Vector rotation around X-axis can be represented as:

Rx = $$\begin{bmatrix}1&0&0\\0&cos(\alpha)&sin(\alpha)\\0&-sin(\alpha)&cos(\alpha)\end{bmatrix}$$

Vector rotation around Y-axis can be represented as:

Ry = $$\begin{bmatrix}cos(\beta)&0&-sin(\beta)\\0&1&0\\sin(\beta)&0&cos(\beta)\end{bmatrix}$$

Vector rotation around Z-axis can be represented as:

Rz = $$\begin{bmatrix}cos(\gamma)&sin(\gamma)&0\\-sin(\gamma)&cos(\gamma)&0\\0&0&1\end{bmatrix}$$

Now final rotation matrix can be calculated and represented as:

R = Rx * Ry * Rz

Now, thrust generated by rotors are used to produce lift. On the other hand force produced by gravity works just opposite. There is one more thing which is force generated by rotor due to its rotation.

For each rotor:

F = $$\begin{bmatrix}Fr*cos(\theta)\\Fr*sin(\theta)\\Fm\end{bmatrix}$$

Here Fm is force produced by that rotor, Fr is force generated due to rotation.

Now summing up, we get:

Ft = $$\begin{bmatrix}\sum_{i=1}^4Fri*cos(\theta)\\sum_{i=1}^4Fri*sin(\theta)\\sum_{i=1}^4Fmi\end{bmatrix}$$

$$\theta$$ - is angle of the rotor at any given time.

All kind of force generated by rotor is taken into consideration in Ft. And orientation effects only Ft, as any given orientation Fg or force produced due to gravity will remain the same.

And, Fg can be written as:

$$\begin{bmatrix}0\\0\\mg\end{bmatrix}$$

Now Fnet = Ft * R – Fg

Is this calculation correct ?

Or does it need more ?