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PUMA机器人正逆运动学推导及运动空间解算求解:①建立坐标系;②给出D-H参数表;③推导正、逆运动学;④编程得工作空间1.建立坐标系根据PUMA机器人运动自由度,在各关节处建立坐标系如图2所示。图1PUMA560机器人坐标系图2.D-H参数表A4A6A5D-H参数表可根据坐标系设定而得出,见表1。(1)i为绕1iZ轴从1iX到iX的角度;(2)1i为绕iX轴从1iZ到iZ的角度;(3)1ia为沿iX轴从1iZ与iX交点到iO的距离;(4)id为沿1iZ轴从1iZ与iX交点到1iO的距离。表1PUMA机器人的杆件参数表连杆i变量i1i1iaid变量范围11909000160~16022002a2d225~453390903a045~2254409004d110~1705509000100~100660006d266~2663.正运动学推导由坐标系图及各杆件参数可得个连杆变换矩阵。111101000001100001csscT22222222122000010001cscascsaTd33333333230001000001cscascsaT444434400000100001csscTd555545000001000001csscT666656600000010001csscTd根据各连杆变换矩阵相乘,可以得到PUMA560的机械手变换矩阵,其矩阵为关节变量的函数。00123456112233445566TTTTTTT将上述变换矩阵逐个依次相乘可以得到06T。060001xxxxyyyyzzzznoapnoapTnoap6514142315236411234651442311523614231446236235452365141423152364112346514423115236xyzxynccssccccssscsccsncccsccssssscccssnsssccsccsoscssccccssccsccsosccsccssssc142314623545236423152351414235123514423152345231223232165141423152314231223231265144231zxyzxycccssoscsccscssaccsssscccacssscsccsaaccsspcacacdsdssscccccscdspsacaccddscsccsc512341232342232365234523zssdsspcdasasdcccss上式中23232323cos,sincs。将13245690,0带入上式检验,得结果22460630100011000001daddTa计算结果与实际结果一致,故上述推导无误。4.逆运动学推导逆运动学推导为由末端位姿,,,noap确定,求解关节变量i。110111000010000001csTsc222221220000010001csascTd333233300010000001csbTsc444344400001000001csdTsc554555000010000001csTsc666656600000010001csscTd1.求1采用逆变换0111T左乘06T,可得:010123451162233445566TTTTTTT1116110000100000010001xxxxyyyyzzzzcsnoapnoapTscnoap1111111116111111110001xyxyxyxyzzzzxyxyxyxycnsncosocasacpspnoapTsncnsocosacaspcp12345223344556646236523452365234523642352345234232265234523323462365234523652345236423523452342322,,,;,,,TTTTTssccsscccsssccccsccscscdsacdcscscacssscscccssscccscsscccssdcas652345233234656446546452645;,,,;0,0,0,1dcccssascsccscccssssddss可得3,3,3,4对应相等,故有1145112645,xyxysacassspcpddss,联立求解,设:226161166cos,sin,,arctanyxxyyxxyyxppadpadpadpadp可以有如下推导:212121121221262122612sin,cos1arctanarctanarctanyyxxddddpaddpadd2.求23,同理有:0102345112633445566,TTTTTT01011261222112221122211222211221212212122121221211111,,,,;,,,;,,xzyxzyxzyxzyxzyxzyxzyxzyyxyxyTTccnnscnsccooscosaccasacsccppsacpscnscnnsscoscoossacsacasscpscppsscnnscoosac1121,;0,0,0,1xyxascpdps2345334455666353453466353453645334543336533456354533466354536343343534336354354656446,,,;,,,;,TTTTcssccccssssscccccscsccsdsacdcscccccsccssssscsccscsscccsscdasdcccsccsccsccc54645645,,;0,0,0,1ssssdss结合前一步骤公式可得:1222214333653345xzyccppsacpsdsacdcsccc122123433635435xzycpscppsscdasdcccsc设2242343,arctandada,22616111361611166,arctanxyxyxyxyzzzzdacadscpspdacadscpsppdapda可得计算公式:23322,2332cossinsincos0a整理得:2222222232243222322322261611122322222226222234arctanarctanarctanarctan4xyxyzzaadaadacadscpspapdaaa3.求4为求解2,做如下变换:010454123465566,,,TTTT010312346144231141423423144231141423423144231141423423224323222332322232414142314423,,,,,,yxzyxzyxzxyTTncsccsnssccccnsocsccsossccccosacsccsasscccacccacaccasassdspsscccpcscc142323123123231231232312312342322232232123123142314411234423142314411234423;,,,;,,zzxyzxyzxyzxyyxzyxzyscpscncnsnsscocosossacacsassdcpacsacscpspssncccssncsccsnssocccssocsccsossac1423144112344232214231443232223323222341123442423,;0,0,0,1xzyxzccssacsccsasspcccsssacaccasasspcsccscdpss5656565655655645556666000001cccssdscsssccdTTsc联立求解得:1423144112344230yxzacccssacsccsass11423112323arctanyxyxzacasacsaccas4.求56,0105512345666,,,,TTT可得:15234235145152342351455234523142314411234423614231441123442360xyzyxzyxzocccccssssoscsccscssocccssncccssncsccsnsssocccssocsccsossc得:141423142314423512312323arctanxxyyzxyzossoccsosccocsocsocsossoc1423144112344236142314411234423arctanyxzyxzncccssncsccsnssocccssocsccsoss从计算过程,我们可以看出其运动反解存在多组解,此时需要根据机构自身运动范围及运动的优化进行选择。5.工作空间使用MATLAB绘制其工作空间如图2-图5。此处采用正解方式进行工作空间求解,理论上经过6层循环可以,加之运动范围限制可以得出机器人工作空间,实际求解过程中,为保证运算速度能够适应,将工作空间进行简化,考虑4、5、6部分主要影响末端小范围,对总体工作空间影响较小,故将4、5、6轴运动固定,仅对123,,进行组循环计算
本文标题:PUMA机器人正逆运动学推导及运动空间解算-机器人技术大作业
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