ABB工業(yè)機(jī)器人運(yùn)動(dòng)學(xué)研究報(bào)告

::00:..【關(guān)鍵字】構(gòu)造名目

ABBIRB6600爭(zhēng)論報(bào)告機(jī)器人構(gòu)造簡(jiǎn)介ABB工業(yè)機(jī)器人可以用于實(shí)現(xiàn)噴霧、涂膠、物料搬運(yùn)、點(diǎn)焊等多種功能，是典型的機(jī)械臂，在網(wǎng)絡(luò)上可以查找到較多的相關(guān)資料。本次作業(yè)就選取ABBIRB6600機(jī)器人作為爭(zhēng)論對(duì)象，首先對(duì)其構(gòu)造進(jìn)展簡(jiǎn)潔簡(jiǎn)介。12ABBIRB6600是六自由度機(jī)器人，具有六個(gè)旋轉(zhuǎn)關(guān)節(jié)，底座固定，通過(guò)各關(guān)節(jié)的旋轉(zhuǎn)可以完成三維空間內(nèi)的運(yùn)動(dòng)。圖1ABBIRB6600機(jī)器人的照片及工作范圍圖，圖2是其構(gòu)造簡(jiǎn)圖和各軸的轉(zhuǎn)動(dòng)的參數(shù)。機(jī)器人的運(yùn)動(dòng)學(xué)ABBIRB6600D-H建模并求出對(duì)應(yīng)的轉(zhuǎn)換JacobianJacobian變換矩陣。、機(jī)器人正運(yùn)動(dòng)學(xué)為了計(jì)算便利把機(jī)器人各關(guān)節(jié)前后兩連桿共線作為初始狀態(tài)，畫(huà)出構(gòu)造簡(jiǎn)圖如圖3圖337實(shí)際上是末端執(zhí)行機(jī)構(gòu)。運(yùn)用學(xué)過(guò)的D-H建模方法建立模型，建模過(guò)程中為了便利畫(huà)出各關(guān)節(jié)坐標(biāo)系，將局部連桿進(jìn)展了拉長(zhǎng)，且由于局部關(guān)節(jié)坐目標(biāo)Z軸垂直于紙面，所以用XY軸畫(huà)出坐標(biāo)系，用右手定則既得到對(duì)應(yīng)的Z軸。最終建立::1:..4：圖44D-H參數(shù)表：i變量范圍100[-180,180]2900[-65，80]300[-180，60]40-90[-300，300]5-900[-120，120]60-90[-300，300]7〔末端〕00——由此算出各關(guān)節(jié)變換矩陣：將這些關(guān)節(jié)坐標(biāo)變換矩陣連乘就得到了由基坐標(biāo)系到末端的坐標(biāo)變換矩陣：但是由于矩陣規(guī)模較大，不便用矩陣形式寫(xiě)出，所以把malab計(jì)算得出的矩陣用分項(xiàng)的形式寫(xiě)出：sin(θ1)*sin(θ6)+cos(θ6)*(cos(θ5)*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3))+sin(θ5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2)))cos(θ6)*sin(θ1)-sin(θ6)*(cos(θ5)*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3))+sin(θ5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2)))sin(θ5)*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3))-cos(θ5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2)),h5’*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3))-h6*(cos(θ5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2))-sin(θ5)*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3)))-h7*(cos(θ5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2))-sin(θ5)*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3)))+h2’*cos(θ1)-(h4+h5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2))+h7’*(sin(θ1)*sin(θ6)+cos(θ6)*(cos(θ5)*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3))+sin(θ5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2))))-h3*cos(θ1)*sin(θ2)cos(θ6)*(cos(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1))+sin(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2)))-cos(θ1)*sin(θ6),-cos(θ1)*cos(θ6)-sin(θ6)*(cos(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1))+sin(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2)))sin(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1))-cos(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2)),sin(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1))-cos(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2)),h5’*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1))-h7’*(cos(θ1)*sin(θ6)-cos(θ6)*(cos(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1))+sin(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))))-h6*(cos(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))-sin(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1)))-h7*(cos(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))-sin(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1)))+h2’*sin(θ1)-(h4+h5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))-h3*sin(θ1)*sin(θ2)-cos(θ6)*(cