How do I solve this robot arm movement?

Question

I am moving my robot arm's tool pitch. In this example, the tool is moving pitch by 80 degrees (lime green number).

The red lines are the current position of the arm.

The blue lines are the position the robot needs to be moved to.

The 3 dark green numbers are the angles of the red arm.

I need to find the 3 pink angles. How do I do this?

Answer 1

I got the distance the arm needed to move in the X direction and the Y direction first.

TPitch

Then I moved in the X and Y directions.

X

Y

Answer 2

Labelling the joints as $A$ (bottom right), $B$ (top), $C$ (left), $D$ (end-effector). We can write the coordinates of the end-effector for the initial position as follows

$D = 100(\cos(100^\circ), \sin(100^\circ)) + 90(\cos((100-180-80)^\circ, \sin(100-180-80)^\circ) + 5 (\cos((-160 + 180 - 120)^\circ, \sin((-160+180-120)^\circ) ) $

And this simplifes to

$D = 100 (\cos(100^\circ), \sin(100^\circ) ) + 90 (\cos(-160^\circ), \sin(-160^\circ) ) + 5 (\cos(-100^\circ), \sin(-100^\circ) ) \hspace{25pt}(1)$

Now using the pink angles $t_1, t_2, t_3$, we can write the coordinates of the tip as follows

$D = 100( \cos(\pi - t_1),\sin(\pi - t_1)) + 90 (\cos(-t_1 - t_2), \sin(-t_1 - t_2)) + 5 (\cos( \pi - t_1 - t_2 + t_3), \sin(\pi - t_1 - t_2 + t_3 ) )\hspace{25pt}(2)$

In addition, we know that the difference in direction between the directions of the last link between the initial configuration and the final configuration is $80^\circ$, and therefore, we can write

$ -100^\circ - (180^\circ - t_1 - t_2 + t_3) = 80^\circ $

From which,

$ t_1 + t_2 - t_3 = 80^\circ + 280^\circ = 360^\circ = 0^\circ $

Thus, $t_3 = t_1 + t_2 $

Now, from equations $(1)$ and $(2)$, we get two equations for the two components $x$ and $y$ of the tip $D$, and using trigonometric identities,

$-100 \cos(t_1) + 90 \cos(t_1 + t_2) = 100 \cos(100^\circ) + 90 \cos(160^\circ) + 5 \cos(100^\circ) + 5\hspace{25pt}(3) $

$ 100 \sin(t_1) - 90 \sin(t_1 + t_2) = 100 \sin(100^\circ) - 90 \sin(160^\circ) - 5 \sin(100^\circ) \hspace{25pt}(4)$

Equations $(3),(4)$ are of the form

$ - a \cos(t_1) + b \cos(t_1 + t_2) = c_1 \hspace{25pt}(5) $

$ a \sin(t_1) - b \sin(t_1 + t_2) = c_2 \hspace{25pt}(6)$

where $a = 100, b = 90$ and

$ c_1 = 100 \cos(100^\circ) + 90 \cos(160^\circ) + 5 \cos(100^\circ) + 5$

$ c_2 = 100 \sin(100^\circ) - 90 \sin(160^\circ) - 5 \sin(100^\circ) $

Squaring equations $(5), (6)$ and adding results in

$ a^2 + b^2 - 2 a b \cos(t_2) = c_1^2 + c_2^2 $

So that,

$ t_2 = \cos^{-1} \bigg( \dfrac{ a^2 + b^2 - c_1^2 - c_2^2 }{2 a b } \bigg) $

Using the values of $a, b, c_1, c_2$, we get the following numerical value for $t_2$

$t_2 = 1.31275 \text{ radian} = 75.215^\circ $

Now equations $(5), (6)$ can be expanded as follows

$ - a \cos(t_1) + b (\cos(t_1)\cos(t_2) - \sin(t_1)\sin(t_2) ) = c_1 \hspace{25pt}(7) $

$ a \sin(t_1) - b (\sin(t_1)\cos(t_2) + \cos(t_1)\sin(t_2) = c_2 \hspace{25pt}(8)$

Which are of the form

$ \begin{bmatrix} -a + b \cos(t_2) && -b \sin(t_2) \\ - b \sin(t_2) && a - b \cos(t_2) \end{bmatrix} \begin{bmatrix} \cos(t_1) \\ \sin(t_1) \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} \hspace{25pt} (9)$

Now we can solve for $\cos(t_1) $ and $\sin(t_1)$, and we get

$ \cos(t_1) =0.1533733473, \hspace{25pt} \sin(t_1) = 0.9881683137 $

From which $ t_1 = 1.4168152 \text{ radian} = 81.177^\circ $

And finally,

$t_3 = t_1 + t_2 = 2.72956769 \text{ radian} = 156.3927^\circ $

Both configurations, initial and final are plotted below, with the red lines indicating the initial configuration, and the blue lines the final configuration.