機(jī)器人學(xué)第二章運(yùn)動(dòng)學(xué)

1、1PAzyxApppP2(cylindrical) 兩個(gè)線性平移運(yùn)動(dòng)和一個(gè)旋轉(zhuǎn)運(yùn)動(dòng)兩個(gè)線性平移運(yùn)動(dòng)和一個(gè)旋轉(zhuǎn)運(yùn)動(dòng)spherical) 一個(gè)線性平移運(yùn)動(dòng)和兩個(gè)旋轉(zhuǎn)運(yùn)動(dòng)一個(gè)線性平移運(yùn)動(dòng)和兩個(gè)旋轉(zhuǎn)運(yùn)動(dòng)3xAyzwpxwpyPzwpww輊輊犏犏犏犏犏犏=犏犏犏犏犏犏犏臌臌,xyzxyzpppwww=4Bx AAAABBBBRxyz 333231232221131211rrrrrrrrrRABByBz00001,Rnoa01xxxyyyzzznoaRnoanoa輊犏犏=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輊輊鬃犏犏犏犏=鬃犏犏犏犏鬃犏犏臌臌5 0001xxxxyyyyzzzznoaPnoaPFnoaP輊犏犏犏=犏犏犏臌6 : 表示坐標(biāo)系表示坐標(biāo)系 BB主軸方向的單位矢量主軸方向的單位矢量. . : : 相對(duì)于坐標(biāo)系相對(duì)于坐標(biāo)系 AA的描述的描述. . 將這些單位矢量組成一個(gè)將這些單位矢量組成一個(gè) 3 33 3的矩陣，按照的矩陣，按照的順序的順序 . . 旋轉(zhuǎn)矩陣旋轉(zhuǎn)矩陣: : 標(biāo)量標(biāo)量 可用每個(gè)矢量在其參考坐標(biāo)系中單位方向上的投可用每個(gè)矢量在其參考坐標(biāo)系中單位方向上的投影的分

3、量來表示。影的分量來表示。 111213212223313233AAAABBBBrrrRXYZrrrrrr,BBBX Y Z,AAABBBXYZ,AAABBBXYZijr7cossin0sincos0001),(xRcos0sin010sin0cos),(yR1000cossin0sincos),(zRRAB重要！重要！8 ,ABABABxxAByyzzPPPPPPPPcossinsincosABBABBABxxyyxyzzPPPPPPPPcs0sc0001ABABABxxyyzzPPPPPP9 可用每個(gè)矢量在其參考坐標(biāo)系中單位方向上的投影的分量來表示: 的各個(gè)分量可用一對(duì)單位矢量的點(diǎn)積來表示

4、 為了簡單，上式的前置上標(biāo)被省略。 由兩個(gè)單位矢量的點(diǎn)積可得到二者之間的余弦，因此可以理解為什么旋轉(zhuǎn)矩陣的各分量常被稱作為方向余弦。components of rotation matrices are often referred to as direction cosinesABRBABABAAAAABBBBBABABABABABAXXYXZXRXYZXYYYZYXZYZZZijrPAPB=|PA|PB|cos 10 進(jìn)一步觀察 ，可以看出矩陣的行是單位矢量 A在 B中的描述. 因?yàn)?為坐標(biāo)系A(chǔ)相對(duì)于 B的描述 由轉(zhuǎn)置得到這表明旋轉(zhuǎn)矩陣的逆矩陣等于它的轉(zhuǎn)置 BATABRR1ABBTBAAR

5、RRBARBTAAAAABTBBBBABTAXRXYZYZ 3ATBAT AATAAABBBBBBATBXRRYXYZIZ111noaRAB1BABABABABABAzzyyxx0BABABABABABAxzzyyxRABTABABRR11RABnoaPRPBABARAB12 用 和 來描述坐標(biāo)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=創(chuàng)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 用于坐標(biāo)系間點(diǎn)的映射的通用數(shù)學(xué)表達(dá)式被稱為算子包括點(diǎn)的平移算子、矢量旋轉(zhuǎn)算子和平移加旋轉(zhuǎn)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 運(yùn)算的結(jié)果得到一個(gè)新的矢量，計(jì)算如下： 用矩陣算子寫出平移變換 where q is the signed magnitude of the translation along the v

11、ector direction .21AAAPPQ21( )AAQPD qPQ22 算子 可以被看成是一種特殊形式的齊次變換: 式中 是平移矢量 Q 的分量 通過定義B相對(duì)于A的位置， (用 ) , 我們使得這兩個(gè)描述具有相同的數(shù)學(xué)表達(dá)式?，F(xiàn)在引入了 ，我們可以用它來描述坐標(biāo)系和映射。100010( )0010001xyQzqqDqqQD,xyzq q q222xyzqqqqQDABORGP23 2) 旋轉(zhuǎn)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) 變換算子（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) 齊次變換陣是坐標(biāo)系的描述. describes the frame B relative to the frame A. （description of a frame） (2)齊次變換陣是變換映射. maps .（） (3)齊次變換陣是變換算子. 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輊-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，我們都采用了坐標(biāo)系的圖形表示法，即用一個(gè)坐標(biāo)系的原點(diǎn)指向另一個(gè)坐標(biāo)系的原點(diǎn)的箭頭來表示。將箭頭串聯(lián)起來，通過

