Proof: F-35 has good sustained rate of turn. Same A/B time

The F-35 compared with other modern jets.
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gta4

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Unread post23 Mar 2017, 16:25

Performance standard: 60% internal fuel (5000 kg), 2 aim-120 missiles, 15000 ft, M0.8.
viewtopic.php?f=55&t=52510
Threshold: 4.6G at M0.8
Demonstrated performance: 4.95G at M0.8 (F-35 240-3 configuration)
http://warships1discussionboards.yuku.c ... NPjFNUrKUk

4.95G at M0.8 corresponds to 11 deg/sec. Not a good looking figure?
This is because F-35 is carrying too much fuel. That is not fair.

Note that Mig-29 uses only 1700 kg as performance standard (1700=3400*50%). For the same afterburner time, a F-35 needs about 2000 kg fuel (fuel consumption = SFC*thrust*time. Since most military turbofan engines have similar SFC of about 1.9, fuel consumption is proportional to total thrust.)

With 3000 kg weight reduction, the total flying weight gives a factor of 16000/19000=0.84. Since the normal load factor is inverse-proportional to weight, the turn rate will be increased by 19%. Which is 13.1 deg/sec at M0.8, 15000 ft.
F-35 is already among the top performer even we don’t consider the turn rate increase from M0.9 to M0.8. We haven’t considered the fact that 2 aim-120s are definitely heavier than 2 infrared missiles neither.
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It is safe to assume that F-35 could out-turn its major adversary, Su-27/30 families, for the fuel of same afterburner duration
Last edited by gta4 on 23 Mar 2017, 21:57, edited 1 time in total.
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sferrin

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Unread post23 Mar 2017, 18:48

"Since all turbofan engines have similar SFC of about 1.9, fuel consumption is proportional to total thrust."

Really? So a TF30 is as fuel efficient as an F414?
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Unread post23 Mar 2017, 22:00

sferrin wrote:"Since all turbofan engines have similar SFC of about 1.9, fuel consumption is proportional to total thrust."

Really? So a TF30 is as fuel efficient as an F414?


SFC for some typical engines:

AL-31F: 1.96
F110-GE129: 1.9
F110-GE400:1.98
F100-PW229:2.05
RD-33: 1.85

TF30 is close to turbojet. Its SFC is about 2.78. https://www.flightglobal.com/FlightPDFA ... 200882.PDF
F135 is a high by-pass ratio engine and is close to F110 instead of TF30.
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garrya

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Unread post24 Mar 2017, 06:56

gta4 wrote:
With 3000 kg weight reduction, the total flying weight gives a factor of 16000/19000=0.84. Since the normal load factor is inverse-proportional to weight, the turn rate will be increased by 19%. Which is 13.1 deg/sec at M0.8, 15000 ft.

We are talking about sustain turn rate here so i dont think it is that simple. Sustained G is affected by drag more than weight AFAIK
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sprstdlyscottsmn

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Unread post24 Mar 2017, 08:15

garrya wrote:
gta4 wrote:
With 3000 kg weight reduction, the total flying weight gives a factor of 16000/19000=0.84. Since the normal load factor is inverse-proportional to weight, the turn rate will be increased by 19%. Which is 13.1 deg/sec at M0.8, 15000 ft.

We are talking about sustain turn rate here so i dont think it is that simple. Sustained G is affected by drag more than weight AFAIK

So I've been thinking about this too. To sustain the speed the thrust and drag must equal. So if the speed is constant we can ignore form drag here. That means induced drag, the drag due to lift must also be the same. In order for the drag due to lift to be the same then the lift coefficient has to be the same. This means the lift force is the same. This means the G does indeed scale directly with weight. The only assumption that needs to be made is that the C.G. does not move significantly as that may impact trim drag.
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Unread post24 Mar 2017, 14:49

sprstdlyscottsmn wrote:
garrya wrote:
gta4 wrote:
With 3000 kg weight reduction, the total flying weight gives a factor of 16000/19000=0.84. Since the normal load factor is inverse-proportional to weight, the turn rate will be increased by 19%. Which is 13.1 deg/sec at M0.8, 15000 ft.

