Integrals of Motion for Discrete-Time Optimal Control Problems
Strapdown Inertial Navigation Integration Algorithm Design Part Velocity and Position Algorithms
Received July 7, 1997; revision received Oct. 9, 1997; accepted for publication Oct. 9, 1997. Copyright °c 1997 by Strapdown Associates, Inc. Published by the American Institute of Aeronautics and Astronautics, Inc.,
I. Introduction
A STRAPDOWN inertial navigation system (INS) is typically composed of an orthogonal three-axis set of inertial angular rate sensors and accelerometers providing data to the INS computer. The inertial sensors are directly mounted (strapdown) to the INS chassis structure in contrast with original INS technology that utilized an active multiaxis gimbal isolation mounting assembly
= direction cosine matrix that transforms a vector from its A2 frame projection form to its A1 frame projection form
= identity matrix = column matrix with elements equal to the projection
gtsam 常用因子
gtsam 常用因子English answers:IMU Preintegration.Inertial Measurement Unit (IMU) preintegration is a technique used to integrate the IMU measurements over a period of time, typically between two keyframes. This reduces the computational cost of integrating the measurements online and allows for more efficient optimization. GTSAM provides a robust and accurate implementation of IMU preintegration.Stereo Vision.Stereo vision is a technique used to estimate the depth of a scene using two or more cameras. GTSAM provides a variety of stereo vision factors, including the pinhole model and the fisheye model. These factors can be used to estimate the pose of the cameras and the depth of the scene.Lidar.Lidar (Light Detection and Ranging) is a remote sensing technology that uses laser pulses to measure the distance to objects. GTSAM provides a variety of lidar factors, including the point-to-plane, point-to-line, and plane-to-plane models. These factors can be used to estimate the pose of the lidar sensor and the location of objects in the scene.GPS.Global Positioning System (GPS) is a satellite-based navigation system that provides location and time information. GTSAM provides a variety of GPS factors, including the position-only model and the velocity-aided model. These factors can be used to estimate the pose of the GPS receiver and the velocity of the vehicle.Odomery.Odometry is a technique used to estimate the pose of a vehicle using the measurements from its wheel encoders. GTSAM provides a variety of odometry factors, including the differential drive model and the unicycle model. These factors can be used to estimate the pose of the vehicle and the velocity of the wheels.Chinese answers:IMU预积分。
定积分的英文专用名词
定积分的英文专用名词Definite Integral: A Deep Dive into Its Concepts and Applications.In the realm of mathematics, the definite integral occupies a pivotal position, bridging the gap between discrete and continuous functions. It represents a fundamental tool for understanding the properties of functions and for solving a wide range of practical problems. This article aims to delve into the essence of the definite integral, exploring its definitions, properties, applications, and the underlying theories that govern it.1. Introduction to Definite Integrals.The definite integral is a mathematical operation that assigns a real number to a given function on a closed interval. It is denoted as ∫[a,b]f(x)dx, where [a,b] is the closed interval and f(x) is the function beingintegrated. This notation represents the accumulation of infinitesimal changes in the function over the specified interval.The definite integral is closely related to the indefinite integral or the antiderivative. While the indefinite integral represents a family of functions that differ by a constant, the definite integral provides a specific numerical value for a given function over a specific interval.2. Properties of Definite Integrals.The definite integral possesses several important properties that govern its behavior and