Model of a rotary joint driven by a linear motor [Part 2]

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:

Angle as a function of length

Let's calculate the angle $ \alpha $ as a function of the motor length $ | \overrightarrow{AB} | $.

Let's start by calculating the length $ | \overrightarrow{AC} | $. Since the triangle $ABC$ is rectangle in $B$, let's write the Pythagorean theorem:

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

To get the rotation angle of the joint ($ \alpha $), let use the law of cosines in the triangle $ OAC $:

$$ | \overrightarrow{AC} |^2 = | \overrightarrow{OA} |^2 + | \overrightarrow{OC} |^2 - 2 | \overrightarrow{OA} | | \overrightarrow{OC} | \cos(\alpha) $$

Let's replace $ | \overrightarrow{AC} |^2 $ by the previous equation:

$$ | \overrightarrow{BC} |^2 + | \overrightarrow{BA} |^2 = | \overrightarrow{OA} |^2 + | \overrightarrow{OC} |^2 - 2 | \overrightarrow{OA} | | \overrightarrow{OC} | \cos(\alpha) $$

We can rewrite this equation:

$$ \cos(\alpha) = \dfrac{| \overrightarrow{OA} |^2 + | \overrightarrow{OC} |^2 - | \overrightarrow{BC} |^2 - | \overrightarrow{BA} |^2 }{2 | \overrightarrow{OA} | | \overrightarrow{OC} |} $$

The angle $\alpha$ is given by:

$$ \alpha = \arccos \left( \dfrac{| \overrightarrow{OA} |^2 + | \overrightarrow{OC} |^2 - | \overrightarrow{BC} |^2 - | \overrightarrow{BA} |^2 }{2 | \overrightarrow{OA} | | \overrightarrow{OC} ) |} \right) $$

Part 1 Summary Part 3

Model of a rotary joint driven by a linear motor [Part 2]

Angle as a function of length

See also