How Do You Calculate Robot Arm Motion to Reach Specific Coordinates?

[Kernul]

Homework Statement

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:
$$P(t) = \left(x_p(t) = 40 + 5 cos\left(\frac{2 \pi}{60t + \pi}\right), y_p(t) = 11 + 5 sin\left(\frac{2 \pi t}{60 + \pi}\right)\right)$$

Homework Equations

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?

 

[kuruman]

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

 

[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?

 

[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.

 

[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?

 

[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 (x_f, y_f) and have its tip at (x_p, y_p) both of which you know.

 

[Kernul]

The internal angles of a any closed four-sided figure add to 2π, always.

Oh I get it.

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.

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

 

[kuruman]

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

 

[Kernul]

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

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 $sin(2\pi - \alpha_3)$?

 

[kuruman]

So I still have to find the distance between these two points to get the hypotenuse and then draw a right triangle, right?

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

 

[Scott Kotzur]

I am brushing up on some equations of motion stuff for a project I am working on at work and I stumbled across this old problem. If it's more appropriate for me to start a new thread, please say so and I will do that instead. I figured since this was never fully resolved that it may still be appropriate for me to post here. I apologize in advance for not quite knowing the proper formatting procedure. Any pointers on that and how to insert proper formulas would be appreciated. Also, if any of my wordy descriptions require diagrams I'm more than happy to make them.

I believe I have properly solved for the specific case when (X_p, Y_p)=(35,11). I am curious, however, whether or not there is a more eloquent way of solving it. It becomes cumbersome when written out on paper and when implementing something like this into code.

Starting where this previously left off, I calculated the value of α₃ using the coordinates of F and P. Specifically, α₃=atan((Y_p-Y_f)/(X_p-X_f)

From that I determined (X₃, Y₃) using more right triangle trigonometry, specifically (X₃, Y₃) = ((X_p,Y_p)+L₃(cos(α₃+π),sin(α₃+π))

Next, between (X_b, Y_b) and (X₃, Y₃) I created another vector and which I used to find α₁. α₁ would be the sum of the angles formed by 2,b,3 which I called α_x and the angle formed between 3b and a horizontal line, which I called α_w.

α_w was found using right angle trig, specifically α_w = atan((X₃-X_b)/(Y₃-Y_b))

The length of the vector B3 was also calculated B3=√((X₃-X_b)²+(Y₃-Y_b)²)

α_x was found using the law of cosines, being given L₁, L₂ and having calculated B3. α_x = acos((L₂²-L₁²-B3²)/(-2L₁B3))

α₁ =α_x+α_w

The inside angle at 2 was also found using the law of cosines on the same triangle as the one above. I'll call this α_ia2

α₂ was found by forming a right triangle using points B and 2 and subtracting the inside angle formed at 2 with this right triangle from the inside angle found at 2 in the immediately preceding step. The end result after some cancelling and whatnot is:
α₂ = α_ia2-π+α₁

I am fairly confident that this is all correct. For the specific instance where P= (35,11) I have calculated the values for α₁, α₂ and α₃ to be 2.059, 0.641, and -0.876 respectively. As stated at the beginning of the post, this feels "clunky" and perhaps I'm missing a more eloquent way of doing it.

Beyond that, the first major snag that I run into is using the equations for P(t). Specifically X_P(t)= 40+5cos(2π/(60t+π)). Can someone please confirm that the t does not belong in the denominator of the cosine function? It would appear as if there may have been a typo and the equation should read:
X_P(t)= 40+5cos(2πt/(60+π))

 

[kuruman]

Please post this on a new thread, with a new diagram of your project and the questions you wish to ask. The fact that the original poster did not confirm that the issue has been resolved, does not necessarily mean that it has not as far as OP is concerned. As for the 60t +π in the denominator, I agree. That looks like a typo. Note that points $x_P$ and $y_P$ lie on a circle whose center is offset from the origin. After a complete revolution, the values of $x_P$ and $y_P$ must be the same. The $t$ in the denominator does not allow that.