机器人学第二章运动学

1、1PAzyxApppP2(cylindrical) 两个线性平移运动和一个旋转运动两个线性平移运动和一个旋转运动spherical) 一个线性平移运动和两个旋转运动一个线性平移运动和两个旋转运动3xAyzwpxwpyPzwpww轾轾犏犏犏犏犏犏=犏犏犏犏犏犏犏臌臌,xyzxyzpppwww=4Bx AAAABBBBRxyz 333231232221131211rrrrrrrrrRABByBz00001,Rnoa01xxxyyyzzznoaRnoanoa轾犏犏=犏犏犏臌01cos( , ) cos( , ) cos( , )cos( , ) cos( , ) cos( , )cos( , )co

2、s( , )cos( , )nxoxaxn xo xa xRnyoyayn yo ya ynzozazn zo za z轾轾鬃犏犏犏犏=鬃犏犏犏犏鬃犏犏臌臌5 0001xxxxyyyyzzzznoaPnoaPFnoaP轾犏犏犏=犏犏犏臌6 : 表示坐标系表示坐标系 BB主轴方向的单位矢量主轴方向的单位矢量. . : : 相对于坐标系相对于坐标系 AA的描述的描述. . 将这些单位矢量组成一个将这些单位矢量组成一个 3 33 3的矩阵，按照的矩阵，按照的顺序的顺序 . . 旋转矩阵旋转矩阵: : 标量标量 可用每个矢量在其参考坐标系中单位方向上的投可用每个矢量在其参考坐标系中单位方向上的投影的分

3、量来表示。影的分量来表示。 111213212223313233AAAABBBBrrrRXYZrrrrrr,BBBX Y Z,AAABBBXYZ,AAABBBXYZijr7cossin0sincos0001),(xRcos0sin010sin0cos),(yR1000cossin0sincos),(zRRAB重要！重要！8 ,ABABABxxAByyzzPPPPPPPPcossinsincosABBABBABxxyyxyzzPPPPPPPPcs0sc0001ABABABxxyyzzPPPPPP9 可用每个矢量在其参考坐标系中单位方向上的投影的分量来表示: 的各个分量可用一对单位矢量的点积来表示

4、 为了简单，上式的前置上标被省略。 由两个单位矢量的点积可得到二者之间的余弦，因此可以理解为什么旋转矩阵的各分量常被称作为方向余弦。components of rotation matrices are often referred to as direction cosinesABRBABABAAAAABBBBBABABABABABAXXYXZXRXYZXYYYZYXZYZZZijrPAPB=|PA|PB|cos 10 进一步观察 ，可以看出矩阵的行是单位矢量 A在 B中的描述. 因为 为坐标系A相对于 B的描述 由转置得到这表明旋转矩阵的逆矩阵等于它的转置 BATABRR1ABBTBAAR

5、RRBARBTAAAAABTBBBBABTAXRXYZYZ 3ATBAT AATAAABBBBBBATBXRRYXYZIZ111noaRAB1BABABABABABAzzyyxx0BABABABABABAxzzyyxRABTABABRR11RABnoaPRPBABARAB12 用 和 来描述坐标系 ,AABBORGBRPBARABORGP B13BAPPBPA0ABABPPP=+14PAPBPRPBABARABRBA1BAATABBRRRRAB15 Example: Frame B is rotated relative to frame A about Z by 30 degrees. He

6、re Z is pointing out of the page. Writing the unit vectors of B in terms of A and stacking them as the columns of the rotation matrix: The original vector P is not changed, we compute a new description relative to another frame.cossin00.8660.5000.000sincos00.5000.8660.0000010.0000.0001.000ABR0.02.00

7、.0BP1.0001.7320.000AABBPR P16AABABBORGPR PP17110001PPRPBBAABApB10010100033RpIpRABBABAABpApBpTpBABA18a123 , , l l lb123( , )( , , )( , )ABRRot yTrans l l lRot xba=创19 Example: Frame B is rotated relative to frame A about Z by 30 degrees, translated 10 units in , and translated 5 unit in . Find , wher

8、e . The definition of frame B is We use the definition of B just given a transformation:0.8660.5000.00010.00.5000.8660.0005.00.0000.0001.0000.00001ABT9.09812.5620.000AABBPT PAXAYAP3.0 7.0 0.0BTP 20 用于坐标系间点的映射的通用数学表达式被称为算子包括点的平移算子、矢量旋转算子和平移加旋转算子。 1) 平移算子（Translational operators） A translation moves a

