機(jī)器人位姿運(yùn)動(dòng)學(xué)2017

1、KINEMATICS OF ROBOTS: POSITION ANALYSIS工業(yè)機(jī)器人是多自由度機(jī)構(gòu)，必須知道每個(gè)關(guān)節(jié)變量才能確定機(jī)器人手機(jī)器人手的位置。Robot Reference FrameslWorld Reference FramelJoint Reference FramelTool Reference FramexyzWorld Reference FrameJoint Reference FramexyznoayznoaTool Reference Framex【機(jī)器人的參考坐標(biāo)系機(jī)器人的參考坐標(biāo)系】nRepresentation of a Point in SpacelA

2、point P in space can be represented by its three coordinates relative to a reference frame as: zxyaxczbyPxyzPabcijk【空間點(diǎn)的表示空間點(diǎn)的表示】1. 1. 機(jī)器人運(yùn)動(dòng)學(xué)的矩陣表示機(jī)器人運(yùn)動(dòng)學(xué)的矩陣表示nRepresentation of a Vector in SpaceA vector can be represented by three coordinates of its tail and its head. If the vector starts at point A

3、and ends at point B, then it can be represented by: ()()()ABxxyyzzBABABAPijkzxyaxczbyPxyzabcP【空間向量的表示空間向量的表示】-【矩陣矩陣】Application of a scale factorlMakes the matrix 4 by 1lAllows for introducing directional vectorsxyzPPPwP, yxxyPPabww為什么要引入比例因子？為什么要引入比例因子？nThe n-o-a Frame designationlaonApproach, Orie

4、ntation, Normal directionsyznoaxaonF,運(yùn)動(dòng)坐標(biāo)系運(yùn)動(dòng)坐標(biāo)系z(mì)yxF,全局參考坐標(biāo)系全局參考坐標(biāo)系z(mì)zzyyyxxxaonaonaonF【坐標(biāo)系在參考坐標(biāo)系原點(diǎn)的表示坐標(biāo)系在參考坐標(biāo)系原點(diǎn)的表示】方向余弦？方向余弦？nRepresentation of a Frame Relative to a Fixed Reference Framezyxnoap0001xxxxyyyyzzzznoapnoapFnoapaonF,運(yùn)動(dòng)坐標(biāo)系z(mì)yxF,全局參考坐標(biāo)系【坐標(biāo)系在參考坐標(biāo)系的表示坐標(biāo)系在參考坐標(biāo)系的表示】nRepresentation of a Rigid Bo

5、dy 0001xxxxyyyyobjectzzzznoapnoapFnoapzyxnoap【剛體的表示剛體的表示】1000zzzzyyyyxxxxpaonpaonpaonPaonTFrame representation Requirementslthe three unit vectors n, o, a are mutually perpendicularleach unit vectors length, represented by its directional cosines, must be equal to 1lThese constraints translate into

6、the following six constraint equations: (the dot-product of n and o vectors must be zero) (the magnitude of the length of the vector must be 1) and1n1o1a0on0an0oalThe same can be achieved by:n oa上式包含了正確的右手法則關(guān)系右手法則關(guān)系，所以一般使用這個(gè)等式判斷3個(gè)向量之間的關(guān)系。Homogeneous Transformation Matricesl4 by 4 matrices:lCan be pr

7、e- or post-multipliedlEasy to find inverse of the matrixlRepresents both orientation and position information, including directional vectors0001xxxxyyyyzzzznoapnoapFnoap【齊次變換矩陣齊次變換矩陣】2. 2. 齊次齊次( (變換變換) )矩陣矩陣Representation of TransformationsA transformation may be in one of the following forms:lA pur

8、e translationlA pure rotation about an axislA combination of translations and/or rotations3. 3. 變換的表示變換的表示當(dāng)空間的坐標(biāo)系（向量、物體或運(yùn)動(dòng)坐標(biāo)系）相對(duì)于固定的參考坐標(biāo)系運(yùn)動(dòng)時(shí)，這一運(yùn)動(dòng)可以用類似于表示坐標(biāo)系的方式來表示。nRepresentation of a Pure Translation zyxpnoanoad【純平移變換的表示純平移變換的表示】1000100010001zyxdddTnRepresentation of a Pure Translation zyxpnoanoad1

9、00010001000100010001xxxxxxxxxxyyyyyyyyyynewzzzzzzzzzzdnoapnoapddnoapnoapdFdnoapnoapdFnew = Trans (dx ,dy ,dz ) Fold 相對(duì)于固定坐標(biāo)系，新坐標(biāo)系位置可通過在原坐標(biāo)系矩陣前面左乘左乘變換矩陣變換矩陣得到。nRepresentation of a Pure Rotation about an Axis yzoapl1l2l3l4pzpapopapopy1234cossinsincosxnyoazoapppllpppllpp1000cossin0sincosxnyozapppppp( ,

