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年 月 日
國際工程與技術(shù)雜志2卷10號，十月，2012
自動(dòng)封口機(jī)的逆運(yùn)動(dòng)學(xué)分析
Akinola A. Adeniyi 1, Abubakar Mohammed 2, Aladeniyi Kehinde 3
機(jī)械工程1，大學(xué)管理，管理，尼日利亞
機(jī)械工程系，聯(lián)邦技術(shù)大學(xué)，明娜，尼日利亞
科學(xué)實(shí)驗(yàn)室技術(shù)系，魯弗斯性理工，尼日利亞
摘要
自動(dòng)封口機(jī)主要用于自動(dòng)化生產(chǎn)線。正確封裝一樣物品的時(shí)間基本取決于電機(jī)控制器是否能精確地判斷出這樣物品在不在最佳的位置上。活塞頭的伸縮度很大程度上取決于紅外傳感器，由活塞頭的伸縮程度，該連接所需的角位移要用逆運(yùn)動(dòng)學(xué)的方法來測定。僅為了展示這種方法同時(shí)減少方程的復(fù)雜程度，連桿的重力作用和慣性將忽略不計(jì)。
關(guān)鍵詞：正向運(yùn)動(dòng)學(xué) 逆運(yùn)動(dòng)學(xué) 自動(dòng)化 封口機(jī)
1簡介
一個(gè)自動(dòng)化的工廠通常使用大量的電子控制的機(jī)械來完成生產(chǎn)，工廠的自動(dòng)化對于其管理具有非常多的戰(zhàn)略意義上的好處。標(biāo)準(zhǔn)的機(jī)械連接通常由電機(jī)、氣動(dòng)系統(tǒng)或電磁閥組成。在手動(dòng)操作的機(jī)器上，人們通過目視來檢測，其它方面的檢測則是機(jī)械化地重復(fù)。這項(xiàng)研究的重點(diǎn)集中在一個(gè)品牌生產(chǎn)線中假想的用于壓制和簽名的封口機(jī)器，逆運(yùn)動(dòng)學(xué)的分析方法將使我們確定鏈接的角位移。運(yùn)動(dòng)學(xué)僅考慮物體的運(yùn)動(dòng)而忽略驅(qū)動(dòng)力，而通過逆運(yùn)動(dòng)學(xué)的方法，則可以確定達(dá)到在最終效應(yīng)下的指定目標(biāo)所需要的鏈接角度和端部執(zhí)行器的位置。
Nagchaudhuri進(jìn)行了一個(gè)關(guān)于曲柄滑塊機(jī)構(gòu)用于PID控制器使用的可行性研究，然而其中卻忽視了偏移量。Tolani通過審查將逆運(yùn)動(dòng)學(xué)的求解問題分為七類，該些方法分別是牛頓迭代方法及其變體，還有雅可比矩陣和與偽逆的變體（也稱為穆爾彭羅斯逆）方或非方形矩陣，以及其它基于控制理論和優(yōu)化技術(shù)的方法。一些研究者提出了一些求解IK問題的算法，但不包括神經(jīng)網(wǎng)絡(luò)算法，循環(huán)坐標(biāo)下降停止算法和不精確的策略算法，但像所有其他技術(shù)一樣，對于一個(gè)給定問題的方法選擇取決于問題的細(xì)節(jié)。Buss討論了雅可比矩陣的轉(zhuǎn)置，穆爾彭羅斯和阻尼最小二乘法，雅克比轉(zhuǎn)置在計(jì)算量上較小然而在基于機(jī)器人的結(jié)構(gòu)前提上卻表現(xiàn)欠佳，在這項(xiàng)研究上，雅克比矩陣轉(zhuǎn)置表現(xiàn)不如人意，但是其逆方法確實(shí)更加合適的，因?yàn)樗鼘⑦@個(gè)問題簡化成了簡單二維平面上四個(gè)自由度的問題。
2 自動(dòng)聯(lián)接部分
