第一部 Inverse kinematics for a 2-joint robot arm using geometry <<
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第二部 Inverse kinematics for a 2-joint robot arm using algebra
Inverse kinematics for a 2-joint robot arm using algebra
使用 2 軸的機器人手臂的逆運動學
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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鏈接機器人，但這次我們要
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，
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
transformation which represents the pose of the robots endefector and we looked at this
代表機器人 endefector 姿勢的轉換，我們查看了這個
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 endefector 姿勢寫為一系列基本齊次
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
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
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
matrix representing the pose of the robot's endefector.
表示機器人尾端姿態的矩陣。
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兩連桿機器人，我們只對它的位置感興趣
endefector, it's X and Y co-ordinate and they are these two elements within the homogeneous
endefector，它是 X 和 Y 坐標，它們是同質中的這兩個元素
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
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
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
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。
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
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
going to make some substations, so where ever I had cos Q2, I'm going to write C2 and where
打算做一些變電站，所以我在哪裡有 cos Q2，我要寫 C2 和哪裡
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
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
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
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
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
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
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
is the equivalent of theta in this particular case.
在這種特殊情況下相當於 theta。
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
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
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
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
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。
第一部 Inverse kinematics for a 2-joint robot arm using geometry <<
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