Interesting problem Vara...
Let's put it in nbr
let's say that your speed is 400kph.
if the guy in your 6 is 20kph faster it equate to 5% diff in frd speed.
When you'll start to maneuver E will be the most important thing.
With 95% less speed, your Kin E would be 0.95*0.95=0.90 less than that of the other guy (kin E = 0.5 x m x Speed²) with m the mass of the planes.
No you'll say to me that your hurri is heavier than a 109 thx to her heavy plywood and tube structure and that masse at speed equate Kin E.
humm let's think about that...
Mass = lift at level flight. hence the more mass to carry, the more lift you'll have to produce. As lift = drag in the sense that the more lift at a given speed generate more drag, hence you'll see that given you are at the same speed but with an heavier mass, your plane will generate more drag. And then less E. So let's put this aside as it complicate the prob without giving any advantage o your hurri (but let's keep it in mind when we will need to firce him to bypass our plane
).
So your hurri has 10% ((1.-0.90)*100 in %) less kin E than the 109 in your 6 and you generate more drag. You'll understand now that even that "little" 20kph disadvantage is uncomfortable whan you'll start to move the plane in the vertical plane or in a high G situation.
Why in high G ?
let's have a look
let's say you pull hard in the vertical at 5g for a brief moment. In a 5G verical turn (not a 360, or a full loop - you'll soon see that it is impossible), your plane will have a corresponding mass of 5 time heavier. Hence it will need 5 time more lift to stay on the same horizontal plane with your wings level.
hence your drag will increase, let's say linearly by a factor of 5.
What will counteract the drag in that situation ? You'd say so : engine power and you E state.
How your Engine power will contribute ?
Drag increase basically with the square of the speed. Has the power necessary to sustain a given speed is the product of the forward speed AND the force opposite to the speed vector, the power of the drag forces will be :
the fraction of the mass x by the G x by the forward speed.
As you see, basically pulling 5 G increase 5 time your drag and necessitate 5 time more power than in a level flight
Let's say that at 400kph your hurri will need 80% of it's eng power. Hence at 5g it would have need 0.8 x 5 = 4 time more power. A total amount that your engine can not produce. This is were your kin E will start to play it's role : being a source of E to keep your global E state approximatively cte (your hurri is not a conservative system !).
How will that kin E play in this situation ?
humm let's have a look. Kin E is a variable of the square of the speed. It has to compensate for the lack of the power of your engine. But the power is the product of a force by a forward speed and the drag force itself evolve in function of the square of the speed !
You'd see : speed x speed for the drag and drag x speed for the power means speed x speed x speed for he drag power
Hence your kin E will proportionally contribute only as the square root of itself to compensate for the lack of power.... that is 3 time what your eng can deliver at it's peak power (4 -1 x engPower)
Humm let's put it together :
you have 95% of the 109 speed
hence you'll get only 0.95x0.95x0.95 = 85 % of what the 109 will have for power to balance the 5G pull up
during the pul up your 90% less Kin E will transform some E to help you to sustain your forward speed acting only proportionally to the root square of the speed (90%) = 0.95 = 95%
SO if we sum the overall lack of power to sustain your frwd speed we will have (1-0.85) + 3x(1-0.95) = 30% !
30% less power at 5 G means that your speed (considering that drag btw your hurri and the 109 are equal - what we hve seen above is not !) will drop by P = DragForce x Speed => Speed = P/DragForce what means that the drop will be more than 30% of that of the 109 (more since your plane is heavier and have to generate more lift hence face more drag) !
And this is only for the first move !
As you've seen (if you are still there) 5% diff in forward speed means a lot less of manoevrability in term of E and speed after each move (30%).
Let's imagine that after your brief 5G move your speed has dropped down to 300kph. The 109 will still hev a speed of 300 x (1+0.30) = 390 kph hence a 90kph difference vs 20 at the begining of the move.
~S!
PS : Humm humm now you have said that he was 50m behind you (what is really close for two fighters). 50m at 390kph means (390-300)/3.6 = 25 m/s means that the 109 pilot has 2 seconds to do something to stay behind you. Holly Cow ! you are just ready to make an evasive break
PPS: I kown that Vara know all about the above. I just took his remark as a
pretexte for a short refreshing cursus