9、 point in space a finite distance along a given vector direction. Only one coordinate system need be involved. It turns out that translating the point in space is accomplished with the same mathematics as mapping the point to a second frame. The distinction is: when a vector is moved “forward” relat

10、ive to a frame, we may consider either that the vector moved forward or that the frame moved backword. The mathematics involved in the two cases is identical, only our view of the situation is different.21 运算的结果得到一个新的矢量，计算如下： 用矩阵算子写出平移变换 where q is the signed magnitude of the translation along the v

11、ector direction .21AAAPPQ21( )AAQPD qPQ22 算子 可以被看成是一种特殊形式的齐次变换: 式中 是平移矢量 Q 的分量 通过定义B相对于A的位置， (用 ) , 我们使得这两个描述具有相同的数学表达式。现在引入了 ，我们可以用它来描述坐标系和映射。100010( )0010001xyQzqqDqqQD,xyzq q q222xyzqqqqQDABORGP23 2) 旋转算子（Rotational operators） Another interpretation of a rotation matrix is as a rotational operato

12、r that operates on a vector and changes that vector to a new vector, , by means of a rotation, R. When a rotation matrix is shown as an operator, no sub- or superscripts appear, because it is not viewed as relating two frame. We may write: Again, the mathematics is the same, only our interpretation

13、is different. How to obtain rotational matrices that are to be used as operators: The rotation matrix that rotates vectors through some rotation, R, is the same as the rotation matrix that describes a frame rotated by R relative to the refrence frame. 21AAPR P2AP1AP24 Although a rotation matrix is e

14、asily viewed as an operator, we can also define another notation for a rotational operator that clearly indicates which axis is being rotated about: is a rotational operator that performs a rotation about the axis direction by degrees. For example:21( )AAKPRPcossin00sincos00( )00100001ZRK( )KR25 Exa

15、mple: Figure shows a vector . We wish to compute the vector obtained by rotating this vector about Z by 30 degrees. Call the new vector . The rotation matrix that rotates vectors by 30 degrees about Z is the same as the rotation matrix that describes a frame rotated 30 degrees about Z relative to th

16、e reference frame. Thus, the correct rotational operator is 10.02.00.0TAP 211.000(30.0)1.7320.000AAZPRP1AP2APcossin00.8660.5000.000(30.0)sincos00.5000.8660.0000010.0000.0001.000ZR26 3) 变换算子（Transformation operators） As with vectors and rotation matrices, a frame has another interpretation as a trans

17、formation operator. In the interpretation, only one coordinate system is involved, and so the symbol T is used without sub- or superscripts. How to obtain homogeneous transform that are to be used as operators: The transform that rotates by R and translated by Q is the same as the transform that des

18、cribes a frame rotated by R and translated by Q relative to the refrence frame. 21AAPT P27 Example: Figure shows vector . We wish to rotate it about Z by 30 degrees and translate it 10 units in and 5 units in . Find ,where . The operator T, which performs the translation and rotation: 1APAXAY2AP13.0

19、7.00.0TAP 0.8660.5000.00010.00.5000.8660.0005.00.0000.0001.0000.00001T219.09812.5620.000AAPT P13.07.00.0AP28 Summary of interpretations (1) 齐次变换阵是坐标系的描述. describes the frame B relative to the frame A. （description of a frame） (2)齐次变换阵是变换映射. maps .（） (3)齐次变换阵是变换算子. T operates on to create . From this

20、 point on, the terms frame and transform will both be used to refer to a position vector plus an orientation. Frame is the term favored in speaking of a description, Transform is used most frequently when function as a mapping or operator is implied. Note that transformation are generalizations of (

21、and subsume) translations and rotations; we will often use the term transform when speaking of a pure rotation (or translation). BAPPABTABT1AP2AP29pTpCBCBpTTpTpCBCABBABA1000BACBABBCABBCABACppRRRTTTTABABBARRR1000BATABBABAABpRpRp001ATAT ABBBBARRpT轾-犏=犏犏臌1 TTABBA30 Example: Frame B is rotated relative

22、to frame A about by 30 degrees and translated four units in and three units in . Thus, we have a description of . Find . The frame defining B is: AYZABTAXBAT0.8660.5000.0004.00.5000.8660.0003.00.0000.0001.0000.00001ABT|0 0 0 |10.8660.5000.0004.9640.5000.8660.0000.5980.0000.0001.0000.00001ATAT ABBBBO

