Roll angle for banked turn of an aircraft on a tilted plane It is known that for a banked turn of an aircraft on a plane parallel to the ground, the roll angle $\Theta$ computes as
$\Theta=\arctan(\frac{a}{g})$, with $a=\frac{v^2}{r}$ the centripetal acceleration and $g$ the gravitational field strength.
But what if the turn is not made parallel to the ground, but an a plane tilted with a angle $\phi$? How is the roll angle and the pitch angle computed? 
I suppose this is a question of computing some transformation matrices, but I am stuck.
The airplane is flying along the red line in this image:

 A: Let's start with an airplane making a circle level to the ground at a consant speed. Its position is given by
$$\vec{x}_1(t) = r\begin{bmatrix}
\cos(\omega t) \\
\sin(\omega t) \\
0
\end{bmatrix}$$
where $r$ is the radius of the turn, $\omega$ is the angular velocity (i.e., $v/r$), and $t$ is time. The $x$- and $y$-coordinates are parallel to the ground and $z$ is the elevation of the plane. I'm assigning an elevation of $0$ to the center of the airplane's circular path for convenience.
Since the plane is keeping a constant velocity, all we have to do is tilt the plane of the airplane's flight. I'm going to do this around the $y$-axis.
$$\vec{x}_2(t) = \begin{bmatrix}
cos(\phi)  &   0   &   sin(\phi) \\
0          &   1   &   0 \\
-sin(\phi) &   0   &   cos(\phi)
\end{bmatrix}
r\begin{bmatrix}
\cos(\omega t) \\
\sin(\omega t) \\
0
\end{bmatrix}
=
r\begin{bmatrix}
\cos(\omega t)cos(\phi) \\
\sin(\omega t) \\
-\cos(\omega t)sin(\phi)
\end{bmatrix}$$
The total force needed to sustain this path is simply the mass of the airplane times the acceleration ($F=ma$), and the acceleration is given by the second derivative of the position. Luckily, the angle $\phi$ is constant, so the derivatives are rather straightforwards.
$$\vec{a}_2 = \frac{d\vec{x}_2}{dt} = 
-\omega^2r\begin{bmatrix}
\cos(\omega t)cos(\phi) \\
\sin(\omega t) \\
-\cos(\omega t)sin(\phi)
\end{bmatrix}$$
There are four forces on the airplane: thrust, drag, weight, and lift. 
Weight is a constant
$$\vec{F}_w = m\vec{g} = mg\begin{bmatrix}0\\0\\-1\end{bmatrix}$$ where $g$ is the acceleration due to gravity.
To keep the plane at a constant speed, the thrust and drag of the airplane must balance the component of gravity in the same direction as the plane. First, the plane's velocity is given by
$$\vec{v}_2 = r\omega\begin{bmatrix}
-\sin(\omega t)cos(\phi) \\
\cos(\omega t) \\
\sin(\omega t)sin(\phi)
\end{bmatrix}
=
v\begin{bmatrix}
-\sin(\omega t)cos(\phi) \\
\cos(\omega t) \\
\sin(\omega t)sin(\phi)
\end{bmatrix}
$$
\begin{array}{r l}
\vec{F}_T + \vec{F}_d&=  -\left(\hat{v}_2 \cdot \vec{F}_g\right)\hat{v}_2 \\
 &= mg\sin(\omega t)\sin(\phi)\hat{v}_2 \\
 &= mg\sin(\omega t)\sin(\phi)\begin{bmatrix}
-\sin(\omega t)cos(\phi) \\
\cos(\omega t) \\
\sin(\omega t)sin(\phi)
\end{bmatrix}
\end{array}
Now, all that's left is to plug all this in to $\Sigma \vec{F} = m\vec{a}$ and see what's left for lift to do:
$$\vec{F}_T + \vec{F}_d + \vec{F}_g + \vec{F}_L = m\vec{a}_2$$
$$\vec{F}_L = m\vec{a}_2 - \vec{F}_T - \vec{F}_d - \vec{F}_g$$
$$\vec{F}_L = -m\omega^2r\begin{bmatrix}
\cos(\omega t)cos(\phi) \\
\sin(\omega t) \\
-\cos(\omega t)sin(\phi)
\end{bmatrix} - mg\sin(\omega t)\sin(\phi)\begin{bmatrix}
-\sin(\omega t)cos(\phi) \\
\cos(\omega t) \\
\sin(\omega t)sin(\phi)
\end{bmatrix} + mg\begin{bmatrix}0\\0\\1\end{bmatrix}$$
So, umm, yeah. The roll and pitch angles are not that simple for a banked circle. This isn't one of those problems where everything collapses into a neat formula.
A: We know the lift force is perpendicular to the airplane. Roll angle (about longitudinal axis) is such that the lift force vertical component equals the weight of the plane $mg$and the lateral component equals the centrifugal force requirement $ma$. 
If the plane tilts, then the lift force has three components: vertical, lateral and longitudinal. If pitch angle is $\phi$, roll angle is $\Theta$, the three components are, 
$$F_{longitudinal}=F_{lift}\sin \phi$$ 
$$F_{vertical}=F_{lift}\cos \phi \cos \Theta =mg$$
$$F_{lateral}=F_{lift}\cos \phi \sin \Theta =ma$$
Still, roll angle is $$\arctan(\frac ga)$$
However, the plane needs more thrust to gain more speed and thus more lift force. Otherwise, the plane will fall. And it is decelerating if pitch is positive. Interesting topic!
