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:

- Part 1. Introduction
- Part 2. Angle as a function of length
- Part 3. Coordinates of motor
- Part 4. Torque

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:

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

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Last update : 05/20/2022