I'm currently studying flight mechanics on a subject at my engineering degree and I'm trying to solve some standard case exercises with the general equations and diferent reference frames. I've been able to solve multiple exercises on different looping manoeuvres enclosed in the vertical plane but I can't deduce the general equations for a horizontal plane. $$ T \cos\varepsilon\cos\nu - D - mg\sin\gamma-m\dot{V}=0\\ T\cos\epsilon\sin\nu-Q+mg\cos\gamma\sin\mu+mV(\dot{\gamma}\sin\mu-\dot\chi\cos\gamma\cos\mu)=0\\ -T\sin\varepsilon-L+mg\cos\gamma\cos\mu+mV(\dot\gamma\cos\mu+\dot\chi\cos\gamma\sin\mu)=0 $$

So starting with the above and taking into account some simplifications I get to an intermediate state which is quite similar to the solution for the problem which is the following:

$$T\cos\nu-D=0\\ L\sin\mu+T\sin\nu\cos\mu=\dfrac{W}{g}\dfrac{V^2}{R} \\ L\cos\mu-T\sin\nu\sin\mu-W=0$$

My question is: How is it possible that the Lift term appears in the second equation if the starting equation didn't consider it? And also, why the trigonometric relations for the thrust change? (This results are for a non-symmetric manoeuvre). I can't get in my mind why the general equations used in all the books can't give in a direct way the desired result without adding new terms.

Reference frame for the equations is wind-axis. The angles used are:

$\nu =$ Thrust side-slip angle

$\varepsilon=$ Thrust angle of atack

As we are using wind-aixis reference frame:

$\mu=$ velocity roll angle

$\gamma=$ velocity pitching angle

$\chi=$ velocity yaw angle

$W=mg$, $T=$ thrust, $D=$ drag, and $Q=$ lateral aerodynamic forces

  • 2
    $\begingroup$ Your angular notations are different from the western industry-standard. You should explain first what these angles refer to. And what do you mean by non-symmetric turn? $\endgroup$
    – JZYL
    Commented Oct 27, 2019 at 13:15
  • $\begingroup$ I've edited the post with the references for the angles. The non-symmetric turn means (from what I understood from my books) that the forces on the two wings are not balanced basically the lift, also, the thrust side-slip angle is different from 0. $\endgroup$ Commented Oct 27, 2019 at 18:20
  • $\begingroup$ I edited your equations to MathJax. Please check if I made any mistakes in transcribing your equations. Remember to use \cos and \sin rather than plain cos and sin, so it is typeset as a function rather than plain variables. $\endgroup$
    – Sanchises
    Commented Oct 28, 2019 at 9:24
  • $\begingroup$ Checked! Now everything is correct. :) $\endgroup$ Commented Oct 28, 2019 at 10:09

1 Answer 1


Your equations seem to be in wind-axis, and look a little weird from my literature (referencing Stevens and Lewis, Aircraft Control and Simulation). I'm more used to specifying mechanics from the body-axis.

Legend: $\theta$ is Euler pitch angle, $\phi$ is bank angle, $\psi$ is yaw angle, $p$ is body roll rate, $q$ is body pitch rate, $r$ is body yaw rate, $\alpha$ is angle of attack, $Y$ is aerodynamic side force (in stability-axis), $u$ is body forward speed, $v$ is body lateral speed, $w$ is body vertical speed.

A. Thrust aligned with body-axis

I will assume:

  1. Thrust is aligned with the body-axis.
  2. All aerodynamic forces are aligned with the stability-axis (see this question for the rationale).
  3. No wind, so that the velocity in body-axis is the relative air velocity ($\alpha=\tan^{-1}(w/u)$,$\beta=\sin^{-1}(v/\sqrt{u^2+v^2+w^2})=\sin^{-1}(v/V)$).

For the general 6-dof, we have:

$$T-D\cos\alpha+L\sin\alpha-mg\sin\theta=m(\dot{u}+qw-rv)$$ $$Y+mg\cos\theta\sin\phi=m(\dot{v}+ru-pw)$$ $$D\sin\alpha-L\cos\alpha+mg\cos\theta\cos\phi=m(\dot{w}+pv-qu)$$

In a coordinated turn, the ball is centered, which means that the gravity component cancels out the centripetal force, and the aerodynamic forces in the y-axis must be zero.

