During a discussion on start-and-stop vehicle technology some bloke began pushing the point that re-starting the car uses stored energy from the battery, which needs to be replenished by increased fuel usage once the engine is moving. Well, this is obvious, the question is that of significance of such additional usage, especially in comparison with savings from reduced idle. Common sense tells me that this is splitting hairs: if ICE efficiency outlays put all accessory usage at 2 to 3% of overall energy consumption, which includes AC, and all electricals, then it can’t be that much.

More over, this dissertation explores idle-reduction technologies for long-haul commercial trucks, and one of the systems exampled is battery-powered (p. 14) that stores enough energy to power AC or heater overnight, and takes about six hours of charge while driving. Author acknowledges increased fuel consumption due to necessity of higher-amperage alternator, although there are no specific numbers provided. However, the mere existence of such commercial application available on the market leads me to believe that such added load is still better than idling.

But, in the interest of science, I need some hard numbers (besides, the fellow just wouldn’t go away). I have some ideas on what to consider, but I’m not well versed in electrical and mechanical engineering, so I do not think I can account for most major factors.

Energy use of the starter motor can be calculated by using the amount of current used per start (which would be 2 to 3 seconds) by the motor itself and the solenoid that engages the starter gear onto the flywheel. Both current and power demand can be found in the starter’s specifications, but I am not sure as to how reliable those numbers would be in the real-world application.

Then there is fuel consumption of the start itself which is estimated at 10 to 15 seconds worth of idling (and Florida chapter of ASME even calculated six seconds for 6-cylinder engine (in a simple, non-rigorous field experiment), but the original link is broken).

Now, how to calculate the increased consumption due to the charging of depleted battery from the starter motor itself, and, additionally, to account for all the accessories that were running while the engine was stopped? Is it a simple matter of using the same number calculated for energy usage of starter motor, and arriving at it by calculating the amount of fuel needed to produce that much extra energy given the losses in the engine itself and in the charging circuit? And, ultimately, how significant are those considerations in the bigger picture?


6 Answers 6


First, consider the case with negligible auxiliary loads (no air conditioning).

For a Civic-sized engine (1.8 liters), this US DOE worksheet estimates about 0.3 US gallon/hour fuel consumption at idle.

Here is a conservative starter calculation:

  • The Civic starter is rated at 1.0 kW (83A$\times$12V). A 3 second start therefore produces 3 kJ. Assume an additional 25% in battery internal dissipation that must be replaced.
  • As you note, this energy must be replenished by the ICE (internal combustion engine). Max ICE efficiency is only 30%. The incremental efficiency, which is what matters for this small additional load, is no doubt higher, but I’ll use 25% as a conservative estimate.
  • Alternator efficiency is not great either; I’ll use a conservative 50%.

    With these values, it requires 3.0 kJ$\times$(1.25 / 0.25 / 0.5) = 30 kJ worth of fuel to recharge the battery (Note the overall charging efficiency is only 10%!).

Now, the energy density of gasoline is 120 MJ per US gallon (42.4 MJ/kg), so the amount of fuel required to recharge the battery, including all the inefficiencies, is 30 kJ $\div$ 120 MJ/gal = 0.00025 US gallon.

So, the “crossover” idle time in this case, above which it is more efficient to stop and restart, is 0.00025 gal $\div$ 0.3 gal/hour $\approxeq$ 8.3 $\times10^{-4}$ hours, or about 3 seconds.

Now suppose an air conditioner (PDF) is consuming 1 kW of electrical power.

  • With the engine running, the A/C requires (via the alternator) an additional engine fuel consumption equivalent to 1 kW / 0.5 / 0.25 = 8 kW, or 29 MJ/hour, or 0.24 gal/hour of gasoline. For a duration $t$, the total fuel consumption with the engine running is (0.3 + 0.24)$t$ = 0.54$t$ (with $t$ in hours).
  • With the engine stopped, the A/C still consumes 1 kW, or 3.6 MJ per hour. With that low 10% charging efficiency, it requires 36 MJ worth of fuel (or 0.3 gal) to recharge an hour’s worth of A/C operation. Adding in the starter contribution, the total fuel requirement is 0.00025 + 0.3$t$ (with $t$ again in hours).

Equating these two new fuel requirements, the crossover time with the A/C on increases, but only to about 4 seconds.

Although the battery charging efficiency is low, the waste of the idling fuel consumption dominates the calculation.

Note that I don’t have a reference for the 25% battery re-charge inefficiency. Unfortunately, that’s an important number when running an A/C, since it reduces the advantage of shutting off the engine. At some high load level (in the neighborhood or 4 kW) that disadvantage outweighs the advantage of turning off the engine.

