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第一部

中文：

我們在前面看到了這個簡單的雙連桿機器人關於正向運動學的講座。描述了該機器人的工具提示位姿簡單地由兩個數字，坐標 x 和 y 相對於世界坐標系。所以，這裡的問題是，給定 x 和 y，我們要確定連接的角度 Q1 和 Q2。我們將遵循的解決方案在這個特定的部分是一個幾何一。我們將從一個簡單的作品開始的建設。我們將覆蓋紅色三角形在我們的機器人之上。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。並且，使用畢達哥拉斯定理，我們可以寫出 r 平方等於 x 平方加上 y 平方。到目前為止，很容易。現在，我們要看看這個三角形此處以紅色突出顯示，我們想確定角度α。為了做到這一點，我們需要使用餘弦規則。而且，如果你對余弦有點生疏規則，這裡有一點複習。我們有一個任意三角形。我們不必有任何直角它，我們將標記長度這條邊作為 A 和相反的角度邊緣，我們將標記為小 a。而且，我們對這條邊和這條邊做同樣的事情角度，還有這條邊和這個角。所以，總而言之，雙方都被標記為首都 A、B 和 C，角度標記為小 a、小 b 和小 c。所以，餘弦規則就是這種關係這裡。這有點像畢達哥拉斯定理，除了對於這個額外的術語，最後的 cos 一個在裡面。現在，讓我們將餘弦規則應用於我們看了一會兒特定的三角形前。寫下來很簡單這種特殊的關係。我們可以隔離術語 cos alpha，它給出我們感興趣的角度α 在。而且，它是根據常數定義的鏈接長度，A1 和 A2 以及位置末端執行器，x 和 y。我們可以寫出這個簡單的關係角α和 Q2。而且，我們從餘弦的形狀知道 Q2 的 cos 必須等於負的函數 cos 阿爾法。這一次，我們只寫一個表達式對於連接角 Q2 的餘弦。現在，我們要再畫一個紅色三角形，我們將應用一些簡單的三角函數在這裡。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。我們可以把這個關係寫成正弦的連接角 Q2。現在，我們可以考慮這個更大的三角形其角為β，此邊長為三角形在這里以藍色給出。並且，三角形另一邊的長度這是。所以，現在我們可以寫一個表達式此處的這些參數的角度β。回到我們畫的紅色三角形之前，我們可以建立之間的關係 Q1 和角β。介紹另一個角度，這個伽馬我們可以寫出之間的關係角度伽馬和工具提示坐標 x 和 y。現在，我們可以寫出一個簡單的關係我們構建的角度，伽馬和 beta 以及我們感興趣的連接角其中是 Q1。而且，總的關係看起來有些東西像這樣。相當複雜的關係，它給了我們連接的角度，即 Q1 末端執行器坐標 y 和 x，以及一堆常量，a1 和 a2，它是也是第二關節角度的函數， Q2。所以，讓我們總結一下我們有什麼派生於此。我們有 Q2 的餘弦表達式我們有 Q1 的表達式。現在，餘弦函數關於 0. 所以，如果我們知道餘弦值 Q2，那麼有兩種可能的解決方案，一個正角和一個負角。我們將明確選擇積極的角度，這意味著我可以寫出這個表達式這裡。現在，我們有了我們所說的逆這個雙連桿機器人的運動學解決方案。我們有兩個連接角的表達式， Q1 和 Q2 在末端執行器姿勢方面 x 和 y，以及一堆常量。你注意到這兩個方程不是獨立的。事實上，Q1 的方程取決於 Q2 的解決方案。在這種情況下，Q2 是負數，我們將用負數寫出 Q2 的解符號在反餘弦前面。現在，我們需要解決 Q1，所以我們要介紹這個特殊的紅色三角形，我們之前求解的角度β，以及用術語定義的角度伽馬 y 和 x。現在，我們寫一個稍微不同的關係在 Q1、gamma 和 beta 之間，與什麼不同我們以前有過。涉及到符號的變化。然後，我們可以替換之前的所有方程並提出這個表達式對於 Q1。同樣，這裡的符號發生了變化。以前，這是一個負面信號。而且，這裡是總結形式的解決方案對於我們的雙連桿的逆運動學當機器人處於這種特定配置時，其中 Q2 為負。讓我們比較這兩種解決方案，t

英文：

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

The tooltip pose of this robot is described
simply by two numbers, the coordinates x and

y with respect to the world coordinate frame.

So, the problem here is that given x and y,
we want to determine the joined angles, Q1

and Q2.

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

one.

We’re going to start with a simple piece
of construction.

We’re going to overlay the red triangle
on top of our robot.

We know that the end point coordinate is x,
y, so the vertical height of the triangle

is y, the horizontal width is x.

And, using Pythagoras theorem, we can write
r squared equals x squared plus y squared.

So far, so easy.

Now, we’re going to look at this triangle
highlighted here in red and we want to determine

the angle alpha.

In order to do that, we need to use the cosine
rule.

And, if you’re a little rusty on the cosine
rule, here is a bit of a refresher.

We have an arbitrary triangle.

We don’t have to 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.

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.

So, the cosine rule is simply this relationship
here.

It’s a bit like Pythagoras’ theorem except
for this extra term on the end with the cos

a in it.

Now, let’s apply the cosine rule to the
particular triangle we looked at a moment

ago.

It’s pretty straightforward to write down
this particular relationship.

We can isolate the term cos alpha which gives
us the angle alpha that we’re interested

in.

And, it’s defined in terms of the constant
link lengths, A1 and A2 and the position of

the end effector, x and y.

We can write this simple relationship between
the angles alpha and Q2.

And, we know from the shape of the cosine
function that cos of Q2 must be equal to negative

of cos alpha.

This time, let’s just write an expression
for the cosine of the joined angle Q2.

Now, we’re going to draw yet another red
triangle and we’re going apply some simple

trigonometry here.

If we know Q2, then we know this length and
this length of the red triangle.

We can write this relationship for the sine
of the joined angle Q2.

Now, we can consider this bigger triangle
whose angle is beta and this side length of

the triangle is given here in blue.

And, the length of the other side of the triangle
is this.

So, now we can write an expression for the
angle beta in terms of these parameters here.

Going back to the red triangle that we drew
earlier, we can establish a relationship between

Q1 and the angle beta.

Introduce yet another angle, this one gamma
and we can write a relationship between the

angle gamma and the tooltip coordinates x
and y.

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.

And, the total relationship looks something
like this.

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.

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

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

Now, the cosine function is symmetrical about
0.

So, if we know the value of the cosine of
Q2, then there are two possible solutions,

a positive angle and a negative angle.

We’re going to explicitly choose the positive
angle, which means that I can write this expression

here.

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

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.

You notice that the two equations are not
independent.

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

In this case, Q2 is negative and we’re going
to write the solution for Q2 with a negative

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.

Now, we write a slightly different relationship
between Q1, gamma and beta, different to what

we had before.

There’s a change of sign involved.

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

for Q1.

Again, there is a change of sign here.

Previously, this was a negative sign.

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.

Let’s compare the two solutions, the case
where Q2 is positive and the case where Q2

is negative.

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