Phantom Kinematics
Phantom Kinematics Model

Forward Kinematics

To define the end point of the manipulator, the position $P(x1,y1)$ was need in terms of the angles $\theta_{2}$ and $\theta_{3}$. This can be done with the formula:

\begin{equation} P_{1}=O^{1}+R_{2}^{1}P^{2} \end{equation}

where $P^{1}$ is te position of $P(x,y)$ using the (x1,y1) coordinae system as the global coordinates.

$O^{1}=\left[\begin{array}{c} L_{1}cos(\theta_{2})\\ L_{2}sin(\theta_{2})\end{array}\right]$ , is the orientation of the end point of the first handle about its own coordinate system (x,y).

$R_{2}^{1}=\left[\begin{array}{cc} cos(\theta_{2}) & -sin(\theta_{2})\\ sin(\theta_{2}) & cos(\theta_{2})\end{array}\right]$, is the transofrmation matrix of the (x2,y2) coordinate system about (x1,y1)

$P^{2}=\left[\begin{array}{c} L_{2}sin(\theta_{3}-\theta_{2})\\ -L_{2}cos(\theta_{3}-\theta_{2})\end{array}\right]$ i the orientation of the end point of the second handle about its own coordinate system (x2,y2)

Backward Kinematics

To find backward kinematics, I have used just simple trigonometric structure and functions:

\begin{array} {c} cos(\theta_{2})L_{1}+sin(\theta_{3})L_{2}=P(x)\\ sin(\theta_{2})L_{1}-cos(\theta_{3})L_{2}=P(y) \end{array}

now using the equation $sin^{2}(\alpha)+cos^{2}(\alpha)=1$ I have eliminated the $sin(\theta_{3})$ which led to the equation : $L^{2}_{2}=(x-cos(\theta_{2})L1)^{2}+(sin(\theta_{2})L1-y)^{2}$ which led to $(x^{2}+y^{2})sin^{2}(\theta_{2})-2bysin(\theta_{2})+b^{2}-x^{2}=0$ where $b = \frac{L_{2}^{2}-x^{2}-y^{2}-L^{2}_{1}}{-2L_{1}}$ gives me the $sin(\theta_{2})$

Wheels Ratio

The big wheel is 115mm diameter. The motor hub is 9mm diameter.
Therefore the ratio between the circumference of the big wheel to the small wheel is 12.7.
This means that for every rotation of the big wheen, the small one makes 12.7 rotations.
Therefore an angle ($\theta_{3}$ or $\theta_{2}$) equals $\theta = ((number of rotations of the motor)*360^\circ)/12.7$

Copyright (c) 2009