oldtimesake

There is no proof this paper is about the production F-16 model (or maybe a prototype), especially given the low lift curve slope. At 10deg AOA the untrimmed CL is less then 0.7 while the flight test value is 0.85.

I'm glad that you get my point to some extent. The drag curve above Mach 0.9 is not classified (published in AGARD-242 test report). The drag at Mach 0.8 can be found in a text book, but below Mach 0.8 we can not find a reliable drag polar. I think the dev is trying some curve fitting which poses big challenge.

The tacview can display both lateral G force and longitudinal G force. In this video only the total G force is displayed. If it is body frame the total G is normal G * cos 72deg + longitudinal G * sin 18 deg. The lift is defined in wind frame. Whether my calculation holds depends on whether the displayed G has an orientation significantly deviates from that of the lift. If the deviation is small, that's good; if the deviation is big, the normal G (perpendicular to wind) is even smaller than the displayed G, make the lift coefficient even smaller than 0.86, that won't affect my conclusion.

Sure, but this thread has nothing to do with drag. I am planning to post another thread on ps loss, that has something to do with drag, but not now.

Either way, that won't affect much. The reduction of lift is 0% or 5%, that won't explain the difference between 0.86 and 1.23

The paper I posted reads Sref = 300 square feet and the corresponding lift coefficient at 17.8deg, mach 0.9 is 1.23. If you can't read sorry for wasting your time.

You are free to select a trimmed state frame by your standard, and calculate the lift coefficient accordingly. I am pretty sure you won't get anything near the test value. My source, and almost any source, claim wing area = 300 square feet = 27..87 square meters.

Start from 11:24

If you look at the video you will notice the AOA is almost stable and it is very close to trimmed state. Or you can average the AOA among 1 sec and I am sure you will get something close

1kg = 2.204lbs. So 26888lbs = 26888/2.204kg=12199kg. Sure, that's my typo.

1) Weight has nothing to do with lift curve slope 2) All other elements you mentioned (antenna, tail...) has very few aerodynamic impact. Besides, bigger tail plane reduces required trim deflection, it actually gives better trim performance and less trim drag, and more available lift at a given AOA. Proof: F-16C-block50 at 26000 lbs can pull 9G at 433.6 knots. With the same calculation it is easy to verify that the lift coefficient at 15 deg is about 1.2 and exceeds that of F-16A. Proof2: YF-16 and F-16A share almost the same aerodynamic curve The shape deference between F-16A and F-16C is significantly less than that between YF-16 and F-16A. At least the reference area remains unchanged, unlike YF-16 and F-16.

A model and C model share the same aerodynamics

You are wrong again. the main problem for F-16 flight model is not its peak sustained turn rate, is its lower sustained turn rate below mach 0.5 and higher energy bleed rate when the turn rate exceeds sustained turn rate. This is induced by much higher AOA required to pull only a small amount of G. One more thing: I think you have a wrong idea about instantaneous turn rate. Even you can get the instantaneous turn rate correct in the future, the energy bleed rate is still incorrect. As tested in this thread: At 22000lbs, 191.4m/s and 7.7G pull the tested ps loss is greater than 391feet/sec, while on the manual it is less than 200feet/sec At 22000lbs, Mach 0.42 you are supposed to sustain 19.5deg/sec but the tested number is only 18.5deg/sec

At 16756 feet, DCS F-16 (flying weight=26888lbs=12199kg) needs 17.8deg AOA to pull 7.9G at true air speed of 637.8 knots (328.1m/s). Lets calculate: Lift = 7.9 * weight * g = 7.9 * 12199 * 9.8 = 944446 Newton Lift = 0.5 * Lift Coefficient * air density * speed ^2 * wing area = 937478 Newton where air density = 0.727613 kg / m^3 at this altitude, wing area = 300 square feet = 27.87 m^2 One can easily calculate that Lift Coefficient = 944446 / 0.5 / 0.727613 / 328.1^2 / 27.87 = 0.86 (I haven't counted the thrust contribution, which will make the Lift Coefficient even lower than 0.86) At AOA=17.8deg around mach 0.9 the lift coefficient is lower than 0.86? What? The true value is 1.23 according to AGARD-242 test report As I mentioned before, the main problem for F-16 flight model is not its peak sustained turn rate, is its lower sustained turn rate below mach 0.5 and higher energy bleed rate when the turn rate exceeds sustained turn rate. This is induced by much higher AOA required to pull only a small amount of G. Hope this time BIGNEWY gets the correct idea about what is wrong.