25、簡單的變換方程就可得到混合坐標(biāo)系。箭頭的方向指明了坐標(biāo)系定義的方式。如果有一個(gè)箭頭的方向與串聯(lián)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執(zhí)行器的坐標(biāo)系 , 它是相對(duì)于操作臂基座的坐標(biāo)系B定義的，又已知工作臺(tái)相對(duì)于操作臂基座的空間位置 , 并且已知工作臺(tái)上螺栓的坐標(biāo)系相對(duì)于工作臺(tái)坐標(biāo)系的位置 計(jì)算螺栓相對(duì)于操作手的位姿： TGTSGT1TBBSGTSGTTT TBSTBTT34 Problem: 能否用少于九個(gè)數(shù)字來表示一個(gè)姿態(tài)? 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的旋轉(zhuǎn)矩陣都可用三個(gè)參量確定.000zyzxyxSSSSSSS(,)xyzSSS133() ()RISIS35 顯然，旋轉(zhuǎn)矩陣的九個(gè)分量線性相關(guān)。實(shí)際上，對(duì)于一個(gè)旋轉(zhuǎn)矩陣R很容易 寫出六個(gè)

27、線性無關(guān)的分量。假定R為三列： 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: 是否能找到一種姿態(tài)表示法，用三個(gè)參量就能簡便進(jìn)行表達(dá)? 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: 考慮兩個(gè)軸旋轉(zhuǎn)，一個(gè)繞Z轉(zhuǎn)30度，另一個(gè)繞X軸轉(zhuǎn)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:固連在坐標(biāo)系固連在坐標(biāo)系BB上的點(diǎn)上的點(diǎn) （1 1）繞）

30、繞z z軸旋轉(zhuǎn)軸旋轉(zhuǎn)9090度度: : （1 1）繞）繞z z軸旋轉(zhuǎn)軸旋轉(zhuǎn)9090度；度； （2 2）然后繞）然后繞y y軸轉(zhuǎn)軸轉(zhuǎn)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）然后繞）然后繞y y軸轉(zhuǎn)軸轉(zhuǎn)9090度。度。(90.0)ZR(90.0)YR(4, 3,7)Trans(7,3,2)TP38 1) X-Y-Z 固定角坐標(biāo)系（fixed angles） 下面介紹描述坐標(biāo)系B姿態(tài)的另一種方法: 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 . 每個(gè)旋轉(zhuǎn)都是繞著固定參考坐標(biāo)系A(chǔ)的軸。我們規(guī)定這種姿態(tài)的表示法為X-Y-Z固定角坐標(biāo)系?！肮潭ā币辉~是指旋轉(zhuǎn)是在固定（即不運(yùn)動(dòng)的）參考坐標(biāo)系中確定的。有時(shí)把它們定義為回轉(zhuǎn)角、俯仰角和偏轉(zhuǎn)角。AXAYAZ39 可以直接推導(dǎo)等價(jià)旋轉(zhuǎn)矩陣，因?yàn)樗械男D(zhuǎn)都是繞著參考坐標(biāo)系各軸的， where is shorthand for , for . 最重要的是搞清楚上式中的旋轉(zhuǎn)順序. 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 . 常常使人感興趣的是逆解問題，即從一個(gè)旋轉(zhuǎn)矩陣等價(jià)推出X-Y-Z固定角坐標(biāo)系。逆解取決于求解一組超越方程；如果方程相當(dāng)于一個(gè)已知的旋轉(zhuǎn)矩陣，那么就有九個(gè)方程和三個(gè)未知量。在這九個(gè)方程中有六個(gè)方程是相關(guān)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 歐拉角（Euler angles） 坐標(biāo)系 B的另一種表示法如下: 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: 注意這個(gè)結(jié)果與以相反順序繞固定軸旋轉(zhuǎn)三次得到的結(jié)果完全相同！總之，這是一個(gè)不太直觀的結(jié)果：三次繞固定軸旋轉(zhuǎn)的最終姿態(tài)和以相反順序三次繞運(yùn)動(dòng)坐標(biāo)軸旋轉(zhuǎn)的最終姿態(tài)相同。 因?yàn)榈葍r(jià)，所以無需通過旋轉(zhuǎn)矩陣的反復(fù)計(jì)算去求Z-Y-X的歐拉角。. 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) 其它角坐標(biāo)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. 每個(gè)表示法均需要按一定順序進(jìn)行三次繞主軸的旋轉(zhuǎn)。這些表示法是24種表示法中的典型方法，且都被稱作角坐標(biāo)系表示法。其中，12種為固定角坐標(biāo)系法，另12種為歐拉角坐標(biāo)系法。注意到由于二者之間的對(duì)偶性，對(duì)于繞主軸連續(xù)旋轉(zhuǎn)的旋轉(zhuǎn)矩陣實(shí)際上只有12種唯一的參數(shù)表示方法。 感興趣的同學(xué)可以參考本書附錄B 45 5) 等效軸角坐標(biāo)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