applications. Some of the key properties include:Linearity: The definite integral distributes over addition and scalar multiplication, allowing us to integrate complex expressions by breaking them down into simpler parts.Interval Additivity: If an interval [a,b] is divided into two subintervals [a,c] and [c,b], then the integral over the entire interval is equal to the sum of the integrals over the subintervals.Constant Factor Rule: If a constant k is multiplied with a function f(x) within the integral, then the integral of kf(x) is equal to k times the integral of f(x).Comparison Theorem: If f(x) ≤ g(x) for all x in the interval [a,b], then the integral of f(x) over [a,b] is less than or equal to the integral of g(x) over the same interval.These properties provide a solid foundation for solving complex integration problems and for understanding the behavior of functions under integration.3. Applications of Definite Integrals.The definite integral finds applications in various fields of mathematics, science, and engineering. Some ofits key applications include:Area Calculation: The most fundamental application of the definite integral is in calculating the area under a curve between two given points. This is achieved by integrating the function representing the curve over the corresponding interval.Volume Calculation: The definite integral can be extended to calculate volumes of three-dimensional objects, such as solids of revolution and regions between surfaces.Length of Curves: The integral can be used to compute the length of a curve in the plane or in space, by integrating the square root of the sum of squares of the derivatives of the curve's coordinates.Physics and Engineering: The definite integral plays a crucial role in physics and engineering, particularly in areas such as mechanics, electromagnetism, and fluid dynamics. It is used to calculate quantities like displacement, velocity, acceleration, force, work, energy,and momentum.Probability and Statistics: In probability theory and statistics, the definite integral is employed to calculate probabilities, expected values, and other statistical measures.4. Theories Underlying Definite Integrals.The theory of definite integrals is rooted in the fundamental theorem of calculus, which establishes a link between differentiation and integration. This theorem states that if F(x) is an antiderivative of f(x), then the integral of f(x) from a to b is equal to F(b) F(a). This theorem provides a practical method for evaluating definite integrals by finding antiderivatives of the integrated functions.In addition to the fundamental theorem, other important theories include the convergence theorems for infinite series and improper integrals, which extend the concept of definite integrals to infinite intervals and infinite sums.These theories provide a rigorous foundation for understanding the behavior of integrals and for solving complex integration problems.5. Conclusion.The definite integral is a powerful tool that lies at the heart of mathematics and its applications. Its ability to measure areas, volumes, lengths, and other quantities has made it indispensable in various fields ranging from physics and engineering to economics and finance. By delving into its definitions, properties, applications, and underlying theories, we can gain a deeper understanding of this fundamental concept and appreciate its value in addressing real-world problems.。
英文有限元方法Finite element method讲义 (1)
