Hello to all, After reading comments about F-35’s performance compared to legacy aircraft, I have decided to make some *rough* analysis of F-35’s turn, acceleration and excess power, and compare it with the current F-16 (the F-35 was supposed to have same maneuverability of F-16 doesn’t it?). While the calculations themselves are precise, there are many unknowns so I had to use many assumptions (thats why I call it rough).
Disclaimer#1: This post will likely to turn into a small fluid dynamics course and likely to be boring. Feel free to skip to the graphs –and comments- at the end.
Disclaimer#2: I had to use generic data from fluid dynamics and advanced aerodynamics books and my own guesstimates to fill in the gaps:
-To be perfectly honest, this little “project” all started to answer “what if F-35 used same airfoil as F-16” question. So first and most obvious of my assumptions that I take F-35 uses same airfoil as F-16 (64A204). It may or it may not, but after seeing the final data, I’ve hoped I can speculate whether thicker airfoils would work best or not. As for Cl graph I can read the data, and modify it for presence of slats. However on a FBW aircraft making high AOA turns at extremely slow speed; TE flaps also come online at continiously variable angles. I assumed TE flaps have 0,2 improvement on Clmax on maneuvering condition, and slowly fading at about M0,5. (F-16’s lift graph behaves as such, but IDK if they would be the same. Can wider TE flaps of F-35 may improve this further? IDK)
– for Cx0 (drag of non lifting area) vs Mach number , I had to use generic data from “area ruled aircraft vs non area ruled” graph; It starts at Cx0 = 0,02 ; starts incresing at transonic, tops out at 0,05 at M1.0 then slowly drops down. For F-15, Soviet spy data estimate its 0,022 to 0,045, M2000= 0,018 to 0,04. MiG-29 aerodynamic booklet says its 0,025 at M0,3 and 0,05 at M1,1 then drops afterwards. I think F-35 has less drag coef. than F-15 because its single engined, but its fat and have big nose so it should be higher than M2k. In line with those, I believe generic info is more or less usable.
-Inlet drag coefficent is also generic data, for subsonic, transonic and supersonic regimes its 0,02 0,028 and 0,015 respectively.
-Dynamic engine thrust on a fixed inlet is accepted as 85% of static at 0 airspeed and 100% at design point. I’ve declared the design point to be M0,8. Both are assumptions, inlet may have differen efficiency or design point can be different.
After having stated the fact each of these points contribute to errors, now how my analysis works;
First, instantenious turn rates; This is simple and highly accurate. By using and modifying the lift formula like I’ve said in the Su-27 topic;
G load = [1/2 * (density of air) * (Wing Area) * (Lift Coefficient) * (airspeed) ^2] / (aircraft weight).
This is the overall G load aircraft pulls. As 1G will be “wasted” to put the aircraft in level flight; only the lateral G can be used to pull the aircraft into the turn. As turn is horizontal and gravity is vertical; simple (G load)^2 = 1+(G lateral)^2
By simply using “acceleration = velocity^2/radius” formula; and velocity itself will complete the circumference of the circle (found by using the radius of turn),
Turn rate = velocity*360/[velocity^2/(lateralG*9,81)*2*PI] where turn rate is in deg/s velocity is in m/s.
On excel calculating turn rate for M0,05 intervals from M0,2 to M1,8; there we have the instantenious turn graph.
Now the slightly trickier, in line acceleration and specific excess power or climb rate:
“Cx0 * frontal area (excluding wings and inlet) + cd inlet * inlet area + Cd * wing area “ gives us “total CdA”. From the drag formula we can find the drag force.
Thrust * (thrust modifier) – drag will give us our excess thrust.
***Cd will be calculated in similar way explained in STR calculation for enough Cl for level flight.
Acceleration is simple; a=F/mass; as I was iterating in M0,05 intervals, I assumed linear acceleration between two airspeeds, with average of two “a” vaules found.
As SEP or climb rate is the amount of potential energy aircraft can is ABLE to gain with its excess thrust. On differantial forms; Potential energy change is dPE/dt = m*g*dH/dt and amount of energy change by the applied force is dE/dt = Force*Velocity. Equating dE = dPE will give us
dH/dt = Velocity * Excess Power / (9,81 * mass in kg)
dH/dt slope, as the name suggests is the the instant amount of change in height with respect to time; or put it in english, it is the “instantenious climb rate” =)
Before anyone brags about inconsistancy, I would like to state a “slight error”. This calculation ignores the fact whether aircraft can actually fly in level or not. For example if you look at the 30k feet graph below, at M0,3 aircraft has 0 deg/s ITR, because our F-35 model can only generate 0,874 Gs, insufficient for level flight let alone turn. However while finding “total CdA” my Cd calculation algorithm merely takes the Cdmax if aircraft cannot generate 1Gs and calculates accordingly –which is why it has positive SEP at M0,3. Theoratically, the calculation is not wrong, it shows at highest non-stalled drag, F-35 will be accelerating; no matter if it will also be falling due to insufficient lift.
Now the last and the hardest part; Sustained turn rates:
After calculating the excess power, a drag formula is applied to find to “counter” that excess power. This “Sustainable Cd” is used to find the “Sustainable Cl” from the wing data. Unfortunalately, excel cannot be made to work like that. So I’ve made a formula from a lagrange polynomial by using various points on Cd graph: At the Cl@0 deg; Cd=Cdmin, at Clmax Cd=Cdmax. This formula forms a fuction that actually “calculates” Cl from the given “Sustainable Cd”, repeatedly taking the maximum Clmax and Cdmax for each Mach number. This is the main reason why SEP graph is off at <1G condition. Also, there are some inaccuracies because the graph is not exact but interpolated. Anyway, after finding the “sustainable Cl” I simply calculate and convert it to Sustained turn rate it in the same way as ITR part, and put in on the graphs.