圖1展示了機(jī)械密封系統(tǒng)的原理圖。加蓋或沖壓由活塞或壓裝頭實(shí)現(xiàn)，P.C是輸送線，蓋子或商標(biāo)由紅外線傳感器感知并放置在正確的位置，S.依靠反饋傳感器來密封或貼印商標(biāo)。如果要印章、封蓋或封裝的物品離開了例子當(dāng)中壓裝頭將要接觸的位置，傳感器將反饋并使壓裝頭縮回。有一系列的反饋給了電機(jī)控制器的話，它也可以不走得那么遠(yuǎn)。M.如果它是手動(dòng)的，這種控制系統(tǒng)就類似于一個(gè)人工操作員,使用傳感器和快速響應(yīng)電動(dòng)機(jī)控制器將使這個(gè)假設(shè)的機(jī)器成為一個(gè)在工廠中執(zhí)行平常任務(wù)非常有用的工具。這個(gè)工廠支線由一個(gè)簡單的曲柄滑塊機(jī)構(gòu)與致動(dòng)器臂A.組成。用更清楚的話來說，如果蓋子和容器在一條線上，那么指令將會(huì)發(fā)出使得活塞壓下來完成密封；如果堵塞則扭轉(zhuǎn)活塞；如果容器或蓋子不在活塞頭將不會(huì)壓下；如果由磨損或撕裂所造成的密封長度短于預(yù)期長度則會(huì)加深壓裝力度。這清楚地表明，活塞確定鏈接的角度、方向和電機(jī)的運(yùn)動(dòng)。這就是一個(gè)逆運(yùn)動(dòng)學(xué)問題，傳感器反饋部分是一個(gè)很復(fù)雜的控制工程問題，就不在此考慮了。
圖1:自動(dòng)封口機(jī)示意圖
（A：致動(dòng)器臂 C：輸送線 M：電機(jī)控制器 P：壓裝頭 S：傳感器 O：需要封口或封蓋的物品）
3 分析
圖2是一個(gè)代表性的曲柄滑塊機(jī)構(gòu)，從活塞軸到電機(jī)軸有一個(gè)偏移量f，O1、O2是軸的活塞移動(dòng)的坐標(biāo)(x,y)。電機(jī)相對O1順時(shí)針或逆時(shí)針方向旋轉(zhuǎn)，如果曲柄使位移Δs在活塞平面，它相當(dāng)于一個(gè)Δex和Δey的運(yùn)動(dòng)。Δα這個(gè)運(yùn)動(dòng)是由曲柄的順時(shí)針或逆時(shí)針運(yùn)動(dòng)產(chǎn)生的，Δβ為連桿與曲柄之間夾角，Δλ也表示連桿與活塞平面之間的夾角。
圖2:偏移滑塊曲柄(笛卡爾坐標(biāo)系)
在電腦的仿真中設(shè)計(jì)這些，角度會(huì)有明確的要求以防止連接不發(fā)生“物理分離”；對于一個(gè)真實(shí)物理連接環(huán)節(jié)，電機(jī)控制器需要控制移動(dòng)的就只有曲柄。
3.1 坐標(biāo)系
在這里采用笛卡爾坐標(biāo)系，順時(shí)針為正、向右移動(dòng)為正、向上為正。上死點(diǎn)（TDC）的計(jì)算式如下，設(shè)曲柄半徑為r,連桿長度為l,則有
其中，fm是指基于幾何上的最大偏移量。下死點(diǎn)的計(jì)算如下
上死點(diǎn)與下死點(diǎn)以及偏移量如圖3所示。
圖3 上死點(diǎn)與下死點(diǎn)
活塞被限制為只在平面方向移動(dòng)，為向量，在這個(gè)研究中，方向向量上有，使其平面在45°角至水平方向。
3.2 正運(yùn)動(dòng)學(xué)
電機(jī)順時(shí)針方向移動(dòng)產(chǎn)生的位移的位置以圖2字母表示反映在（1）式中。在下標(biāo)中以（i，f）分別表示起始和最終值，f的位置是在現(xiàn)實(shí)中平穩(wěn)轉(zhuǎn)動(dòng)曲柄達(dá)到的，平滑度可以用數(shù)值方法細(xì)致漸進(jìn)地達(dá)到要求。在這個(gè)增量的最后，最終的位移目標(biāo)以一個(gè)角度參數(shù)的函數(shù)方式給出：
角度的線性關(guān)系在這個(gè)問題中可以幫助減少在方程（1）計(jì)算中的自由度數(shù)量，由可以得出。利用三角法，可以計(jì)算出圖2中任何時(shí)刻的活塞位置，見式（2）、（3）。
由雅克比矩陣，給出了方程（4）同時(shí)將其簡化為方程（5）。
計(jì)算新的活塞位置涉及求解方程(1)，第一級下的泰勒級數(shù)下活塞的新坐標(biāo)見方程（6）。θ是相關(guān)連接的位移矢量角，數(shù)學(xué)上，，這里，我們設(shè)，由此活塞或壓頭的位置大約可由方程（6）得出。應(yīng)該指出的是,β可以取水平位置來進(jìn)一步降低方程集，在這里可以將其稱為β0.