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))),sin(θ6)*(cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))),cos(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-sin(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2)),h1+h2+(h4+h5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-h5’*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+h3*cos(θ2)+h6*(cos(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-sin(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2)))+h7*(cos(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-sin(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2)))-h7’*cos(θ6)*(cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3)))用各關(guān)節(jié)轉(zhuǎn)角的初值來(lái)檢查變換矩陣內(nèi)的正確性：設(shè)：代入各關(guān)節(jié)變換矩陣可以求出各矩陣初值，由于中沒(méi)有關(guān)節(jié)變量，所以保持不變記為矩陣g圖5將圖5中的初值矩陣連乘可以得到基坐標(biāo)系到末端坐標(biāo)系的變換矩陣：6：圖6觀看圖6中的矩陣，檢查初始狀態(tài)末端坐標(biāo)系在基坐標(biāo)系的位姿，很簡(jiǎn)潔看出：在中的坐標(biāo)為，并且與反向，與反向，與同向，這與D-H建模圖〔圖4〕中所得到的結(jié)果一樣，可以確定計(jì)算過(guò)程是正確的。、機(jī)器人逆運(yùn)動(dòng)學(xué)Jacobian法進(jìn)展機(jī)器人逆運(yùn)動(dòng)學(xué)的求解。首先已經(jīng)列出的，可以依次求出matlab計(jì)算過(guò)程中：把依次記作a，b，c，d，e，f把記作〔sitai〕計(jì)算結(jié)果如下:6T由于2 矩陣元素表達(dá)式較簡(jiǎn)潔，不便列出整個(gè)矩陣，所以逐列寫(xiě)出：第一列：其次列：第三列：第四列：6T 6T由于1 比2 的矩陣元素表達(dá)式更為簡(jiǎn)潔，所以逐項(xiàng)列出：-cos(θ6)*(cos(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-sin(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))),sin(θ6)*(cos(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-sin(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))),-cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))-sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3)),h2’-(h4+h5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))-h5’*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-h3*sin(θ2)-h6*(cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3)))21a -sin(θ6),2122a -cos(θ6),22-cos(θ6)*(cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3)))sin(θ6)*(cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))),cos(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-sin(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2)),(h4+h5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-h5’*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+h3*cos(θ2)+h6*(cos(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-sin(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2)))066T與正運(yùn)動(dòng)學(xué)計(jì)算00時(shí)求得的7T除了最終一列之外完全一樣，下面寫(xiě)出最終一列的元素：0h5’*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3))-h6*(cos(θ5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2))-sin(θ5)*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3)))+h2’*cos(θ1)-(h4+h5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2))-h3*cos(θ1)*sin(θ2)h5’*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1))-h6*(cos(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))-sin(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1)))+h2’*sin(θ1)-(h4+h5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))-h3*sin(θ1)*sin(θ2)H1+h2+(h4+h5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-h5’*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+h3*cos(θ2)+h6*(cos(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-sin(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2)))利用求出各關(guān)節(jié)到末端的坐標(biāo)變換矩陣，可以求出Jacobian矩陣各列。Jacobian矩陣第iJ的計(jì)算方法如下：in o ax x

