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第二部

國文：

這裡我們有與我們相同的兩個鏈接機器人只是看著，但這次我們要使用分析方法解決它，即我們是否會更多地依賴代數，特定的線性代數而不是幾何。我們有一個表達式 E，它是齊次的表示姿勢的變換機器人終結者，我們看了這個在上一課中，我們可以編寫 endefector 構成一個基本齊次序列轉換。 Q1 的旋轉，平移沿 X 方向 A1，旋轉 Q2 然後在 X 方向上平移通過 A2。如果我展開這個，乘以所有一起轉換，我得到了表達顯示在這裡；三乘三同構變換表示機器人姿態的矩陣終結者。現在對於這個特殊的兩連桿機器人，我們只對它的位置感興趣 endefector，它是 X 和 Y 坐標，它們這兩個元素是同質的嗎變換矩陣，所以我要復制那些出來。所以這裡又是我們的表達對於 X 和 Y 我們要做的是一個相當常見的技巧，我們要平方並添加這兩個方程，我得到了一個關係看起來像這樣。現在我可以解決根據 endefector 的關節角度 Q2 姿勢 X 和 Y 以及機器人的常數 A1 和 A2。現在我要做的是應用總和的角度身份。我要擴展這些項，Q1 的正弦加 Q2 或 Q1 的餘弦加 Q2 為了讓生活更輕鬆一點，我打算建造一些變電站，所以無論在哪裡我有 cos Q2，我要寫 C2 以及在哪裡曾經我有正弦 Q2，我要寫 S2。這是一個相當常見的速記，當人們正在研究機器人運動學方程。這是製作後的方程式替代品。看這兩個方程，我可以看到他們掉進了一個很好的井裡已知的形式，對於這種形式，有一個非常眾所周知的解決方案。所以我要考慮只是其中一個方程，方程為 Y 並使用我們眾所周知的身份解決方案，我可以確定變量小 a、小 b 和小 c 一旦我確定了這些，我就可以只需寫下 Q1 的解決方案，即在這個特定的情況下相當於 theta 案件。這裡再次是我們對 Q1 的表達式，複製從上一張幻燈片結束，我們可能還記得從我們早期的工作中我們確定這種特殊的關係； X 平方加 Y 平方等於這個特定的複數表達。所以我可以用和代替它做一些簡化，我最終得到這個 Q1 的表達式稍微簡單一些。和這是我得到的相同表達上一節中的幾何方法。

英文：

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

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

particular linear algebra rather than geometry.
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. 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. 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.

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. 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. 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.

Now what I'm going to do is apply the sum
of angles identity. 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.
It's a fairly common shorthand when people

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

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

is the equivalent of theta in this particular
case.

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. So I can substitute that in and

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

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

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