$x = 2L.cos(\alpha)$
$y = 4L.sin(\alpha)$
\begin{align} F_{AB} dx - mg dy&=0 \\\ F_{AB}&=mg\frac{dy}{dx} \\\ F_{AB} &=-\frac{2mg}{tan(\alpha)} \\\ \end{align}
$x = BH =L\sqrt{cos(\alpha)^2+9sin(\alpha)^2}=L\sqrt{1+8sin(\alpha)^2}$
$y = 4L.sin(\alpha)$
$dx = 8L\frac{sin(\alpha)cos(\alpha)}{\sqrt{1+8sin(\alpha)^2}}d\alpha$
$dy = 4Lcos(\alpha)d\alpha$
\begin{align} F_{BH}&=mg\frac{dy}{dx} \\\ F_{BH} &=mg\dfrac{\sqrt{1+8sin(\alpha)^2}}{2sin(\alpha)} \\\ \end{align}
Response file to download here : Chap6_ScissorPlatform.mo
It should be noted that the actuator forces for the BH configuration are much lower than for the AB configuration. It is therefore recommended to use the BH configuration to minimize the forces on the actuators.
$R = \rho\dfrac{L}{S} = 1,7e^{-8}\dfrac{350}{3,14.(\frac{0,23e^{-3}}{2})^2} \approx 143 \,\Omega$
A : f
B : a
C : b
D : c
E : d
F : e
You can find the Modelica model here : Chap6_PowerOffBrake.mo
The winding effect is modelled by the electromagnetic transformer (blue/orange box). In the Modelica code of this component you will find :
//converter equations:
V_m = i*N;
// Ampere's law
N*der(Phi) = -v;
// Faraday's law
which represent a transformer that binds the power variables :
and representing the equations : $v=-N\frac{d\varphi}{dt}$ et $\oint \frac{B}{\mu}dl=NI$
The equation representing the Ampere theorem $\oint \frac{B}{\mu}dl=NI$ can be represented on the form of a mesh law with fmm drops (Prduct of Reluctance.Flux as in electricity voltage drop product of resistance.current) :
$\oint \frac{B}{\mu}dl=\varphi \sum R_k = NI$
where the reluctance is calculated with $R = \int \frac{dl}{\mu S} $
For the reluctances b and f the section is constant over the length and so we have:
$R = \frac{l}{\mu S} $
that is found in the Modelica code :
A = pi*(r_o^2 - r_i^2);
G_m = (mu_0*mu_r*A)/l;
where G_m is the inverse of the magnetic reluctance $G_m = 1/R_m$
For the reluctances a and d the section is not constant over the length. It is therefore necessary to integrate the previous expression on the radial geometry (of internal radius $r_i$, external radius $r_o$, thickness l) : $R = \int_{r_i}^{r_o} \frac{dr}{\mu S} = \int_{r_i}^{r_o} \frac{dr}{\mu 2\pi r l} = \frac{ln(r_o/r_i)}{\mu 2\pi l}$
that is found in the Modelica code :
A = l*pi*(r_o + r_i);
// Area at arithmetic mean radius for calculation of average flux density
G_m = 2*pi*mu_0*mu_r*l/Modelica.Math.log(r_o/r_i);
where G_m is the inverse of the magnetic reluctance $G_m = 1/R_m$
An electro-magnetic actuator consisting as here of a coil and a mechanical moving part:
This gives the following energy balance :
$ dW_{elec} = dV_{mag} + dW_{meca}$
If we assume the position of the fixed mechanical effector, $dx=0$ and :
$ dV_{mag} = dW_{elec} = vidt = id\varphi$ or $V_{mag}=\int id\varphi=\frac{1}{2}Li^2$
If we assume the flow in the fixed coil, $d\varphi=0$ and :
$ dW_{elec} = id\varphi = 0 = dV_{mag} + dW_{meca}$
so dW_{meca} = -dV_{mag} or $F=-\frac{\partial V_{mag}}{\partial x}$
but having a constant flow is difficult to verify in practice.
By introducing the notion of magnetic coenergy: $V_{mag} + V_{comag} = i\varphi$
or $dV_{mag} + dV_{comag} = di\varphi + id\varphi $
In case of constant current, we then have :
$ dW_{elec} = id\varphi = dV_{mag} + dW_{meca} = di\varphi + id\varphi - dV_{comag} + dW_{meca}$
or $0 = \varphi di - dV_{comag} + dW_{meca}$
so $F=\frac{\partial V_{comag}}{\partial x}$ where : $V_{comag}=\int \varphi di=\frac{1}{2}Li^2$
so : $F=\frac{1}{2}\frac{\partial L}{\partial x} i^2$
$V_{comag}=\frac{1}{2}Li^2=\frac{1}{2} \varphi i $
At the level of a reluctance of a magnetic circuit :
$\frac{1}{2} \varphi i = \frac{1}{2} \varphi^2 R_m = \frac{1}{2} \varphi^2 / G_m = \frac{1}{2} G_m \varepsilon^2 $
with $G_m$ the opposite of reluctance and $varepsilon$ the fmm. We then have :
$F=\frac{1}{2}\frac{\partial G_m}{\partial x} \varepsilon^2$
f we observe the partial model “PartialForce” of the MSL (Magnetic/FluxTube/Interfaces), we find these equations :
V_m = Phi*R_m;
R_m = 1/G_m;
F_m = 0.5*V_m^2*dGmBydx;
which is finally found in component e of Figure 6.27.
