diff --git a/rotating_frame.html b/rotating_frame.html
index aefefeb..3c30a2c 100644
Binary files a/rotating_frame.html and b/rotating_frame.html differ
diff --git a/rotating_frame.org b/rotating_frame.org
index 9dc95d3..b07e538 100644
--- a/rotating_frame.org
+++ b/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
+ <>
+#+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 |
+| | Light | Heavy |
+|-------------+---------+---------|
+| Neg. Spring | 3.5 N/m | 8.5 N/m |
* 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
<>
+
+#+begin_src matlab :exports none :results silent :noweb yes
+ <>
+#+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