"Simple" math question about length and rotation relations

Question

I'm currently building a robot arm as a hobby, and I'm still in the planning phase. But I've encountered a small problem, where my knowledge doesn't suffice.

This is what I am trying to achieve:

I need to calculate the length of the two hydraulics, for the "hand" to be at a certain position.

Let's say I want to move the hand to the length indicator for the second arm. how would I calculate the corresponding hydraulic lengths ?

This is of course a 2d representation of what will be "3d". but I just need some guidelines, and I should be able to figure the rest out.

Answer 1

Once you have those angles, we can solve this with some trigonometry on the small triangle formed by the pivot point and the points where the hydraulics meet the arms.

Let's call the pivot point $A$, and the hydraulic meeting points $B$ and $C$. Then the law of cosines says $a^{2} = b^{2} + c^{2} - 2bc \cos(A)$. In this case, $b$ and $c$ are the fixed lengths from the pivot point to where the hydraulics are attached to the arms, and the angle at $A$ is just the desired angle for that arm. From there, you can calculate $a$, the necessary length of the hydraulic.

Answer 2

Hint:

Given the situation, say $r_1$ is the first arm, and $r_2$ is the second arm. Then

$$ x = r_1 \cos(\phi_1) + r_2 \cos(\phi_2)\\ y = r_1 \sin(\phi_1) + r_2 \sin(\phi_2) $$

When you solve the angles $\phi_1$ and $\phi_2$ you can find the hydraulics lengths...