【正文】 1 （3）Ti4=1 0 0cosφi1 0 0sinφi100sinφi10 0 cosφi10 0 1 （4）式中:ψi1 即手動(dòng)轉(zhuǎn)軸坐標(biāo)系O4x4y4z4 繞x4 軸的轉(zhuǎn)角。Ti5= cosθi1 0 0 1 sinθi1 0 0 0sinθi1 0 0 0 cosθi1 0 0 1 （5）式中:θi1即兩調(diào)整螺桿坐標(biāo)系由于高度差所形成的夾角。下面應(yīng)用解析法分別求出各坐標(biāo)系在基礎(chǔ)坐標(biāo)系中的位姿矩陣。散牙坐標(biāo)系O5x5y5z5 的求解:由于我們已經(jīng)求解出散牙坐標(biāo)系在排牙坐標(biāo)系中的位姿矩陣, 所以只需將之轉(zhuǎn)化到基礎(chǔ)坐標(biāo)系中,即O T5 = OAS Ti1 Ti2 Ti3Ti4Ti5 oAS=10010a50b500001c501 ( 6 )式中: OAS 為排牙坐標(biāo)系到基礎(chǔ)坐標(biāo)系的轉(zhuǎn)化矩陣,參數(shù)abc5 表示的是初始位置時(shí)坐標(biāo)系間x、y、z向的距離,下同。手動(dòng)轉(zhuǎn)軸坐標(biāo)系的O4x4y4z4 :由圖2可知,手動(dòng)轉(zhuǎn)軸坐標(biāo)系各個(gè)坐標(biāo)軸的方向與散牙坐標(biāo)系的各坐標(biāo)軸方向完全相同且無(wú)任何轉(zhuǎn)動(dòng), 所以只需將散牙坐標(biāo)系沿各個(gè)坐標(biāo)軸方向進(jìn)行相應(yīng)的平移,即OT4 = OT55A4 （7）式中: 4A5 為散牙坐標(biāo)系到手動(dòng)轉(zhuǎn)軸坐標(biāo)系的轉(zhuǎn)化矩陣，如圖5所示。 圖5 轉(zhuǎn)化矩陣圖調(diào)整螺桿坐標(biāo)系O2x2y2zO3x3y3z3的求解:當(dāng)坐標(biāo)系O2x2y2z2相對(duì)于坐標(biāo)系O3x3y3z3沿z2 軸多運(yùn)動(dòng)位移d時(shí),帶動(dòng)坐標(biāo)系O5x5y5zO4x4y4z4發(fā)生轉(zhuǎn)動(dòng),轉(zhuǎn)動(dòng)的角度即式( 5)中的θi ,這樣就會(huì)使手動(dòng)轉(zhuǎn)軸坐標(biāo)系O4x4y4z4與調(diào)整螺桿坐標(biāo)系O3x3y3z3的相對(duì)位置發(fā)生改變,如圖5所示。從圖5中我們可以很容易的求解出變量Δx、Δy的數(shù)值,那么將手動(dòng)轉(zhuǎn)軸坐標(biāo)系O4x4y4z4繞x4軸旋轉(zhuǎn)ψi ,然后繞y4軸旋轉(zhuǎn)θi 角,使z4軸與z3軸方向相同,再分別沿x4軸反向平移a3 +Δx,沿y4軸反向平移b3 ,最后沿z4軸反向平移c3Δz,使兩坐標(biāo)系重合，則：OT3 =OT4roz( x4 ,ψi )roz(y4,θi )T(z4 , c3+Δz)T(y4,b3)T(x4,a3Δx) （8）由于兩調(diào)整螺桿坐標(biāo)系間x向距離為定值a2(通過(guò)測(cè)量可得),則將坐標(biāo)系O2x2y2z2沿x2軸反向平移a2,然后沿z2軸正向平移a2tanθi1即oT2=oT3T(x3，a2)T（Z3,a2tanθi） ( 9 )滑動(dòng)板坐標(biāo)系O1x1y1z1的求解:將調(diào)整螺桿坐標(biāo)系O2x2y2z2延z2軸反向平移zi1 (即公式2中的zi1 ) ,使之回到零點(diǎn)處, 然后分別沿x2 軸正向平移a1 ,沿y2 軸正向平移b1 ,最后沿z2 軸反向平移c1 ,即OT1 =O T2 T ( z2 , zi1 ) T ( x2 , a1 ) T ( y2 , b1 ) T ( z2 , c1 ) (10)U簧擬合點(diǎn)坐標(biāo)系Oixiyizi( i = 9 ) 的求解。由于我們求解U簧擬合點(diǎn)坐標(biāo)系的目的是通過(guò)擬合點(diǎn)坐標(biāo)系Oixiyizi( i = 9)的位姿變化量來(lái)推倒出各個(gè)電機(jī)所需的脈沖, 因此我們只需求出各個(gè)擬合點(diǎn)坐標(biāo)系在基礎(chǔ)坐標(biāo)系中X 向和Y向的變化量即可。如圖6所示, 我們先求出各個(gè)擬合點(diǎn)所對(duì)應(yīng)的彈簧板上的點(diǎn)的坐標(biāo), 然后通過(guò)兩點(diǎn)間距離公式對(duì)各個(gè)擬合點(diǎn)的坐標(biāo)進(jìn)行求解。如圖6所示。圖6　擬合曲線與彈簧板曲線位置關(guān)系由排牙經(jīng)驗(yàn)我們可知,在各個(gè)散牙位姿確定的情況下,牙弓曲線的方程為[ 7～9 ]y =αxβ式中:參數(shù)α和β是由牙弓曲線的弧長(zhǎng)、弓長(zhǎng)和弓寬決定。而彈簧板的曲線方程是將牙弓曲線沿法線方向等距偏移20 mm得到的。對(duì)于給定的平面參數(shù)曲線c(t),設(shè)其有切線, c(t)的等距為d的等距曲線方程為[10]c0(t) = c(t)+dN(t) (11)式中: N(t)為曲線c(t)在參數(shù)t處的單位法向量?！? （12）將牙弓曲線方程寫(xiě)成平面參數(shù)方程的形式有 t ∈ (0, 1) (13) 　冪函數(shù)處處有切線,這里,等距曲線在牙弓曲線的外側(cè),其距離d = 20 mm,則彈簧板曲線c0 (t)的方程為 (14)將x′(t) = 1, y′(t) =α βxβ 1代入式(12)得N ( t) =1/{1 +α2β2t2β 2[ αβtβ 1 , 1 ]}　　彈簧板曲線方程的參數(shù)形式為 (15)式中: 參數(shù)t是指牙弓曲線上某點(diǎn)的橫坐標(biāo)。圖6中弧長(zhǎng)SS2 已知,如果利用弧微分公式S=0x1+f(x)39。dx,那么未知量x是積分上限且曲線方程為參數(shù)形式,不好求解。因此我們用MATLAB編寫(xiě)一個(gè)循環(huán)程序,將曲線分成很多小微段,求出每一小段的弧長(zhǎng),一點(diǎn)一點(diǎn)累加,判斷每一次累加的結(jié)果是否超過(guò)弧長(zhǎng)S,如果超出則程序停止,并輸出此時(shí)點(diǎn)的坐標(biāo)。求出各個(gè)擬合點(diǎn)所對(duì)應(yīng)的彈簧板上點(diǎn)的坐標(biāo)后, 由兩點(diǎn)間距離公式以及這兩點(diǎn)都在過(guò)A點(diǎn)的法線上,那么可得出各個(gè)擬合點(diǎn)的坐標(biāo)。算例通過(guò)上述算法,我們?nèi)∠伦笱罏槔?