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Here is a simple way to think of this problem: Without thrust from its engine, the airplane begins instead to "coast downhill" like someone on a bike- and the pilot can choose whatever "slope" he or she wishes to follow, consistent with the need to keep the wings generating lift and the control surfaces to generate directional control forces. In a ...


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The weight $W$ of the glider has two components, $Wt$ and Wn. $Wt$ is in the same direction as $V$ and $Wn$ is perpendicular to $V$. Aerodynamic $F$ also has two components $L$ and $D$, where $L$ is perpendicular to $V$ and $D$ is parallel to $V$. When $L = Wn$: $Wt> D$, the linear acceleration of the glider is positive; $Wt <D$, the linear ...


3

Ejection of the tipjet exhaust streams in different directions creates a reactive torque on the rotor. This causes the rotor to spin up until aerodynamic drag exactly opposes the jet reaction. So in a way you can say that tipjet helicopters cancel the rotor torque through aerodynamic drag. But no torque is applied to the rotor through the rotor hub. Its ...


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Think of 2 scenarios: in one you are holding a rocket shaped projectile that is inert and you throw this projectile using your arm. In the second scenario you hold a real rocket, which is then launched. In the first scenario you throw the projectile, and you have to counter the twisting motion of your arm because your arm is imparting a force. In the ...


32

This is a basic physics question, involving Newton's third law of motion (For every action, there is an equal and opposite reaction.) When a centrally mounted engine applies force to turn the rotor, the equal and opposite reaction creates torque on the fuselage. With a tip jet, the force is applied by the jet shooting its exhaust perpendicular to the blade,...


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Think of the aircraft engine as one isolated system and the rotor as another isolated system. In its simplest terms, torque is the force required to move mass in a circular motion. Torque is caused by the engine providing power to the rotor shaft which moves the mass of the blades. The torque is the interaction of the stationary engine trying to move the ...


1

The weight $W$ of the glider has two components, $Wt$ and Wn. $Wt$ is in the same direction as $V$ and $Wn$ is perpendicular to $V$. Aerodynamic $F$ also has two components $L$ and $D$, where $L$ is perpendicular to $V$ and $D$ is parallel to $V$. When $L = Wn$: $Wt> D$, the linear acceleration of the glider is positive; $Wt <D$, the linear ...


2

To estimate whether the dutch-roll mode goes unstable, we can assess the real-part of the dutch-roll eigenvalue: negative is stable and positive is unstable. From Etkins, Dynamics of Flight, one illustrative estimate (although not necessarily numerically accurate) is as follows: $$n_{DR}\approx\frac{1}{2} \left[ \frac{Y_v}{m} + \frac{N_r}{I'_z} + \frac{I'...


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Computers in these days do not care about "significant decimal digits" for computation as they use binary IEEE-754 representation internally. The typical number sizes are float (most often 32 bits in these days) and double (most often 64 bit). C offers 80-bit extended precision (long double) and sometimes 128 bit double is available. You can read about sizes ...


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From your description, it's not an issue with the aerodynamic coefficients. If as you have described previously that you're modeling F-18, then the airframe should be unstable longitudinally. A 1e-10 deviation (assuming it applies to body rates, body speeds, etc. as well) is well-within typical trim error bound; in fact, a deviation is unavoidable when ...


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I am not a flight dynamics expert, but let me give an extremely general answer on verification of a non-linear dynamic system. consider that you have the non-linear system $\dot x =f(x)$. This system may have multiple equilibrium points, which are the $x$ values such that $f(x)=0$. You have checked that one of the desired equilibrium points is actually ...


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It's very prudent of you to be concerned about the axis frames in which different forces/motions are defined. It's a common mistake to ignore the fact that aerodynamic coefficients are normally defined in wind axes. The standard approach is to calculate Equations of Motions in the body frame; this is how they are usually defined. For this, you'll indeed ...


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