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Algebra Translation

Inverse kinematics for a 2-joint robot arm using algebra 翻譯

0:04

Here we have the same two link robot as we just looked at but this time we're going tosolve it using an analytical approach , that is we're going to rely much more on algebra ,particular linear algebra rather than geometry .

在這裡，我們有與前一篇相同的兩連桿機械手臂，但這次我們將使用分析方法來解決它，意味著我們將更多地依賴代數，特別是使用線性代數而不是幾何。

0:16

We have an expression E , which is the homogeneous transformation which represents the pose of the robots endefector and we looked at this in the last lecture , we can write the endefector pose as a sequence of elementary homogeneous transformations .

我們有一個表達式 E，它是表示機械手臂最終位置的齊次變換式，我們在上一課中看到了這一點，我們可以將最終位置寫為一系列基本齊次變換式。

0:33

A rotation by Q1 , a translation along the X direction by A1 , a rotation by Q2 and then a translation in the X direction by A2 .

Q1 旋轉，A1 沿 X 方向平移，Q2 旋轉，然後 A2 沿 X 方向平移。

0:43

If I expand this out , multiply all the transformations together , I get the expression shown here ; a three by three homogeneous transformation matrix representing the pose of the robot's endefector.

如果我將其展開，將所有變換相乘，就會得到此處所示的表達式 E ;它是一個三乘三的齊次變換矩陣，表示機械手臂的最終的位置。

0:56

Now for this particular two link robot , we are only interested in the position of its endefector , it's X and Y co - ordinate and they are these two elements within the homogeneous transformation matrix , so I'm going to copy those out .

現在對於這個特殊的兩連桿機器人，我們只對它的最終位置感興趣，與它有關的是 X 和 Y 坐標，它們是齊次變換矩陣中的這兩個元素，所以我將把它們複製出來。

1:10

So here again is our expression for X and Y and what we're going to do is a fairly common trick , we're going to square and add these two equations and I get a relationship that looks like this .

所以這裡又是我們對 X 和 Y 的表達式，我們要做的是一個相當常見的技巧，我們要將它平方並將這兩個方程相加，我得到一個看起來像這樣的關係式。

1:22

Now I can solve for the joint angle Q2 in terms of the endefector pose X and Y and the robot's constants A1 and A2.

現在我可以根據最終位置X和Y以及機械手臂的常數A1及A2來求出關節角度Q2。

1:32

Now what I'm going to do is apply the sum of angles identity .

現在我要做的是應用角度之和的特性。

1:35

I'm going to expand these terms, sine of Q1 plus Q2 or cos of Q1 plus Q2 and to make life a little bit easier, I'm going to make some substations, so where ever I had cos Q2, I'm going to write C2 and where ever I had sine Q2, I'm going to write S2.

我將展開這些項，Q1 的正弦加 Q2 或 Q1 的餘弦加 Q2，為了讓生活更輕鬆一點，我將建立一些分部，所以只要有 cos Q2，我會將其取代為C2並且在有正弦Q2的地方，我會將其取代為S2。

1:51

It's a fairly common shorthand when people are looking at robot kinematic equations.

當人們解決機械手臂運動學方程式時，這是一個相當常見的快速記法。

1:57

And here are the equations after making those substitutions.

這是進行替換後的方程式。

2:00

Looking at these two equations, I can see that they fall into a very well known form and for that form there is a very well known solution.

看看這兩個方程式，我可以看到它們屬於一個眾所周知的形式，對於這種形式，有一個淺顯易見的解決方案。

2:09

So I'm going to consider just one of the equations, the equation for Y and using our well known identity and it's solution, I can determine the values for the variables little a, little b and little c and once l've determined those, then I can just write down the solution for Q1, which x is the equivalent of theta in this particular case.

所以我將只考慮其中一個方程，Y 的方程，並使用我們眾所周知的恆等式和它的解，我可以確定變量小 a、小 b 和小 c 的值，一旦我確定了這些，然後我可以寫下 Q1 的解決方程，在這種特殊情況下，x 相當於 θ。

2:31

Here again is our expression for Q1, copied over from the previous slide and we may remember from earlier in our workings that we determined this particular relationship; X squared plus Y squared is equal to this particular complex expression.

這裡再次是我們對 Q1 的表達，從上一張幻燈片複製過來，我們可能還記得在我們早期的說明中，確定了這種特殊關係； X 平方加 Y 平方等於這個特定的複雜表達式。

2:45

So I can substitute that in and do some simplification and I end up with this slightly less complex expression for Q1.

因此，我可以將其替換並進行一些簡化，最終得到 Q1 的這個稍微不那麼複雜的表達式。

2:53

And it is the same expression that I got following the geometric approach in the previous section.

這與我在上一節中遵循幾何方法求出的表達式一模一樣。

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