MSc in Mechanical Engineering Design MSc in Structural Engineering LECTURER: Dr. K. DAVEY(P/C10)Week LectureThursday(11.00am)SB/C53LectureFriday(2.00pm)Mill/B19Tut/Example/Seminar/Lecture ClassFriday(3.00pm)Mill/B192nd Sem. Lab.Wed(9am)Friday(11am)GB/B7DeadlineforReports1 DiscreteSystems DiscreteSystems DiscreteSystems2 Discrete Systems. Discrete Systems. Tutorials/Example I.Meshing I.Deadline 3 Discrete Systems Discrete Systems Tutorials/Example IIStart4 Discrete Systems. Discrete Systems. DiscreteSystems.5 Continuous Systems Continuous Systems Tutorials/Example II. Mini Project6 Continuous Systems Continuous Systems Tutorials/Example7 Continuous Systems Continuous Systems Special elements8 Special elements Special elements Tutorials/Example III.Composite IIDeadline *9 Special elements Special elements Tutorials/Example10 Vibration Analysis Vibration Analysis Vibration Analysis III Deadline11 Vibration Analysis Vibration Analysis Tutorials/Example12 VibrationAnalysis Tutorials/Example Tutorials/Example13 Examination Period Examination Period14 Examination Period Examination Period15 Examination Period Examination Period*Week 9 is after the Easter vacation Assignment I submission (Box in GB by 3pm on the next workingday following the lab.) Assignment II and III submissions (Box in GB by 3pm on Wed.)CONTENTS OF LECTURE COURSEPrinciple of virtual work; minimum potential energy.Discrete spring systems, stiffness matrices, properties.Discretisation of a continuous system.Elements, shape functions; integration (Gauss-Legendre).Assembly of element equations and application of boundary conditions.Beams, rods and shafts.Variational calculus; Hamilton’s principleMass matrices (lumped and consistent)Modal shapes and time-steppingLarge deformation and special elements.ASSESSMENT: May examination (70%); Short Lab – Holed Plate (5%); Long Lab – Compositebeam (10%); Mini Project – Notched component (15%).COURSE BOOKSBuchanan, G R (1995), Schaum’s Outline Series: Finite Element Analysis, McGraw-Hill.Hughes, T J R (2000), The Finite Element Method, Dover.Astley, R. J., (1992), Finite Elements in Solids and Structures: An Introduction, Chapman &HallZienkiewicz, O.C. and Morgan, K., (2000), Finite Elements and Approximation, DoverZienkiewicz, O C and Taylor, R L, (2000), The Finite Element Method: Solid Mechanics,Butterworth-Heinemann.IntroductionThe finite element method (FEM) is a numerical technique that can be applied to solve a range of physical problems. The method involves the discretisation of the body (domain) of interest into subregions, which are known as elements. This enables a continuum problem to be described by a finite system of equations. In the field of solid mechanics the FEM is undoubtedly the solver of choice and its use has revolutionised design and analysis approaches. Many commercial FE codes are available for many types of analyses such as stress analysis, fluid flow, electromagnetism, etc. In fact if a physical phenomena can be described by differential or integral equations, then the FE approach can be used. Many universities, research centres and commercial software houses are involved in writing software. The differences between using and creating code are outlined below:(A) To create FE software1. Confirm nature of physical problem: solid mechanics; fluid dynamics; electromagnetic; heat transfer; 1-D, 2-D, 3-D; Linear; non-linear; etc.2. Describe mathematically: governing equations; loading conditions.3. Derive element equations: convert governing equations into algebraic form; select trial functions; prepare integrals for numerical evaluation.4. Assembly and solve: assemble system of equations; application of loads; solution of equations.5. Compute:6. Process output: select type of data; generate related data; display meaningfully and attractively.(B) To use FE software1. Define a specific problem: geometry; physical properties; loads.2. Input data to program: geometry of domain, mesh generation; physical properties; loads-interior and boundary.3. Compute:4. Process output: select type of data; generate related data; display meaningfully and attractively.DISCRETE SYSTEMSSTATICSThe finite element involves the transformation of a continuous system (infinite degrees of freedom) into a discrete system (finite degrees of freedom). It is instructive therefore to examine the behaviour of simple discrete systems and associated variational methods as this provides real insight and understanding into the more complicated systems arising