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Full untrimmed Flight Dynamics model of an Airplane is separated into two separate systems: Longitudinal and Lateral.

When the untrimmed model in Longitudinal direction, which is six dimensional, is trimmed (f(x,u) = 0), what is the dimension of the equilibrium space? Is it two dimensional? If yes, then why?

x = state vector for longitudinal direction u = control vector in longitudinal direction

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    $\begingroup$ When you say 6-dim, I assume you have included the elevator actuator and/or engine as separate states? If it's pure surface driven, then it would be 4-dim only. $\endgroup$
    – JZYL
    Feb 3, 2020 at 17:57
  • $\begingroup$ Yes, Elevator and Thrust are different states. Here, x = [V, alpha, q, theta] and u = [Delta_Elevator, Delta_Thrust]. So, a total of 6 dimensions. I want to know what will be the dimension of equilibrium space. $\endgroup$
    – Pavan
    Feb 4, 2020 at 7:13

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I never heard of the term "Equilibirum Space" in flight dynamics and I can't find references in a quick google search.

However, I assume you are talking about describing the flight dynamics of an aircraft as a linear state-space model. When linearising, you can indeed separate the equations of motion in the plane symmetry (longitudinal) from the assymetric (lateral) motions.

$\dot{ \vec{x}}=\mathbf{A} \vec{x} + \mathbf{B} \vec{u} $

The dimension of the state vector $\vec{x}$ is usually 4 (velocity, flight path angle, pitch angle and pitch rate), in your case the dimension of the control vector $\vec{u}$ is 2 (thrust and elevator deflection).

You linearise the equations of motion around an equilibrium point; the state and control vectors describe the deviation from that equilibrium point. The dimension of the equilibrium points is 6 (4 state variable, 2 control variables).

One could say that the equilibrium point is in the space spanned by the 6 independent state and control variables. If that is what you mean, then dimension of the equilibrium space is 6.

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  • $\begingroup$ I found about equilibrium space when I was reading a paper (DOI: 10.1109/ICCA.2019.8899603) on the control of an impaired aircraft. In the paper, a straight level flight trim plot had only 2 dimensional. I want to know why the dimension is 2. $\endgroup$
    – Pavan
    Feb 5, 2020 at 13:31
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The longitudinal differential system is:

$$\dot{\vec{\bf{x}}}=f(\vec{\bf{x}},\vec{\bf{u}})$$

where $\vec{\bf{x}}=[u,w,q,\theta]$, $\vec{\bf{u}}=[\delta_e,T]$

As far as linear space is concerned, control inputs $\vec{\bf{u}}$ are not states. So unless you have actuator models that construct internal actuator states, the longitudinal differential system is 4-dimensional.

As far as the equilibrium point is concerned, you set all the derivatives to zero, and impose a further restriction on $q=0$, $u=u_0$, then you have four unknowns (two state variables and two control variables) with four non-homogeneous algebraic equations. With linear aerodynamics, you have one unique solution, and it is not a vector space.

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