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Subtitle translation:

Title:Inverse kinematics for a 2-joint robot arm using geometry

標題:反向運動學對使用幾何體的2個關節機械臂

We saw this simple two-link robot in the previous lecture about forward kinematics

我們在上一堂關於正向運動學的課程中看到了這個簡單的雙連桿機構

The tooltip pose of this robot is described simply by two numbers, the coordinates x and y with respect to the world coordinate frame.
該機構的姿態由兩個簡單的數字描述，即相對於絕對坐標系的坐標 x 和 y

So,the problem hwre us that given x and y.we want to determine the joined angles,Q1 and q2.
所以，給我們 x 和 y 的問題是我們想要確定連接角 Q1 和 Q2是多少

The solution that we’re going to follow in this particular section is a geometric one.

我們將用一些幾何的技巧來解決這個問題

We’re going to start with a simple piece of construction.
我們將從一個簡單的結構開始

We’re going to overlay the red triangle on top of our robot.
我們將在機構上覆蓋紅色三角形

We know that the end point coordinate is x, y, so the vertical height of the triangle is y, the horizontal width is x
我們知道終點坐標是x，y，所以三角形的垂直高度是y，水平寬度是x

And, using Pythagoras theorem, we can write r squared equals x squared plus y squared.
並且，使用畢氏定理，我們可以寫出 r 平方等於 x 平方加上 y 平方

So far,so easy
到目前為止都很容易

Now, w’re going to look at this triangle highlighted here in red and we want to determine the angle alpha.
現在，我們將查看此處紅色突出顯示的三角形，我們要決定α角

In order to do that, we need to use the cosine rule.
為了做到這一點，我們需要使用餘弦定理

And, if you’re a little rusty on the cosine rule, here is a bit of a refresher.
而且，如果你對餘弦定理有點生疏，這裡有一點複習

We have an arbitrary triangle.
我們這裡有一個任意三角形

We don’t have any right angles in it and we’re going to label the length of this edge as A and the angle opposite that edge, we’re going to label as little a

我們沒有任何直角，我們將這條邊的長度標記為 A，而該邊的對面角，我們將標記為 a

And, we do the same for this edge and this angle, and this edge and this angle,So, all together, the sides are labelled capitals A, B and C, and the angles are labelled little a, little b, and little c
並且，我們對這條邊和這個角做同樣的事情，這條邊和這個角，所以，所有的邊都被標記為大寫的A、B和C，而這些角被標記小寫的a、b和c

So, the cosine rule is simply this relationship here.
所以，餘弦定理就是這種關係
It’s a bit like Pythagoras’ theorem except for this extra term on the end with the cos a in it.
它有點像畢氏定理，除了末尾有 cos a 的這個額外項

Now, let’s apply the cosine rule to the particular triangle we looked at a moment ago
現在，讓我們將餘弦定理應用於我們剛才看到的特定三角形

It’s pretty straightforward to write down this particular relationship.
寫下這種特殊關係非常簡單

We can isolate the term cos alpha which gives us the angle alpha that we’re interested in.
我們可分離出 cosα ，只需要得到我要的α角

And, it’s defined in terms of the constant link lengths, A1 and A2 and the position of the end effector, x and y.
而且，它是根據恆定連桿長度 A1 和 A2 以及末端的位置 x 和 y 去定義

We can write this simple relationship between the angles alpha and Q2
我們可以寫出α角和 Q2 之間的這種簡單關係

And, we know from the shape of the cosine function that cos of q2 must be equal to negative of cos alpha
而且，我們從餘弦函數的形狀知道，q2 的 cos 必須等於 cos α 的負數

This time, let’s just write an expression for the cosine of the joined angle q2
這一次，我們只寫一個關於連接角 q2 的餘弦的表達式

Now, we’re going to draw yet another red triangle and we’re going apply some simple trigonometry here
現在，我們將繪製另一個紅色三角形，並在此處應用一些簡單的三角函數

If we know Q2, then we know this length and this length of the red triangle
如果我們知道 Q2，那麼我們就知道這個長度和這個紅色三角形的長度

We can write this relationship for the sine of the joined angle q2
我們可以為連接角 Q2 的正弦寫出這種關係

