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A geometrical inverse kinematics method for hyper-redundant manipulators

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200810th Intl.Conf.on Control,Automation,Robotics and VisionHanoi,Vietnam,17–20December2008A Geometrical Inverse Kinematics Method forHyper-Redundant ManipulatorsSamer Yahya, Haider A. F. Mohamed, M. Moghavvemi, and S. S. YangCentre for Research in Applied Electronics (CRAE)University of Malaya, Kuala Lumpur, WP 50603, Malaysiahaider@.myAbstract— Hyper-redundant manipulators have large numberof kinematic degrees of freedom, thus processing someunconventional features such as the ability to enter a narrowspace while avoiding obstacles. To solve the problem of multi-solution caused by redundancy, a geometrical method ispresented in this paper. This proposed method finds one optimalsolution to the inverse kinematics of redundant or hyperredundant manipulators from these infinite solutions with fewercomputations. This method can be used for any planar n-serialmanipulators. Experiments are conducted 3-links redundantmanipulator and 10-links hyper-redundant manipulator todemonstrate the effectiveness of this proposed method.Keywords— hyper redundant manipulator, inversekinematics, geometrical method.I.I NTRODUCTIONThe term “hyper-redundancy” refers to redundantmanipulators with a very large number of degrees of freedom[1]. It is well known that redundant and hyper-redundantmanipulators have some advantages when compared withclassical arms because they allow the trajectory optimization,both on the free space and on the presence of obstacles, and theresolution of singularities [2].Many algorithms for hyper-redundant manipulator inverse kinematics were proposed and these algorithms generally were divided into three kinds [3]: the algebraic approach [4], the iterative approach, such as neural networks [5],[6], genetics algorithm [7], and the geometric one [3],[8].Most of the techniques for solving the inverse kinematics of hyper-redundant manipulators have many disadvantages when executing, like the complexity in computations, high time-consuming computation, and the difficulty in finding the optimal solution.This paper proposed a method that can be used to find the optimal solution with less computation.II.THE GEOMETRICAL METHOD Consider a manipulator with n dof whose joint variables are denoted by q= [q1,q2,.. ., q n], as shown in Fig. 1.Figure 1. A planar manipulator with n-linksHere,q i denotes the i-th joint's variable, j i denotes the i-th joints, and l i denotes the i-th link. For simplicity, consider that all the links have the same lengths.To understand the proposed method, suppose that (tp) is the target point. The steps below explain this method:1.Draw a circle with a radius equal to the link length and thecenter of which is at (tp).2.Draw a straight line from the center of this circle to thejoint (jn), Fig. 2 shows these steps. Now we can get the new position of the link (ln), where the tip of this link is on the target point (tp) and the other end (the new joint (jn)) is on the intersection between the circle and the straight line. Therefore, the joint angle of the last link (qn) is found.3. A new circle with the same radius is drawn, its center is atthe new joint (jn), and Fig. 3 shows this step.1954978-1-4244-2287-6/08/$25.00c 2008IEEE ICARCV20084.Now draw another straight line from the center of the lastcircle to the joint (jn-1), Fig. 3. Now we can get the new position of the link (ln-1), where the tip of this link is on the center of the last circle, and the other end (the new joint (jn-1)) is on the intersection between this circle and the straight line as shown in Figs. 4 and 5. Therefore, the joint angle of the link (qn-1) is found.Figure 4. Finding the new state of the link (l n-1)n-15.Find the new state for the other links by the same wayuntil touching the new joint (j3).6.To find the new state of the first two links (l1 and l2), twocircles with the radius equal to the link length must be1955drawn, the center of the first one is at the origin point (j1) and the center of the second circle is on the new joint (j3) as shown in Fig. 6.Figure 6. The two circles of step (6)7.Draw a straight line from the new joint (j3) to theintersection point of the two circles (the new joint (j2));this straight line represents the new position of the link (l2). By the same way the straight line between the intersection of the two circles and the origin point (j1) represents the new position of the link (l1). It is noticed that there are two possible states for the two links (l1), and (l2), because there are two intersection points between the circles. Fig. 7 illustrates these possible positions.8.Choose the intersection point which is nearer to thecurrent joint (j2) as shown in Fig. 8.To summarize this algorithm, the flow chart depicted in Fig. 9 is constructed. By following these steps as in the flow chart, the inverse kinematics of each redundant or hyper redundant manipulator can be found, except there is a special case, when the intersection point between the circle and the straight line is out of the reachable area of the current joint. This case will be explained