from the finite element method.Work and Strain energyFLuxConsider a metal bar of uniform cross section, A , fixed at one end (unrestrained laterally) and subjected to an axial force, F , at the other.Small deflection theory is assumed to apply unless otherwise stated.The work done, W , by the applied force F is .a ()∫′′=uau d u F WIt is worth mentioning at this early stage that it is not always possible to express work in this manner for various reasons associated with reversibility and irreversibility. (To be discussed later)The work done, W , by the internal forces, denoted strain energy , is se22200se ku 21u L EA 2121EAL d EAL d AL W ==ε=ε′ε′=ε′σ=∫∫εεwhere ε=u L and stiffness k EA L=.The principle of virtual workThe principle of virtual work states that the variation in strain energy is equal to the variation in the work done by applied forces , i.e.()u F u u d u F du d W u ku u ku 21du d ku 21W u0a 22se δ=δ⎟⎟⎠⎞⎜⎜⎝⎛′′=δ=δ=δ⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛δ=δ∫()0u F ku =δ−⇒Note that use has been made of the relationship δf dfduu =δ where f is an arbitrary functional of u . In general displacement u is a function of position (x say) and it is understood that ()x u δ means a change in ()u x with xfixed. Appreciate that varies with from zero to ()'u F 'u ()u F F = in the above integral.Bearing in mind that δ is an arbitrary variation; then this equation is satisfied if and only if F , which is as expected. Before going on to apply the principle of virtual work to a continuous system it is worth investigating discrete systems further. This is because the finite element formulation involves the transformation of a continuous system into a discrete one. u ku =Spring systemsConsider a single spring with stiffness independent of deflection. Then, 2F21u1F1u2k()()⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡−−=−=2121212se u u k k k k u u 21u u k 21W()()()⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡−−δδ=δ−δ−=δ21211212se u u k k k k u u u u u u k W()⎟⎟⎠⎞⎜⎜⎝⎛δδ=δ+δ=δ21212211a F F u u u F u F W , where ()111u F F = and ()222u F F =.Note here that use has been made of the relationship δ∂∂δ∂∂δf f u u f u u =+1122, where f is an arbitrary functional of and . Observe that in this case is a functional of 1u u 2W se u u u 2121=−, so()()(121212*********se se u u u u k u u ku 21du d u du dW W δ−δ−=−δ⎟⎠⎞⎜⎝⎛=δ=δ).The principle of virtual work provides,()()()()0F u u k u F u u k u 0W W 21221121a se =−−δ+−−−δ⇒=δ−δand since δ and δ are arbitrary we have. u 1u 2F ku ku 11=−2u 2 F ku k 21=−+represented in matrix form,u F K u u k k k k F F 2121=⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡−−=⎟⎟⎠⎞⎜⎜⎝⎛=where K is known as the stiffness matrix . Note that this matrix is singular (det K k k =−=220) andsymmetric (K K T=). The symmetry is a result of the fact that a unit deflection at node 1 results in a force at node 2 which is the same in magnitude at node 1 if node 2 is moved by the same amount.Could also have arrived at equation above via()⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡−−=⎟⎟⎠⎞⎜⎜⎝⎛⇒=⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡−−−⎟⎟⎠⎞⎜⎜⎝⎛δδ=δ−δ2121211121a se u u k k k k F F 0u u k k k k F F u u W WBoundary conditionsWith the finite element method the application of displacement constraint boundary conditions is performed after the equations are assembled. It is an interest to examine the implications of applying and not applying the displacement boundary constraints prior to applying the principle of virtual work. Consider then the single spring element above but fixed at node 1, i.e. 0u 1=. Ignoring the constraint initially gives()212se u u k 21W −=, ()()1212se u u u u k W δ−δ−=δ and 2211a u F u F W δ+δ=δ.The principle of virtual work gives 2211ku ku ku F −=−= and 2212ku ku ku F =+−=, on applicationof the constraint. Note that is the force required at node 1 to prevent the node moving and is the reaction force.21ku F −=21ku F =−Applying the constraint straightaway gives 22se