We are talking about sustain turn rate here so i dont think it is that simple. Sustained G is affected by drag more than weight AFAIK

So I've been thinking about this too. To sustain the speed the thrust and drag must equal. So if the speed is constant we can ignore form drag here. That means induced drag, the drag due to lift must also be the same. In order for the drag due to lift to be the same then the lift coefficient has to be the same. This means the lift force is the same. This means the G does indeed scale directly with weight. The only assumption that needs to be made is that the C.G. does not move significantly as that may impact trim drag.


Wouldn't that mean that it's the sustained G that increases by 19%, not the sustained turn rate? What's the turn rate for 5.89G at that speed?
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castlebravo

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Unread post24 Mar 2017, 15:20

One other consideration is that the F-35 KPP for sustained turn rate is for 0.8 Mach, but that chart of other fighters it at 0.9 Mach. Also, the F-35A KPP is assuming an end of life engine with less thrust.

I think the sustained turn performance is close enough to other fighters that in real combat, any advantage another jet might have in the metric is going to be a fourth-order factor behind the F-35's non-kinematic qualities, the skill of the pilot, and kinematic metrics other than STR which are more important in modern combat.
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Unread post24 Mar 2017, 15:43

castlebravo, you are very much correct. The SA and networking advantage more than nullify any marginal STR issue.

mikemag, 5.89G at 0.8M at 15,000ft gives 12.7 degrees per second for a turn where neither speed or altitude are lost.
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castlebravo

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Unread post24 Mar 2017, 22:03

One thing I just noticed is that in the OP's chart, the F-16A's STR and T/W is inconsistent with the following image of an F-16A with 50% fuel. Does anyone have any details on the exact "combat weight" for all the aircraft in the OP's image? I think some of them including the F-16A might have had significantly less than 50% fuel.

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basher54321

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Unread post24 Mar 2017, 23:29

Ha - well deduct 2000 lbs of fuel and use that 23K static thrust rating.

or

deduct 1000 lbs fuel and use the flight manual static thrust rating (25K)

or

deduct 1000 lbs to get Block 5 weight and use the flight manual static thrust rating.

:P
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castlebravo

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Unread post25 Mar 2017, 02:37

Well, since the block 5 F-16A was such a hotrod and isn't really equivalent to anything in use today, I took a stab at comparing the Block 50 to the F-35A instead. Rather than adjust the F-35A into something that matches the 22,000lb 15k' E-M chart in the HAF manual supplement, I'll add weight to the F-16C instead.

28.6k thrust / 43k thrust x 60% fuel x 18,498lbs = 7,381 lbs of magic gas that somehow fits into a Viper (assume a spherical cow)

20,200 lbs basic weight
700lbs for a pair of slammers
30 lbs additional to swap AIM-9 launchers to AIM-120
287lbs for 20mm
48 lbs for chaff/flares

28,646 lbs / 22,000 lbs gross weight = a 1.3/1 ratio

13.8°/s now becomes 10.6°/s for the F-16C compared to the F-35A's 11°/s
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inst

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Unread post25 Mar 2017, 03:21

castlebravo wrote:One other consideration is that the F-35 KPP for sustained turn rate is for 0.8 Mach, but that chart of other fighters it at 0.9 Mach. Also, the F-35A KPP is assuming an end of life engine with less thrust.

I think the sustained turn performance is close enough to other fighters that in real combat, any advantage another jet might have in the metric is going to be a fourth-order factor behind the F-35's non-kinematic qualities, the skill of the pilot, and kinematic metrics other than STR which are more important in modern combat.


