Robot Arm Controller
Let us look at another example of a problem in which the matrix will be useful. The robot arm problem is a good choice, since we’ve already discussed it. Thus, should be a good idea for us to discuss it together, since you could meet some problems. Then we can examine how to solve it using matrices. OK, so, first let’s review the problem. You will see the problem now. The first thing we’re going to do is solve it. Then, after that, we’re going to talk about how there’s a matrix hidden in this problem. OK, so here it is. Here’s a robot arm with two joints. Here’s a joint at the origin, and then there’s a joint over here. And in between them there’s a rod of length L, and you can make L stretch. And then there’s another joint over here. There’s an angle theta between them, and this has length of square root of 2, and at the end of it is xy, a fingertip of the robot. The first part of the problem was to find xy in terms of L and theta, and everyone did fine on that. So here is your formula: x equals L plus square root 2 cosine of theta and y equals square root 2 sine. Now, if you assume that L equals 1 and theta equals pi over 4, then you can find out that x is 2 comma The question is, how can we adjust L and theta? So that xy is 2 comma 1.1. In other words, we’d like to lift the tip straight up a distance 0.1. As a matter of fact, a question will come now, which is the most common question that people have trouble with So evidently, everybody recognized that we want to use linear approximation to solve this problem, since the algebra would be too complicated to do it exactly. The question is: What function in what variables should we approximate with a linear function? One option is to approximate x as a function of L and theta. Another is to approximate L and theta as a function of x and y. A third option is to approximate some other function f, which we have to figure out what it is could approximate that. Think of that as a function of x and y. Who should we think of as a function of who? [Graph] (x,y)=(L+sqrt2cosθ, sqrt2sinθ) initialy L=1,θ=pi/4 =>(x,y)=(2,1) How adjust L,θ so(x,y)=(2,1.1) [Graph] [Graph]