Further (experimental) data to confirm the above estimates can be found here: http://www.iwilltry.org/b/projects/how-many-seconds-of-idling-is-equivalent-to-starting-your-engine/

In my case it consumes about the same amount of fuel as 7 seconds of idling. However, the additional fuel consumption observed seems almost entirely due to a faster idle speed setting for the first 20 seconds after starting. Any good driver would start moving within 1-2 seconds after starting, which would effectively eliminate the fast idle losses. If you can begin extracting useful work from your engine within 1 second after starting the engine then it appears starting the engine consumes fuel equivalent to about 0.2 seconds of idling.

  • 2
    $\begingroup$ Most A/C units require the engine to run the compressor. I don't know of any that run off the battery completely, but perhaps they exist. Thanks for finishing up the engine efficiency question for me, I couldn't find that data. $\endgroup$
    – user6972
    Commented Aug 29, 2013 at 5:54
  • $\begingroup$ @user6972: My Prius stays cool at stop lights... (I think I read that was one of its major innovations.) $\endgroup$
    – Art Brown
    Commented Aug 29, 2013 at 6:01
  • $\begingroup$ The prius is probably alone in that respect. Looks like 2010+ models have a 3-phase high voltage (201V) motor driven Air Conditioning (A/C) compressor. They don't publish any numbers on it, but they say it will only run for 3 minutes then shut down. FYI in 12V systems 1 A-Hr = 43.2KJ so our numbers are quite similar. $\endgroup$
    – user6972
    Commented Aug 29, 2013 at 7:31
  • $\begingroup$ @user6972: I'm glad to hear our numbers are close. $\endgroup$
    – Art Brown
    Commented Aug 29, 2013 at 7:49
  • 1
    $\begingroup$ That was me. The calculations, actually, stayed the same, save for the 1.8 L correction — they were exactly as you now updated. :) Mind if I re-edit fancy formulae and SI conversions? $\endgroup$
    – theUg
    Commented Sep 1, 2013 at 3:43

Firstly, the proof is the the pudding. While I was in Germany I was in a car (a Smart) that automatically turned off the engine when you were stopped and held the brake down, and then re-started the engine when you pushed the gas. It did this so quickly that you didn’t really notice. I assume this is the kind of technology you are referring to. The fact that this is done in a commercial car and apparently improves efficiency shows that, overall, turning off the engine at traffic lights does in fact improve efficiency. For your hard numbers, all you’d need to do is look up the fuel consumption specifications for such a car with and without the engine start/stop feature enabled.

Secondly, with respect to the amount of energy used from the battery by the starter, it’s easy to put an upper bound on the amount of electrical energy used while starting the car using the battery ratings. One of the ratings printed on batteries is the number of cold cranking amps (CCA). This is the amount of current the battery can put out while starting the car at 0 °C. A typical number is 700 CCA for the battery in a four-cylinder. Assuming the starter uses the maximum current, equal to the number of CCA (it doesn’t) and it takes 10 s to start the car (it doesn’t), gives an upper bound of 700 A $\times$ 12 V $\times$ 10 s $=$ 84 kJ of energy use by the starter.

84 kW $\approx$ 112 HP is on the lower end of the maximum power a four-cylinder can put out. 84 kW $\times$ 1 s $=$ 84 kJ, which means that our upper-bound estimate of the energy used by the starter is equivalent to about 1 s of a four-cylinder going full-out. Engines do in fact put out this kind of power in practice; that’s what you get when you’re on the highway in your power band and you floor it, which might be a reasonable thing to do when getting onto the highway in a small car. If you want to know how much idle time this is equivalent to, you’d have to look up the idling fuel consumption, which isn’t really physics and I don’t know off the top of my head.

Finally, all of the energy stored in the battery is less than the battery capacity in amp-hours multiplied by the battery voltage in volts. I say “less than” because the voltage will drop somewhat before the battery goes dead. For a 70 Ah battery at 12 V this is 840 Wh $\approx$ 3 MJ, which is still only equivalent to 30 s of a 100 kW engine.

In general, the batteries we encounter in day-to-day life store a puny amount of energy when compared with fossil fuels. That is why it has been so hard to make a competitive electric car. Tell that to the bloke.

You can find references for the numbers in this post (1) in a vehicle owner’s manual, and (2) printed on a car battery.

  • 1
    $\begingroup$ Why do you (and everybody else) use such huge numbers for the time it takes to start a car? A car that takes a whole second to start its engine is not in good shape; taking 10 seconds means it is on the verge of dying. $\endgroup$ Commented Nov 5, 2018 at 14:37

On average a 4 or 6 cylinder car will draw about 250 amps for 3 seconds to start. That works out to be 0.21 Amp-Hrs.

