10. Linearization Intuition 3
C3 L3 08 Linearization Intuition 3 V3
| Quantity | Value at Hover |
|---|---|
| u_2 | 0 |
| u_1 | mg |
| \phi | 0 |
If you're interested in a more mathematically rigorous explanation of why the above values are true, we'll present that here. If you're not interested, feel free to continue with the lesson!
Let's start with the equations of motion:
\begin{aligned}
\\
\ddot{y} &= \frac{u_1}{m}\sin\phi
\\
\ddot{z} &= g - \frac{u_1}{m}\cos\phi
\\
\ddot{\phi} &= \frac{u_2}{I_{xx}}
\end{aligned}
We chose to analyze the hover state because it's an equilibrium where the second derivatives are all 0.
\begin{aligned}
\\
\\
\frac{u_1}{m}\sin\phi &= 0
\\
\\
\frac{u_1}{m}\cos\phi &= g
\\
\\
\frac{u_2}{I_{xx}}&= 0
\end{aligned}
From the third equation we can see that u_2 =0.
We can square the remaining two equations and then add them together, which leaves us with the following:
\begin{aligned}
\frac{u_1^2}{m^2} \sin^2 \phi &= 0
\\
+ \ \ \frac{u_1^2}{m^2} \cos^2 \phi &= g^2
\\
\\
\frac{u_1^2}{m^2}(\sin^2 \phi + \cos^2 \phi) &= g^2
\end{aligned}
Remembering that \sin^2x + cos^2x = 1…
\begin{aligned}
\frac{u_1^2}{m^2} &= g^2
\\u_1 &= mg
\end{aligned}
And we can plug this into our equation for \ddot{y}=0 to prove that \phi=0.