23、RGARRPT31 Figure indicates a situation in which a frame D can be expressed as products of transformations in two different ways: We can set these two descriptions of equal to construct a transform equation: Transform equations can be used to solve for transforms in the case of n unknown transforms a

24、nd n transform equations. UUAUUBCDADDBCDTT TorTT T TUAUBCADBCDT TT T TUDT32 Consider in the case that all transforms are known except . Here, we have one transform equation and one unknown transform, hence, we easily find its solution: 注意：在所有的途中，我们都采用了坐标系的图形表示法，即用一个坐标系的原点指向另一个坐标系的原点的箭头来表示。将箭头串联起来，通过

25、简单的变换方程就可得到混合坐标系。箭头的方向指明了坐标系定义的方式。如果有一个箭头的方向与串联的方向相反，就先求出它的逆 。 11BUUACCBADDTTT T TBCTUAUBCADBCDT TT T T33 Example: 假定已知操作臂末端执行器的坐标系 , 它是相对于操作臂基座的坐标系B定义的，又已知工作台相对于操作臂基座的空间位置 , 并且已知工作台上螺栓的坐标系相对于工作台坐标系的位置 计算螺栓相对于操作手的位姿： TGTSGT1TBBSGTSGTTT TBSTBTT34 Problem: 能否用少于九个数字来表示一个姿态? A result from linear algebr

26、a (known as Cayleys formula): for any proper orthonormal matrix R, there exists a skew-symmetric matrix (S=-ST) S such that: a skew-symmetric matrix of dimension 3 is specified by three parameters as: 任何 33的旋转矩阵都可用三个参量确定.000zyzxyxSSSSSSS(,)xyzSSS133() ()RISIS35 显然，旋转矩阵的九个分量线性相关。实际上，对于一个旋转矩阵R很容易 写出六个

27、线性无关的分量。假定R为三列： These three vectors are the unit axes of some frame writtern in terms of the refrence frame. Each is a unit vector, and all three must be mutually perpendicular, so we see that there are six constrains on the nine parameters: 是否能找到一种姿态表示法，用三个参量就能简便进行表达? 1 ,1 ,10 ,0 ,0XYZX YX ZY Z ()R

28、X Y Z36 Whereas translations along three mutually perpendicular axes are quite easy to visualize, rotations seem less intuitive. Unfortunately people have a hard time describing and specifying orientation in three-dimensional space. One difficulty is that rotations dont generally commute. That is: E

29、xample: 考虑两个轴旋转，一个绕Z转30度，另一个绕X轴转30度。: ABBABCCBR RR R0.8660.5000.000(30.0)0.5000.8660.0000.0000.0001.000ZR1.0000.0000.000(30.0)0.0000.8660.5000.0000.5000.866XR0.870.430.250.870.500.00(30.0)(30.0)0.500.750.43(30.0)(30.0)0.430.750.50.000.500.870.250.430.87ZXXZRRRR37 Example:固连在坐标系固连在坐标系BB上的点上的点 （1 1）绕）

30、绕z z轴旋转轴旋转9090度度: : （1 1）绕）绕z z轴旋转轴旋转9090度；度； （2 2）然后绕）然后绕y y轴转轴转9090度；度； （2 2）再平移）再平移44，-3-3，77； （3 3）最后再平移）最后再平移44，-3-3，77。 （3 3）然后绕）然后绕y y轴转轴转9090度。度。(90.0)ZR(90.0)YR(4, 3,7)Trans(7,3,2)TP38 1) X-Y-Z 固定角坐标系（fixed angles） 下面介绍描述坐标系B姿态的另一种方法: Start with the frame coincident with a known refrence fr

31、ame A. Rotate B first about by an angle , then about by an angle , and, finally, about by an angle . 每个旋转都是绕着固定参考坐标系A的轴。我们规定这种姿态的表示法为X-Y-Z固定角坐标系。“固定”一词是指旋转是在固定（即不运动的）参考坐标系中确定的。有时把它们定义为回转角、俯仰角和偏转角。AXAYAZ39 可以直接推导等价旋转矩阵，因为所有的旋转都是绕着参考坐标系各轴的， where is shorthand for , for . 最重要的是搞清楚上式中的旋转顺序. Equation abo

