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**: ,

for **3-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[1][4]

y = H[2][4]

z = H[3][4]

That’s it. It is not difficult to calculate the direct kinematics for an n-link manipulator.

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