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oldtimesake

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  1. 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.
  2. 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.
  3. 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 conc
  4. 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.
  5. 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
  6. 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.
  7. 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.
  8. 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
  9. 1kg = 2.204lbs. So 26888lbs = 26888/2.204kg=12199kg. Sure, that's my typo.
  10. 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 shap
  11. Hi, Xavven. I discussed with you the other day. I found it interesting to measure the ps loss. However, our current method takes lots of efforts.

    I was a flight control engineer, and I know aerodynamics and flight dynamics. I know a simple way to measure ps loss.

    However I don'y have DCS installed on my current machine, so I would like to ask you some questions about Tacview:

    1) Is it possible to read the acceleration of aircraft (how fast/slow it's gaining speed/losing speed)?

    2) Is it possible to read the thrust and drag in real time?

    1. Xavven

      Xavven

      Hi, oldtimesake! I only recently got Tacview so I'm not very well versed in it, and it's still in its trial period on my computer. It doesn't look like it has acceleration, thrust, or drag recorded anywhere, unfortunately. I thought I'd include a screenshot with every window open I can find. That appears to be all it offers.

       

      image.png

    2. oldtimesake

      oldtimesake

      That's it! I saw "longitudinal G-force" and that is the acceleration!

      In a tight turn, just tell me the aircraft weight, the true air speed, the lateral G-force, and the longitudinal G-force, I can calculate the corresponding ps loss and compare it with the manual.

       

      We can start with a turn near 22.5 deg/sec between Mach 0.6-0.7, just like the one we discussed.

  12. 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 lo
  13. 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 mak
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