Model of a rotary joint driven by a linear motor [Part 3]

This page is part of an article about rotary joint driven by a linear motor. Please, start by reading the introduction.

This article is splitted in three parts:

Coordinates of motor

Coordinates of A

To compute the torque as a function of the motor length, we'll need the expression of the vector $ \overrightarrow{AB} $. Let's start by calculating the coordinates of $A=(x_A, y_A)$:

$$ \begin{split} x_A = | \overrightarrow{OA} | \times \cos ( \pi - \alpha ) \\ y_A = | \overrightarrow{OA} | \times \sin ( \pi - \alpha ) \end{split} $$

Since $ \cos(\pi-\theta) = -\cos(\theta) $ and $ \sin(\pi-\theta) = \sin(\theta) $, the previous expression can be simplified:

$$ \begin{split} x_A &=& -| \overrightarrow{OA} | \times \cos ( \alpha ) \\ y_A &=& | \overrightarrow{OA} | \times \sin ( \alpha ) \end{split} $$

Coordinates of B

To get the coordinates of point $B$, we will calculate the intersection of circle $C_1$ and $C_2$:

$C_1$ is a circle of center $A$ and radius $ | \overrightarrow{AB} | $
$C_2$ is a circle of center $C$ and radius $ | \overrightarrow{CB} | $

You'll find a detailed demonstration of the intersections calculation on this page: How to calculate the intersection points of two circles?.

The distance between centers $ | AC | $ is given by:

$$ d = | \overrightarrow{AC} | = \sqrt {| \overrightarrow{BC} |^2 + | \overrightarrow{BA} |^2 } $$

if $ d > | \overrightarrow{BA} | + | \overrightarrow{BC} | $ the circles are too far apart and do not intersect;
if $ d < | | \overrightarrow{BA} | - | \overrightarrow{BC} | | $ one circle is inside the other and do not intersect;
if $ d = 0 $ and $ \left| \overrightarrow{BA} | = | \overrightarrow{BC} \right| $ the circles are merged and there are an infinite number of points of intersection;
if $ d = | \overrightarrow{BA} | + | \overrightarrow{BC} | $ there is a single intersection point;
if $ d < | \overrightarrow{BA} | + | \overrightarrow{BC} | $ there are two intersection points.

In the three first cases, there is no solution. The coordonates of point $B$ is given by:

$$ \begin{split} x_B &=& ~ x_1 ~\pm~ \dfrac{h}{d}(y_C - y_A) \\ y_B &=& ~ y_1 ~\pm~ \dfrac{h}{d}(x_C - x_A) \end{split} $$

where:

$ a = \dfrac{| \overrightarrow{AB} |^2 - | \overrightarrow{BC} |^2 + d^2 }{2d} $
$ h = \sqrt{ | \overrightarrow{AB} |^2 - a^2 } $
$ x_1 = x_A + \dfrac{a}{d} \times ( x_C - x_A ) $
$ y_1 = y_A + \dfrac{a}{d} \times ( y_C - y_A ) $

Part 2 Summary Part 4

Model of a rotary joint driven by a linear motor [Part 3]

Coordinates of motor

Coordinates of A

Coordinates of B

See also