Is it because aircraft can get more "instant" power for go-around, with throttle full-forward?

Or is it for protecting the aircraft's engine from high RPM, low PROP stress in case of go-around?


3 Answers 3


It's both, but mostly the first one. You need maximum HP for go-around, which means you need the propeller governor regulating RPM at redline when you call for max power.

You could leave the prop control at a cruise RPM setting, but you would have to remember to move the prop RPM to max before the throttle if you have to go around. Easier to save time and just move the RPM to max while on approach so it's ready when you need it. On a piston twin, where you need every fraction of a HP from the good engine if an engine quits on a go around, this can mean the difference between making it and not.

If you leave the prop at a low RPM setting and forget about it when you need to go around, and just shove the throttle forward, you will be at max manifold pressure but with RPM limited to the governor setting, and you won't be getting maximum HP for the engine, and there is a possibility of inducing detonation from the extreme "oversquare" operation, oversquare being RPM lower than MP, especially on a supercharged or turbocharged engine (you won't ruin a Contintental O-470 doing that, but shove the throttle on a supercharged R-985 to get redline MP of 36" while leaving the RPM down at, say, 1200, and you'll probably blow a jug or two).

  • $\begingroup$ I would slightly reword this, as engine max rpm is not synonymous for max pwr rpm. $\endgroup$
    – Jpe61
    Apr 20, 2022 at 17:09
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    $\begingroup$ I know what you're getting at, but that generally applies to engines whose redlines are a function of valve train limits, and the engine is able to wind up well beyond the torque peak, so HP drops off toward redline as torque drops off faster than the RPM rise. This doesn't really apply to direct drive a/c engines that are propeller limited, and where the published HP limit is an arbitrary value reached at the published redline. In racing, the engines will be allowed to wind up past normal redline for more HP, and they will use a shorter prop. $\endgroup$
    – John K
    Apr 20, 2022 at 19:01
  • $\begingroup$ @Jpe61, for a constant speed propeller in practice they are. Given the choice, which you have with a constant speed prop, there is no reason to ever select higher rpm than max power, so the governor will always be set up so that the max selectable rpm is at or below max power rpm of the engine, and is therefore what gives you most available power. $\endgroup$
    – Jan Hudec
    Apr 24, 2022 at 0:53
  • $\begingroup$ Yes, but as you wrote, "...at or below max power rpm", latter of which might be the case in which propeller max rpm is the limit, not engine rpm. I know this is probably overly complex for this specific question, but anyway... $\endgroup$
    – Jpe61
    Apr 24, 2022 at 6:18
  • $\begingroup$ @Jpe61 the max power rpm might be actually higher for the engine alone, but for the propulsion system as a whole the max rpm allowed by the governor is the max power rpm. $\endgroup$
    – Jan Hudec
    Apr 24, 2022 at 14:31

It's the former.

Having your props taking smaller bites of air means that you get faster response from increased throttle, allowing you to reconfigure from landing to missed with less flailing about. There's an additional benefit at lower throttle settings: you get more drag since your prop is the literal opposite of feathered, so if the air it hitting the prop faster than the engine would be pulling it, you get a bit of a free braking effect, which can be useful for landing as well.


When the propeller pitch is set for high rpm (propeller lever full forward), the engine is operating at (or very close to) its maximum power rpm. Therefore if a go around is initiated, the engine will give its best performance.

If the propeller lever was set at a lower rpm, the full power of the engine could not be utilized, as power is simply torque x rpm. For example for horsepower the equation is torque (in lbft) x rpm / 5252.

  • 3
    $\begingroup$ Adding to your answer, power comes from rate of fuel burned (fuel flow in gal/sec). The more fuel being converted to HP each second, the more power you have. So the higher the rpm the more fuel-air mixture is going through the engine (engine displacement in cubic inches times 1/4 RPM for 4-cycle engine). $\endgroup$ Apr 19, 2022 at 15:23
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    $\begingroup$ @CharlesBretana not really. Fuel burned equates to energy, not output power. Some energy is lost (heat) and some arrives at the crankshaft output. If there is no torque on the shaft, there is no output power, just losses. For instance a car engine with gear in neutral does not produce power. Increase the fuel, the RPM goes up untill losses equals fuel burned. Clutch the engine, a torque is produced, together with the RPM there is now an effective output power. Torque is merely the balance between effort (pressure on top of the pistons) and resistance (propellor on a plane). $\endgroup$
    – Tim
    Apr 20, 2022 at 15:02
  • $\begingroup$ @Tim, please read more carefully (no offense (grin)), but I said Rate of fuel burned, not quantity. I included the parenthetical (fuel flow in gal/sec) to make that as clear as possible. But you are right that fuel quantity equates to energy... $\endgroup$ Apr 20, 2022 at 17:52
  • $\begingroup$ To be absolutely specific, the energy content of the fuel burned is related to the quantity of the fuel in mass, not volume 🙃 $\endgroup$
    – Jpe61
    Apr 24, 2022 at 6:21

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