…posted in the india-russia-israel thread
Here are some of my calculations on the F7 performance which is based on the numbers you have provided. Lets keep this as a MiG 21 / F7 PG comparison thread.
This could turn out to be interesting, and members are requested to keep on topic – that means post here with your opinions only if you can back it up with numbers and actual references – NO could have / would have / maybe / speculation posts please!
First the post I am replying to
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You’re free to make your calculation here. I got these figures from CFF of the ACIG forum, and this is one thorough guy when it comes to research (I know since I also argued with him before.)
As for F-7MG:
Aspect Ratio 2.76
Wings: 24.88 m^2
Span: 8.32m
Length: 14.66m
Height 4.1m
Weight: 5292kg empty; 7540kg normal take-off; 9100kg max take-off
Power: WP-13F, max thrust 6600kg, military thrust 4500kg
Max Speed: 2 Mach
Max level speed: 648 kt
Climb rate: 11700 m/min.
Service ceiling: 17500m
Max instantaneous turn rate: 25.2 degree/s
Sustained turn rate at 1000m: 16 degree/s
Sustained turn rate at 5000m: 11 degree/s
Sustained turn rate at 8000m: 8 degree/s
Sustained turn rate at sea level, Mach 0.7: 14.7 degree/s
Sustained turn rate at 5000m, Mach 0.8: 9.5 degree/s
g limits: +8/-3
Fuel: Internal 2041kg
Operational radius: air superiority with two AAMs and three 500L drop tanks, 5 min with afterburner: 850km
Operational radius: air to ground with two bombs and two 500 L drop tanks: 550km.
Ferry range: 2200km
Weapon: 4 hardpoints. 2x 30mm cannon with 126 rounds.
Could carry PL-2, PL-5, PL-7, or PL-9 AAM, or rocket pod,
bombs.
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As you can see, you vastly underestimate the engine power, to assume that the F7’s TW is only achieved at 6,200kg, where in fact, the engine produces 6,600kg. At 6,200kg, the TW is actually in excess of unity. With 6,600kg thrust engine, .97 TW can be achieved roughly at just under 7,000kg. Note that plane only has an empty weight of 5,200kg.
Sustained turns are not very good however. This means that the plane is very good in making quick sudden maneuvers, but is not good in sustaining them, where the sustained performance falls roughly back to a standard MiG-21.
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Now my reply:
lets do the t/w calculations for the F7 PG again:
t/w ratio = engine thrust / weight of plane (at any particular instance)
We will use newtons N as our standard unit of measurement:
Weights: kg (N)
Empty: 5,292 (51,897)
Normal t/o: 7,540 (73,942)
MTOW: 9,100 (89,241)
Engine Thrust from the CATIC.com.cn site referenced by you in the earlier thread.
Military: 42,700 N = 4,354 kgf
Afterburner: 65,900 N = 6,719 kgf
Now lets calculate the t/w at normal t/o and MTOW using both military thrust and with afterburners
Normal t/o weight: 73,942 N
t/w with military thrust = 42,700 N / 73,942 N = .58
t/w with afterburner = 65,900 N / 73,942 N = .89
Therefore your claim of .97 with normal weight is ridiculous. The equation does not balance. Its a very simple calculation. Use the figures from the official CATIC site provided by you, and you will arrive at the above.
A t/w of .97 can be acheived by the plane PROVIDED the weight is lowered so lets calculate at what weight it will achieve .97 t/w ratio
with military thrust 42,700 N
we know that
t/w = Thrust / weight (N)
t/w x weight (N) = thrust
weight (N) = Thrust / tw
weight (N) = 42,700 (N) / .97
weight (N) = 44,021 N
weight (kg) = 4,489 kg = less than the empty weight (5,292 kg) of the aircraft!!!
Therefore we can conclude the F7 PG can NEVER achieve a t/w of .97 with military thrust
with afterburner thrust 65,900 N
we know that
t/w = Thrust / weight (N)
t/w x weight (N) = thrust
weight (N) = Thrust / tw
weight (N) = 65,900 (N) / .97
weight (N) = 67,938 N
weight (kg) = 6,928 kg
Based on the above it can be concluded to achieve a t/w of .97 a F7 would have to weigh ~6,900 kg. and it has to use its afterburners to do so. Please show to me how it would acheive a t/w of > 1. Don’t speculate; calculate and let us all know.
Lets move to the other calculations:
>Max instantaneous turn rate: 25.2 degree/s
>Sustained turn rate at 1000m: 16 degree/s
>Sustained turn rate at 5000m: 11 degree/s
>Sustained turn rate at 8000m: 8 degree/s
>Sustained turn rate at sea level, Mach 0.7: 14.7 degree/s
>Sustained turn rate at 5000m, Mach 0.8: 9.5 degree/s
Turn rates are a function of bank angle and speed, which determine the raidus of turn and therefrom the turn rate. The g loads effecting the plane can also be calculated. So lets work out the various radii and from the STRs. We will also claculate the g forces acting on the plane at those turn rates…
The formula to calculate turn radius is V^2 / g x tan(b)
V = velocity
g = 9.80665 m/sec^2
tan(b) = tan of the bank angle
b = bank angle
From this turn radius we can calculate the time required to turn one degree and thereby calculate the turn rate per second.
t = pi V / 180 x g x tan(b) = time for 1 degree of turn
Since we have numbers for 5,000 m at mach 0.8 we will work on this first. => >Sustained turn rate at 5000m, Mach 0.8: 9.5 degree/s
For this excercise, we will use the ICAO standard atmosphere table for 5,000 m for some of our calculations.