in the next section.Figure 7. The two possible positions of the links (l1) and (l2) Figure 8. The initial and expectative states of the manipulator.1956Figure 9. The flow chart for finding the inverse solution of hyper-redundant manipulator.After implementing these steps, an n-series of lines will be plotted. These lines are the new states for the links of the manipulator.III.SPECIAL CASEWhen the new position of any joint (i-th joint) is specified by finding the intersection point of the circle and the straight line as explained previously, if()()¦−=>+1122ijjiilyxthat is this intersection point is out of the reachable area of the i-th joint. Where x i and y i are the x-axis and y-axis respectively of the intersection point. In this case another circle must be drawn, the center of which is at the origin (j1) and the radius is equal to the reachable area of the i-th joint, Fig. 10 shows this case.To further explain this case, the following example based on Fig. 10 is considered. To find the new joint (j n), a circle must be drawn, the center of which is at the target point (tp) and the radius of this circle is equal to the link length. Then a straight line must be drawn from the target point (tp) to the joint (j n). The intersection point (A) between them is the new joint (j n). Now if the distance between this new joint (j n) which is presented by (A) and the origin (j1) is more than (l1+l2+…..+l n-1), this means that the point (A) is out the reachable area of the joint (j n). Therefore, the circle (C R) is 1957drawn. The center of which is at the origin point (j1) and the radius of the circle is equal to (l1+l2+…..+l n-1). One of the intersection points (C) and (D) between the two circles is to be the new position of the joint (j n). The nearer one of the points (C) or (D) to the current position of the joint (j n) is the new position of the joint (j n). in this example, (B) is the new position of the joint (jn).IV.SIMULATION RESULTSCase1:The first simulation case is a 3-links redundant manipulator; each link is 3 units length. The initial state of every joint is known as q=[80°,60°,0°]. The target point is tp=[5,3] units length. Fig. 11 shows the initial and final states of the manipulator.Figure 11. The first simulation caseAfter implementing the proposed method the joint variables values are q=[80.87°,41.54°,-40.59°].Case2:The second simulation case is a 10-links hyper-redundant manipulator; each link is 2 units length. The initial state of every joint is known as q=[80°,60°,40°,30°,20°,5°,-5°,10°,15°,25°]. The target point is tp=[10,2] units length. Fig. 12 shows the initial and final states of the manipulator.By implementing the proposed method, the joint variables values are found to be: q=[ 86.128°, 55.277°, 40.484°, 30.476°, 20.774°, 5.419°, -2.749°, -40.449°, -109.251°,-128.330°].V.C ONCLUSIONThis paper modeled the inverse kinematics of hyper-redundant manipulator using a simple geometrical method; this method can be used for any planar n-serial manipulator. A flow chart that can be readily used to solve the hyper-redundant inverse kinematics was constructed in this paper.Two cases were simulated in this paper, and the joint variables of each case were calculated using simple calculations. This method can be used for finding the path planning of the end-effecter of the hyper-redundant manipulator.xyFigure 12. The second simulation caseVI.R EFERENCES[1] Chau-Chang W., Vijay K., and Guay-Ming C.," A motion control andobstacle avoidance algorithm for hyper-redundant manipulators", Proceedings of the 1998 International Symposium on UnderwaterTechnology, pp. 466 - 471.[2] Fernando B.M., and Tenreiro J.A., "Kinematic optimization of redundantand hyper-redundant robot trajectories", 1998 IEEE InternationalConference. Electronics, Circuits and Systems, vol. 2, pp. 467 - 470. [3] Li Sheng, Wang Yiqing, Chen Qingwei and Hu Weili, "A newGeometrical Method for the Inverse Kinematics of the Hyper-Redundant Manipulators", 2006 IEEE International Conference.Robotics and Biomimetics, pp. 1356 – 1359.[4] Alba Perez and J. Michael McCarthy, " Sizing a Serial Chain to Fit aTask Trajectory Using Clifford Algebra Exponentials", Proceedings ofthe 2005 IEEE International Conference on Robotics and Automation,pp. 4709-4715.[5] Youshen Xia and Jun Wang, " A dual neural network for kinematiccontrol of redundant robot manipulators", IEEE Transactions. Systems,Man, and Cybernetics. Volume 31, Issue 1, pp. 147 – 154. Feb 2001. [6] Samy F.M. Assal, Keigo Watanabe and Kiyotaka Izumi, "Neuralnetwork learning from hint for the inverse kinematics problem ofredundant arm subject to joint limits", 2005 IEEE/RSJ InternationalConference. Intelligent Robots and Systems, pp. 1477 – 1482.[7] ANDREAS C. NEARCHOU, "Solving the inverse kinematics problemof redundant robots operating in complex environments via a modifiedgenetic algorithm", Mechanism and Machine Theory, Volume 33, Issue3, April 1998, pp. 273-292.[8] Zeki Y. BAYRAKTAROGLU, Fabienne BUTEL, Pierre BLAZEVIC,Viviane PASQUI" A geometrical approach to the trajectory planning ofa snake-like mechanism", Proceedings, 1999 IEEE/RSJ InternationalConference on Intelligent Robots and Systems, vol.3, pp. 1322 – 1327.1958。