10、 )xyznoapRot xpcossin0sincos0001),Rot(x【繞軸純旋轉(zhuǎn)變換的表示繞軸純旋轉(zhuǎn)變換的表示】Rotation Matrices100( , )00Rot xCSSC0( , )0100CSRot ySC0( , )0001CSRot zSC1000010000cossin00sincos,10000cos0sin00100sin0cos,10000cossin00sincos00001,zRotyRotxRot齊次變換矩陣？齊次變換矩陣？nRepresentation of Combined Transformations lExample: lRotation

11、of degrees about the x-axis,lFollowed by a translation of l1,l2,l3 (relative to the x-, y-, and z-axes respectively),lFollowed by a rotation of degrees about the y-axis.1.Pre-multiply by each matrix:1, =( ,)xyznoapRot xp2,1231,123( , , )( , , )( , )xyzxyznoapTrans l l lpTrans l l lRot xp3,2,123( ,)(

12、 ,)( , , )( , )xyzxyzxyznoappRot ypRot yTrans l l lRot xp相對(duì)于固定的參考坐標(biāo)系的每次變換，變換矩陣都是左乘的?！緩?fù)合變換的表示復(fù)合變換的表示】nTransformations Relative to the Rotating (current) Frame lIn this case, matrices representing each transformation are post-multiplied.lIf transformations are relative to both the Universe frame and t

13、he current frame, each matrix is accordingly multiplied, either pre- or post-.當(dāng)進(jìn)行相對(duì)于運(yùn)動(dòng)坐標(biāo)系或當(dāng)前坐標(biāo)系的軸的變換時(shí)：為計(jì)算當(dāng)前坐標(biāo)系中點(diǎn)的坐標(biāo)相當(dāng)于參考坐標(biāo)系的變化，這時(shí)需要右乘變換矩陣右乘變換矩陣而不是左乘。【相對(duì)于旋轉(zhuǎn)坐標(biāo)系（當(dāng)前坐標(biāo)系相對(duì)于旋轉(zhuǎn)坐標(biāo)系（當(dāng)前坐標(biāo)系/運(yùn)動(dòng)坐標(biāo)系）的變換運(yùn)動(dòng)坐標(biāo)系）的變換】Inverse of MatriceslThe following steps must be taken to calculate the inverse of a matrix:lCalculate

14、the determinant of the matrix.lTranspose the matrix.lReplace each element of the transposed matrix by its own minor (adjoint matrix).lDivide the converted matrix by the determinant.4. 4. 變換矩陣的逆變換矩陣的逆所謂逆變換就是將被變換的坐標(biāo)系返回到原來的坐標(biāo)系。所謂逆變換就是將被變換的坐標(biāo)系返回到原來的坐標(biāo)系。nInverse of Rotation MatriceslThe inverse of a rota

15、tion matrix is its transpose because rotation matrices are “unitary”. Txx),(Rot),(Rot1【旋轉(zhuǎn)矩陣的逆旋轉(zhuǎn)矩陣的逆】nInverse of Transformation MatriceslThe inverse of a transformation (or a frame) matrix is the following:l1. Transpose the rotation portion of the matrix.l2. Take the negative of the dot-product of th

16、e P and n, P and o, and P and a vectors.lThe scale factors remain the same.1 and 00010001xxxxxyzyyyyxyzzzzzxyznoapnnnnoapoooTTnoapaaa p np op a【變換矩陣的逆變換矩陣的逆】Forward and Inverse Kinematic EquationslForward kinematics includes substituting the known joint values into the equations to find the location

17、 and orientationlInverse kinematics includes finding an equation that results in joint values if the desired position and orientation are specified.5. 5. 機(jī)器人的正逆運(yùn)動(dòng)學(xué)機(jī)器人的正逆運(yùn)動(dòng)學(xué)23關(guān)節(jié)與連桿l 在機(jī)器人中，通常有兩類關(guān)節(jié)：轉(zhuǎn)動(dòng)關(guān) 節(jié)和移動(dòng)關(guān)節(jié)l 自由度：物體能夠相對(duì)于坐標(biāo)系進(jìn)行獨(dú)立運(yùn)動(dòng)的數(shù)目l 不同于人類的關(guān)節(jié)，一般機(jī)器人關(guān)節(jié)為一個(gè)自由度的關(guān)節(jié)，其目的是為了簡(jiǎn)化力學(xué)、運(yùn)動(dòng)學(xué)和機(jī)器人的控制l 轉(zhuǎn)動(dòng)關(guān)節(jié)提供了一個(gè)轉(zhuǎn)動(dòng)自由度，移動(dòng)關(guān)節(jié)

