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°).
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.
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
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
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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
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
On the graph 6.4 at 500 km/h Nx=0. The aircraft will have a constant speed
You haven’t taken into account the contribution of thrust to lift yet Fip=5 degrees
