If you wanted a 737 to have the same design as a single prop light aircraft (i.e. just one prop on the nose, no turbines, APUs or other engines) - how big (and how fast) would the propeller have to be to provide enough thrust to provide lift and enable flight?

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    $\begingroup$ Just a note on the OP’s question: if you have a single propeller - no turbines, APUs or anything else to drive it - what you have is a speedbrake. $\endgroup$ Commented Nov 27, 2023 at 11:43
  • $\begingroup$ @Romeo_4808N I'm talking about if it was a large 912UL or similar $\endgroup$
    – Cloud
    Commented Nov 27, 2023 at 12:39
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    $\begingroup$ Obviously it's hypothetical, so we don't need to be concerned about what happens during takeoff and landing (probably history's greatest prop strike). But it might help to clarify if you are referring to just staying aloft in level flight at minimum speed, or flying at normal turboprop speed and altitude. $\endgroup$ Commented Nov 27, 2023 at 15:29

1 Answer 1


As usual, we can use the simple momentum theory to quickly estimate the thrust $T$ generated by a propeller:

$ T = \sqrt[3]{2 k \rho A P^2} $

where $A$ is the propeller area, $\rho$ is the density and $P$ the power needed to spin the propeller. This equation is strictly valid only at zero forward speed but being the highest thrust normally required at takeoff, then it's just perfect for our purpose. $k$ is a coefficient taking into account the limitations of the momentum theory: a value of $0.5$ can be used in a conservative way.

So, the two CFM LEAP-1B of a 737 Max generate a total thrust $T$ of 260 kN. Substituting we get:

$AP²=\frac{T³}{2k\rho}=\frac{260'000³}{2 \times 0.5 \times 1.125}=1.5e16 [m²W²]$

Note that we can only estimate the product of the propeller area with the needed power. We can now decide to either use a very powerful engine with a small propeller or a less thirsty engine and a big propeller.

Taking into account the height of the nose (some 3m) as limit for the propeller radius, we get a propeller area of $A=\pi 3²=28m²$ and therefore a needed power of $P=\sqrt{1.5e16/28}=24MW$. Three of these would be just perfect to power the propeller.

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    $\begingroup$ Thrust requirement for a twin engine is different than for a single engine. A twin will need to do with either engine down most of the stuff a single can do with it's only engine, so for this single prop 737 we do not need to match the full rated power of both of the regular 737's engines 😉 $\endgroup$
    – Jpe61
    Commented Nov 27, 2023 at 18:45
  • $\begingroup$ 260 kN thrust is about 58,500 lbs, a bit more than the MTOW of a Chinook, with two 3-blade 60 ft diameter rotors. So a single (subsonic) rotor, ahem, prop, would be about 90 ft. Smaller if you add more blades, as with even heavier helicopters. $\endgroup$ Commented Nov 27, 2023 at 19:27
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    $\begingroup$ With a giant prop, more thrust is created at lower airspeeds, less at higher airspeeds than the LEAP. Takeoff performance is a selling point for turbo props. >20,000 hp will create 58,500 lbs (260 kN) of thrust at 0.2 Mach, especially if the "giant prop" is more efficient. The idea was tried here. $\endgroup$ Commented Nov 28, 2023 at 9:59
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    $\begingroup$ It's 23 foot prop swung at 545 RPM. $\endgroup$ Commented Nov 28, 2023 at 10:13

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