ku 21W =, 22se u ku W δ=δ and 22a u F W δ=δ. The principle of virtual work gives with no information about the reaction force at node 1.22ku F =Exam Standard Question:The spring-mass system depicted in the Figure consists of three massless springs, which are attached to fixed boundaries by means of pin-joints at nodes 1, 3 and 5. The springs are connected to a rigid bar by means of pin-joints at nodes 2 and 4. The rigid bar is free to rotate about pivot A. Nodes 2 and 4 are distances and below pivot A, respectively. Each spring has the same stiffness k. Node 2 is subjected to an external horizontal force F 2/l 4/l 2. All deflections can be assumed to be small.(i) Write expressions for the extension of each spring in terms of the displacement of node 2.(ii) In terms of the degrees of freedom at node 2, write expressions for the total strain energy W of the spring-mass system. In addition, specify the variation in work done se a W δ resulting from the application of the force.2F (iii) Use Use the principle of virtual work to find a relationship between the magnitude of and the horizontal components of displacement at node 2.2F (iv) Use the principle of virtual work to show that the net vertical force imposed by the springs on the rigid-bar at node 2 is zero.Solution:(i) Directional vectors for springs are: 2112e 21e 23e +=, 2132e 21e 23e +−= and 145e e =. Extensions for bottom springs are: 221212u 23u e =⋅=δ, 223232u 23u e −=⋅=δ.Note that 2u u 24=, so 2u245−=δ.(ii)()2222222245232212se ku 87u 212323k 21k 21W =⎟⎟⎠⎞⎜⎜⎝⎛⎟⎠⎞⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛−+⎟⎟⎠⎞⎜⎜⎝⎛=δ+δ+δ=, 222u F W δ=δ(iii) 2222a 22se ku 47F u F W u ku 47W =⇒δ=δ=δ=δ(iv) Need additional displacement degree of freedom at node 2. Let 22122e v e u u += and note that2221212v 21u 23u e +=⋅=δ and 2223232v 21u 23u e +−=⋅=δ.()⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛+−+⎟⎟⎠⎞⎜⎜⎝⎛+=δ+δ=222222232212se v 21u 23v 21u 23k 21k 21W ⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛δ+δ−⎟⎟⎠⎞⎜⎜⎝⎛+−+⎟⎟⎠⎞⎜⎜⎝⎛δ+δ⎟⎟⎠⎞⎜⎜⎝⎛+=δ22222222se v 21u 23v 21u 23v 21u 23v 21u 23k W Setting and gives0v 2=0u 2=δ2vert 222222se v F v 0v 21u 23v 21u 23k W δ=δ=⎟⎟⎠⎞⎜⎜⎝⎛⎟⎠⎞⎜⎝⎛δ⎟⎟⎠⎞⎜⎜⎝⎛−+⎟⎠⎞⎜⎝⎛δ⎟⎟⎠⎞⎜⎜⎝⎛=δ hence . 0F vert 2=Method of Minimum PotentialConsider the expression,()()F u u u TT 21212121c se K 21F F u u u u k k k k u u 21W W P −=⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡−−=−=where W F and can be considered as a work term with independent of . u F u c =+1122F i u iThe approach of minimising P is known as the method of minimum potential .Note that,()()u F 0F -u u =F u u u +u u K K K K 21W W P T T T T c se =⇒=δδ−δδ=δ−δ=δwhere use has been made of the fact that δδu u =u u T TK K as a result of K 's symmetry.It is useful at this stage to consider the minimisation of an arbitrary functional ()u P where()()3T T O H 21P P u u u u u δ+δδ+∇δ=δand the gradient ∇=P P u i i ∂∂, and the Hessian matrix coefficients H P u u ij i j=∂∂∂2.A stationary point requires that ∇=, i.e.P 0∂∂Pu i=0.Moreover, a minimum point requires that δδu u TH >0 for all δu ≠0 and matrices that possess this property are known as positive definite .Setting P W W K se c T=−=−12u u u F T provides ∇=−=P K u F 0 and H K =.It is a simple matter to check that with u 10= (to prevent rigid body movement) that K is positive definite and this is a property commonly associated with FE stiffness matrices.Exam Standard Question:The spring system depicted in the Figure consists of four massless unstretched springs, which are attached to fixed boundaries by means of pin-joints at nodes 1 to 4. The springs are connected to a slider at node 5. Theslider is constrained to move in a frictionless channel whose axis is to the horizontal. Each spring has the same stiffness k. The slider is subjected to an external force F 0453 whose direction is along the axis of the frictionless channel.