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构造地质学参考文献

构造地质学参考文献

构造地质学参考文献现阶段国内流行多个种版本的《构造地质学》教材：（1）朱志澄、宋鸿林主编，马杏垣、单文琅主审的《构造地质学》，中国地质大学出版社出版；（2）徐开礼、朱志澄主编的《构造地质学》，地质出版社出版；（3）朱志澄主编《构造地质学》，中国地质大学出版社，1990年第1版，1999年第2版，2008年第三版（第三版是2008年度国家精品课程、中国地质大学（武汉）地学类系列精品教材）；（4）普通高等教育“十一五”国家级规划教材《构造地质学》，李忠权、刘顺主编，地质出版社，2010年第三版；（5）长安大学地质工程国家级特色专业建设系列教材《构造地质学》，地质出版社，2011；（6）高职高专院校资源勘查类专业“十一五”规划教材《构造地质学》，地质出版社，2007。

考虑到学生的专业特点，本课程计划选用朱志澄主编的《构造地质学》（2008年第三版）作为主导教材，同时辅以朱志澄主编的《构造地质学》（1999年第二版）和朱志澄、宋鸿林主编的《构造地质学》（中国地质大学出版社1990年出版）；徐开礼、朱志澄主编《构造地质学》（地质出版社1989年第2版）和地质系于在平等人主编的《构造地质学》CAI多媒体教材。

二、主要参考文献构造地质学教学过程中用到的参考文献主要来自国内外有关期刊。

国内目前还没有专门的构造地质学类期刊，构造地质学类的文献主要见于综合性的学术期刊。

国外关于构造地质学的英文期刊主要有两个：《Journal of Structural Geology》和《Tectonics》，这两个英文期刊上的文献基本能够反映构造地质学的主要研究方向和新进展，建议构造地质学学习者尽可能多地浏览这两个期刊。

这里列出部分参考文献。

邓洪菱，张长厚，李海龙，邹云。

褶皱相关断裂构造及其地质意义。

自然科学进展，2009， 19（3）： 285-296。

黄德志。

构造地质学实践教学与地质专业学生专业素质培养。

创新与创业教育，2012，3（4）：65-67。

学术英语综合季佩玉四单元翻译

学术英语综合季佩玉四单元翻译

Unit 4 翻译环保的当今时代是约半个世纪之久。

那段时间意识不断壮大，我们所面临的挑战知识增加，而重要的实践已经取得了进展，例如在一些减少各种污染，并在建立保护区。

我们是，但是，仍然由协调人什么我们的星球能提供可持续的要求很远。

人与自然之间的不平衡的后果是出现在改变地球的气候，动物和植物的势头，并在关键的资源，包括野生鱼类资源，淡水和土壤的枯竭的大灭绝。

而这些环境压力不是一成不变的。

他们不断升级，随着我国人口的增长和国家继续为更多的经济增长的不懈追求。

如果我们要避免这些趋势的最严重后果则毫无疑问是较为迅速的进展将需要比迄今取得的，但我们在这里应该关注我们的努力？什么可能是在未来的半个世纪行动的优先领域？这让我感到眼前的主要挑战主要不是相对于良好的信息，更好的技术和良好的政策思路。

这些东西是至关重要的当然的，但所有这些东西都已经可用。

我们知道如何让清洁电力，节约资源，培育生物多样性。

我们知道如何规范污染，防止损坏的生态系统，如果我们想。

我们有这些能力的事实是不够的。

如果我们继续前进的决定性方式的争论它需要被重新定义。

我们需要从“做正确的事'上移动，谈到风险管理，促进抗灾能力。

要查看关爱地球的自然系统为某种道德选择的是完全误解了危机，我们都在这个挑战是关于人类社会的未来，而不是一些可选的慈善事业，我们可以留给慈善事业慷慨解囊，做社会改良。