The kinetic energy of the system is : $E_{c} = \frac{1}{2} M_{eq} \dot{x}^{2} + \frac{1}{2} M_u V_{g}^{2} + \frac{1}{2} J_{G} \dot{\theta}^{2}$
with :
$M_{eq} = M + 2 \frac{J_r + J_m N^2}{R^2}$
$V_{eq}^2 = \dot{x}^{2} + 2 L_{G} \dot{\theta} \dot{x} \cos\left(\theta + \beta\right) + L_{G}^{2} \dot{\theta}^{2}$
The potential energy of the system is : $E_{p} = M_{u} g L_{G} \cos\left(\theta + \beta\right)$
A Lagrangian is being written : \begin{equation} \mathcal{L} = E_{c} - E_{p} \end{equation}
with :
\begin{equation} \mathcal{L} = \frac{1}{2} M_{eq} \dot{x}^{2} + \frac{1}{2} M_u \left( \dot{x}^{2} + 2 L_{G} \dot{\theta} \dot{x} \cos\left(\theta + \beta\right) + L_{G}^{2} \dot{\theta}^{2} \right) + \frac{1}{2} J_{G} \dot{\theta}^{2} - M_{u} g L_{G} \cos\left(\theta + \beta\right) \end{equation}
The Lagrange equation on the degree of freedom x is defined as : \begin{equation} \frac{d}{dt} \left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) - \frac{\partial \mathcal{L}}{\partial x} = F_{ext,x} \end{equation}
with :
\begin{align}
\frac{\partial \mathcal{L}}{\partial \dot{x}} & = M_{eq} \dot{x} + M_u \dot{x} + M_u L_G \dot{\theta} \cos\left(\theta\right) + M_{u} L_{G} \dot{\theta} \cos \left(\theta + \beta\right)
\frac{d}{dt} \left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) & = M_{eq} \ddot{x} + M_u \ddot{x} + M_u L_G \ddot{\theta} \cos\left(\theta +\beta\right) - M_{u} l_{G} \dot{\theta}^{2} \sin\left(\theta + \beta\right)
\frac{\partial \mathcal{L}}{\partial x} & = 0
\end{align}
So :
\begin{align}
\left(M_{eq}+M_{u}\right) \ddot{x} + M_{u} L_{G} \cos\left(\theta + \beta \right) \ddot{\theta} - M_{u} L_{G} \sin\left(\theta + \beta\right) \dot{\theta}^{2} = F_{ext,x}
\end{align}
The Lagrange equation according to $\theta$ gives : \begin{equation} \frac{d}{dt} \left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right) - \frac{\partial \mathcal{L}}{\partial \theta} = F_{ext,\theta} \end{equation}
With :
\begin{align}
\frac{\partial \mathcal{L}}{\partial \dot{\theta}} & = M_u L_G \dot{x} \cos\left(\theta + \beta\right) + M_u L_G^{2} \dot{\theta} + J_{G} \dot{\theta}
\frac{d}{dt} \left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right) & = M_u L_G \ddot{x} \cos\left(\theta + \beta\right) - M_u L_G \dot{x} \dot{\theta} \sin \left(\theta + \beta\right) + M_u L_G^{2} \ddot{\theta} + J_{G} \ddot{\theta}
\frac{\partial \mathcal{L}}{\partial \theta} & = - M_u L_G \dot{\theta} \dot{x} \sin\left(\theta +\beta\right) + M_u g L_G \sin \left(\theta + \beta\right)
F_{ext,\theta} & = 0
\end{align}
So finally :
\begin{align}
\left(J_{G} + M_{u} L_{G}^{2}\right) \ddot{\theta} + M_{u} L_{G} \cos\left(\theta + \beta\right) \ddot{x} & = M_{G} L_{G} g \sin\left(\theta + \beta\right)
\end{align}
The external actions are (with $C_m$ the motor torque) :
\begin{align}
F_{ext,z} & = F_{m} - F_{rr} - F_{aero}
F_{m} & = \frac{2 C_{m}}{R}
F_{rr} & = \left(M_{u}+M\right) g C_{rr}
F_{aero} & = \frac{1}{2} \rho C_{d} S \dot{x}^{2}
\end{align}
Here you will find a Modelica file with the implementation and use of these equations : Chap6_Segway.mo