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Algebra Translation

00:03

Here we have the same two link robot as we just looked at but this time we're going to

在這裡，我們有剛剛看到的相同的兩個雙聯桿件，但這次我們要

00:07

solve it using an analytical approach, that is we're going to rely much more on algebra,

使用分析方法解決它，也就是說我們將更多地依賴代數，

00:12

particular linear algebra rather than geometry. We have an expression E, which is the homogeneous

特定的線性代數而不是幾何。我們有一個表達式 E，它是齊次的

00:19

transformation which represents the pose of the robots endefector and we looked at this

代表機器人末端執行器的轉換，我們查看了這個

00:24

in the last lecture, we can write the endefector pose as a sequence of elementary homogeneous

在上一課中，我們可以將末端執行器寫為一系列基本齊次

00:31

transformations. A rotation by Q1, a translation along the X direction by A1, a rotation by

轉換。旋轉 Q1，沿 X 方向平移 A1，旋轉

00:38

Q2 and then a translation in the X direction by A2. If I expand this out, multiply all

Q2，然後由 A2 在 X 方向進行平移。如果我展開這個，乘以所有

00:45

the transformations together, I get the expression shown here; a three by three homogeneous transformation

一起轉換，我得到這裡顯示的表達式；三乘三同構變換

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matrix representing the pose of the robot's endefector.

表示桿件尾端的矩陣。

00:56

Now for this particular two link robot, we are only interested in the position of its

現在對於這個特殊的雙連桿件，我們只對它的位置感興趣

01:00

endefector, it's X and Y co-ordinate and they are these two elements within the homogeneous

末端執行器，它是 X 和 Y 坐標，它們是同質中的這兩個元素

01:06

transformation matrix, so I'm going to copy those out. So here again is our expression

轉換矩陣，所以我要把它們複製出來。所以這裡又是我們的表達

01:11

for X and Y and what we're going to do is a fairly common trick, we're going to square

對於 X 和 Y 我們要做的是一個相當常見的技巧，我們要平方

01:16

and add these two equations and I get a relationship that looks like this. Now I can solve for

並添加這兩個方程，我得到一個看起來像這樣的關係。現在我可以解決

01:23

the joint angle Q2 in terms of the endefector pose X and Y and the robot's constants A1 and A2.

關節角度 Q2 根據 endefector 姿勢 X 和 Y 以及機器人的常數 A1 和 A2。

01:31

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

現在我要做的是應用角度恆等式。我要擴展這些

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terms, sine of Q1 plus Q2 or cos of Q1 plus Q2 and to make life a little bit easier, I'm

條款，Q1 的正弦加 Q2 或 Q1 的 cos 加 Q2 並讓生活更輕鬆一點，我是

01:43

going to make some substations, so where ever I had cos Q2, I'm going to write C2 and where

打算做一些變化，所以我在哪裡有 cos Q2，我要寫 C2 和哪裡

01:48

ever I had sine Q2, I’m going to write S2. It's a fairly common shorthand when people

曾經我有正弦 Q2，我要寫 S2。這是一個相當常見的速記，當人們

01:54

are looking at robot kinematic equations. And here are the equations after making those

正在研究機器人運動學方程。這是製作後的方程式

01:58

substitutions. Looking at these two equations, I can see that they fall into a very well

替代品。看這兩個方程，我可以看出它們落入了一個很好的

02:04

known form and for that form there is a very well known solution. So I'm going to consider

已知形式，對於該形式，有一個眾所周知的解決方案。所以我要考慮

02:10

just one of the equations, the equation for Y and using our well known identity and it's

只是其中一個方程，Y 的方程並使用我們眾所周知的身份，它是

02:15

solution, I can determine the values for the variables little a, little b and little c

解決方案，我可以確定變量小 a、小 b 和小 c 的值

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and once I've determined those, then I can just write down the solution for Q1, which

一旦我確定了這些，那麼我就可以寫下 Q1 的解決方案，即

02:27

is the equivalent of theta in this particular case.

在這種特殊情況下相當於特殊案例。

02:31

Here again is our expression for Q1, copied over from the previous slide and we may remember

這又是我們對 Q1 的表達，從上一張幻燈片複製過來，我們可能還記得

02:36

from earlier in our workings that we determined this particular relationship; X squared plus

在我們工作的早期，我們確定了這種特殊的關係； X平方加

02:41

Y squared is equal to this particular complex expression. So I can substitute that in and

Y 平方等於這個特定的複數表達式。所以我可以用和代替它

02:47

do some simplification and I end up with this slightly less complex expression for Q1. And

做一些簡化，我最終得到了 Q1 的這個稍微不那麼複雜的表達式。和

02:53

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

它與我在上一節中遵循幾何方法得到的表達式相同。

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