嵌入了使我们从保护自然的人们保护自然为人们的叙述是这样的重新规划的重要组成部分。

我们正处在一个时期的后果，世界必须知道，健康的本质不是一些可选的精密而是一组不可缺少的物质资产。

如果这样的叙述是为了获得实际效果再想找性质后必须立即被看作不仅是一个环境的挑战，也是一个经济问题。

只要我们继续滑向两个方向行进，一方面是促进环保目标的同时，对其他直接矛盾与措施，以实现更多的经济增长，我们不再将无法取得真正的进展。

当涉及到经济学和生态学有大量的好思已经完成。

例如，在不同的经济措施，以取代成功与失败，生态税的改革更为现实的措施，收入转移的负担，污染，生态系统服务补偿，补贴的重定向和如何调动国内生产总值的原油措施融资以扩大清洁行业。

invariants and K-spectrums of local theta lifts

invariants and K-spectrums of local theta lifts


arXiv:1302.1031v2 [math.n 1.1. Let WR be a finite dimensional symplectic real vector space. Throughout this paper (G, G′) will denote a type I irreducible reductive dual pair in Sp(WR ). Such dual pairs are listed in Table 1 in Section 2.1. We follow the notation in [15] closely. Let Sp(WR ) be the metaplectic double cover of Sp(WR ). For any subgroup E of Sp(WR ), let E denote its inverse image in Sp(WR ). We choose a maximal compact subgroup U of Sp(WR ) such that K := G ∩ U and K ′ := G′ ∩ U are maximal compact subgroups of G and G′ respectively. Hence K and K ′ are maximal compact subgroups of G and G′ respectively. The choice of U determines a unique complex structure on WR with the following property: there is a positive definite Hermitian form , on the resulting complex vector space W so that the imaginary part of , coincides with the symplectic form on WR , and U coincides with the unitary group attached to (W, , ). We choose the oscillator representation of Sp(WR ) whose Fock model Y is realized as the space C [W ] of complex polynomials on W with the U action as described in Appendix A.1. Let ς denote the minimal U-type of Y . It is a one dimensional representation of U acting on the space of constant functions in C [W ]. Let ς |E denote the restriction of ς to E for any subgroup E of U. Let g = k ⊕ p denote the complexified Cartan decomposition of the Lie algebra of G corresponding to the maximal compact subgroup K . Likewise we define g′ = k′ ⊕ p′ for G′ . Let ρ be an irreducible admissible genuine (g, K )-module. By (2.5) in [15], Y /(∩ψ∈Homg,K (Y ,ρ) ker ψ ) ≃ ρ ⊗ Θ(ρ) where Θ(ρ) is a (g′ , K ′ )-module called the full (local) theta lift of ρ. Theorem 2.1 in [15] states that if Θ(ρ) = 0, then Θ(ρ) is a (g′ , K ′ )-module of finite length with an infinitesimal character and it has a unique irreducible quotient θ(ρ) called the (local) theta lift of ρ.
Uncovering the Origin of Metallurgy

Uncovering the Origin of Metallurgy

Uncovering the Origin of MetallurgyMetallurgy is the science and technology of extracting metals from their ores, refining them, and preparing them for use. It is a crucial aspect of human civilization and has played a significant role in shaping our world. The origin of metallurgy is a topic of great interest to historians, archaeologists, and scientists alike. In this essay, we will explore the different perspectives on the origin of metallurgy and try to uncover the truth behind this fascinating subject. One perspective on the origin of metallurgy is that it emerged independently in different parts of the world. This theory suggests that humans discovered the properties of metals through trial and error and developed the necessarytechniques to extract and refine them. This is supported by the fact that early metallurgy was found in several regions, including the Middle East, Asia, andSouth America. For example, the ancient Egyptians were known to have used copper and gold for decorative purposes, while the people of the Andes used copper to create intricate jewelry and ornaments. Another perspective on the origin of metallurgy is that it was developed through cultural diffusion. This theory suggests that metallurgy was first developed in one region and then spread toother parts of the world through trade and migration. For example, some historians believe that metallurgy was first developed in the Middle East and then spread to Europe and Asia. This is supported by the fact that the earliest known metal objects were found in the Middle East, such as the copper beads from the 6th millennium BCE found in the area now known as Iraq. A third perspective on the origin of metallurgy is that it was developed as a result of environmental factors. This theory suggests that the availability of certain metals in a particularregion led to the development of metallurgy in that area. For example, the abundance of copper in the Middle East and the Andes may have led to the development of early metallurgy in those regions. Similarly, the availability of iron in certain parts of Africa may have led to the development of ironworking in those areas. Regardless of the origin of metallurgy, it is clear that it played a crucial role in the development of human civilization. The ability to extract and refine metals allowed humans to create tools and weapons that were stronger and more durable than those made from stone or wood. This, in turn, led to thedevelopment of agriculture, trade, and the rise of civilizations. In conclusion, the origin of metallurgy is a fascinating subject that has captured the attention of historians, archaeologists, and scientists for centuries. While there are different perspectives on how metallurgy first emerged, it is clear that it played a crucial role in shaping our world. The ability to extract and refine metals allowed humans to create tools and weapons that were stronger and more durable, leading to the development of agriculture, trade, and the rise of civilizations. As we continue to uncover new evidence about the origin of metallurgy, we will gain a deeper understanding of our shared history and the remarkable achievements of our ancestors.。