3.3 逆運(yùn)動(dòng)學(xué)
問題不在于給出Xi和θ來求解Xf，而是由給定Xi和Xf來求解θ。由迭代實(shí)得出了活塞的位移目標(biāo)為，活塞的向量位移可表示為。因?yàn)檫@是一個(gè)沒有其他方向位移的平面問題，所以可以將其簡化為。為了減少可能的顛簸或跳動(dòng)影響，可以逐步使用比基礎(chǔ)上的r和L更直觀的系數(shù)μ，前提是μ≤1且為逆雅克比矩陣。該算法可以檢查是否已經(jīng)達(dá)到目標(biāo)，當(dāng)求解少于預(yù)先確定的誤差等級或達(dá)到最大數(shù)量迭代時(shí)，此迭代停止，這就是實(shí)時(shí)應(yīng)用的一個(gè)關(guān)鍵部分。
4 結(jié)果和討論
考慮活塞頭在位置P時(shí)，機(jī)械臂在若干任意位置的方向，假設(shè)傳感器系統(tǒng)需要活塞移動(dòng)到新的位置P2，完成了幾個(gè)任意起始位置的曲柄仿真所得出的結(jié)果都是同想要達(dá)到的目標(biāo)接近的。如果曲柄當(dāng)前方向與曲柄角之間為-5°，且有一個(gè)指令從傳感器發(fā)出使活塞壓頭收回其曲柄臂0.1倍的長度。模擬指示曲柄繼續(xù)逆時(shí)針方向走15.58°,這對應(yīng)于β0增加到19.26°，相應(yīng)的，λ減少到86.32°。圖4顯示了模擬進(jìn)展的活塞頭從當(dāng)前位置P1到新目標(biāo)P2和完成迭代次數(shù)。
圖4:曲柄位置和迭代與雅可比矩陣求逆
該技術(shù)使用的是逆雅克比技術(shù)，雅克比行列式的轉(zhuǎn)置方法在這種情況下對于同樣的問題是無法預(yù)測的，解決方法只有落定到一個(gè)角度內(nèi)的局部最小值，但其收斂速度更快，見圖5。
圖5 使用逆向和轉(zhuǎn)置的雅克比矩陣得出的曲柄位置
如果要求計(jì)算一個(gè)無法在實(shí)際上達(dá)到的目標(biāo)，例如超過上死點(diǎn)或下死點(diǎn)位置的點(diǎn)P3，模擬運(yùn)行并且在達(dá)到最大迭代數(shù)量時(shí)停止，亦可能是雅克比矩陣變?yōu)椴豢赡?，見圖6。
圖6 不可實(shí)現(xiàn)目標(biāo)情況
5 小結(jié)
本文的重點(diǎn)是應(yīng)用逆運(yùn)動(dòng)學(xué)技術(shù)分析機(jī)器人鏈接如獲得封閉的自動(dòng)化工廠，不考慮慣性效應(yīng)的影響。反求技術(shù)的雅可比行列式,如上文獻(xiàn),在這個(gè)應(yīng)用程序更可靠。雅可比行列式的方法是不可逆的轉(zhuǎn)置。本文簡要介紹了一個(gè)簡單的應(yīng)用了逆運(yùn)動(dòng)學(xué)的自動(dòng)封口機(jī)；以活塞頭收回0.1單位作為測試實(shí)例。新的曲柄角度可以由雅克比逆向方法相對雅克比行列式轉(zhuǎn)置方法更精確地得出。這個(gè)問題還可以擴(kuò)展到包括動(dòng)力學(xué)對于最優(yōu)轉(zhuǎn)矩或電動(dòng)馬達(dá)驅(qū)動(dòng)部件的可能性選擇上。
International Journal of Engineering and Technology Volume 2 No. 10, October, 2012
An Inverse Kinematic Analysis of a Robotic Sealer
Akinola A. Adeniyi 1, Abubakar Mohammed 2, Aladeniyi Kehinde 3
1Department of Mechanical Engineering, University of Ilorin, Ilorin, Nigeria
2Department of Mechanical Engineering, Federal University of Technology, Minna, Nigeria
3Department of Science Laboratory Technology, Rufus Giwa Polytechnic, Owo, Nigeria
ABSTRACT
A planar robotic sealing or brand stamping machine is presented for an automated factory line. The appropriate time to seal or to stamp an object is basically determined by a motor controller which relies critically on whether or not the object is in the best position. The extent of protraction and retraction of the piston head is largely dictated by an infrared sensor. Given the extent to protract or retract the piston head, the angular displacements of the link required are determined using the Inverse Kinematic (IK) techniques. The inertia and gravity effects of the links have been ignored to reduce the complexity of the equations and to demonstrate the technique.