Furthermore, in the inertial frame, we should only see a non-zero yaw rate ($\dot{\psi}$) for a steady turn. So translating this back into body-axis, we have:

$$p=-\dot{\psi}\sin\theta$$ $$q=\dot{\psi}\cos\theta\sin\phi$$ $$r=\dot{\psi}\cos\theta\cos\phi$$

To make the matter somewhat clearer, we can choose to make small angle assumptions on $\alpha$ and $\theta$, and $u\approx V$, and eliminate all second order terms. This brings us to:

$$T-D+L\alpha-mg\theta=m(\alpha V\dot{\psi}\sin\phi- \beta V\dot{\psi}\cos\phi)$$ $$mg\sin\phi=mV\dot{\psi}\cos\phi$$ $$-L+mg\cos\phi=-mV\dot{\psi}\sin\phi$$

From geometry, a particle tracing out a constant circular path with a rate of $\dot{\psi}$ will have a radius (i.e. turn radius) of: $R=V/\dot{\psi}$. So the second equation is now:


We can define a stability-axis load factor ($n_{z_s}=\frac{L}{mg}$), which would closely approximate the body load factor for small AOA. The third equation can be expressed as:


At this point, we've completely reproduced the performance equations relating load factor, bank angle and turn radius from the 6-dof equations.

B. Thrust asymmetric with body axis

Let's say that the thrust vector is angled with the body axis by a pitching angle of $\epsilon$ and a yaw angle of $\nu$.

So in the body-axis, the thrust vector is:

$$\boldsymbol{T_b}=\begin{bmatrix} \cos\epsilon\cos\nu & -\cos\epsilon\sin\nu & \sin\epsilon \\ \sin\nu & \cos\nu & 0 \\ -\sin\epsilon\cos\nu & \sin\epsilon\sin\nu & \cos\epsilon \end{bmatrix} \begin{bmatrix}T \\ 0 \\ 0 \end{bmatrix}$$

We now have modified 6-dof:

$$T\cos\epsilon\cos\nu-D\cos\alpha+L\sin\alpha-mg\sin\theta=m(\dot{u}+qw-rv)$$ $$T\sin\nu + Y+mg\cos\theta\sin\phi=m(\dot{v}+ru-pw)$$ $$-T\sin\epsilon\cos\nu + D\sin\alpha-L\cos\alpha+mg\cos\theta\cos\phi=m(\dot{w}+pv-qu)$$

Again assume that the ball is centered, thus $T\sin\nu = -Y$. The conclusion from the second equation (relating bank angle to yaw rate and turn radius) would be the same as before.

The third equation can be reduced to:


  • $\begingroup$ Thanks for your elaborated answer, I really appreciate it. But, still, I see what you are doing with this on the body-axis, as getting back to my initial question you have from start on the x-axis equation the terms for Thrust, Drag and Lift, and the same for the z-axis (Lift, Drag). In my case when starting with the equations referenced in (3.5) lift doesn't appear in the y-axis but it's in the answer for the exercise. I can imagine that, while in the turn and with the plane in a banking angle its normal that some of the lift is also produced in the Y axis from a wind-fram point of view. $\endgroup$ Commented Oct 27, 2019 at 18:30
  • $\begingroup$ But I don't know how to incorporate it to the wind-axis equations and what is the procedure to get there and solve it using the wind-axis. Thank you again for taking the time to answer. $\endgroup$ Commented Oct 27, 2019 at 18:35
  • $\begingroup$ @RogerVallès Sorry, I prefer the body-axis :) See my modified answer and see if it helps any. $\endgroup$
    – JZYL
    Commented Oct 28, 2019 at 0:27
  • $\begingroup$ Yes, with the new part that you added I've been able to deduce what I was missing. I draw the thrust and the lift vectors on the Fh plane and then moved them to the wind-axis with the transformation matrix and got the final result. I've been thinking about it for a while and it looks more logical to use the body-axis for most of the calculations, but I'm not in a position to choice what axis to use as that's what's on my book for the subject and all the exercises are usually with wind-axis. Basically I think my teacher likes the wind-axis a lot more hahaha $\endgroup$ Commented Oct 28, 2019 at 5:49

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