Nope. Only the peak sustained turn rate is accurate. Sustained turn rate below mach 0.5 is 1deg/sec lower than manual: One more thing: I think you have a wrong idea about instantaneous turn rate. Even you can get the instantaneous turn rate correct in the future, the energy bleed rate is still incorrect, as tested in this thread: At 22000lbs, 191.4m/s and 7.7G pull the tested ps loss is greater than 391feet/sec, while on the manual it is less than 200feet/sec At 22000lbs, Mach 0.42 you are supposed to sustain 19.5deg/sec but the tested number is only 18.5deg/sec

Can you try deploy the trailing edge flap and check if that helps the turn rate?

4:01:45 to 4:02:04 is somewhere between 19s and 20s depending on the "sub-sec" accuracy, so the average turn rate should be 18-18.95 deg/sec. That is 0.5~1.5 deg/sec below manual (19.5deg/sec). Based on the speed gain and altitude loss that is almost balanced. It is safe to assume it is still about 0.5~1.5 deg/sec lower than manual.

We can conclude from your test that a DCS F-16 (weight = 22000lbs ) can only sustain 17.4deg / sec at air speed slightly higher than Mach 0.4, which is 2.1 deg/sec lower than the manual. Don't you find it ridiculous?

Let's calculate the speed of sound in flight manual: pick a point from flight manual: 9G, 0.7Mach, 21.5deg/sec. speed = centripetal acceleration / angular vilocity = square root(9^2 - 1) * 9.8 / (21.5 / 57.3) = 87.653 / 0.3752 = 233.6167 m / s = 0.7 mach speed of sound in manual = 233.6167 / 0.7 = 333.73 m/s speed of sound in DCS = 145.07 / 0.42 = 345.4 m/s

The only explanation is that the speed of sound in manual is slightly different from that in DCS, making the mach number slightly different. Only the true air speed (TAS) is reliable. Let's calculate the speed of sound in flight manual: pick a point from flight manual: 9G, 0.7Mach, 21.5deg/sec. speed = centripetal acceleration / angular vilocity = square root(9^2 - 1) * 9.8 / (21.5 / 57.3) = 87.653 / 0.3752 = 233.6167 m / s = 0.7 mach speed of sound in manual = 233.6167 / 0.7 = 333.73 m/s speed of sound in DCS = 145.07 / 0.42 = 345.4 m/s

That won't explain the difference of 19.5deg/sec (manual value) and 17.4deg/sec (test value). The only explanation is that the speed of sound in manual is slightly different from that in DCS, making the mach number slightly different. Only the true air speed (TAS) is reliable.

Oh boy you are again reading directly the indicated turn rate... True air speed 282knots = 145.07m/s Centripedal acceleration = square root (4.6^2 - 1)G = 44.001m/s^2 Horizontal turn rate = 44.001 / 145.07 * 57.3 = 17.37deg/sec From the chart it should be around 19.5deg/sec. That is not close at all. Even the direct turn rate reading is 18.5deg/s and that is still far from being checked out.

As I said, the indicated turn rate is not horizontal turn rate. Confusing the two results in big error. I never read the indicated turn rate and I calculate the equivalent horizontal turn rate with airspeed and G loading.

OK, so basically I need to repeat my test but try to get 24.9 deg./sec. of ITR without banking more than 90 degrees, correct? ---------------- For now I don't think it is necessary. a) We are more interested in the ps bleed around ps=0 curve which is more frequently used in air combat. b) With the G load reading you don't need to maintain a horizontal turning attitude. The equivalent horizontal turn rate can be calculated. c) PS bleed rate test at Mach 0.4 or 0.45 is a good start, and I suggest you turn on the AOA indicator, read the G load only, ignore the turn rate