Now, we can consider this bigger triangle whose angle is beta and this side length of this triangle is given here in blue
現在，我們可以考慮這個更大的三角形，它的角是β，這個三角形的邊長在這裡用藍色給出

And, the length of the other side of the triangle is this
而且，三角形的另一邊的長度是這個

So, now we can write an expression for the angle beta in terms of these parameters here
所以，現在我們可以在此處根據這些參數編寫角度 beta 的表達式

Going back to the red triangle that we drew earlier, we can establish a relationship between q1 and the angle beta
回到我們之前畫的紅色三角形，我們可以建立 Q1 和角度 beta 之間的關係

Introduce yet another angle, this one gamma and we can write a relationship between the angle gamma and the tooltip coordinates x and y
引入另一個角度，這個ɣ，我們可以寫出角度ɣ和工具提示坐標 x 和 y 之間的關係

Now, we can write a simple relationship between the angles that we’ve constructed, gamma and beta and the joined angle we’re interested in which is q1
現在，我們可以在我們構建的角度 ɣ 和 β與我們感興趣的連接角度 Q1 之間寫出一個簡單的關係

And, the total relationship looks something like this
而且，整個關係看起來像這樣

Quite a complex relationship, it gives us the angle of joined one, that’s q1 in terms of the end effector coordinates y and x, and a bunch of constants, a1 and a2, and it’s also a function of the second joint angle, Q2

一個相當複雜的關係，它給了我們連接的角度，即末端執行器坐標 y 和 x 的 q1，以及一堆常數 a1 和 a2，它也是第二個關節角度 Q2 的函數

So, let’s summarize what it is that we have derived here

所以，總結我們在這裡得出什麼

We have an expression for the cosine of Q2 and we have an expression for Q1

我們有 Q2 的餘弦表達方程式和 Q1 的表達方程式

Now, the cosine function is symmetrical about 0

現在，餘弦函數在0對稱

So, if we know the value of the cosine of Q2, then there are two possible solutions a positive angle and a negative angle.

所以我們知道cos Q2的值可能為正角和負角。

We’re going to explicitly choose the positive angle. Which means that I can write this expression here.

我們選擇正角，表示我可以寫出此表達式。

And now, we have what we call the inverse kinematic solution for this two-link robot.

現在我們能用逆向運動學來解決雙連桿機構

We have an expression for the two joined angles, Q1 and Q2 in terms of the end effector pose x and y, and a bunch of constants.

我們有兩個角 Q1 和 Q2 的表達式，根據最終執行的x 和 y 以及一系列常數

You notice that the two equations are not independent

你注意到這兩個方程不是獨立的

The equation for Q1, in fact, depends on the solution for Q2

實際上，Q1 的方程取決於 Q2 的解

In this case, Q2 is negative and we’re going to write the solution for Q2 with a negative sign in front of the inverse cosine

在這種情況下，Q2 是負數，我們將用負號在反餘弦前寫出 Q2 的解

Now, we need to solve for Q1, so we’re going to introduce this particular red triangle, the angle beta that we solved previously, and the angle gamma which is defined in terms of y and x

現在，我們需要求解 Q1，因此我們將介紹這個特定的紅色三角形、我們之前求解的β角，以及根據 y 和 x 定義的ɣ角

Now, we write a slightly different relationship between q1, gamma and beta, different to what we had before

現在，我們在 q1、ɣ 和 β之間寫出略有不同的關係，與我們之前的不同

There’s a change of sign involved

這涉及符號的變化

Then, we can substitute all that previous equation and come up with this expression for Q1

然後，我們可以替換之前的所有等式並得出 Q1 的這個表達式

Again, there is a change of sign here

同樣，這裡的符號發生了變化

Previously, this was a negative sign

先前，這是個負號

And, here in summary form is the solution for the inverse kinematics of our two-link robot when it is in this particular configuration, where Q2 is negative

並且，這裡總結的形式是我們的雙連桿機構在這種特定配置下的逆向運動學的解決方案，其中 Q2 為負

Let’s compare the two solutions, the case where q2 is positive and the case where q2 is negative.

讓我們比較兩個解決方案，Q2為正和Q2為負的情況

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