Proof: F-35 can out-accelerate Su-27/35 in subsonic region

The F-35 compared with other modern jets.
  • Author
  • Message
Offline

gta4

Forum Veteran

Forum Veteran

  • Posts: 828
  • Joined: 17 Oct 2010, 19:10

Unread post21 Nov 2016, 03:05

This is from the note of an aerospace engineer. I know there are other engineers in this forum so feel free to review it.

We have read from a TsAGI report that a Su-27 could accelerate from 600km/h to 1100km/h in 15 seconds, on 1000m, with 18920kg flying weight:

https://s14.postimg.org/nkknvh56p/Su_27 ... ration.jpg

The average acceleration is 9.25m/s2 from 600-1100km/h at 1000m.
We have also read from F-35 240-4.2 configuration report that F-35 could accelerate from 0.6-0.95 mach (696km/h-1102km/h) in 17.9 seconds, under Maneuver Weight at 15000 ft (4527 m):

download/file.php?id=18000&mode=view

The Maneuver Weight is defined as follows (60% internal fuel, about 5000kg):
https://s15.postimg.org/n9x5n6wyz/Tactical_MW.jpg

(The 540NM combat radius is almost the radius of full internal fuel with JDAMs loaded at take-off (but launched afterwards). I will prove that later in the appendix)

The question is: how to convert F-35’s performance at 4527m to 1000m under the same standard? Let’s do it.

Calculation standard: Both aircrafts carry the fuel allowance for the same afterburner time.
This standard is justified as follows: If we adopt some conventional standards, such as 50% internal fuel, this will be unfair for the aircraft with very high internal fuel or low fuel consumption. We can obtain that under 18920kg flying weight, Su-27 has only about 2000kg (4400lb) internal fuel, because a Su-27 with 5270kg fuel, 2xR-27 and 2xR-73 missiles, has a total weight of 23430 kg:
http://www.sukhoi.org/eng/planes/military/su27sk/lth/
So we need to know how much fuel is needed for F35 to have the same afterburner time as a Su-27 with 2000kg fuel. This leads to Prerequisite 1:

Prerequisite 1: It’s fair to let F-35 carry only 1560kg fuel, for allowance of the same afterburner time as a Su-27 with 2000kg fuel.

Explanation: The fuel consumption is proportional to engine thrust:
Fuel Consumption=SFC*Thrust*Time.
Modern fighter jet engines all have a SFC (specific fuel consumption) of about 1.9 (this approximation could be easily verified with published engine data), therefore, since the afterburner thrust of a F135 is 78% of that of two AL-31s, the fuel consumption of F135 is also 78% of that of two AL-31s. The result is 2000kg*78%=1560kg.
Compared to the Maneuver Weight (19000 kg) of F-35, this new fuel standard yeilds a total flying weight of 15600kg, which is a 18% reduction. (15600/19000=82%)
We are facing a new problem: How is engine thrust at 1000m compared to that at 4572m? This leads to Prerequisite 2:

Prerequisite 2: At a given airspeed (subsonic) and from medium to low altitude, engine thrust is proportional to air density.

Explanation: Theoretically, this is because the volumetric flow rate is almost constant at a given speed (at subsonic) and the air density is to be multiplexed with. This could be easily verified with published engine data. This is a very good approximation. You are welcome to use published data (i.e., RD-33 or AL-31 engine performance curves) to verify this Prerequisite.
So, at a given airspeed (for instance, 600km/h), the thrust at 1000m is 1.44 times as big as that at 4572m, because the ratio of air density is 1.44.
We know the comparison of thrust. What about the drag? This leads to Prerequisite 3:

Prerequisite 3: At a given airspeed (subsonic), the drag at 1000m is less than 1.44 times as big as that at 4527m.

Explanation: The drag is given by:
Drag=½*Drag Coefficient*air density*speed^2*wing area.
Let's compare 1000m and 4527m. The speed is fixed because it is given, and wing area remains unchanged. The air density gives a factor of 1.44. The drag coefficient is almost the same but slightly smaller, because the zero lift drag coefficient is identical, but the lift coefficient required to maintain level flight is smaller due to higher air density, which yields a smaller induced drag coefficient. This concludes the explanation.

Prerequisite 4: At subsonic acceleration, the speed-time curve is a convex function, or in other words, the faster you fly, the harder you accelerate.

Explanation: Theoretically, this is because the drag increases so rapidly as you accelerate. This could also be easily verified with published data of some aircrafts, such as:
http://forum.keypublishing.com/attachme ... 1389782706
http://forum.keypublishing.com/attachme ... 1419959707

Prerequisite 4 tells us that, If we have two aircrafts A and B, and aircraft B has the same or better average acceleration in a faster speed interval, then it’s safe to say B can out-accelerate A.

Thank you for your patience for reading through the theories. Now here comes the essential part: performance conversion from 4527m to 1000m.

