# What is the Dimension of Equilibrium Space in Longitudinal Space?

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

• 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.
– JZYL
Feb 3 '20 at 17:57
• 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. Feb 4 '20 at 7:13

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.

• 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. Feb 5 '20 at 13:31

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.