The alternator on a car can easily restore that amount of energy in 30 seconds at about 40% efficiency. Which means about 0.52 Amp-Hrs are needed from the alternator to recover the battery.

Most alternators put out about 60 Amp-hrs at idle or 100+ Amp-hrs when running at higher rpm.

Thus starting a car takes about 1% of the the available power coming from the alternator at idle. or about 0.65% when driving. So this shows the actual load on the battery and the alternator are insignificant.

To compare the difference in electrical energy wasted to the fuel wasted in starting a car or idling a car is dramatic. The motor is using a ton of power every second it is running on the order of 100 kwatts. Turning it off even for a second is going to save a dramatic difference in overall power consumption. If you ran the engine for just 1 second you're talking about 2.3 Amp-Hrs (assuming all that power is compared to a 12V battery producing 100 kwatts).

Or if you compare it to 3 seconds of starting that's 6.95 Amp-hrs (engine running for 3 seconds) vs 0.52 Amp-hrs to recharge after starting.

Stopping the car is a no-brainer.

EDIT: Using Art's 0.4 gal/hour at idle for a 2.4L, that is 1,111 A-Hr worth of energy in a 12V system. By comparison that's 11 standard 12 volt automotive batteries worth of power every hour. Now consider you only need about 0.52 A-Hrs to recharge after each start you can see that we are talking about 0.045% of a power drain which equates to about 2 seconds of the engine's fuel consumption at idle, much less if you are driving. (1,111 A-hr = 0.31 A-seconds)


  • 4 cylinder engine hp/watts (I used a low estimate of 100 kwatts or 134 HP.)
  • Alternator current vs rpm vs efficiency (see first graph Figure 3.)
  • Most small car starters are 1.4KW or ~116 amps or 0.097 Amp-hrs at 3 seconds, I roughly doubled the number to 250 amps to cover just about any type of small engine. For example the popular Honda Civic uses a Denso 280-0324 starter that is only rated at 1.0 kW or 83.3 amps or 0.069 Amp-Hrs for 3 seconds.
  • 1 a-hr @ 12V = 43.2KJ, 1 gal = 120MJ => 2.78 kA-Hr @ 12V.
  • $\begingroup$ Good edit. Let’s see how people vote on that during the week, and if there are other comments or answers. Although there’s one more thing — what about consumption from lights, radio, electromechanical AC that would run while the engine is stopped? Would they be on the same level of insignificance as the starter’s use? $\endgroup$
    – theUg
    Commented Aug 25, 2013 at 21:34
  • $\begingroup$ There's really no comparison between the amount of energy used in a running engine versus anything else in a car unless you are running the headlights or a pure electrical AC system for an extended period of time off the battery (multiple days). Most standard car batteries only have 100 Amp Hours of capacity (1200 watts @ 12V), which is ~ 1/83 the power expended in this 134HP example engine. $\endgroup$
    – user6972
    Commented Aug 26, 2013 at 0:47
  • $\begingroup$ Not sure about multiple days. Headlights left on can kill the car in a few hours (still hours, so not a problem for start-stop). And what about any efficiency losses? For instance, 100 kW motor consumes so many joules per second — say, it makes 25 kW at idle, then it ideally consumes 25 kJ of energy per second, but how many Joules of fuel is consumed to produce those 25 kW? $\endgroup$
    – theUg
    Commented Aug 26, 2013 at 4:44
  • $\begingroup$ It would take multiple days 24/7 of lights/AC running to compare to the engines power output. I didn't say it was possible with a 100 amp-hour battery, that would be silly. The power we are talking about for the engine is output power, not power consumed/used. Every engine has a different efficiency ratings so it is hard to give a specific example. Do you have a specific engine in mind? $\endgroup$
    – user6972
    Commented Aug 26, 2013 at 19:03
  • $\begingroup$ Not really, no. But a good example with ubiquitous engine would be helpful. That very Civic, for instance. And I think important part in the calculations is to keep the engine power output realistic — nobody is going to sit on the red light with throttle fully depressed producing 100 kW. And I hope I don’t discourage you with my comments — I’m not being picky, but trying to crystallize the best answer it could possibly be. $\endgroup$
    – theUg
    Commented Aug 28, 2013 at 18:31

Thanks everybody for the good answers here. I got here by searching some data about the problem I have with my old 2.0TD Toyota Avensis, which we use several times per week for 3-5minutes rides. Especially in winter, whatever battery I put in, it always ends discharged/empty in few weeks.

