oldtimesake Posted April 16, 2021 Share Posted April 16, 2021 (edited) 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. Edited April 18, 2021 by oldtimesake 4 Link to comment Share on other sites More sharing options...
TobiasA Posted April 17, 2021 Share Posted April 17, 2021 What's the sense in opening another thread about what is known to be fixed in a later update? It is also the A model, not the C. We all know that the Viper is slightly underperforming below its corner speed and high AOAs. They know it and will fix it. Like... We have 4 threads about that specific issue with the same stuff in it. Let's just pause it at this point and return to it after the FM rework. 1 Link to comment Share on other sites More sharing options...
oldtimesake Posted April 17, 2021 Author Share Posted April 17, 2021 A model and C model share the same aerodynamics Link to comment Share on other sites More sharing options...
Dragon1-1 Posted April 17, 2021 Share Posted April 17, 2021 No, they don't. A model has a different, smaller tail, a different antenna configuration and the earliest models also had no chin hardpoints, further decreasing drag. They were lighter (which influences aerodynamics, too), and had different engines. They look close at first glance, but they're different. 5 Link to comment Share on other sites More sharing options...
oldtimesake Posted April 17, 2021 Author Share Posted April 17, 2021 (edited) 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. Edited April 17, 2021 by oldtimesake 2 Link to comment Share on other sites More sharing options...
Machalot Posted April 17, 2021 Share Posted April 17, 2021 (edited) 20 hours ago, oldtimesake said: At 16756 feet, DCS F-16 (flying weight=26888lbs=12109kg) 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 * 12109 * 9.8 = 937478 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 = 27.87 m^2 One can easily calculates that Lift Coefficient = 937478 / 0.5 / 0.727613 / 328.1^2 / 27.87 = 0.85 (I haven't counted the thrust contribution, which will make the Lift Coefficient even lower than 0.85) At AOA=17.8deg around mach 0.9 the lift coefficient is lower than 0.85? 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. I thought I read somewhere that Tacview is not a reliable source of AoA. Did you get it from Tacview or in game? Also make sure it's degrees and not "AoA units". Also note that the charts you posted are likely for quasi-steady state maneuvers with control surfaces trimmed, whereas you are looking at a transient flight condition to make your calculations. In my experience that can make a huge difference because the control surfaces are not trimmed. Edited April 17, 2021 by Machalot 1 "Subsonic is below Mach 1, supersonic is up to Mach 5. Above Mach 5 is hypersonic. And reentry from space, well, that's like Mach a lot." Link to comment Share on other sites More sharing options...
Snappy Posted April 17, 2021 Share Posted April 17, 2021 (edited) 21 hours ago, oldtimesake said: At 16756 feet, DCS F-16 (flying weight=26888lbs=12109kg) 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 * 12109 * 9.8 = 937478 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 = 27.87 m^2 One can easily calculates that Lift Coefficient = 937478 / 0.5 / 0.727613 / 328.1^2 / 27.87 = 0.85 (I haven't counted the thrust contribution, which will make the Lift Coefficient even lower than 0.85) At AOA=17.8deg around mach 0.9 the lift coefficient is lower than 0.85? 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. Unless it’s a typing error regarding the baseline numbers on your part, already your first lbs to kg conversion is seemingly wrong , by some 87kg too light. 26888 lbs is 12196 (rounded down) kg , not 12109 kg Not much in the grand scheme of things, however my point is, if you expect to have this taken serious by the dev team, probably better to have this basic stuff correct. regards, Snappy Edited April 17, 2021 by Snappy Link to comment Share on other sites More sharing options...
oldtimesake Posted April 17, 2021 Author Share Posted April 17, 2021 (edited) 1kg = 2.204lbs. So 26888lbs = 26888/2.204kg=12199kg. Sure, that's my typo. Edited April 17, 2021 by oldtimesake Link to comment Share on other sites More sharing options...
oldtimesake Posted April 17, 2021 Author Share Posted April 17, 2021 1 hour ago, Machalot said: I thought I read somewhere that Tacview is not a reliable source of AoA. Did you get it from Tacview or in game? Also make sure it's degrees and not "AoA units". Also note that the charts you posted are likely for quasi-steady state maneuvers with control surfaces trimmed, whereas you are looking at a transient flight condition to make your calculations. In my experience that can make a huge difference because the control surfaces are not trimmed. 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 Link to comment Share on other sites More sharing options...