32、ve is correct only for rotations performed in the order: about by an angle , then about by an angle , and, finally, about by an angle . 常常使人感兴趣的是逆解问题，即从一个旋转矩阵等价推出X-Y-Z固定角坐标系。逆解取决于求解一组超越方程；如果方程相当于一个已知的旋转矩阵，那么就有九个方程和三个未知量。在这九个方程中有六个方程是相关的。 00100( ,)0010000100ABXYZcscsRsccsscscc cc s cs cc s cs ss cs s

33、 sc cs s cc ssc sc c ccosssinAXAYAZ40 Let: In summary: Although a second solution exists, by using the positive square root in the formula for , we always compute the single solution for which . This is usually a good practice. If , the solution degenerates. In those cases, one possible convention i

34、s to choose . 111213212223313233( ,)ABXYZc cc s cs cc s cs srrrRs cs s sc cs s cc srrrsc sc crrr 2231112121113233tan2(,)tan2(/,/)tan2(/,/)ArrrArcrcArcrc009090090 001222900tan 2(,)Arr01222900tan 2(,)Arr 41 2) Z-Y-X 欧拉角（Euler angles） 坐标系 B的另一种表示法如下: Start with the frame coincident with a known refrenc

35、e frame A. Rotate B first about by an angle , then about by an angle , and, finally, about by an angle . In this representation, each rotation is performed about an axis of the moving system B rather than one of the fixed refrence A. Such sets of three rotations are called Euler angles. Note that ea

36、ch rotations takes place about an axis whose location depends upon the preceding rotations.BZBYBX42 We can write: 注意这个结果与以相反顺序绕固定轴旋转三次得到的结果完全相同！总之，这是一个不太直观的结果：三次绕固定轴旋转的最终姿态和以相反顺序三次绕运动坐标轴旋转的最终姿态相同。 因为等价，所以无需通过旋转矩阵的反复计算去求Z-Y-X的欧拉角。. 00100( )( )( )0010000100ABZ Y XZYXcscsRRRRsccsscscc cc s cs cc s cs s

37、s cs s sc cs s cc ssc sc c 43 3) Z-Y-Z Euler angles Describing the orientation of a frame B as follow: Start with the frame coincident with a known refrence frame A. Rotate B first about by an angle , then about by an angle , and, finally, about by an angle . Extracting: 111213 212223313233( , )ABZ

38、Y Zc cc s cs cc s cs srrrRs cs s sc cs s cc srrrsc sc crrr BZBYBZ2231323323133231tan2(,)tan2(/,/)tan2(/,/)ArrrArsrsArsrs121100tan 2(,)Arr012111800tan 2(,)Arr44 4) 其它角坐标系的表示法 In the preceding subsections we have seen three conventions for specifying orientation: X-Y-Z fixed angles, Z-Y-X Euler angles

39、, and Z-Y-Z Euler angles. 每个表示法均需要按一定顺序进行三次绕主轴的旋转。这些表示法是24种表示法中的典型方法，且都被称作角坐标系表示法。其中，12种为固定角坐标系法，另12种为欧拉角坐标系法。注意到由于二者之间的对偶性，对于绕主轴连续旋转的旋转矩阵实际上只有12种唯一的参数表示方法。 感兴趣的同学可以参考本书附录B 45 5) 等效轴角坐标系表示法 With the notation we give the description of an orientation by giving an axis, X, and an angle 30 degrees. Thi

40、s is an example of an equivalent angle-axis representation. If the axis is a general direction (rather than one of the unit directions) any orientation may be obtained through proper axis and angle selection. Describing the orientation of a frame B as follow: Start with the frame coincident with a k

41、nown refrence frame A. Then Rotate B first about the vector by an angle according to the right-hand rule. Vector is called the equivalent axis of a finite rotation. AKAK(30)XR46 A general orientation of B relative to A may be written as or . The specification of the vector requires only two paramete

42、rs, because its length is always taken to be one. The angle specifies a third parameter. The equivalent rotation matrix is: where , and . The sign of is determined by the right-hand rule, with the thumb pointing along the positive sense of . The inverse problem:1122333223133121121cos()212sinrrrArrKrrrr(, )ABR K( )KRAK1 cosv ,ATxyzKk kkAK111213212223313233( )xxxyzxzyABKxyzyyyzxxzyyzxzzk k vck k vk sk k vk srrrRk k vk sk k vck k vk srrrk k vk sk k vk sk k vcrrr47 Example: A Frame B is described as initia