px pn

pnn o a pyyyy

z x

y x6T y yi1 n o a

p,并且

poz

po x y

poy xz z z z

pa pa pa0 0 0 1

z x y y xpnpoz zpa 那么:Ji n z z o az z z依據(jù)此公式可以算出Jacobian矩陣的各列:sin(θ6)*(cos(θ3)*sin(θ5)+cos(θ5)*sin(θ3))*(h3+h6*(cos(θ3)*cos(θ5)-sin(θ3)*sin(θ5))+cos(θ3)*(h4+h5)-h5’*sin(θ3))+cos(θ6)*(cos(θ3)*sin(θ5)+cos(θ5)*sin(θ3))*(h6*(cos(θ3)*sin(θ5)+cos(θ5)*sin(θ3))+sin(θ3)*(h4+h5)+h5’*cos(θ3))-sin(θ6)*(cos(θ3)*cos(θ5)-sin(θ3)*sin(θ5))*(h3+h6*(cos(θ3)*cos(θ5)-sin(θ3)*sin(θ5))+cos(θ3)*(h4+h5)-h5’*sin(θ3))-cos(θ6)*(cos(θ3)*cos(θ5)-sin(θ3)*sin(θ5))*(h6*(cos(θ3)*sin(θ5)+cos(θ5)*sin(θ3))+sin(θ3)*(h4+h5)+h5’*cos(θ3)),cos(θ6)*(h3+h6*(cos(θ3)*cos(θ5)-sin(θ3)*sin(θ5))+cos(θ3)*(h4+h5)-h5’*sin(θ3))-sin(θ6)*(h6*(cos(θ3)*sin(θ5)+cos(θ5)*sin(θ3))+sin(θ3)*(h4+h5)+h5’*cos(θ3))]cos(θ3)*cos(θ5)-sin(θ3)*sin(θ5)cos(θ3)*sin(θ5)+cos(θ5)*sin(θ3),0,-sin(θ6)*(cos(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-sin(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2)))*((h4+h5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))-h2’+h5’*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))+h3*sin(θ2)+h6*(cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3)))),cos(θ6)*((h4+h5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))-h2’+h5’*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))+h3*sin(θ2)+h6*(cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3)))),-sin(θ6)*(cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3)))*((h4+h5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))-h2’+h5’*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))+h3*sin(θ2)+h6*(cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))))-cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))-sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3)),0,cos(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3))-sin(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2)),(sin(θ1)*sin(θ6)+cos(θ6)*(cos(θ5)*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3))+sin(θ5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2))))*(h6*(cos(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))-sin(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1)))-h5’*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1))-h2’*sin(θ1)+(h4+h5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))+h3*sin(θ1)*sin(θ2))-(cos(θ6)*sin(θ1)-sin(θ6)*(cos(θ5)*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3))+sin(θ5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2))))*(h6*(cos(θ5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2))-sin(θ5)*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3)))-h5’*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3))-h2’*cos(θ1)+(h4+h5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2))+h3*cos(θ1)*sin(θ2)),(cos(θ1)*cos(θ6)+sin(θ6)*(cos(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1))+sin(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))))*(h6*(cos(θ5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2))-sin(θ5)*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3)))-h5’*(cos(θ1)*sin(θ2)*sin(θ3)-cos(θ1)*cos(θ2)*cos(θ3))-h2’*cos(θ1)+(h4+h5)*(cos(θ1)*cos(θ2)*sin(θ3)+cos(θ1)*cos(θ3)*sin(θ2))+h3*cos(θ1)*sin(θ2))-(cos(θ1)*sin(θ6)-cos(θ6)*(cos(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1))+sin(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))))*(h6*(cos(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))-sin(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1)))-h5’*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ3)*sin(θ1))-h2’*sin(θ1)+(h4+h5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))+h3*sin(θ1)*sin(θ2)),-cos(θ6)*(cos(θ5)*(cos(θ2)*sin(θ3)+cos(θ3)*sin(θ2))+sin(θ5)*(cos(θ2)*cos(θ3)-sin(θ2)*sin(θ3)))*(h6*(cos(θ5)*(cos(θ2)*sin(θ1)*sin(θ3)+cos(θ3)*sin(θ1)*sin(θ2))-sin(θ5)*(sin(θ1)*sin(θ2)*sin(θ3)-cos(θ2)*cos(θ