In the last article I have already described which question answers direct kinematics, and now we will look at examples of how to do this.
How to find the end position for 2-links manipulator? Geometry will tell! Obviously:
All right, but the same for 3-links ?
Also kind of understandable. Even noticeable logic.
Will you do for a 6-links ? – Nonono, David Blaine. Thanks, I’ll see.
And this is a real industrial manipulator. Guess what.
Lazy people like me come up with such a thing as rotation matricies. You don’t even have to think with them where cosines are, where are sinuses. Look,
for 2-links: ,
The main thing is not to get lost in the coordinate axes. Place the robot in the axial position (as drawn 3 and 6 links manipulators) and write down in turn for joint and for link.
And now let’s figure it out for a 6-links:
And that’s it!
So, stop, stop. And what is it? Where are cosines, where are x’s?
The matrix looks like (for the convenience of multiplication immediately bring to the size of 4×4):
And matrices T:
With matrices sorted out, but what do we get? What is H?
H – homogeneous transformation matrix. As you might guess it consists of rotation matrix R(3×3) and translation T(3×1) and added to 4×4.
To get the answer x, y, z, you need to take the corresponding elements:
x = H
y = H
z = H
That’s it. It is not difficult to calculate the direct kinematics for an n-link manipulator.