- #1
jumbo1985
- 19
- 1
I have a particle that moves along a circular arc centered at origin in 2D plane.
I have the following angular displacement function for time T0 through T3 and the following acceleration and velocity constraints in Cartesian coordinates. At t = T0, theta = 0, velocity = 0. At t = T3, velocity = 0.
I want to calculate the maximum A_T and V_T in terms of the linear A_max and V_max quantities (A_max and V_max put a limit on A_T and V_T). I want to eventually be able to calculate T1, T2, T3.
[tex]
\forall t \in [T_0, T_1]
\begin{cases}
\frac{d^2\theta}{dt^2}(t) = A_T \\
\frac{d\theta}{dt}(t) = \dot{\theta}_{0} + A_T(t-T_0) \\
\theta(t) = \theta_{0} + \dot{\theta}_{0}(t-T_0)+\frac{1}{2}A_T(t-T_0)^2 \\
\end{cases}
\\
\forall t \in [T_1, T_2]
\begin{cases}
\frac{d^2\theta}{dt^2}(t) = 0 \\
\frac{d\theta}{dt}(t) = V_T \\
\theta(t) = \theta(T_1)+V_T(t-T_1) \\
\end{cases}
\\
\forall t \in [T_2, T_3]
\begin{cases}
\frac{d^2\theta}{dt^2}(t) = -A_T \\
\frac{d\theta}{dt}(t) = \dot{\theta}(T_2) - A_T(t-T_2) \\
\theta(t) = \theta(T_2) + \dot{\theta}(T_2)(t-T_2) - \frac{1}{2}A_T(t-T_2)^2 \\
\end{cases}
\\
\\
\vec{A_{max}} = \begin{pmatrix}
A_{max,x} \\
A_{max,y} \\
\end{pmatrix}
\\
\vec{V_{max}} = \begin{pmatrix}
V_{max,x} \\
V_{max,y} \\
\end{pmatrix}
\\
\vec{Position} =
\begin{pmatrix}
x \\
y \\
\end{pmatrix} =
\begin{pmatrix}
\cos(\theta(t)) \\
\sin(\theta(t)) \\
\end{pmatrix}
[/tex]
At first I figured I could solve
[tex]\vec A =
\begin{pmatrix}
\frac{d^2}{dt}[\cos(\theta(t))] \\
\frac{d^2}{dt}[\sin(\theta(t))]\\
\end{pmatrix}
[/tex]
for A_T at t = T0 with all of the initial conditions applied but I'm afraid this is not correct.
How do I approach this problem?
Thanks
I have the following angular displacement function for time T0 through T3 and the following acceleration and velocity constraints in Cartesian coordinates. At t = T0, theta = 0, velocity = 0. At t = T3, velocity = 0.
I want to calculate the maximum A_T and V_T in terms of the linear A_max and V_max quantities (A_max and V_max put a limit on A_T and V_T). I want to eventually be able to calculate T1, T2, T3.
[tex]
\forall t \in [T_0, T_1]
\begin{cases}
\frac{d^2\theta}{dt^2}(t) = A_T \\
\frac{d\theta}{dt}(t) = \dot{\theta}_{0} + A_T(t-T_0) \\
\theta(t) = \theta_{0} + \dot{\theta}_{0}(t-T_0)+\frac{1}{2}A_T(t-T_0)^2 \\
\end{cases}
\\
\forall t \in [T_1, T_2]
\begin{cases}
\frac{d^2\theta}{dt^2}(t) = 0 \\
\frac{d\theta}{dt}(t) = V_T \\
\theta(t) = \theta(T_1)+V_T(t-T_1) \\
\end{cases}
\\
\forall t \in [T_2, T_3]
\begin{cases}
\frac{d^2\theta}{dt^2}(t) = -A_T \\
\frac{d\theta}{dt}(t) = \dot{\theta}(T_2) - A_T(t-T_2) \\
\theta(t) = \theta(T_2) + \dot{\theta}(T_2)(t-T_2) - \frac{1}{2}A_T(t-T_2)^2 \\
\end{cases}
\\
\\
\vec{A_{max}} = \begin{pmatrix}
A_{max,x} \\
A_{max,y} \\
\end{pmatrix}
\\
\vec{V_{max}} = \begin{pmatrix}
V_{max,x} \\
V_{max,y} \\
\end{pmatrix}
\\
\vec{Position} =
\begin{pmatrix}
x \\
y \\
\end{pmatrix} =
\begin{pmatrix}
\cos(\theta(t)) \\
\sin(\theta(t)) \\
\end{pmatrix}
[/tex]
At first I figured I could solve
[tex]\vec A =
\begin{pmatrix}
\frac{d^2}{dt}[\cos(\theta(t))] \\
\frac{d^2}{dt}[\sin(\theta(t))]\\
\end{pmatrix}
[/tex]
for A_T at t = T0 with all of the initial conditions applied but I'm afraid this is not correct.
How do I approach this problem?
Thanks