它的坐標(biāo)原點(diǎn)與排牙參考坐標(biāo)系的原點(diǎn)重合,已知它的位姿以及牙弓曲線的方程,來(lái)確定各個(gè)關(guān)節(jié)的位姿,其中散牙在排牙坐標(biāo)系中的參數(shù)如下表1所示,各坐標(biāo)系的轉(zhuǎn)化矩陣的參數(shù)如表2所示。表1 散牙位姿的參數(shù)xi1yi1zi1Φi1(176。)ψi1(176。)θi1(176。)牙弓曲線方程00181015Y=表2 各坐標(biāo)系的轉(zhuǎn)化矩陣參數(shù)序號(hào)ai（mm）bi（mm）ci（mm）i=14300i=21200i=30i=4i=50計(jì)算結(jié)果如下：oT5= 731oT4= oT3= 0 0 00 0 1 0 1oT2= 0 0 00 0 1 0 1oT1= 0 0 00 0 1 0 1o6=( ) o7=( )o8 = ( ) o9 = ( )結(jié)論本文通過(guò)對(duì)排牙機(jī)器人機(jī)構(gòu)的分析,取單操作機(jī)為研究對(duì)象建立了運(yùn)動(dòng)學(xué)方程,應(yīng)用解析法和矩陣法,提出了一套對(duì)串并聯(lián)機(jī)器人運(yùn)動(dòng)學(xué)逆問(wèn)題的求解方法。當(dāng)我們已知患者的牙弓曲線和各散牙的末端位姿時(shí),便可通過(guò)以上算法求得各個(gè)關(guān)節(jié)坐標(biāo)系的位置和姿態(tài)。實(shí)驗(yàn)證明,此方法計(jì)算準(zhǔn)確、合理,符合實(shí)際,為下一步排牙機(jī)器人14個(gè)操作機(jī)整體運(yùn)動(dòng)以及電機(jī)的控制做好了準(zhǔn)備工作。附錄2Kinematic Analysis of ToothArrangement Robot with SerialParallel JointsAbstractBased on the characteristics of serial joints and parallel joints, we put forward a tootharrangement robot posed of a kind of 5 degrees of freedom mechanism of serial structure and parallel structure. Through the analysis of the tootharrangement robot mechanism, the equation of kinematics has been built. We analyze the question of inverse kinematics and deduce the matrix of position and gesture of each joint by using analytic method. Finally, we take downleft tooth as an example, on the condition of knowing the equation of dental arch curve and the parameters of single tooth in the coordinate system, and the matrix of position and gesture of each joint has been solved. The result shows that the arithmetic is reasonable and efficient and partly solve the problem of teeth arrangement robot kinematics.I. INTRODUCTIONParallel joints robot has higher rigidity and precision。higher bearing capacity and ratio of the intensity, and can realize the quite plex movement with very simple mechanism than traditional serial joints robot. .Moreover,it actuatesconveniently[1]. Simultaneously, in position solution,The parallel mechanism countersolution is extremely the robot online and realtime solution must calculates the countersolution which is extremely advantageous to parallel joints. However its working space is often small, it is not suit direct application for industry robot with bigger working space. In a word, the serial mechanism and the parallel mechanism all have its good and bad far, the dynamics analysis of mechanism has been matured. However, in robot kinematics aspect, nearly all the researches on dynamics analysis of robot concentrates on the the serial and parallel joint robot, the research on the serialparallel robot are few, Because the pound robot has close structure and openchain structure, its dynamics modeling and solution is extremely plex. In this paper, we put forward a tootharrangement robot of serial structure and parallel structure. Through the analysis of the tootharrangement robot mechanism, the equation of kinematics has been built and the problem of inverse kinematics also have been partly solved. Finally, we take downleft tooth as an example, The result shows that the arithmetic is reasonable and effic