Update some typography

index.org

@@ -61,7 +61,7 @@ As the translation stage is rotating around the Z axis due to the spindle, the f
 The measurement is either the $x-y$ displacement of the object located on top of the translation stage or the $u-v$ displacement of the sample with respect to a fixed reference frame.
 
 #+name: fig:rotating_frame_2dof
-#+caption: Schematic of the mecanical system
+#+caption: Schematic of the mechanical system
 [[./figs/rotating_frame_2dof.png]]
 
 In the following block diagram:
@@ -79,7 +79,7 @@ Indices $u$ and $v$ corresponds to signals in the rotating reference frame ($\ve
 - $\epsilon_u$ and $\epsilon_v$ are position error of the sample along $\vec{i}_u$ and $\vec{i}_v$
 
 ** Equations
-  <<sec:equations>>
+<<sec:equations>>
 Based on the figure [[fig:rotating_frame_2dof]], we can write the equations of motion of the system.
 
 Let's express the kinetic energy $T$ and the potential energy $V$ of the mass $m$:
@@ -111,7 +111,7 @@ F_{\text{ext}, x} &= F_u \cos{\theta} - F_v \sin{\theta}\\
 F_{\text{ext}, y} &= F_u \sin{\theta} + F_v \cos{\theta}
 \end{align*}
 
-By appling the Lagrangian equations, we obtain:
+By applying the Lagrangian equations, we obtain:
 \begin{align}
 m\ddot{x} + kx = F_u \cos{\theta} - F_v \sin{\theta}\\
 m\ddot{y} + ky = F_u \sin{\theta} + F_v \cos{\theta}
@@ -125,6 +125,11 @@ y & = d_u \sin{\theta} + d_v \cos{\theta}
 
 We obtain:
 \begin{align*}
+\dot{x} & = \dot{d_u} \cos{\theta} - d_u\dot{\theta}\sin{\theta} - \dot{d_v}\sin{\theta} - d_v\dot{\theta} \cos{\theta} \\
+\dot{y} & = \dot{d_u} \sin{\theta} + d_u\dot{\theta}\cos{\theta} + \dot{d_v}\cos{\theta} - d_v\dot{\theta} \sin{\theta}
+\end{align*}
+and:
+\begin{align*}
 \ddot{x} & = \ddot{d_u} \cos{\theta} - 2\dot{d_u}\dot{\theta}\sin{\theta} - d_u\ddot{\theta}\sin{\theta} - d_u\dot{\theta}^2 \cos{\theta}
            - \ddot{d_v} \sin{\theta} - 2\dot{d_v}\dot{\theta}\cos{\theta} - d_v\ddot{\theta}\cos{\theta} + d_v\dot{\theta}^2 \sin{\theta} \\
 \ddot{y} & = \ddot{d_u} \sin{\theta} + 2\dot{d_u}\dot{\theta}\cos{\theta} + d_u\ddot{\theta}\cos{\theta} - d_u\dot{\theta}^2 \sin{\theta}
@@ -133,21 +138,26 @@ We obtain:
 
 By injecting the previous result into the Lagrangian equation, we obtain:
 \begin{align*}
- m \ddot{d_u} \cos{\theta} - 2m\dot{d_u}\dot{\theta}\sin{\theta} - m d_u\ddot{\theta}\sin{\theta} - m d_u\dot{\theta}^2 \cos{\theta}
--m \ddot{d_v} \sin{\theta} - 2m\dot{d_v}\dot{\theta}\cos{\theta} - m d_v\ddot{\theta}\cos{\theta} + m d_v\dot{\theta}^2 \sin{\theta}
+  m \ddot{d_u} \cos{\theta} - 2m\dot{d_u}\dot{\theta}\sin{\theta} - m d_u\ddot{\theta}\sin{\theta} - m d_u\dot{\theta}^2 \cos{\theta}
+- m \ddot{d_v} \sin{\theta} - 2m\dot{d_v}\dot{\theta}\cos{\theta} - m d_v\ddot{\theta}\cos{\theta} + m d_v\dot{\theta}^2 \sin{\theta}
++ c \dot{d_u} \cos{\theta} - c d_u\dot{\theta}\sin{\theta} - c \dot{d_v}\sin{\theta} - c d_v\dot{\theta} \cos{\theta}
 + k d_u \cos{\theta} - k d_v \sin{\theta} = F_u \cos{\theta} - F_v \sin{\theta} \\
   m \ddot{d_u} \sin{\theta} + 2m\dot{d_u}\dot{\theta}\cos{\theta} + m d_u\ddot{\theta}\cos{\theta} - m d_u\dot{\theta}^2 \sin{\theta}
 + m \ddot{d_v} \cos{\theta} - 2m\dot{d_v}\dot{\theta}\sin{\theta} - m d_v\ddot{\theta}\sin{\theta} - m d_v\dot{\theta}^2 \cos{\theta}
++ c \dot{d_u} \sin{\theta} + c d_u\dot{\theta}\cos{\theta} + c \dot{d_v}\cos{\theta} - c d_v\dot{\theta} \sin{\theta}
 + k d_u \sin{\theta} + k d_v \cos{\theta} = F_u \sin{\theta} + F_v \cos{\theta}
 \end{align*}
 