18、提供一個(gè)移動(dòng)自由度，各關(guān)節(jié)間是以固定桿件相連接的24連桿參數(shù) 連桿長度：兩個(gè)關(guān)節(jié)的關(guān)節(jié)軸線 Ji與 Ji+1 的公垂線距離為連桿長度，記為ai。 連桿扭轉(zhuǎn)角：由Ji與公垂線組成平面P，Ji+1與平面P的夾角為連桿扭轉(zhuǎn)角，記為i 。連桿偏移量：除第一和最后連桿外，中間的連桿的兩個(gè)關(guān)節(jié)軸線Ji與Ji+1 都有一條公垂線ai，一個(gè)關(guān)節(jié)的相鄰兩條公垂線 ai與ai-1的距離為連桿偏移量，記為di。關(guān)節(jié)角：關(guān)節(jié)Ji的相鄰兩條公垂線ai與ai-1在以Ji為法線的平面上的投影的夾角為關(guān)節(jié)角，記為i。ai、i、di、i這組參數(shù)稱為Denavit-Hartenberg(D-H)參數(shù)。Denavit-Harten

19、berg (DH) Representation of Forward Kinematic Equations of Robots lMay be used for any configuration, whether specific coordinates or not.lCan include joint offset, twist angles, multi-variable joints, and so on.lVery common.lMany other equations are based on this methodology6. 6. 機(jī)器人正運(yùn)動(dòng)學(xué)方程的機(jī)器人正運(yùn)動(dòng)學(xué)方

20、程的D-HD-H表示表示nD-H RepresentationlZ-axes along the joint motion. represents joint rotation. ld is joint linear displacement or distance between common normals.l is the twist angle between z-axes.la is the length of the common normal.D-H RepresentationlAssign z-axes to each joint along linear motion or

21、 revolute axis.lAssign x-axes along the common normal between successive z-axes.lNo need for y-axes.lIf z-axes coincide, x-axis is perpendicular to both.lIf z-axes are parallel, x-axes can be anywhere.D-H Representation連桿本身的參數(shù)連桿長度連桿兩個(gè)軸的公垂線距離（x方向）連桿扭轉(zhuǎn)角連桿兩個(gè)軸的夾角（x軸的扭轉(zhuǎn)角）連桿之間的參數(shù)連桿之間的距離相連兩連桿公垂線距離（z方向平移距）連

22、桿之間的夾角相連兩連桿公垂線的夾角（z軸旋轉(zhuǎn)角）lFour transformations are necessary to go from one frame to the next:D-H RepresentationlRotate about the zn-axis an angle of . This will make xn and xn+1 parallel to each other. This is true because an and an+1 are both perpendicular to zn, and rotating zn an angle of will ma

23、ke them parallel (and thus, coplanar).lTranslate along the zn-axis a distance of dn+1 to make xn and xn+1 colinear. Since xn and xn+1 were already parallel and normal to zn, moving along zn will lay them over each other.lTranslate along the (already rotated) xn-axis a distance of an+1 to bring the o

24、rigins of xn and xn+1 together. At this point, the origins of the two reference frames will be at the same location.lRotate zn-axis about xn+1-axis an angle of to align zn-axis with zn+1-axis. At this point, frames n and n+1 will be exactly the same, and we have transformed from one to the next.D-H

25、Representation1n1n1nlA transformation matrix can be formed by:1111111111111110010001001000000100010000001000100100000100010001( ,)(0,0,)(,0,0)( ,)nnnnnnnnnnnnnnnnCSaSCCSdSTARot zTransdTrans aRot x 100001nCD-H RepresentationlAn A-matrix is:D-H Representation11111111111111111100001nnnnnnnnnnnnnnnnnnCS

26、CSSaCSCCCSaSASCdp在機(jī)器人的在機(jī)器人的基座基座與與手手之間的總變換為：之間的總變換為：yznoaTool Reference FramexnAAAA321n1 -n32211RHRTTTTTlA parameters table may look like:D-H RepresentationlExample: A simple 6-axis robot zyx123z0 x0z1x1z2x2z3x4x3z4z5x5z6x6a2a3a44D-H RepresentationlFind a set of equations that allow determination of

27、joint values from desired position and orientation information.lEach robot has a different solution.lIt may be necessary to use different approaches for each robot.lThis usually requires pre-multiplication of transformation matrices by inverse of individual A matrices, squaring of terms, divisions, and so on.Inverse Kinematic EquationslFor the shown example, the following may be found:where and In