(i)The deflection of node 5 can be represented by the vector 25155v u e e u +=, where and areunit orthogonal vectors which are shown in the Figure. Write the components of deflection and in terms of , where is the magnitude of , i.e. e 1e 25u 5v 5U 5U 5u 25U 5u =. Show that the extensions of eachspring, in terms of , are: 5U ()22/31U 515+=δ, ()22/31U 525−=δ, and2/U 54535−=δ−=δ.(ii) In terms of k and write expressions for the total strain energy W of the spring-mass system. Inaddition, specify the variation in work done 5U se a W δ resulting from the application of the force . 5F (iii) Use the principal of virtual work to find a relationship between the magnitude of and thedisplacement at node 5.5F 5U (iv) Use the principal of virtual work to determine an expression for the force imposed by the frictionless channel on the slider.(v)Form a potential energy function for the spring system. Assume here that nodes 1, 3 and 4 are fixed and node 5 is restricted to move in the channel. Use this function to determine the reaction force at node 2.Solution:(i) Directional vectors for springs and channel are: ()2115e e 321e +=, ()2125e e 321e +−=, 135e e −=, 45e e = and (21c 5e e 21e +=). Deflection c 555e U u =, so 2U v u 555==. Extensions springs are: ()3122U u e 551515+=⋅=δ, ()3122U u e 552525−=⋅=δ, 2Uu e 553535−=⋅=δand 2Uu e 554545=⋅=δ(ii)()()()252522245235225215se kU U 83131k 8121k 21W =⎟⎠⎞⎜⎝⎛+−++=δ+δ+δ+δ=, 55a U F W δ=δ(iii)5555a 55se kU 2F U F W U kU 2W =⇒δ=δ=δ=δ(iv) Need additional displacement degree of freedom at node 3. A unit vector perpendicular to the channel is(21p 5e e 21e +−=) and let p 55c 555e V e U u += and note that()()3122V3122U u e 5551515−++=⋅=δ and ()()3122V3122U u e 5552525++−=⋅=δ, 2V 2U u e 5553535+−=⋅=δ and 2V 2U u e 5554545−=⋅=δ()()()()()()()()⎟⎠⎞⎜⎝⎛−+++−+−++=δ+δ+δ+δ=255255255245235225215se V U 831V 31U 31V 31U k 8121k 21W ()()()()()555se V 0V 831313131kU 81W δ=δ−+−+−+=δ, where variation is onlyconsidered and is set to zero. Principle of virtual work .5V δ3V 0F V F V 0W p 55p 55se =⇒δ=δ=δ(v)()3122U u e 551515+=⋅=δ, ()()5552525V 3122Uu u e −−=−⋅=δ, where 2522e V u =. ()()()223333223232233245235225215V F U F U V 3122U 3122U k 21V F U F k 21P −−⎟⎟⎠⎞⎜⎜⎝⎛+⎥⎦⎤⎢⎣⎡−−+⎥⎦⎤⎢⎣⎡+=−−δ+δ+δ+δ=and ()0F V 3122U k V P 2232=−⎥⎦⎤⎢⎣⎡−−−=∂∂, which on setting 0V 2= gives ()⎥⎦⎤⎢⎣⎡−−=3122U k F 32.The reaction is .2F −System AssemblyConsider the following three-spring system 2F 21u 1F 1u 2kF 3F 4u 3u 4k 1k2334()()()234322322121se u u k 21u u k 21u u k 21W −+−+−=,()()()()()()343432323212121se u u u u k u u u u k u u u u k W δ−δ−+δ−δ−+δ−δ−=δ,44332211a u F u F u F u F W δ+δ+δ+δ=δ,and δδ implies that,W W se a −=0u F K u u u u k k 0k k k k 00k k k k 00k k F FF F 43213333222211114321=⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−+−−+−−=⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛=where again it is apparent that K is symmetric but also it is banded, i.e. the non-zero coefficients are located around the principal diagonal. This is a property commonly associated with assembled FE stiffness matrices and depends on node connectivity. Note also that the summation of coefficients in individual rows or columns gives zero. The matrix is singular and 0K det =.Note that element stiffness matrices are: , and where on examination of K it is apparent how these are assembled to form K .⎥⎦⎤⎢⎣⎡−−1111k k k k ⎥⎦⎤⎢⎣⎡−−2222k k k k ⎥⎦⎤⎢⎣⎡−−3333k k k kIf a boundary constraint is imposed then row one is removed to give:0u 1=u F K u u u k k 0k k k k 0k k k F F F 432333322221432=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−+−−+=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛=. If however a boundary constraint (say) is imposed then row one is again removed but a somewhatdifferent answer is obtained: 1u 1=u F K u u u k k 0k k k k 0k k k F F k F 4323333222214312=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−+−−+=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛+=)Direct FormulationIt is possible to formulate the stiffness matrix directly by moving one node and keeping the others fixed and noting the reactions.The above system can be solved for u , once