Keywords: Forward Kinematics, Inverse Kinematics, Robotics, Sealer
1. INTRODUCTION
An automated factory uses a number of mechanical links electronically controlled to achieve tasks. The benefits of factory automation are many and of strategic importance to management [1]. Standard mechanical links are usually powered with electrical motors, pneumatic systems or solenoids. In a manually operated machine, the human performs visual checks and other standard checks that are to be replicated by automation. The interest of this work is centered on a hypothetical sealing machine which is used for stamping some signatures and logos as done in a branding factory line. Inverse kinematic analysis is applied to enable us determine angular displacements of the link. Kinematics involves the study of motion without consideration for the actuating forces. Inverse Kinematics (IK) is a method for determining the joint angles and desired position of the end-effectors given a desired goal to reach by the end effectors [1].
A feasibility of using a PID controller was studied by Nagchaudhuri [2] for a slider crank mechanism but without an offset. Tolani et al [3] reviewed and grouped the techniques of solving inverse kinematics problems into seven. The techniques are the Newton-Raphson’s method and its other variants. There are the Jacobian and the variants with pseudo-inverse (otherwise known as the Moore-Penrose inverse) for square or non-square Jacobian. Other methods are the control-theory based and the optimisation techniques. A number of authors [1, 4-7] have proposed algorithms for solving IK problems which include but not limited to Neural Network algorithm, Cyclic Coordinate Descent closure and Inexact strategy, but like every other techniques for a given problem the choice of method depends on the specifics of the problem.
Buss [8] discussed the Jacobian transpose, the Moore-Penrose and the Damped Least Squares techniques. In terms of computational cost, the Jacobian transpose method is the cheap but can perform poorly based on the robot configurations. In this work the Jacobian transpose technique ill-performed but the Jacobian Inverse technique is suitable and more so it is a simple 2D planar representation of the problem with only 4 degrees of freedom.
2. OPERATIONS OF THE ROBOTIC LINK
Fig. 1 shows the schematic diagram of the robotic sealing system. The capping or stamping is achieved with the piston or ram head, P. C is the conveyor line. The caps or the branding heads are placed in position and sensed by an infrared sensor, S. The instruction to seal or brand is dependent on feedback from the sensor. If the item to be branded, capped or stamped is out of place at the instance when the ram head was going to touch, the sensor feedback will be to retract the head. It can also be to not go too far. There can be a range of feedback to the motor controller, M. This kind of control system is similar to what a human operator would do if it were manually operated. The use of sensors and fast responding motor controller will make this hypothetical machine a very useful tool in a factory performing this kind of mundane task. This factory sub-line is a simple slider-crank mechanism with actuator arm A.
In clearer terms, the instructions would be to press the piston ram to seal if the cap and the container are in line; to reverse the piston in case of a jam; to not press the piston ram if either the container or the cap is absent; to press further if the seal length is shorter than expected as may be caused by wear and tear. This clearly shows that the piston determines the angle of the link or the direction or action of the motor. This is an inverse kinematics problem. The sensor feedback part is much of a control engineering problem, not considered in this paper.
Fig. 1: The robotic sealing rig schematic
3. ANALYSIS
Fig. 2 is a representation of the slider-crank mechanism. There is an offset, f, of the piston axis from the motor axis, O1. O2 is the axis of the piston with moving coordinates (x,y). The motor rotates clockwise or counter clockwise about O1. If the crank makes displacement Δs on the piston plane, it is equivalent to a motion of Δex and Δey. This motion is caused by the crank making an angular motion clockwise or counter-clockwise, . The angle between the connecting rod and crank makes an angular displacement of, . This also means the angular shift of is made between the connecting rod and the piston or ram plane.
Fig. 2: The offset slider crank (Cartesian coordinate world)
In a computer game application for these, the angles would be explicitly required so that the links do not “physically disjoint”; for a physically connected link, the motor controller only would need the instruction to move only the crank.
