As I'm not sure I understand the implications of your simultaneous equations, may I ask a question?
Cn(total) = Cn(beta)*beta + Cn(rudder)*rudder + Cn(OEI) = 0
It seems to me that Cn(OEI) does not depend on longitudinal c.g. position, but the other two coefficients do. It also seems to me that it's important that Cn(beta) is less than Cn(rudder) for the wings level case - much less, in fact, as Cn(rudder) has to oppose the other two put together in the wings level case.
If Cn(OEI) is independent of longitudinal c.g. (independent of whatever longitudinal reference point you choose to use), and Cn(rudder) is the dominant of the other two expressions, doesn't this mean that the longitudinal reference point is important? Otherwise we could pick the point at the point of origin of force beta, making Cn(beta) zero. Or we could put the reference point far ahead of the aircraft, and Cn(beta) would approximate Cn(rudder). Either way the equation would cease to equal zero, simply by our apparently arbitrary choice of longitudinal reference point. In turn, this implies a fault somewhere in the reasoning.
None of the terms in that equation are dependent upon the c.g. position. That obviously seems unintuitive, because it's causing problems in another thread, so let's look at it in more detail.
When I measure the forces and moments on an aircraft ina wind tunnel, I use an arbitrary reference point for all forces and moemnts. I don't use the c.g. position, because I don't KNOW the cg.
When a simulation model of an aircraft is produced, specifically the "flight model" which is the six degree of freedom model of the aerodynamic and other forces on the aircraft, all the forces and moments are STILL referenced to the (same) arbitary reference point.
The cg (inertial) effects are added separately.
To consider moving the reference point. Suppose I use the nominal' reference point for forces and moments - 25% mean aerodynamic chord is typical. I then construct my yawing moment balance about that reference point.
Cn_25=Cn_25(beta)*beta + Cn_25(rudder)*rudder + Cn_25(OEI);
where "_25" indicates we're using the 25% reference point.
If I now wish to calculate the yawing moment about the aircraft c.g., I have to translate the point of action of the forces and moments from my 25% point to the cg; let's assume it's at 15%.
Cn_15 = Cn_25 + CY_25 * (25%-15%)*chord
I can then dimensionalise the Cn_15 term into an actual moment, N_15, and then determine the acceleration of the aircraft about the yaw axis:
yaw accel=N_15/Iyy
Now, cg mattered in that lot only because of the transfer of the sideforce term. But we assumed in that particular case that the aircraft wasn't banked, so the residual sideforce on the aircraft is zero. Therefore CY_25 was zero. Therefore the Cn_15=0 condition for trim is identical to the Cn_25=0 condition. Unless there is a lateral force component (which occurs in trimmed flight only with a steady bank angle) then the cg has no effect in determining the trimmed condition in the yawing axis.
Of course, if the yaw accel is non zero - i.e. the aircraft is still moving about the yawing axis - the c.g. matters to determine the motion. But if the aircraft is not accelerating, then the cg has no effect for the special zero bank case.
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It's extremely important when discussing the equations of motion and using these to determine the behaviour of the aircraft to be as precise as possible about axes etc.; the impossible can be made to appear intuitive if the definitions are carelessly used.
SR_71:
The only thing missing is the rolling moment due to rudder, Cl(rudder) and that\'s because I\'ve chosen to ignore the rolling axis for simplicity. Imagine the rudder point of action to lie on the aircraft stability x-axis if you wish a practical case.
Cn(rudder)*rudder is the yawing moment produced by the rudder, and is the only rudder term needed in the yawing moment equation.
CY(rudder) is the rudder sideforce derivative; when that term is non-zero then the rudder produces sideforce and yawing moment (i.e. is not producing a pure moment) and in that case we find that to "solve" the sideforce equation for the equilibirum case another source of sideforce is required. Sideslip, through the CY(beta)*beta terms, is the example used.