Circular Motion - Artificial Intelligence for Robotics Udacity English Japanese

Circular Motion - Artificial Intelligence for RoboticsYouTube↗ IA↗(long shot)
Education 17.2K views 19 likes 2 dislikes 23 Feb 2015

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Here's our programming assignment.

次の課題はプログラミングです

In class you already programmed a particle filter for a really simplistic robot

授業ではごく単純なロボットの粒子フィルタをプログラミングしました

that was able to measure ranges to landmarks and moved pretty much like a trashbin.

ランドマークとの距離の観測が可能でごみ箱によく似た動作をします

Now I'd like replace it with a more interesting robot that's more realistic.

今度はより現実的で面白いロボットに取り替えましょう

In particular, I'd like you to use a car. Here's a car.

具体的には車を使います　これが車です

A car tends to have fixed tires and two steerable in the front.

通常は2本の固定タイヤと前方に2本の可動タイヤがついています

Suppose the location of our car in a coordinate system

座標システムにおける車の位置は

is given by its x-coordinate and its y-coordinate--

x座標とy座標とで表されます

I'm picking the halfway point on the rear axle as the reference point--

基準点として後方の車軸の中間点を選び

and by its heading dire theta.

進行方向のθが得られます

The state will be x, y, and the orientation theta.

状態はx、y、方向のθです

Then this car also has a steering angle that is called the alpha.

この車にはαという操舵角もあります

The question is how is the state effected by driving a certain distance d

初期状態がx、y、θだと仮定すると

with a fixed steering angle, assuming the initial state is x, y, and theta.

固定された操舵角である距離dを運転すると状態はどのような影響を受けるでしょうか？

It turns out to answer this, I also need to know the length of the vehicle,

これに答えるためには車体の長さも知る必要があります

which I will just call L for length that is a constant throughout out consideration.

この長さをLと呼びますこれは定数になります

This is usually called a bicycle model.

いわゆる自転車モデルなので

Obviously, it suffices to look at one pair of tires because the other one--

1組のタイヤを見れば十分です

at least in approximation--runs pretty much parallel.

少なくとももう1組は平行に動くからです

If we look at the robot locally where we have a steering angle, alpha,

ロボットを局所的に見れば操舵角とα そしてロボットの長さのLがあり

robot length L, and we're driving the rear tire forward by distance D,

距離のDによって後方のタイヤを前に運転します

then the robot will turn to the left, and it's turning angle, beta, is proportional

ロボットは左折して回転角度のβは

to the distance that the rear tire moves forward divided by the length of the vehicle

車両の長さに回転角度を掛けたもので割られて

times the tangent of the steering angle.

後方のタイヤで動く距離のDと釣り合います

Let's now compute the changes of x, y, and theta in the local coordinate system.

次はx、y、θの変化をローカル座標システムで計算しましょう

Realize the turning radius R of this robot

ロボットの回転半径Rが

is simply the distance that we drive forward divided by beta.

ただβで割られた前へ運転する距離だと理解してください

It's relatively simple math, which I don't want to explain in detail.

比較的に単純な数学なので詳しは説明しません

This means that the robot is turning around a point over here at cx and cy.

これはロボットがcxとcyのポイントを曲がることを意味します

After the turn, the vehicle is located somewhere over here.

曲がったあとの車の位置はこの辺りです

In global coordinates, here is the way we describe this.

グローバル座標では以下のように説明します

Cx is the x coordinate of the robot x minus--now we go to the left--

左へ動くならcxはロボットxのx座標から

the sine of the robot orientation before motion times radius R.

動く前のロボットの方向の正弦に半径Rを掛けたものを引きます

Similarly, cy is this expression over here--y plus cosine of the orientation times R.

同じようにcyもこのような式で yに方向の余弦にRを掛けたものを足します

Then after motion, we can go back from cx to cy to a new state over here

そして動いたあとはcxとcyへ戻り回転半径のβを追加することによって

simply by adding in the turning radius beta.

ここにある新しい状態へ移動します

That is, our new x is cx plus sine of theta plus beta times radius.

つまり新しいxはcxに θとβを足した正弦に半径を掛けたものを足します

Notice how this parallels this guy over here, except for two changes.

この式は2つの変化を除いて似ていることに気づいてください

What we previously subtracted we're now adding,

前に引かれたものをここでは足して

and we're adding beta to the argument of the sine.

正弦の引数にβを加えます

The same with y, and the orientation is just increased by beta--modulo 2π.

yでも同じようにして方向にはただβに加えます mod 2πです

This all works if the robot is actually turning.

これはロボットが実際に曲がっている場合にはうまくいきます

If the robot were to go straight, then R would become infinity by this division over here.

ロボットが真っすぐ進むなら Rはこの割り算で無限になります

For small betas--that's smaller than 0.001--we can approximate this all as straight motion.