Altitude (m): 5,000
Air Density (kg/m^3): 0.7361
Pressure (N/m^2): 54020
Temprature K: 255.65
Speed of Sound (m/s): 320.53
Kinematic viscosity: 2.212
V = mach 0.8 = 320.53 m/s x 0.8 = 256.42 m/s
9.5 deg / sec = 0.1053 sec / deg
First we calculate the angle of bank using the formula for turn rate:
0.1053 = 3.1416 x 256.42 / 180 x 9.80665 x tan(b)
tan(b) x 0.1053 = 3.1416 x 256.42 / 180 x 9.80665
tan(b) = 3.1416 x 256.42 / 180 x 9.80665 x .1053
tan(b) = 805.57 / 185.87
tan(b) = 4.33
THEREFORE at 5,000 m at mach 0.8 the angle of bank = 77 deg
Next we calculate the turn radius
r = V^2 / g tan(b)
r = 256.42^2 / 9.80665 x 4.33
r = 65,751.22 / 42.46
r = 1,548 m
So at 5000 m at ~77 deg bank at mach 0.8 the F7 PG has a turn radius of ~1.5 km
Doing the same calculations for mach 0.7 at sea level (>Sustained turn rate at sea level, Mach 0.7: 14.7 degree/s). we arrive at the following using the ICAO numbers as under:
Altitude (m): 0
Density (kg/m^3): 1.2250
Pressure (N/m^2): 101325
Temprature K: 288.15
Speed of Sound (m/s): 340.29
Kinematic viscosity: 1.461
tan(b) = 6.23
angle of bank = 80.88 deg (have to use the decimals here and you will see the significance of the decimal places in the g load calculations later)
r = 928.69 m = 0.93 km
The Load Factor in a turn (g load) is calculated using the formula
LF = 1 / cos(b)
Therefore we get the following g loads for the F7PG
0.8mach @ 5,000 m = 1 / cos(77) = 1 / 0.2250 = ~4.44 g
AT 80 deg bank angle
0.7mach @ 0 m = 1 / cos(80) = 1 / 0.1736 = ~5.76 g
BUT AT 80.88 deg bank
0.7 mach @ 0 m = 1 / cos(80.88) = 1 / 0.1585 = ~6.31g
As you can see less than a degree of difference increases g load significantly at those bank angles. Now do you understand why I take the usefullness of the claimed ITR of 22.5 deg / sec for the F7 with a pinch of salt? At 0.8 mach @ 5,000 m altitude it is already pulling 4.44 gs for a 9.5 deg/sec turn rate!
For your claimed ITR we can confirm the actual g load, turn radius and turn rate if you provide numbers such as altitude, velocity and turn rate (which we already have @ a claimed 22.5 deg/sec).
We can also calculate the actual g load on the plane at the instance of the turn @ 22.5 deg based on some assumptions
We will use the ICAO table for 5,000m (numbers above):
We will also assume the following
velocity to be 0.8 mach = 320.53 m/s x 0.8 = 256.42 m/s
22.5 deg / sec = 0.0444 sec / deg
Therefore the bank angle and g load numbers are:
tan(b) = 10.28
angle of bank = 84.44 deg
g load = 1 / cos(84.44) = 10.32 g!!!
This g load is clearly beyond the ability of the plane. Instantaneous turn or not. First it is more than 2 gs above its rated limit – which by the way are not meant to be reached often. Second how will a pilot stand 10.32 g and stay concious? It is too much to expect every pilot to remain concious at this g load.
A-ha you say it is not meant to be carried out at mach 0.8! ITR is only quoted for lower velocities – so lets work backwards based on the maximum g load of the plane and calculate the velocity to carry out a 22.5 deg turn at the max g rating of the F7 = 8 g
We will still use the same numbers
22.5 deg / sec = 0.0444 sec / deg
also we will keep the g load @ 8 g
Working backwards we calculate the angle of bank based on 8 g
g = 1 / cos(b)
cos(b) = 1 / g
cos(b) = 1 / 8
cos(b) = 0.125
bank angle = 82.81 deg
tan(b) = 7.94
Calculate the velocity at which this turn rate can be achieved using the turn rate formula:
t = pi V / 180 x g x tan(b)
0.0444 = 3.1416 x V / 180 x 9.80665 x 7.94
0.0444 x 180 x 9.80665 x 7.94 = 3.1416 x V
V = 0.0444 x 180 x 9.80665 x 7.94 / 3.1416
V = 622.30 / 3.1416
V = 198.06 m/s = 713 km/hr = ~ mach 0.61
So what we have here is a ITR of 22.5 deg @ 8 g @ 0.61 mach.
Now I can ask the question – How useful is the ITR of 22.5 deg/sec in a dogfight when your velocity HAS TO BE mach 0.61? => certainly slower than your opponents, unless you want to assume your opponent is going to slow down and wait for you to do your ITR of 22.5 deg/sec? 😀
I think your opponents will certainly be on their burners going as fast as they can trying to gain and retain the maximum energy. The F7 which slows down to do its ITRs is going be a juicy target in someones HUD. Even if it does not get shot down, it has lost its momemtum, which has to be built up again by going on burners. It is still going to be a target till it regains its energy.
Manufacturer quoted specification only show off the best in a/c., and depending on the state of the a/c. those numbers can be higher or lower than the opponent.
Since you also claim these numbers are better than the MiG 21. Please show us the data points you have used for your comparisons. Preferably the 21bis since this is the closest to the F7 PG.
PS: Sorry for the long-long post…
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May 10, 2002 at 10:15 pm