Snappy Posted April 17, 2021 Share Posted April 17, 2021 (edited) 16 minutes ago, oldtimesake said: 1kg = 2.204lbs. So 26888lbs = 26888/2.204kg=12199kg. Sure, that's my typo. Ok then , I simply used more decimal points in the conversion..Thats why I ended on a 6. Regards, Snappy Edited April 17, 2021 by Snappy Link to comment Share on other sites More sharing options...
Machalot Posted April 17, 2021 Share Posted April 17, 2021 7 minutes ago, oldtimesake said: 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 Can you tell me what timestamp in that 22 minute video? "Subsonic is below Mach 1, supersonic is up to Mach 5. Above Mach 5 is hypersonic. And reentry from space, well, that's like Mach a lot." Link to comment Share on other sites More sharing options...
oldtimesake Posted April 17, 2021 Author Share Posted April 17, 2021 1 minute ago, Machalot said: Can you tell me what timestamp in that 22 minute video? Start from 11:24 Link to comment Share on other sites More sharing options...
Machalot Posted April 18, 2021 Share Posted April 18, 2021 (edited) 4 hours ago, oldtimesake said: Start from 11:24 You go from steady 2.8 deg AoA to ~17 deg and then steady ~19 deg in less than a second at 14:55:52. Hard to know if that's realistic. The bigger issue is that the sample frame you chose, ~17 deg, is at the tail end of the transient so the jet still has a lot of alpha rate, and the FCS hasn't deflected the surfaces to arrest that rate yet, or is in the middle of doing so, so they are likely not in a trim configuration. Edited April 18, 2021 by Machalot "Subsonic is below Mach 1, supersonic is up to Mach 5. Above Mach 5 is hypersonic. And reentry from space, well, that's like Mach a lot." Link to comment Share on other sites More sharing options...
sLYFa Posted April 18, 2021 Share Posted April 18, 2021 On 4/17/2021 at 12:29 AM, oldtimesake said: wing area = 27.87 m^2 Does your source specifically state that the CL values use that exact wing area?`If not then the wing area used to normalize the lift could be quite different from the 27.87 square meters you assumed (e.g. with/without LEX, some equivalent straight wing area). CL/CD/CM values depend heavily on the definition of the area they are normalized with, which is not always the physical area of the wing. i5-8600k @4.9Ghz, 2080ti , 32GB@2666Mhz, 512GB SSD Link to comment Share on other sites More sharing options...
oldtimesake Posted April 18, 2021 Author Share Posted April 18, 2021 21 minutes ago, Machalot said: You go from steady 2.8 deg AoA to ~17 deg and then steady ~19 deg in less than a second at 14:55:52. Hard to know if that's realistic. The bigger issue is that the sample frame you chose, ~17 deg, is at the tail end of the transient so the jet still has a lot of alpha rate, and the FCS hasn't deflected the surfaces to arrest that rate yet, or is in the middle of doing so, so they are likely not in a trim configuration. 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. 1 minute ago, sLYFa said: Does your source specifically state that the CL values use that exact wing area?`If not then the wing area used to normalize the lift could be quite different from the 27.87 square meters you assumed (e.g. with/without LEX, some equivalent straight wing area). CL/CD/CM values depend heavily on the definition of the area they are normalized with, which is not always the physical area of the wing. My source, and almost any source, claim wing area = 300 square feet = 27..87 square meters. Link to comment Share on other sites More sharing options...
sLYFa Posted April 18, 2021 Share Posted April 18, 2021 You didn't understand the question apparently, sorry for wasting your time i5-8600k @4.9Ghz, 2080ti , 32GB@2666Mhz, 512GB SSD Link to comment Share on other sites More sharing options...
oldtimesake Posted April 18, 2021 Author Share Posted April 18, 2021 (edited) 16 minutes ago, sLYFa said: You didn't understand the question apparently, sorry for wasting your time 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. Edited April 18, 2021 by oldtimesake Link to comment Share on other sites More sharing options...
