In other words, in a piston engine, the best-power mixture is richer than that producing the EGT peak that signals stoichiometric combustion.

Why is that, physically?

Lycoming representation of EGT/CHT versus power

  • $\begingroup$ Welcome to Av.SE! $\endgroup$
    – Ralph J
    Sep 1, 2018 at 0:53
  • $\begingroup$ Is it actually true that peak EGT corresponds to a stoichiometric ratio? Bear in mind the indicated EGT depends on the speed of combustion as well as the chemical energy released, since it's sensed at a particular point in the tailpipe. $\endgroup$ Sep 28, 2019 at 8:10
  • $\begingroup$ Related, on space.SE: Pro/cons of burning propellant in stochiometric ratio? $\endgroup$
    – Manu H
    Sep 28, 2019 at 10:52

4 Answers 4


It's because, although the Best Power mixture dumps more fuel than can be consumed by the available oxygen, the extra fuel increases the combustion speed in the charge, increasing the buildup up of pressure in the combustion chamber, and this increases the total potential energy in the air/fuel charge that can be converted to work, even though not all of fuel is burned. You could say that the excess fuel has kind of a catalytic effect, or maybe a turbocharging effect... sort of... up to a point.

Beyond a certain point, adding more excess fuel becomes counterproductive, so Best Power mixture is the sweet spot where the speed of combustion and pressures generated are maximized and the cylinder makes the most horsepower it can make (the highest possible combustion pressure) with the available displacement and airflow.

  • 1
    $\begingroup$ I don't know if by turbocharging effect you are referring this but: evaporation of the fuel cools down the air charge on its way to the cylinder. More fuel, cooler air charge, more air going to the cylinder. Up to a certain point (max power) this might be of significance. Might, because I don't remember and can't find any sources about this to confirm if the difference is significant between max power and stoich in gasoline engines. When comparing gasoline and ethanol engines, the difference is significant. $\endgroup$
    – Jpe61
    Sep 28, 2019 at 7:24
  • $\begingroup$ By difference you mean the amount of evaporative temperature drop? To the extent it does, I would think it's only a factor in carbureted engines where the droplet stream has a long way to go. $\endgroup$
    – John K
    Sep 28, 2019 at 17:27
  • $\begingroup$ Yes. That might very well be the case. Difference in the amount of gasoline between stoich and max power is not that big after all. $\endgroup$
    – Jpe61
    Sep 28, 2019 at 18:00
  • $\begingroup$ I am pretty sure the rich mixture "decreases" combustion speed (increases RON), allowing spark advance and favouring any forced induction boost. Water can do the same. Takeoff power was traditionally hard on pistons, which are really only ring carriers. The heat and stress is considerably higher. A rich mixture is less "knocky" as the combustion speed slows. $\endgroup$
    – mckenzm
    Dec 1, 2019 at 9:03

in practical terms, running rich means that despite the unevenness of the fuel-air mix both in a single cylinder firing and across all cylinders firing and across all firings in the engine, all cylinders are guaranteed to NOT be running lean on ANY firing, which means all of them will be pulling best power on EVERY firing.

In addition, there is an effect that I know about from motorcycle engines which is probably true of airplane engines and that is the "quality" of combustion as a function of richness: If you run rich, preignition and lean misfire (both of which reduce combustion efficiency) are prevented, and the delivered power is smoother. You can hear this: a motorcycle engine sounds "happier" when running rich than it does running lean- and it pulls harder, even though the mix is nonstochiometric.

  • $\begingroup$ This is by far a more correct answer, in my opinion. Because the fuel and air is mixed at the carburetor, not all cylinders get the same amount of fuel. The front valves get more fuel than the rear valves do. In a fuel injected engine, the fuel is introduced just before the cylinders and each gets approximately the same amount of fuel. On fuel injected engines, with individual cylinder head temperatures it is possible to run lean of peak and get some pretty incredible performance. $\endgroup$
    – wbeard52
    Mar 9, 2021 at 2:12
  • $\begingroup$ @wbeard52- agree, and especially if the fuel injectors have been individually tuned to behave exactly the same. there is an (expensive) way to accomplish this which is reputed to furnish outstanding performance at several GPH less than standard for the power settings used. $\endgroup$ Mar 9, 2021 at 5:56

This is simply due to the difference between theory and practice. It turns out that even in a well-balanced charge of fuel and air, there are highly localized “pockets” of varying mixtures at the local, molecular level. By “local level” you should think of a bunch of little fuel molecules huddling together over here and over there in different places inside the cylinder as the piston is rising up towards top dead center and starting down. Some of these “pockets” may be so lean, or so rich, that they won’t burn at all. Some may be in the combustible range, but not really contribute to the combustion event, and some may be perfectly mixed, ready to go, so to speak.

In theory, the ideal mixture for our engines is about 15 parts air and 1 part fuel (by weight), which should result in no oxygen and no unburned fuel molecules going out the exhaust pipe - IF it was a truly homogenous mixture, but, it’s not. It’s actually highly dynamic, varying greatly from cycle to cycle. Each combustion event is comprised of those little pockets of varying fuel-air mixtures I just described, different every time. At a stoichiometric mixture, after the spark occurs and the flame front propagates, some of the oxygen molecules just do not “find” fuel molecules quickly enough to burn, and they remain unused or unburned at the ideal ratio. By supplying just a bit more fuel, there is a greater opportunity for the lonely oxygen molecules to react with the hydrocarbons, meaning more total fuel is burned, a bit more heat is generated, and less oxygen escapes out the exhaust pipe without having had a chance to mate.

This is why we’ve long observed that a slightly richer mixture would produce slightly more power.

  • 2
    $\begingroup$ Like the answer! In the 2nd paragraph, last 2 sentences, why do extra fuel molecules for the O2 to find, work better than extra O2 molecules for the fuel to find? IOW, why wouldn't swapping "fuel" with "oxygen" in those 2 sentences be an equally accurate explanation, since there unburned molecules of each "hanging around" as described. $\endgroup$
    – Ralph J
    Mar 8, 2021 at 2:46

Peak EGT does not mean max power for a simple reason: you have extra heat that was not used mechanically.[1] It wasn't used because the rate of energy release was not fast enough for complete combustion before the exhaust valve opened, but sufficient to affect the EGT sensor downstream.

When more fuel is added up to a level, more oxygen is burned, releasing more energy at a higher rate, and power increases.

1: Bush, M. "Understanding CHT and EGT." Cessna Pilots Association Magazine (2009).


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