I agree here; I don't see the F-35's STR as important to its function, modern HOBS missiles mean that dogfighting is more likely to be a series of attritional exchanges, barring countermeasure systems. The only real combat scenario where the F-35's STR would come into play would be if all combatants ran out of missiles, and for whatever silly reason, they opted to go guns only instead of returning to base. The more common STR scenario, on the other hand, would be in training exercises, not real operations. Consequently, I don't see the F-35 as putting emphasis on STR; the 28-degree pedal turn, for instance, is an example of exceptional ITR, which matters more since the higher ITR is needed to dodge modern dogfight missiles with more than 45G of maneuverability. The F-35's advantages, in contrast, lie in its ITR, its EODAS system, stealth, and its HOBS missiles.

The discussion of sustained maneuverability matters more in the context of the J-20 in whether it is designed as an air superiority craft or as an interceptor; the latter is not intended to dogfight (even if modern dogfights are often attritional exchanges) and thus does not care about STR.

As to the OP's argument, here's a graph of the Su-27 and F-15's STR. At the same level the OP is inflating the F-35 to 13 degree/sec, the Su-27 and F-15 are showing slightly over 16 deg / sec STR turn rates. As I mentioned earlier; it doesn't matter, in a realistic engagement the Su-27s would be wiped out by AIM-120s from extended range.

https://forums.eagle.ru/showthread.php?t=159251
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inst

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Unread post25 Mar 2017, 11:00

Your calculation is overly generous, which is just short of being dead wrong. The reason 60% fuel is used is because you use 40% fuel to get there, 40% fuel to get back, which leaves you with 20% fuel to do the air combat. For instance, where did you get the 3000 kg figure?

I suppose I could redo the math, but I'm lazy and we can just plain assume the F-35 has 40% fuel capacity, instead of 60% fuel capacity. That leaves the F-35 with 10% less weight, or we go up from 10.9 to 12 deg / sec. And I'm being generous here; I'm using 5G on the F-35A instead of 4.6G, which gives us about 10 degrees / second or 11 deg / sec with 40% fuel capacity.
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castlebravo

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Unread post25 Mar 2017, 15:21

inst wrote:Your calculation is overly generous, which is just short of being dead wrong. The reason 60% fuel is used is because you use 40% fuel to get there, 40% fuel to get back, which leaves you with 20% fuel to do the air combat. For instance, where did you get the 3000 kg figure?



60% fuel is a realistic combat weight for an F-35A, so no argument there, but 50% fuel for Viper isn't. In reality, a Viper pilot flying an air to air mission at the same range from home plate as the F-35 in your example would be at 100% internal fuel at the start of the fight after ditching his EFTs.

60% of the F-35's fuel capacity is ~5,000kg (this is the fuel weight used in the F-35's demonstrated STR), and ~2,000kg is the fuel loaded needed to roughly match the same combat endurance as a MiG-29 at 50% fuel, so ~3,000kg is the weight difference between an F-35 pulling 4.95g at 60% fuel and an F-35 with a combat weight equivalent to the MiG-29 in the STR vs T/W graph.

A similar conversion must have been done by the DoD when they wrote the F-35 KPPs since the requirement for STR is to be Viper-like, but it was still only ~5g at 15k. A 50% fuel F-16C can sustain a lot more than 5G at angels 15. Even an F-16C with 60% internal fuel and a pair of slammers would be around ~12°/s and ~5.5g STR.
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Unread post25 Mar 2017, 15:53

inst wrote:here's a graph of the Su-27 and F-15's STR. At the same level the OP is inflating the F-35 to 13 degree/sec, the Su-27 and F-15 are showing slightly over 16 deg / sec STR turn rates. As I mentioned earlier; it doesn't matter, in a realistic engagement the Su-27s would be wiped out by AIM-120s from extended range.

https://forums.eagle.ru/showthread.php?t=159251


Cheater, you are cheating again: comparing turn rate at 3000 m with that at 4527 m (15000 ft)!

Trt not to fool us when you can not out-smart us.
Last edited by gta4 on 25 Mar 2017, 17:16, edited 1 time in total.
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