Math notation:
We note a the acceleration, v the air speed, T the thrust, D the drag, m the total mass of aircraft. At a given altitude, the acceleration, the thrust and the drag are not constant, but functions of speed, which is equivalent to write a=a(v), T=T(v) and D=D(v). Their relationship is given by:
eq1.png

---equation 1
At 4527m and maneuver weight, F-35 at maneuver weight, accelerates from 0.6-0.95 mach (696km/h-1102km/h) in 17.9 seconds (by injecting equation 1 into it):
eq2.png

---equation 2
At 1000m, notations are changed: we note a1000(v) the acceleration, T1000(v) the thrust, D1000(v) the drag, and m1000 the mass. Prerequisite 1~3 yields:
eq3.png

---equation 3
The time required to accelerate from 696km/h-1102km/h at 1000m is given by (by injecting equation 2 and 3 into it):
eq4.png

---equation 4
Equation 4 tells us that the time required to accelerate is less than 10.19 s. The average acceleration in the interval [696-1102km/h] is more than 11.06m/s2. According to Prequisite 4, it is safe to conclude that F-35 can out-accelerate Su-27 in subsonic region, with a significant margin of more than 19.5%.
Sukhoi has been advertising Su-35’s acceleration for a while. Su-35 has about 8% increases over Su-27, which is still inferior to F-35.
We already know F-35 has some special turning technique which delivers an astonishing 28deg/sec sustained turn:
https://s15.postimg.org/awie5l6aj/35turn_rate.jpg

viewtopic.php?f=22&t=52503&p=356274&sid=b041569202d3b491efee6d62de1e0d1a#p356274
With this turning performance, and coupled with strong subsonic acceleration to recover energy, F-35 will become a potent dogfighter once CLAW is opened up and maneuver restrictions are removed.

Appendix:
Why the 540NM combat radius is almost the radius of full internal fuel with JDAMs loaded at take-off (but launched afterwards)?
F-35 with JSMs and 2 AMRAAMs loaded when taking off, JSMs released during the mission, has a combat radius of 610NM with internal fuel:
download/file.php?id=22482&mode=view
The JSM is very light and is less than 908 lbs:
http://www.dtic.mil/ndia/2014PSAR/albright.pdf
SO, with the much heavier 2000lbs JDAM loaded, the combat radius will drop significantly. 540NM is a reasonable figure.
Last edited by gta4 on 22 Nov 2016, 02:53, edited 6 times in total.
Offline

garrya

Forum Veteran

Forum Veteran

  • Posts: 715
  • Joined: 25 Dec 2015, 12:43

Unread post21 Nov 2016, 04:14

Interesting thread , too bad i dont know enough about aerodynamic to contribute :mrgreen:
Offline

eloise

Elite 1K

Elite 1K

  • Posts: 1683
  • Joined: 27 Mar 2015, 16:05

Unread post21 Nov 2016, 13:21

would you mind if i took this to Keypublishing?
Offline
User avatar

sferrin

Elite 5K

Elite 5K

  • Posts: 5293
  • Joined: 22 Jul 2005, 03:23

Unread post21 Nov 2016, 14:20

eloise wrote:would you mind if i took this to Keypublishing?


Probably get more useful answers by taking it to the playground at the local elementary school.
"There I was. . ."
Online

sprstdlyscottsmn

Elite 4K

Elite 4K

  • Posts: 4228
  • Joined: 10 Mar 2006, 01:24
  • Location: Phoenix, Az

Unread post21 Nov 2016, 14:30

That was not a bad "first-order approximation". It also lines up with much of the anecdotal evidence of "recovers airspeed much better than an F-16", "like flying an F/A-18 with (a turbo)/(four engines)".
"Spurts"

-Pilot
-Aerospace Engineer
-Army Medic
-FMS Systems Engineer
Offline

gta4

Forum Veteran

Forum Veteran

  • Posts: 828
  • Joined: 17 Oct 2010, 19:10

Unread post21 Nov 2016, 17:01

eloise wrote:would you mind if i took this to Keypublishing?


It seems the author doesn't mind. You can do it.

Don't forget the equations. They are in images.
Offline

garrya

Forum Veteran

Forum Veteran

  • Posts: 715
  • Joined: 25 Dec 2015, 12:43

Unread post21 Nov 2016, 18:25

gta4 wrote:Don't forget the equations. They are in images.

Who made all these ? he must be very knowledgeable
Offline

gta4

Forum Veteran

Forum Veteran

  • Posts: 828
  • Joined: 17 Oct 2010, 19:10

Unread post21 Nov 2016, 22:18

garrya wrote:
gta4 wrote:Don't forget the equations. They are in images.

Who made all these ? he must be very knowledgeable


A geek from GeorgTech
Offline

garrya

Forum Veteran

Forum Veteran

  • Posts: 715
  • Joined: 25 Dec 2015, 12:43

Unread post21 Nov 2016, 22:47

gta4 wrote:
A geek from GeorgTech

Can you tell him to do Rafale and Typhoon too ? lol
Offline
User avatar

steve2267

Elite 2K

Elite 2K

  • Posts: 2132
  • Joined: 12 Jun 2016, 17:36

Unread post22 Nov 2016, 00:16

gta4 wrote:Prerequisite 3: At a given airspeed (subsonic), the drag at 1000m is less than 1.44 times as big as that at 4527m.

