Inverse kinematics for a 2-joint robot arm using geometry 翻譯 <<
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Inverse kinematics for a 2-joint robot arm using algebra 翻譯
Inverse kinematics for a 2-joint robot arm using algebra
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,
在這裡，我們有與剛才看到的相同的兩個鏈接機器人，但這次我們將使用分析方法來解決它，也就是說，我們將更多地依賴代數，
particular linear algebra rather than geometry.
特定的線性代數而不是幾何。
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，它是表示機器人 endefector 位姿的齊次變換，我們在上一課中看過這個，我們可以將 endefector 位姿寫為一系列基本齊次變換。
A rotation by Q1, a translation along the X direction by A1,
Q1 的旋轉，A1 沿 X 方向的平移，
a rotation by Q2 and then a translation in the X direction
by A2.
旋轉 Q2，然後沿 X 方向平移通過 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.
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； 三乘三同構變換
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 endefector, it's X and Y co-ordinate and they are these two elements within the homogeneous transformation matrix,
現在對於這個特殊的雙連桿機器人，我們只對它的末端位置感興趣，它是 X 和 Y 坐標，它們是齊次變換矩陣中的這兩個元素，
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.
所以我要把這些複製出來。 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，我們要平方並添加這兩個方程，我得到一個看起來像這樣的關係。
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 endefector 求解關節角度 Q2
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 terms,
現在我要做的是應用角度恆等式。 我將擴展這些術語，
sine of Q1 plus Q2 or cos of Q1 plus Q2 and to make life a little bit easier,
Q1 的正弦加 Q2 或 Q1 的 cos 加 Q2 並使式子簡單，
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,
我打算做一些變電站，所以只要有 cos Q2，我就會寫 C2，只要有正弦 Q2，
I’m going to write S2. It's a fairly common shorthand when people are looking at robot kinematic equations.
我要寫S2。 當人們查看機器人運動學方程時，這是一個相當常見的速記。
And here are the equations after making those substitutions. Looking at these two equations,
這是進行這些替換後的方程式。 看這兩個方程，
I can see that they fall into a very well known form and for that form there is a very well known solution.
我可以看到它們屬於一種眾所周知的形式，對於這種形式，有一個眾所周知的解決方案。
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,
所以我將只考慮其中一個方程，即 Y 的方程，並使用我們眾所周知的恆等式及其解，我可以確定變量小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.
小 b 和小 c，一旦我確定了它們，我就可以
只需寫下 Q1 的解決方案，在這種特殊情況下，它相當於 theta。
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;
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工作的早期，我們確定了這種特殊關係；
X squared plus Y squared is equal to this particular complex expression.
X 平方加 Y 平方等於這個特定的複雜表達式。
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。
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得到的表達式相同。
Inverse kinematics for a 2-joint robot arm using geometry 翻譯 <<
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