Add some analysis

rotating_frame.org

@@ -109,7 +109,7 @@ We obtain:
            + \ddot{d_v} \cos{\theta} - 2\dot{d_v}\dot{\theta}\sin{\theta} - d_v\ddot{\theta}\sin{\theta} - d_v\dot{\theta}^2 \cos{\theta} \\
 \end{align*}
 
-By injecting the previous result into the Lagrangian equation [[eq:lagrangian_eq_inertial]], 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}
@@ -137,7 +137,7 @@ We can then subtract and add the previous equations to obtain the following equa
 \end{align*}
 #+end_important
 
-** Analysis
+** TODO Analysis
 We obtain two differential equations that are coupled through:
 - *Euler forces*: $m d_v \ddot{\theta}$
 - *Coriolis forces*: $2 m \dot{d_v} \dot{\theta}$
@@ -145,12 +145,24 @@ We obtain two differential equations that are coupled through:
 Without the coupling terms, each equation is the equation of a one degree of freedom mass-spring system with mass $m$ and stiffness $k-d_u m\dot{\theta}^2$.
 Thus, the term $-d_u m\dot{\theta}^2$ acts like a negative stiffness (due to *centrifugal forces*).
 
+*** Stiff actuators
+Let's say we use stiff actuators such that $m \ddot{d_u} + (k - m\dot{\theta}^2) d_u \approx k d_u$.
+
+Let's suppose that $F_u + 2 m\dot{d_v}\dot{\theta} + m d_v\ddot{\theta} \approx F_u$.
+
+Then we obtain $d_u = \frac{F_u}{k}$ that we can re inject in the other equation to obtain:
+\[ m \ddot{d_v} + (k - m\dot{\theta}^2) d_v &= F_v - 2 m\frac{\dot{F_u}}{k}\dot{\theta} - m \frac{F_u}{k}\ddot{\theta} \]
+
+*** Negative Stiffness
+If $\max{\dot{\theta}} \ll \sqrt{\frac{k}{m}}$, then the negative spring effect is negligible and $k - m\dot{\theta}^2 \approx k$.
+
 * Analytical Computation of forces for the NASS
 For the NASS, the Euler forces should be less of a problem as $\ddot{\theta}$ should be very small when conducting an experiment.
 
-First we will determine the value for Euler and Coriolis forces during regular experiment.
-
-** Euler and Coriolis forces
+** Parameters
+#+begin_src matlab :exports none :results silent :noweb yes
+  <<matlab-init>>
+#+end_src
 
 Let's define the parameters for the NASS.
 #+begin_src matlab :exports code :results silent
@@ -166,6 +178,9 @@ Let's define the parameters for the NASS.
   ddot = 0.2; % [m/s]
 #+end_src
 
+** Euler and Coriolis forces
+First we will determine the value for Euler and Coriolis forces during regular experiment.
+
 #+begin_src matlab :exports none :results silent
   Felight = mlight*d*wdot;
   Feheavy = mheavy*d*wdot;
@@ -188,25 +203,25 @@ We then compute the corresponding values of the Coriolis and Euler forces, and t
 | Coriolis | 44.0 N | 1.8 N |
 | Euler    | 3.5 N  | 8.5 N |
 
-** Spring Softening Effect
+** Negative Spring Effect
 #+begin_src matlab :exports none :results silent
   Klight = mlight*d*wdot^2;
   Kheavy = mheavy*d*wdot^2;
 #+end_src
 
-The values for the spring softening effect are displayed in table [[tab:spring_softening]].
+The values for the negative spring effect are displayed in table [[tab:negative_spring]].
 This is definitely negligible when using piezoelectric actuators. It may not be the case when using voice coil actuators.
 
 #+begin_src matlab :results value table :exports results :post addhdr(*this*)
-  ans = sprintf(' | Light | Heavy | \n Spring Soft. | %.1f N/m | %.1f N/m', Klight, Kheavy)
+  ans = sprintf(' | Light | Heavy | \n Neg. Spring | %.1f N/m | %.1f N/m', Klight, Kheavy)
 #+end_src
 
-#+NAME: tab:spring_softening
-#+CAPTION: Spring Softening effect
+#+NAME: tab:negative_spring
+#+CAPTION: Negative Spring effect
 #+RESULTS:
 |             | Light   | Heavy   |
-|--------------+---------+---------|
-| Spring Soft. | 3.5 N/m | 8.5 N/m |
+|-------------+---------+---------|
+| Neg. Spring | 3.5 N/m | 8.5 N/m |
 
 * Control Strategies
   <<sec:control_strategies>>
@@ -229,11 +244,10 @@ The block diagram is shown on figure [[fig:control_measure_fixed_2dof]].
 
 The loop gain is then $L = G(\theta) K J(\theta)$.
 
-*** QUESTION Is the loop gain is changing with the angle ?
-Is \[ G(\theta) J(\theta) = G(\theta_0) J(\theta_0) \] ?
+One question we wish to answer is: is $G(\theta) J(\theta) = G(\theta_0) J(\theta_0)$?
 
 ** Measurement in the rotating frame
-Let's consider that the measurement is in the rotating reference frame.
+Let's consider that the measurement is made in the rotating reference frame.
 
 The corresponding block diagram is shown figure [[fig:control_measure_rotating_2dof]]
 
@@ -245,6 +259,11 @@ The loop gain is $L = G K$.
 
 * Effect of the rotating Speed
   <<sec:effect_rot_speed>>
+
+#+begin_src matlab :exports none :results silent :noweb yes
+  <<matlab-init>>
+#+end_src
+
 ** TODO Use realistic parameters for the mass of the sample and stiffness of the X-Y stage
 
 ** TODO Check if the plant is changing a lot when we are not turning to when we are turning at the maximum speed (60rpm)
@@ -262,7 +281,14 @@ The loop gain is $L = G K$.
 #+NAME: matlab-init
 #+BEGIN_SRC matlab :results none :exports none
   clear; close all; clc;
+
+  %% Add path with some functions
   addpath('./src/');
+
+  %% Intialize Laplace variable
+  s = tf('s');
+
+  %% Initialize ans with org-babel
   ans = 0;
 #+END_SRC