Explanation: The drag is given by:
Drag=½*Drag Coefficient*air density*speed^2*wing area.
Let's compare 1000m and 4527m. The speed is fixed because it is given, and wing area remains unchanged. The air density gives a factor of 1.44. The drag coefficient is almost the same but slightly smaller, because the zero lift drag coefficient is identical, but the lift required to maintain level flight is smaller due to higher air density, which yields a smaller induced drag coefficient. This concludes the explanation.


My only critique is that unless the author has figured a way to warp the gravity field, the weight of the two aircraft will be (essentially) identical (neglecting the extremely minute differences from being slightly further from the center of mass of the earth), requiring the lift to be the same. Does that not mean the induced drag coefficient will be the same?

Nevertheless, I do not think this negates the overall analysis.
Last edited by steve2267 on 22 Nov 2016, 00:18, edited 1 time in total.
Take an F-16, stir in A-7, dollop of F-117, gob of F-22, dash of F/A-18, sprinkle with AV-8B, stir well + bake. Whaddya get? F-35.
Offline
User avatar

count_to_10

Elite 3K

Elite 3K

  • Posts: 3281
  • Joined: 10 Mar 2012, 15:38

Unread post22 Nov 2016, 00:17

We are talking about wet thrust, right?
Shouldn't that mean that specific thrust won't be at all related to the dry specific thrusts reported?
Einstein got it backward: one cannot prevent a war without preparing for it.

Uncertainty: Learn it, love it, live it.
Offline

garrya

Forum Veteran

Forum Veteran

  • Posts: 715
  • Joined: 25 Dec 2015, 12:43

Unread post22 Nov 2016, 00:57

steve2267 wrote:
My only critique is that unless the author has figured a way to warp the gravity field, the weight of the two aircraft will be (essentially) identical (neglecting the extremely minute differences from being slightly further from the center of mass of the earth), requiring the lift to be the same. Does that not mean the induced drag coefficient will be the same?

Nevertheless, I do not think this negates the overall analysis.

He means lift coefficients. Thinner air mean higher AoA to get adequate CL at the same speed, which mean higher Cd also.
Offline

gta4

Forum Veteran

Forum Veteran

  • Posts: 828
  • Joined: 17 Oct 2010, 19:10

Unread post22 Nov 2016, 02:44

steve2267 wrote:
gta4 wrote:Prerequisite 3: At a given airspeed (subsonic), the drag at 1000m is less than 1.44 times as big as that at 4527m.

Explanation: The drag is given by:
Drag=½*Drag Coefficient*air density*speed^2*wing area.
Let's compare 1000m and 4527m. The speed is fixed because it is given, and wing area remains unchanged. The air density gives a factor of 1.44. The drag coefficient is almost the same but slightly smaller, because the zero lift drag coefficient is identical, but the lift required to maintain level flight is smaller due to higher air density, which yields a smaller induced drag coefficient. This concludes the explanation.


My only critique is that unless the author has figured a way to warp the gravity field, the weight of the two aircraft will be (essentially) identical (neglecting the extremely minute differences from being slightly further from the center of mass of the earth), requiring the lift to be the same. Does that not mean the induced drag coefficient will be the same?

Nevertheless, I do not think this negates the overall analysis.

Sorry that was my typo. I should write "lift coefficient" instead. Even though the weight of aircraft remains the same, the air density has changed. So the required lift coefficient has changed. So the induced drag coefficient has changed.
Last edited by gta4 on 22 Nov 2016, 02:57, edited 2 times in total.
Offline

gta4

Forum Veteran

Forum Veteran

  • Posts: 828
  • Joined: 17 Oct 2010, 19:10

Unread post22 Nov 2016, 02:47

count_to_10 wrote:We are talking about wet thrust, right?
Shouldn't that mean that specific thrust won't be at all related to the dry specific thrusts reported?


But SFC are still almost indentical.

For all fighter jet engines:

SFC for military power are all around 0.8.

SFC for afterburner are all around 1.9.
Offline

garrya

Forum Veteran

Forum Veteran

  • Posts: 715
  • Joined: 25 Dec 2015, 12:43

Unread post22 Nov 2016, 06:05

Can someone confirm this :
“A set of CFTs carries 50 percent more fuel than the centerline external fuel tank, but has only 12 percent of the drag.” The CFTs are designed for the full F-16 flight envelope – up to 9 g’s, maximum angle of attack and sideslip and maximum roll rate.

http://defense-update.com/products/c/F-16-CFT.htm

Centerline tank has drag index of 18 when fully loaded , centerline pylon has drag index of 7

Image

So drag of CFT is 12*( 18+7)/100 =3 even less drag than a single Aim-120

For visual illustration :
This
Image
is about 7 % more draggy than this
Image
Next

Return to F-35 versus XYZ

Who is online

Users browsing this forum: No registered users and 6 guests