According to your answers, this should not happen. I am aware the question here was different (whether start/stop feature really saves fuel). I just wanted to note that the start/stop system can not be implemented into ANY car and is not suitable for ALL users. For me and my old Toyota, it would not work at all...

Here are my calculations & measurements:

  • let's say the starter takes 200A for 3 seconds. In winter, I usually have to crank it twice, because when temperature goes under 0°C, the first attempt barely moves the stiffened engine. So 200A x 6 seconds = 600A.seconds.
  • I measured the Toyota charging "algorithm" yesterday. My initial battery (unconnected/open) voltage was 13.1V. After starting the engine, the charging voltage was 14.xV (don't remember the exact value) and the current was initially limited to 2A. Then the current decreased to 1.x A: while charging, the battery "open" voltage rises so it does not receive that much current.
  • Note: the 2A current limit & decrease must be there, because otherwise the battery would start boiling. E.g., with 5A charging current, a 12V/62Ah lead acid battery starts boiling almost immediatelly. It boils even with 2.5A CONSTANT charging current (this is another measurement I did yesterday...)
  • If recharing with only 1A, the 600A.seconds used for starter would ideally mean 600 seconds (10 minutes) of charging. However, charging efficiency is far below 100%. Especially in winter, the frozen battery does not receive the charge that well, so...
  • ... the "30 mins of driving to recharge the starter usage" is probably NOT just an anecdote, but a reality, at least for frozen diesel cars.
  • $\begingroup$ This anecdotal evidence has a sample size of 1, which really can't be useful for as a basis for an answer. $\endgroup$
    – Kyle Kanos
    Commented Jan 4, 2016 at 14:47

I sorry to dampen your enthusiasm but I am not a physics student but a commercial electrician but had lots of experience in car stereo installs the first mistake I see of what's been writing is about the output of the alternator which can only produce 10% of its rated output at idle and that's when its brand new its also a mechanical device that wears out and efficiency goes down as it gets older, so in real world driving the alternator is under tremendous strain to recharge a car batter, years ago I was told buy a very good car mechanic he told me these wise words of wisdom it takes at least 30 minutes of driving to recharge what 10 to 20 seconds of starting takes from the battery and alternator technology has improved I have always remembered that lesson also the alternator if it needs to recharge at a higher rate will take more power from the engine

The next problem is as I have had first hand experience and advising on this issue is modern battery technology with battery manufactures demanding ever higher Cold Cranking Outputs of there batteries the result is ever thinning battery plates its result's in more batteries dying of buckled plates and battery manufactures also only give a pro rata guarantee on their batteries but we all know lead is a toxic and undesirable metal but on car batteries until they can come up with a better material we are stuck with using lead Yuasa have gone back in technology and now using medium plate thickness while Bosch have gone for thinner plate material AGM batteries with varying degree's of success then also is your alternator ECU controlled or controlled of the battery another tier of technology in my mind is pointless but also you would also factor in a replacement battery at every 3 years plus wear and tear of the starter motor some are easy to replace and some can be very expensive if you factor in these costs to the equation does stop start work yes and no I think the answer is very much not a mathematical one but depends more one real world driving

Sitting in motorway traffic during the clearance of an accident is some what of a no brainer but being stuck on a motorway what happens if your wheels are constantly moving at 1 or 2 miles an hour, but also other drivers would constantly cut in front of you if you allowed small gaps to develop, being a truck driver for ten years of my life I can personally vouch for the impatience and stupidity of other drivers but in summing up I think its more the situation than just a pure mathematical formula

thanks for reading regards Simon

  • $\begingroup$ Hi simon, I'm not sure you really addressed the energy use estimation aspect of the question and the 30 mins of driving to recharge the starter usage is just an anecdote. Do you have any actual data on this? $\endgroup$ Commented Oct 23, 2015 at 3:37
  • $\begingroup$ If your car takes 10 to 20 seconds to start, then there's something wrong with it. Modern petrol/gasoline engines can start in a fraction of that time. $\endgroup$
    – Simon B
    Commented Dec 1, 2015 at 16:41

The answer chosen as solution already explains everything but it is important to consider also the repair costs.

If a car with rugged start/stop system can do 10 000 starts and the repair costs are about 400 Euro to replace the components of the starter system, it's already 4 cents per start (2.5 ml petrol at 1.6 €/liter).

With the assumptions above, each start does not require 0.00025 US gal but rather 0.00066+0.00025 gallons, therefore 0.00091/0.3= 11 seconds (12 seconds with AC).


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