Machalot Posted April 18, 2021 Share Posted April 18, 2021 1 hour ago, oldtimesake said: 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. I'm not trying to be testy here, although apparently this thread is heading that direction. I'll take a look at things you calculate and post, but I'm not interested in trying to pull and calculate my own results from your video. One other thing to consider is whether you are calculating and measuring the same acceleration quantity, i.e. wind-fixed vs body-fixed Nz. As I'm sure you know, the trig says at 18 deg AoA there is a 5% knockdown of lift and a 31% contribution from drag into body-fixed Nz. Do you know if the G value shown in Tacview is body-fixed or wind-fixed? "Subsonic is below Mach 1, supersonic is up to Mach 5. Above Mach 5 is hypersonic. And reentry from space, well, that's like Mach a lot." Link to comment Share on other sites More sharing options...
oldtimesake Posted April 18, 2021 Author Share Posted April 18, 2021 4 minutes ago, Machalot said: I'm not trying to be testy here, although apparently this thread is heading that direction. I'll take a look at things you calculate and post, but I'm not interested in trying to pull and calculate my own results from your video. One other thing to consider is whether you are calculating and measuring the same acceleration quantity, i.e. wind-fixed vs body-fixed Nz. As I'm sure you know, the trig says at 18 deg AoA there is a 5% knockdown of lift and a 31% contribution from drag into body-fixed Nz. Do you know if the G value shown in Tacview is body-fixed or wind-fixed? 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 Link to comment Share on other sites More sharing options...
Machalot Posted April 18, 2021 Share Posted April 18, 2021 1 minute ago, oldtimesake said: 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 But the 31% contribution of drag might. "Subsonic is below Mach 1, supersonic is up to Mach 5. Above Mach 5 is hypersonic. And reentry from space, well, that's like Mach a lot." Link to comment Share on other sites More sharing options...
oldtimesake Posted April 18, 2021 Author Share Posted April 18, 2021 Just now, Machalot said: But the 31% contribution of drag might. 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. Link to comment Share on other sites More sharing options...
Machalot Posted April 18, 2021 Share Posted April 18, 2021 Just now, oldtimesake said: 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. You seem to have missed my point. The G value measured by the aircraft Z accelerometer will include 95% of the lift and 31% of the drag currently on the airframe. It's a change of coordinates to compare CL and Nz that must include the drag component. "Subsonic is below Mach 1, supersonic is up to Mach 5. Above Mach 5 is hypersonic. And reentry from space, well, that's like Mach a lot." Link to comment Share on other sites More sharing options...
oldtimesake Posted April 18, 2021 Author Share Posted April 18, 2021 (edited) 14 minutes ago, Machalot said: You seem to have missed my point. The G value measured by the aircraft Z accelerometer will include 95% of the lift and 31% of the drag currently on the airframe. It's a change of coordinates to compare CL and Nz that must include the drag component. 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. Edited April 18, 2021 by oldtimesake Link to comment Share on other sites More sharing options...
Machalot Posted April 18, 2021 Share Posted April 18, 2021 (edited) 5 hours ago, oldtimesake said: The tacview can display both lateral G force and longitudinal G force. In this video only the total G force is displayed. After some quick math it's easy to show that it depends on if the longitudinal G is big enough to make the lateral G significantly lower than the total G. I don't think so, since at this altitude the longitudinal G would be significantly lower than 1 while the total G is close to 8. Ok, then my remaining question is whether Tacview AoA is trustworthy. Later on I'll try to find the original post that was the source of my doubts. Update: Couldn't find it. The gist was that a guy and his wingman were flying in close formation pulling the same g's with vastly different AoA in Tacview. When he asked about it in the forums, people told him Tacview AoA can be unreliable based on the way track files are constructed. Edited April 18, 2021 by Machalot "Subsonic is below Mach 1, supersonic is up to Mach 5. Above Mach 5 is hypersonic. And reentry from space, well, that's like Mach a lot." Link to comment Share on other sites More sharing options...
Frederf Posted April 18, 2021 Share Posted April 18, 2021 Tacview only knows object attitude and motion vector. "AOA" is just difference between two. Tacview has no idea of winds. Link to comment Share on other sites More sharing options...
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