1 Goal of this note
+1 Goal of this note
The control objective is to stabilize the position of a rotating object with respect to a non-rotating frame. @@ -335,18 +337,18 @@ We want to compare the two different approach:
2 System
+2 System
-2.1 System description
+2.1 System description
-The system consists of one 2 degree of freedom translation stage on top of a spindle (figure 1). +The system consists of one 2 degree of freedom translation stage on top of a spindle (figure 1).
@@ -358,7 +360,7 @@ The measurement is either the \(x-y\) displacement of the object located on top
-
Figure 1: Schematic of the mecanical system
@@ -392,22 +394,22 @@ Indices \(u\) and \(v\) corresponds to signals in the rotating reference frame (2.2 Equations
+2.2 Equations
-Based on the figure 1, we can write the equations of motion of the system. +Based on the figure 1, 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 \(m\):
\begin{align} -\label{org25e6b9c} +\label{org4d9790f} T & = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) \\ V & = \frac{1}{2} k \left( x^2 + y^2 \right) \end{align} @@ -416,7 +418,7 @@ V & = \frac{1}{2} k \left( x^2 + y^2 \right) The Lagrangian is the kinetic energy minus the potential energy. \begin{equation} -\label{orgb6862e5} +\label{orgb67b4fc} L = T-V = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) - \frac{1}{2} k \left( x^2 + y^2 \right) \end{equation} @@ -425,7 +427,7 @@ L = T-V = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) - \frac{1}{2} k \le The partial derivatives of the Lagrangian with respect to the variables \((x, y)\) are: \begin{align*} -\label{org02c8a0d} +\label{orgcf126c0} \frac{\partial L}{\partial x} & = -kx \\ \frac{\partial L}{\partial y} & = -ky \\ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} & = m\ddot{x} \\ @@ -441,10 +443,9 @@ F_{\text{ext}, y} &= F_u \sin{\theta} + F_v \cos{\theta} \end{align*}-By appling the Lagrangian equations, we obtain equation \eqref{orgc6b5c7e}. +By appling the Lagrangian equations, we obtain:
\begin{align} -\label{orgc6b5c7e} m\ddot{x} + kx = F_u \cos{\theta} - F_v \sin{\theta}\\ m\ddot{y} + ky = F_u \sin{\theta} + F_v \cos{\theta} \end{align} @@ -468,7 +469,7 @@ We obtain: \end{align*}-By injecting the previous result into the Lagrangian equation \eqref{orgc6b5c7e}, 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} @@ -504,8 +505,8 @@ We can then subtract and add the previous equations to obtain the following equa2.3 Analysis
+2.3 TODO Analysis
We obtain two differential equations that are coupled through: @@ -520,23 +521,46 @@ Without the coupling terms, each equation is the equation of a one degree of fre Thus, the term \(-d_u m\dot{\theta}^2\) acts like a negative stiffness (due to centrifugal forces).
2.3.1 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: +\[ m \ddot{d_v} + (k - m\dot{\theta}^2) d_v &= F_v - 2 m\frac{\dot{F_u}}{k}\dot{\theta} - m \frac{F_u}{k}\ddot{\theta} \] +
3 Analytical Computation of forces for the NASS
+2.3.2 Negative Stiffness
++If \(\max{\dot{\theta}} \ll \sqrt{\frac{k}{m}}\), then the negative spring effect is negligible and \(k - m\dot{\theta}^2 \approx k\). +
+3 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. -
3.1 Euler and Coriolis forces
+3.1 Parameters
Let's define the parameters for the NASS. @@ -554,12 +578,21 @@ d = 0.10.2; % [m/s]
3.2 Euler and Coriolis forces
+-We then compute the corresponding values of the Coriolis and Euler forces, and the obtained values are displayed in table 1. +First we will determine the value for Euler and Coriolis forces during regular experiment.
-| Spring Soft. | +Neg. Spring | 3.5 N/m | 8.5 N/m |
