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

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:

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):

The Maneuver Weight is defined as follows (60% internal fuel, about 5000kg):

(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 Theorem 1:

Theorem 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.

Proof: The fuel consumption is proportional to engine thrust:
Fuel Consumption=SFC*Thrust*Time.
Modern fighter jet engines all have a 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 theorem 2:

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

Proof: 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 theorem.
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 theorem 3:

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

Proof: 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 proof.

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

Proof: 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:

Theorem 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 theorem. 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:

—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:

—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. Theorem 1~3 yields:

—equation 3
The time required to accelerate from 696km/h-1102km/h at 1000m is given by:

–equation 4
The average acceleration in the interval [696-1102km/h] is more than 11.06m/s2. According to Theorem 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.

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:

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

We already know F-35 has some special turning technique which delivers an astonishing 28deg/sec sustained turn:

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.