrestrial vehicle

more energy consuming is maintaining the velocity for this to happen, the consequence of Forces implies to car should be zero. the car engine should resist with two major forces:

Providing this forces requires Engine Energy : Energy = Force X Distance

  1. friction
  2. AirDrag

The Friction can be calculated using this formula: friction force = support force * friction coefficient

supporting force will be increasing by Mass, Earth Gravitational Acceleration, additional force on top of the vehicle. friction coefficient is related with Wheels width and materials and their deformation. (which Gearing and speed has no effect on them)

Note: just driving with in very high speed formula 1 car will increase on top force significantly.

so driving with 10km/h and 20km/h has same force. 20km/h would require twice power (joule/second) but will end in half of the time.

so: they are equal in friction resistance energy consumption.

but about Air Drag Resistance Conditions Are Different.

Air Drag Force is related to Shape, Cars Cross sectional Area, Air Velocity.

The Velocity is our interest. which depends on Wind and Car Speed. V = Car Speed - (Wind Speed X Cosine(Wind Angle) )

But Force has Cubic Relation with The Speed means:

make speed twice, needs eight time Power, and we will end our journey in half the time, this results in Four Times Energy Consumption for Air Drag Resistance.

Driving Slower Would Significantly Reduce Fuel Consumption

its get more significant in high speeds (80+ km/h) which 80% of fuel consumption is for air drag resistance and 20% is for friction resistance.

This Rules would be perfectly correct for an electric car.

But a Mechanical Car has to Consume Fuel in 4-stroke Engine to make its power, it needs perfect burning and perfect engine breath.

if RPM is low, Engine cannot Breath its required Oxygen (closed throttle) and raw Fuel releases in smoke.

if RPM is high, Engine cannot Breath its required Oxygen (imperfect burnt) and Carbon-monoxide releases in smoke.

The Above Conditions Are Relative (in both we have raw fuel and carbon-monoxide)

The Economy-Friendly Would be The Maximum Torque RPM (my car is 2800 RPM)

Driving in Maximum Torque RPM Would Result in Fuel Perfect Burning

for example i have calculated my efficient maximum torque RPM in each gears as follow:

gear & speed relation :

1(20 km/h),2(40 km/h),3(60 km/h),4(90 km/h),5(110 km/h)

Note: IN CVT Cars we dont have just a few steps, and Every speed after RPM Reach its Point Would be Fuel-Perfect-Burning.

The Above Rule is True For Injector Based Car, Which ECU Controls The Fuel Flow.

For Old Cars, There is One-State Gasoline Flow Screw. in Those Cars, Gears Mostly For Setting Torque And Driving In Higher Gears Is More Fuel-Efficient.

In Carbrator Cars, Driving in High Gears Are More Fuel Efficient


in Injector based Cars Its More Fuel Efficient to Drive In Maximum Torque, Not Like Old-Cars Which Driving In High Gears Were More Fuel Efficient.

PS: my car is Saipa Pride (injector)

PS: Its Based on My Researches, Internet Searches, Intellectual Toughs, and Calculations. If Some one Thinks I am Wrong, Please Let me Know.

PS: I used to Drive my car in Higher Gear Strategy before, But After Switching to Maximum Torque RPM Strategy, it seems to give me more millage (10% more).for 30 Liter of gasoline it was like 500Km, now it seems like 550km.


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