+
 Which is equivalent to:
 \begin{align*}
- m \ddot{d_u} - 2m\dot{d_u}\dot{\theta}\frac{\sin{\theta}}{\cos{\theta}} - m d_u\ddot{\theta}\frac{\sin{\theta}}{\cos{\theta}} - m d_u\dot{\theta}^2
--m \ddot{d_v} \frac{\sin{\theta}}{\cos{\theta}} - 2m\dot{d_v}\dot{\theta} - m d_v\ddot{\theta} + m d_v\dot{\theta}^2 \frac{\sin{\theta}}{\cos{\theta}}
+  m \ddot{d_u} - 2m\dot{d_u}\dot{\theta}\frac{\sin{\theta}}{\cos{\theta}} - m d_u\ddot{\theta}\frac{\sin{\theta}}{\cos{\theta}} - m d_u\dot{\theta}^2
+- m \ddot{d_v} \frac{\sin{\theta}}{\cos{\theta}} - 2m\dot{d_v}\dot{\theta} - m d_v\ddot{\theta} + m d_v\dot{\theta}^2 \frac{\sin{\theta}}{\cos{\theta}}
++ c \dot{d_u} - c d_u\dot{\theta}\frac{\sin{\theta}}{\cos{\theta}} - c \dot{d_v}\frac{\sin{\theta}}{\cos{\theta}} - c d_v\dot{\theta}
 + k d_u - k d_v \frac{\sin{\theta}}{\cos{\theta}} = F_u - F_v \frac{\sin{\theta}}{\cos{\theta}} \\
   m \ddot{d_u} + 2m\dot{d_u}\dot{\theta}\frac{\cos{\theta}}{\sin{\theta}} + m d_u\ddot{\theta}\frac{\cos{\theta}}{\sin{\theta}} - m d_u\dot{\theta}^2
 + m \ddot{d_v} \frac{\cos{\theta}}{\sin{\theta}} - 2m\dot{d_v}\dot{\theta} - m d_v\ddot{\theta} - m d_v\dot{\theta}^2 \frac{\cos{\theta}}{\sin{\theta}}
++ c \dot{d_u} + c d_u\dot{\theta}\frac{\cos{\theta}}{\sin{\theta}} + c \dot{d_v}\frac{\cos{\theta}}{\sin{\theta}} - c d_v\dot{\theta}
 + k d_u + k d_v \frac{\cos{\theta}}{\sin{\theta}} = F_u + F_v \frac{\cos{\theta}}{\sin{\theta}}
 \end{align*}
 
@@ -155,11 +165,11 @@ We can then subtract and add the previous equations to obtain the following equa
 #+begin_important
 #+NAME: eq:du_coupled
 \begin{equation}
- m \ddot{d_u} + (k - m\dot{\theta}^2) d_u = F_u + 2 m\dot{d_v}\dot{\theta} + m d_v\ddot{\theta}
+  m \ddot{d}_{u} + c \dot{d}_{u} + (k - m\dot{\theta}^2) d_u = F_u + 2 m\dot{d}_v\dot{\theta} + m d_v\ddot{\theta} + c d_{v} \dot{\theta}
 \end{equation}
 #+NAME: eq:dv_coupled
 \begin{equation}
- m \ddot{d_v} + (k - m\dot{\theta}^2) d_v = F_v - 2 m\dot{d_u}\dot{\theta} - m d_u\ddot{\theta}
+  m \ddot{d}_{v} + c \dot{d}_{v} + (k - m\dot{\theta}^2) d_v = F_v - 2 m\dot{d}_u\dot{\theta} - m d_u\ddot{\theta} - c d_{u} \dot{\theta}
 \end{equation}
 #+end_important
 
@@ -272,12 +282,13 @@ This is definitely negligible when using piezoelectric actuators. It may not be
 | Neg. Spring | 1381.7[N/m] | 0.9[N/m] |
 
 ** Limitations due to coupling
+*** Equations
 To simplify, we consider a constant rotating speed $\dot{\theta} = \omega_0$ and thus $\ddot{\theta} = 0$.
 
 From equations [[eq:du_coupled]] and [[eq:dv_coupled]], we obtain:
 \begin{align*}
-  (m s^2 + (k - m{\omega_0}^2)) d_u &= F_u + 2 m {\omega_0} s d_v \\
-  (m s^2 + (k - m{\omega_0}^2)) d_v &= F_v - 2 m {\omega_0} s d_u \\
+  (m s^2 + c s + (k - m{\omega_0}^2)) d_u &= F_u + 2 m \omega_0 s d_v + c \omega_0 d_v \\
+  (m s^2 + c s + (k - m{\omega_0}^2)) d_v &= F_v - 2 m \omega_0 s d_u - c \omega_0 d_v
 \end{align*}
 
 From second equation:
@@ -303,10 +314,10 @@ The two previous equations can be written in a matrix form:
 #+NAME: eq:coupledplant
 \begin{equation}
 \begin{bmatrix} d_u \\ d_v \end{bmatrix} =
-\frac{1}{(m s^2 + (k - m{\omega_0}^2))^2 + (2 m {\omega_0} s)^2}
+\frac{1}{(m s^2 + cs + (k - m{\omega_r}^2))^2 + (2 m {\omega_r} s + c \omega_r)^2}
 \begin{bmatrix}
-  ms^2 + (k-m{\omega_0}^2) & 2 m \omega_0 s \\
-  -2 m \omega_0 s          & ms^2 + (k-m{\omega_0}^2) \\
+  ms^2 + cs + (k-m{\omega_r}^2) & 2 m \omega_r s + c \omega_r \\
+  -2 m \omega_r s + c \omega_r  & ms^2 + cs + (k-m{\omega_r}^2)
 \end{bmatrix}
 \begin{bmatrix} F_u \\ F_v \end{bmatrix}
 \end{equation}
@@ -411,6 +422,7 @@ From this analysis, we can determine the lowest practical stiffness that is poss
 | k min [N/m] |  2199 |    89 |
 
 ** Effect of rotation speed on the plant
+*** Introduction                                                    :ignore:
 As shown in equation [[eq:coupledplant]], the plant changes with the rotation speed $\omega_0$.
 
 Then, we compute the bode plot of the direct term and coupling term for multiple rotating speed.