3.1 The World
Cartesian coordinate system is adopted. Clockwise is positive and motion to right and upwards are positive. The Top Dead Centre (TDC) is attained when the crank, radius r, and the connecting rod, length l, are in line. This is attained when . fm is the maximum variable offset based on the geometry. The Bottom Dead Centre (BDC) is reached when . The TDC and BDC with the variable offset are shown in Fig. 3.
Fig. 3: The Top and Bottom Dead centre
The piston has been constrained to move only in planar direction, on the vector of . In this work, the direction vector is , making the plane at 45° to the horizontal.
3.2 The Forward Kinematics
The displacement caused by the motor moving clockwise from the position in Fig. 2 is represented in equation (1). Where subscripts (i,f) are respectively mean initial and final values. The position at f is reached in reality smoothly for a rotating crank, but the smoothness can be reached in fine incremental steps, in the numerical approach. At the end of the stepped increments, the final displacement to the goal is seen as a function of angular parameters given as:
The linear dependence of the angles, in this problem, can help to reduce the number of degrees of freedom to compute in equation (1). It can be shown that, there by making .Using trigonometry, the instantaneous initial, arbitrary, position of the piston in Fig. 2 is given by Equation (2)(3).
The Jacobian matrix for is given in equation (4) and simplified to equation (5).
Computing the new piston position involves solving equation (1). The new coordinate of the piston by the first term of expansion of the Taylor series can be shown to be given in equation (6). θis the vector of the robot angular displacements for the related links. Mathematically, . Here, we have . Therefore the current position of the piston or the pressing head is approximately given in equation (6). It should be noted thatβcan be measured from the horizontal to further reduce the equation sets, this is referred to as β0 elsewhere in this paper.
3.3 Inverse Kinematics
The problem is not that of solving for Xf given Xi andθbut it is that of solving for θgiven Xi, and the desired Xf. This is iteratively implemented such that the target displacement of the piston is given as .This is a vector of the piston displacement and can be represented as.Since this is a planar problem with no displacements in the other directions, it reduces to a.To smoothen the possible jerk or jumpy effect, this can be stepped using a factor ofμwhich can be selected intuitively based on the ratio of r to L but and is the inverse of Jacobian matrix. The algorithm checks if the target has been reached or not. Iteration is stopped when the solution is within a pre-determined level of error or a maximum number of iterations. The choice of these limiting values should depend on the response time acceptable. This can be critical for a real time application.
4. RESULT AND DISCUSSIONS
Consider a current orientation of the robotic arm at any arbitrary position with the piston head at a position P. Suppose the sensor system requires the piston to move to a target new position P2. The simulation is done for several arbitrary starting positions of the crank and results are similar for reachable targets. Supposing the crank angle is at a current orientation with crank angle of -5°, and there is an instruction from the sensor to retract the piston ram head by 0.1times the crank arm length. The simulation instructs the crank proceeds to counter clockwise by 15.58°, this corresponds to an increase ofβ0 to 19.26°and correspondingly,λreduces to 86.32°. Fig. 4 shows the simulation progress of the piston head from a current position P1 to the new target P2 and the number of iterations done.
Fig. 4: Crank Position and Iteration with the Jacobian Inverse Matrix
The technique used is the Jacobian inverse technique. The Jacobian transpose technique is not predictable for the same problem and in this case, the solution settles to a local minimum for only one of the angles but the convergence rate is faster, see Fig. 5.
Fig. 5: Crank Positions using the Inverse and Transpose of the Jacobian Matrix
If there is a request to a physically unreachable target, such as to a more than the TDC or BDC locations, P3, the simulation runs and stops after the maximum number of iterations or if the Jacobian Matrix becomes un-invertible, Fig. 6.
Fig. 6: Unreachable Target situation
5. CONCLUSION
This paper is focused on the application of the Inverse Kinematics technique to the analysis of a robotic link, such as obtained in a sealer of an automated factory, without consideration for the effects of inertia effects. The Jacobian inverse technique, as mentioned in literatures, is more reliable in this application. The Jacobian transpose approach is not reliable. This paper has demonstrated the application of the inverse kinematics to a simple robotic sealer; the piston is instructed to retract by 0.1 units as a test case. The new crank angle was found more accurately with the Jacobian Inverse technique better that the Jacobian Transpose technique. The problem can be extended to include the dynamics for possible selection of the optimal driving torque or electric motor selection for the driving parts.
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