# Homework Help: Robot arm Equation of Motion

1. Oct 23, 2016

### Kernul

1. The problem statement, all variables and given/known data
My professor posted this problem on his page as a self-assessment for the students and said that you just need some analytical geometry notions and trigonometry to solve it, although I don't quite get how to solve it. This is the problem statement:
Considering the articulated mechanical structure in the picture (cinematic scheme of a robot).
In the hypothesis that:
$x_b = 10; y_b = 5;$ Coordinates of the base of the robot.
$L_1 = 15; L_2 = 20; L_3 = 25;$ Length of the arms.
$x_f = 30; y_f = 17;$ Coordinates of the hole.
Determine the angles $\alpha_1, \alpha_2, \alpha_3$ so that the extremity of the third arm coincides with the point $P$ of coordinates $(35, 11)$.
More in general, write an algorithm that allows to calculate the equations of motion $\alpha_1(t), \alpha_2(t), \alpha_3(t)$ so that you can draw the circumference:
$$P(t) = \left(x_p(t) = 40 + 5 cos\left(\frac{2 \pi}{60t + \pi}\right), y_p(t) = 11 + 5 sin\left(\frac{2 \pi t}{60 + \pi}\right)\right)$$

2. Relevant equations

3. The attempt at a solution
I really do not understand how this should be solved since I've never done this kind of problem.
Could someone explain step by step how it should be solved and what notions need to be used?

2. Oct 23, 2016

### kuruman

This looks like a vector addition problem. Start at the origin and move to the tip at {xP, yP} using the tip of one vector as the starting point of the next. For example, to get to {x2, y2}, you need to add
{xb, yb} + L1{ cosα1, sinα1} and so on.

3. Oct 23, 2016

### Kernul

So something like this?
$$(x_b, y_b) + L_1(cos\alpha_1, sin\alpha_1) + L_2(cos\alpha_2, sin\alpha_2) + L_3(cos\alpha_3, sin\alpha_3) = (10, 5) + 15(cos\alpha_1, sin\alpha_1) + 20(cos\alpha_2, sin\alpha_2) + 25(cos\alpha_3, sin\alpha_3)$$
But how do I get to know the equations for the three alphas? I would just get a vector that points to the hole? Or did I misunderstand?

4. Oct 23, 2016

### kuruman

You did not misunderstand. You know the coordinates of the hole, so you can write two equations for the x and y components of the resultant vector that relate the three alphas. You need a third equation relating the alphas. You get that by observing that the four internal angles of the four-sided figure formed by the three vectors and the resultant add to 2π.

5. Oct 23, 2016

### Kernul

So I would have this:
$$(10 + 15cos\alpha_1 + 20cos\alpha_2 + 25cos\alpha_3, 5 + 15sin\alpha_1 + 20sin\alpha_2 + 25sin\alpha_3) = (30, 17)$$
and these two equations:
$\begin{cases} 10 + 15cos\alpha_1 + 20cos\alpha_2 + 25cos\alpha_3 = 30 \\ 5 + 15sin\alpha_1 + 20sin\alpha_2 + 25sin\alpha_3 = 17 \end{cases}$
But I don't quite get how is it possible that the four angles add to $2\pi$. It doesn't look like that too me.
And, by the way, isn't adding a fourth angle $\alpha_4$ making things impossible since I'll have three equations with four unknowns?

6. Oct 23, 2016

### kuruman

The internal angles of a any closed four-sided figure add to 2π, always. Also, you can find α3 quite easily (I realized later) because L3 must pass through the hole at (xf, yf) and have its tip at (xp, yp) both of which you know.

7. Oct 25, 2016

### Kernul

Oh I get it.

So I multiply $L_3$ to the difference between the point $F$ and the point $P$ in order to get the angle $\alpha_3$?

8. Oct 25, 2016

### kuruman

Not the angle, a trig function of the angle. Draw a right triangle with a piece of L3 as hypotenuse.

9. Oct 26, 2016

### Kernul

Just a piece of $L_3$? The piece between the point $F$ and the point $P$? So I still have to find the distance between these two points to get the hypotenuse and then draw a right triangle, right? Would it be $sin(2\pi - \alpha_3)$?

10. Oct 26, 2016

### kuruman

If you know the coordinates of points P and F you know the distance between them.