So basically I need to take the continuous line, since in game slats are extending
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The first MiG-29s did not have an adaptive wing, i.e. the LE was produced at the maximum angle. In version 9-15, an adaptive wing was made. For a smoother flow
Spoiler
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Done a relatively quick drag calculation (still need to check everything again, so I will post again tomorrow as in Italy it’s past mid night right now) and figures for chart 6.4 unfortunately are not for 13000kg. AoA needed to sustain 5g at 550kph TAS (500kph IAS) at 2000m for that weight is ~21.5º, which corresponds to around 1.325Cy → 0.44 Cx → Over 19000kg of Drag, which is far above the 11834kg (12720kg * cos 21.5º) tangent thrust value.
So unless the aircraft is missing thrust the value for chart 6.4 seem to be calculated for a mass close to empty weight.
That said the figures for chart 6.14 are right for a mass 13000kg (which matches @ZVO_12_INCH tables)
All of this means that, unless aircraft is missing quite a lot of thrust, the MiG-29 is only slightly underperforming (more detail on this tomorrow).
The reason chart 6.4 and 6.14 match well at higher speeds could very well be because mass matters a lot less at higher speeds
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A bit of a mistranslation here.
Graph 2.13 shows the relation of the lift coefficient of a balanced plane (Cy.bal, y-axis) to the drag coefficient (Cx, x-axis). The plane is clean (no ordinance/fuel tank) at H=0 with flaps retracted (not deployed).
The smooth line is for the leading edge retracted.
The dashed line is for the leading edge deployed (slats) at (delta = -20°).
“убраны” - “retracted” (or “removed”, depending on context)
“выпущены” - “deployed”
Graph 2.14 shows the relation of the aerodynamic quality (K, y-axis) to the lift coefficient of a (not necessarily balanced) plane (C_y, x-axis). The plane is clean at M = 0.6.
“Выпуск носков”, in the middle of the graph, means “Slapts deployment”.
For some reason, the lines flip from smooth to dashed as they intersect, but thankfully the slat angle is outlined for both sections.
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Thanks, all much clearer now.
I will check everything again tomorrow and share it all here.
Ended up using the dashed line (which is better for higher Cl) anyway when i calculated the drag value of over 19000kg.
It seems the slats actually improve lift to drag ratio quite a bit on the MiG-29 beyond a certain AoA (which is why the 2 lines cross in the second chart)
Since this polar diagrams were done at H=0 I will do the same calculations for the overloads on the German manual.
Although Cy to Cx ratio should be independent of altitude
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you only calculate the thrust on the X axis, but for some reason you forget to add a component from the thrust P × Sin AoA to lifting force. Grahf 6.4 mass 13000kg
5G= Yp/G or Y+P×sinAoA/G (Lift+P×sinaoa)/G
You don’t understand much about what I’m talking about
This Thread will remain Civilized… Do not know how many times it needs to be said…
Bulling and harassment will NOT be tolerated whatsoever…
You will either Respect fellow members of the community, or… more warnings will be issued… one member already has some time away
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The book indicates the overload of the high-speed coordinate system
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Height 2000m speed IAS 500km/h. Ny=4 AoA=16.7 y=76 degrees.
Speed of sound=332.53 m/s
TAS=550km/h=550/3.6=152,77 m/s 152,77/332.53=0.459 Mach
We find traction through the graph of Zhukovsky curves through the Get Graph program
P=13065kgs
fip=5 degrees
4=1/G (128167.65sin(16.7+5)+Ya); 4=(46956.12+Ya)/G. 4G=47389.57+Ya=510120=
Ya=510120-47389.57=462730.42
q=pV^2/2 =1.006152,77^2/2=11739.35 Pa
Cya=Ya/qS=462730.42/38*11739.35=1.0372
Cxi=0.1868
Сx0=0.0327
Cx=0.0327+0.1868=0.2195
Xa=cxqS=97917,91
nxa=(Pcos(Aoa+fip)-Xa)/G=(117995.5207-97917,91)/G=8655.69/127530=0.1574
As we can see by calculation, Nx is positive. So the plane should pick up speed a little
Perhaps yes, you’re right, graph 6.4 is made for a little less mass.But even for a mass of 13000kg, it is clear that nx is positive. So there will be no loss of speed
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How did you get the AoA result of 16.7 degrees? That’s what I get using 6.14 chart G load at 500kph (4.1G).
At 5G I get 21.63 degrees
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2000m 500 km/h
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The blue function which goes f(sinAoA) → Cl is an interpolation for the values of Sin(AoA) and Cl, and is certainly accurate when AoA result is above 20 degrees
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500km/h IAS at 2000m = 550km/h TAS, which is the speed used in the lift equation
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533 km/h
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In game 500km/h IAS at 2000m is 550 km/h TAS
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ok so I have a mistake now I will correct the calculations
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According the 6.4 chart Ny=5.
Using Ny = 4 I also get a result around 17 degrees AoA
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I recalculated and got Nx=0.15 at V IAS -500 km/h the plane still tends to accelerate
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