Now I’ve explained how all these works, I will also put the F-16 Blk50 data from its supplemental manual.
First graph is for clean, 50% fuelled F-35 at 17495 kg versus drag index=0 F-16 block 50. Standard atmospheric conditions (do matter because I am taking density and thrust modifier accordingly) and at Sea Level:
[ATTACH=CONFIG]225995[/ATTACH]
Unsuprisingly (to the most at least) chances of F-35 beating a F-16 in a gun fight is laughable to none. While F-35 pulls slightly higher maximal ITR on paper (24,9 vs 24,8), F-35’s ITR graph is constistantly 2,5 deg/s below the F-16’s. Again unsuprisingly its also draggy while turning, and its STR graph is approximately 3 deg/s below F-16’s all the time. While F-16 maxes out at 21,4 STR, F-35 can only pull a puny 18,7 deg/s highest STR.
Now the second graph. Same two aircraft at 30000 feet;
[ATTACH=CONFIG]225996[/ATTACH]
Well, this is slightly suprising; at thinner air where drag drops significantly, F-35 can approach the performance of F-16, at least when subsonic. Both aircraft are pretty comperable below M0,75, but at supersonic speeds F-16 shines. As the speed increase, chances of F-35 kinematically beating the F-16 drops to 0.
Third graph; same aircraft, Specific Excess Power in m/s at Sea Level and 30k feet:
[ATTACH=CONFIG]225997[/ATTACH]
At sea level, F-16 puts tremendous power thanks to its minimal drag, clearly a leap ahead of F-35 in acceleration and climb, which can barely exceed 245 m/s. F-16 has no problems reaching 330m/s climb rate. However at 30k feet, F-35 actually has better SEP when subsonic, and mostly comperable at transonic regime. Supersonic is beyond comparison, of course. Translation of this graph is; F-16 will have a lot (read = 35+%) better acceleration and climb performance at S/L, but F-35 will have (very) slightly better subsonic acceleration/climb at 30k feet.
Now let’s move into a more realistic scenario for close air combat; Lets assume both aircraft expanded their BVR missiles and moving into merge with 4x AIM-9s. For F-16
-2x LAU-129 launchers for wingtip (drag index 1) (previously carrying the expended AIM-120s)
-4x LAU-129 launcher+adapters for 2,3 7 and 8 (drag index 6)
-4x AIM-9M missiles on stations 2,3,7 and 8 (drag index 5)
-basic drag index includes two wingtip AIM-9s, removing both -> drag index -8
-F-16 “C” basic drag index = 7
Calculating those will give us drag index = 45. If F-16 was carrying wing EFTs and dropped them, there would be additional +8 drag index for each NNJET pylon, additonal centerline tank pylon (after dropping the tank itself) would add +7 to the drag index. I will simply take the F-16’s data from Drag Index 50 graph.
As F-35 will be carrying AIM-9s internally there will be no drag penalty. Its launcher/adapter mechanism is also integrated, so I will only add 88×4 = 352 kg to the weight.
Comparing armed F-35 armed with 4x internal AIM-9s versus F-16 armed with 4 AIM-9Ms and 2 empty pylons. At sea level;
[ATTACH=CONFIG]225998[/ATTACH]
This was by far the most suprising result to me. I was expecting extra drag would degrade F-16’s performance a little, but not this much; At very slow speeds, F-16 still has slight advantage, but at above M0,6 F-35 actually sustains turns BETTER than F-16 blk50. On ITR part, F-35 gets advantage as the speed increases, topping out at 24,4 deg/s versus F-16’s 22,5 deg/s.
Same aircraft, at 30k feet:
[ATTACH=CONFIG]225999[/ATTACH]
On average; F-35 has 1,2 deg/s superiority to F-16’s Sustained turn performance at subsonic and transonic realm. While supersonic F-16 has better STR. Their ITR is mostly comperable, however at supersonic F-16 enters PhiMax state which degrades ITR performance. While F-35 looks better in theory, I don’t know if similar conditions would affect F-35 too.
Same aircraft SEP graph:
[ATTACH=CONFIG]226000[/ATTACH]
At sea level, they appear to be comperable, however at 30k feet, F-35 gets a clear advantage in terms of climb and acceleration performance.
To summerize; Clean F-35 is clearly inferior to clean F-16 Block 50 in overall performance, with the gap narrowing at higher altitudes. However when armed with 4 missiles, F-35 is equal or better than F-16 Block 50 over most of the flight envelope. With the payload increasing advantage should move to F-35. IMHO, no matter how lightly loaded F-35 is, it wont have enough maneuverability to do any nice tricks at the airshows. For spectators it will always be a flying brick. However it will have sufficient kinematic performance where it matters, at least when compared to legacy fighters.
Also this modelling calculates the maximum speed and acceleration of F-35 as follows;
At Sea level, its M1.1, translates to 1357 km/h
At 30000feet; its M1.67, translates to 1822 km/h
I had made an acceleration comparison graph some time ago for a topic in this forum by using FM datas. Putting my F-35 model’s acceleration into the graph gives;
[ATTACH=CONFIG]226001[/ATTACH]
Its M0,8 to M1,2 time is 34,95 seconds.
Obviously this is simple excel modelling, nothing more but I tried to be as objective and scientific as possible; I am from Turkey, we have a large F-16 fleet, and will have one of the largest F-35A fleet, so there is no reason for me to pimp up one aircraft’s performance. Though I will admit I really dont like F-35, that will not change whatever numbers it generates. Ugly is ugly. Here’s the raw data for further thoughts;
[ATTACH=CONFIG]225994[/ATTACH]
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March 1, 2014 at 10:02 pm