I've been trying to calculate the power a P&W F100 engine on a F-16 is producing using some of the available data on thrust and velocity and I ended up with a number that I want to confirm. Here's how I made the calculations.

First of all, I don't want my calculations to involve any complications due to the altitude and its effect on the engine thrust, so I just chose to make my calculations at sea level. Wikipedia states that the maximum speed of an F-16 at sea level is 1,470 $km/h $ which is ~$408$ $m/s$.

Secondly, I found out that jet engines in general produce different thrust at different speeds. I did some research and found that turbojets produce somewhat constant thrust at subsonic speeds and gain extra thrust at supersonic speeds, while turbofans suffer degrading thrust with speed with a degree depending on the bypass ratio of the turbofan. I found that high bypass ratio turbofans suffer much more thrust degradation at high speeds compared to low bypass ratio turbofans. which means that low BPR turbofans fall somewhere in between the turbojet and high BPR turbofans according to their BPR.

I found that P&W F100 Wiki gives a bypass ratio of $0.36:1$, then looking at this thrust vs speed graph, which confirms what I concluded above, I could determine that the F100 engine with 0.36 BPR can have no more than 10% thrust degradation at mach 1.2. I also think that using the afterburner shifts the curve further up since more air is now entering the core than before without increasing the fan flow and therefore the engine behaves even more like a turbojet than a turbofan. Overall, I think it would be safe to say that the afterburner thrust at Mach 1.2 at sea level will not degrade more than 10% and so the F100 engine thrust will become 0.9×127 kN = $114.3$ $kN$

Now the power is simply $P = F . v$ = $(114.3 × 10^3) × 408$ = $46.6 × 10^6$ $W$ or about $62,000$ $hp$.

So my question now is: Are my calculations acceptable to a certain degree or are they completely off and I didn't take into account another significant factor?

EDIT: I have another question. I'm assuming here that the engine is producing full thrust at the sea level to counter the air drag and achieve the quoted speed. But is this a correct assumption? In other words, could an F-16 fly at its maximum speed at sea level while not using the maximum thrust?

  • 1
    $\begingroup$ In answer to your edit, it is possible for an aircraft to exceed is maximum speed (e.g. a limit speed - Vmo/Mmo) while not using max thrust. In an aircraft designed for supersonic flight like the F-16, the maximum theoretical speed is probably much closer to its design limit speed than it would be in an airliner, but there will be a difference $\endgroup$
    – SSumner
    Mar 2, 2020 at 18:49
  • $\begingroup$ Also, the first question is more of a physics question than an aviation one, while the edited one is more aviation-related $\endgroup$
    – SSumner
    Mar 2, 2020 at 18:50

1 Answer 1


https://en.wikipedia.org/wiki/Pratt_%26_Whitney_F100 gives a maximum afterburning thrust for the F100-PW-220 of 105.7 kN, so you were pretty close there. Power is not listed on wikipedia, but you can get there. It lists "Specific fuel consumption: Military thrust: (0.73 lb/(lbf·h))", so all you need to know is the lower heating value of jet fuel, and you can calculate the power. Off the top of my head, 46 MW seems a little high for that size engine, but it's certainly on that order of magnitude.

  • $\begingroup$ I think the data on the F-16 Wikipedia is about the F100-PW-229. The engine WP says it is used for late model F-16s: en.wikipedia.org/wiki/Pratt_%26_Whitney_F100#F100-PW-229. Or it could be the F110-GE-129: en.wikipedia.org/wiki/General_Electric_F110#F-16 In any case, I think 127 kN is within the possible range for an F-16 and using it at the mentioned speeds wouldn't be incorrect. $\endgroup$ Mar 25, 2019 at 10:49
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    $\begingroup$ If I understood you correctly, I found that the lower heating value of jet fuel to be 43.1 MJ/kg. For the F100-PW-229, the afterburner SFC is 1.94 lb/(lbf·h) which for 127 kN (29,000 pounds-force) would be 6.56 kg/second. This would generate power of 282 MW which is much more than the value I calculated and which shows that not all of the fuel chemical energy is utilized as mechanical work. Please correct me if I'm wrong or if I misunderstood something you said. $\endgroup$ Mar 25, 2019 at 11:37
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    $\begingroup$ @AbanobEbrahim: That points to a thermal efficiency near 17%, which is entirely believable. $\endgroup$
    – MSalters
    Mar 25, 2019 at 16:56

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