0．001より小さいβでは直線的な動きだと見積もることができます

Our new x is the old x times our driven distance pointed in the cosine of the heading direction.

新しいxは古いxに運転した距離dを掛けたもので進んだ方向の余弦です

Similarly for y we go in sine of heading direction, and the heading direction stays the same.

同じようにyも進行方向の正弦へ進みますこの方向は変わりません

You could add beta, which is basically zero, to be slight more precise

より正確にするために基本的にゼロであるβを足したり

or you could just use theta. It doesn't really matter.

θを使ってもどちらでも構いません

In this programming assignment, I'd like you to implement this piece of math over here in our particle filter.

この課題では粒子フィルタにこれらの計算を実装してもらいます.

To make sure we increment it correctly, I will give you some example data that you can check.

正しく増加できるように確認用の例となるデータを与えます

In our first part, I've prepared a lot of software for you,

最初の部分では様々なソフトウェアを用意しました

basically copying the old particle filter software over,

基本的には古い粒子フィルタのソフトウェアをコピーします

and removing the motion and the measurement model for now.

今のところは動作と観測モデルを取り除きました

In this I just want you to practice the motion model.

ここではただ動作モデルを練習してもらいたいのです

We assume a length of the road of 20. We initialize the road with this length parameter,

道路の長さは20であると仮定し長さのパラメータで道路を初期化します

and for this first iteration we assume no steering noise and no distance noise.

最初の反復ではsteering＿noiseと distance＿noiseはありません

I set the robot to (0.0, 0.0, 0.0) in the beginning,

ロボットをまず（0．0、0．0、0．0）とセットし

and then I cache away a number of motions,

いくつかの動作を格納します

The way to read those is this robot is moving by 10 total with the steering angle of zero.

このロボットが操舵角がゼロで合計で10動いているので読み込みます

Then it moves another 10 with a steering angle of pi divided by 6.

そして操舵角π／6でまた10動きます

Then it moves 20, again with a steering angle of zero.

その後はまた操舵角ゼロで20動きます

A pi over 6 is a left steering,

π÷6は左舵なので

so the robot should change its coordinates in the beginning just in the x direction,

ロボットは最初にxの方向だけで座標を変えます

because it's facing an x direction over here.

xの方向を向いているからです

Then turn left a little bit, go forward, and go straight again.

少し左折して進んだあともう一度真っすぐ進みます

Scrolling down a little bit, I also give you the code to run the robot.

このロボットを実行するコードも与えましょう

We've created the robot here. You print the initial coordinate.

ここにロボットを作ったので初期の座標を出力してください

Then for each of the motions in this list over here,

そしてリストにあるすべての動作に

we apply the myrobot equals myrobot.move command,

myrobot＝myrobot．moveのコマンドを適用して

and we print out the successive command.

連続するコマンドを出力します

If you get this right, these are the numbers I would like to see.

正しくできればこのように数字が分かるはずです

Initially, the robot position is 0, 0, 0, 0. That's just the one over here.

最初はロボットの位置はこのように0．0です

It's out first print command. It then moves forward in the x direction by 10.

これは最初の出力コマンドでxの方向に10進みます

The orientation stays 0 and so does y, because there's no steering.

操縦がないので方向もyもゼロのままです

Now we steer. This affects x. It doesn't quite move 10. It only moves 9.86.

操縦するとxに影響が出ます 10まではいきませんが9．86動きます

In the y direction it only moves 1.433, and its new orientation is 0.2886.

yの方向では1．433だけ動いて新しい方向性は0．2866です

Then we move straight again, and now the x coordinate becomes 0.3903,

そしてもう一度真っすぐに動くと x座標は0．3903になります

y coordinate becomes 7.12, orientation 0.28.

y座標は7．12で方向は0．28となります

Your code should output exactly the same values also over here.

これとまったく同じ数値を出力するコードにしてください

Just to give you a second test--this is a sequence of 10 motions

2つ目のテストです　連続する10の動作があり

where the robot moves 10 forward and always turns right but an angle of 0.2 in radians.

ロボットが前に10動くと常に右折し弧度は0．2になります

We look at the outputs we get the following array.

出力を見るとこのような配列になります

You can see that the orientation starting at zero, which is the same as 2π,

方向は2πと同じくゼロで始まって

decreases all the way to 5.26,

5．26まで減少しました

and you can also see that the robot starts running in a circle

ロボットが円を描いて動き始めたことも分かります

whereas initially we add almost 10 to the x direction and almost nothing in the y direction.

最初にx方向に約10足して y方向にはほとんど何も足しません

As you come down here, we subtract quite a bit in the y direction,

ここまでくればぐるぐる回っているので

because now the robot is going in a circle.

yの方向でかなりたくさん減算します

You should look at these numbers over here and see if your code matches these exact numbers that my code outputs.

ここの数値を見て私のコードが出力するものと一致するか確かめてください

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