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I want to know what is the minimum speed that an F-16 can fly at without stalling.
I have searched the web, but found no exact information answering my question.
Most of them say that it depends on many factors, like payload and maneuvering. But, for simplicity, I want to know the stall speed of that plane flying in a straight path at a constant altitude.

If there is no exact answer, I want to know the regular stall speed, just an average value of that speed, for a clean F-16A weighing 18,000 kg flying in a straight & level path.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Federico Sep 11 '19 at 12:53
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I don't have exact data either, but we can easily make an estimate with data from Wikipedia. Note that I explicitly neglect high-lift devices which are present on the F-16, the variable angle of incidence (washout), as well as body lift which probably plays a significant role as well.

I estimate the aircraft weight to be around $150\,\mathrm{kN}$. The wing area is approximately $28\,\mathrm{m}^2$, and the NACA 64A204 airfoil used has a max $C_L$ of 0.8. Fill that in the lift equation $$L=\frac{1}{2}C_L\rho v^2$$

we get a speed of about $100\,\mathrm{m/s}$ which is about 200kts.

At the angle of attack of max $C_L$, the L/D is about 10, so the engine has to deliver $15\,\mathrm{kN}$ of thrust to maintain level flight. This angle of attack is approximately 6 degrees, and due to the probably positive angle of incidence between the fuselage and wings, the thrust angle is likely less than 6 degrees w.r.t. the horizon. This means that the vertical component of thrust is about $\sin(6°)\approx 0.1$ times smaller, or $1.5\,kN$. Given the error bars on my estimates (especially the aircraft weight), this does not warrant a recalculation because most of the weight is indeed carried by the wings.

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    $\begingroup$ 200 kts seems like a really fast stall speed, yes? Aren't most planes around 180kts on approach, and land even slower than that, yet are still above their stall speed? $\endgroup$ – abelenky Jul 8 '19 at 16:24
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    $\begingroup$ @abelenky I agree. Then again, this is a high-performance airplane, and I neglected high-lift devices which significantly delay stall. $\endgroup$ – Sanchises Jul 8 '19 at 16:29
  • $\begingroup$ I does not want the speed to be accurate but an approximated number. just a number. Any one have tested this or near this stall speed, this will be helpful $\endgroup$ – AAEM Jul 9 '19 at 11:22
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    $\begingroup$ @Ahmed If you read my answer more carefully, you'll notice that the answer is highly dependent on weight. This can change significantly based on fuel and weapons carried. There's no "number", just different numbers for different weights, following roughly the relation given in this answer. $\endgroup$ – Sanchises Jul 10 '19 at 6:52
  • $\begingroup$ The F-16 is not a glider, it uses vortex lift from the strakes to lower its minimum speed. Without wind tunnel data it is very hard to calculate its maximum lift coefficient, but I would guess it is around 1.1 using the wing for the reference area and adding tail lift on top of wing and fuselage lift. 85 m/s looks about right to me. $\endgroup$ – Peter Kämpf 2 hours ago
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It depends on the weight of the F-16. If weight is below afterburner thrust the F-16 can fly vertical, without having to worry about any lift that the wings are generating. At airshows it passes by straight and level with the nose pointing almost straight up.

So its stall speed in that situation is Not Defined: even at speed zero there is no stall.

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  • $\begingroup$ I think the "straight flight path" + "constant altitude" part of the question removes the afterburner vertical climb solution. $\endgroup$ – Ralph J 2 hours ago
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The minimum speed is 110 knots in level flight pattern .http://www.dailysabah.com/turkey/2017/01/25/turkish-aerobats-break-slowest-airspeed-record-in-breathtaking-show

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  • $\begingroup$ Are we taking your word for this, or can you provide a link to an authoritative source? Right now, this answer amounts to, "somebody on the internet said..." which by itself isn't useful information. $\endgroup$ – Ralph J 2 hours ago
  • $\begingroup$ Note that a nearly identical answer by the same author was deleted on this question a few days ago. $\endgroup$ – Ralph J 2 hours ago

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