possible rigid body motion is prevented, by setting u (say) to give 10=⇒=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−+−−+=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛=u F K u u u k k 0k k k k 0k k k F F F 432333322221432⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−+−−+=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛−4321333322221432F F F k k 0k k k k 0k k k u u uThe inverse stiffness matrix, K −1, is known as the flexibility matrix and, for this example at least, can be assembled directly by noting the system response to prescribed forces.In practice K −1is never calculated and the system K u F = is solved using a modern numerical linear system solver.It is a simple matter to confirm thatu u K 21u u u u k k 0k k k k 00k k k k 00k k u u u u 21W T 4321333322221111T4321se =⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−+−−+−−⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛= with F u T4321T4321a F F F F u u u u W δ=⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛δδδδ=δThus,()u F F u u K 0K W W Ta se =⇒=−δ=δ−δExample:k1F 2u23k2u3F321With use a direct method to find the assembled stiffness and flexibility matrices.0u 1=Solution:The equations of interest are of the form: 3232222u k u k F += and 3332323u k u k F +=.Consider and equilibrium at nodes 2 and 3. At node 2, 0u 3=()2212u k k F += and at node 3,.223u k F −=Consider and equilibrium at nodes 2 and 3. At node 2, 0u 2=322u k F −= and at node 3, . 323u k F =Thus: , , 2122k k k +=223k k −=232k k −= and 233k k =.For flexibility the equations of interest are of the form: 3232222F c F c u += and . 3332323F c F c u +=Consider and equilibrium at nodes 2 and 3. At node 2, 0F 3=122k F u = and at node 3,1223k F u u ==.Consider and equilibrium at nodes 2 and 3. At node 2, 0F 2=122k F u = and at node 3,()2133k 1k 1F u +=.Thus: 122k 1c =, 123k 1c =, 132k 1c = and 2133k 1k 1c +=.Can check that ⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡+⎥⎦⎤⎢⎣⎡−−+1001k 1k 1k 1k 1k 1k k k k k 2111122221 as required,It should be noted that the direct determination requires boundary constraints to be applied to ensure that the flexibility matrix exists, which requires the stiffness to be non-singular. However, the stiffness matrix always exists, so boundary conditions need not be applied prior to constructing the stiffness matrix with the direct approach.Large deformation theory for spring elementsThus far small deflection theory has been applied where the strains are measured using the Cauchy strainxu11∂∂=ε. A conjugate stress can be obtained by differentiating with respect the expression for strain energy density (energy per unit volume) 11ε211E 21ε=ω, i.e. 111111E ε=ε∂ω∂=σ, where E is Young’s Modulusand is the Cauchy stress (sometimes referred to as the Euler stress). 11σIn the case of large deformation theory we will restrict our attention to hyperelastic materials which are materials that possess an expression for strain energy density Ω (say) that is analytical in strain.The strain used in large deformation theory is Green’s strain (see Appendix II) which for a uniformly loadeduniaxial bar is 211x u 21x u E ⎟⎠⎞⎜⎝⎛∂∂+∂∂=.An expression for strain energy density (energy per unit volume) 211EE 21=Ω and the derived stress is 111111EE E S =∂Ω∂=, where E is Young’s Modulus and is known as the 211S nd Piola-Kirchoff stress . 2F21u1F1u2kBar subject to longitudinal deformationConsider a bar of length L and cross sectional area A represented by a spring element and subject to nodal forces and . 1F 2FThe strain energy is∫∫∫∫⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛∂∂+∂∂==Ω=Ω=212121x x 22x x 211x x V se dx x u 21x u EA 21dx E EA 21dx A dV WConsider further a linear displacement field of the form ()21u L x u L x L x u ⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛−= and note thatL u u xu 12−=∂∂. ()()221212x x 221212se u u L 21u u L EA 21dx L u u 21L u u EA 21W 21⎥⎦⎤⎢⎣⎡−+−=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛−+−=∫ ()()()⎥⎦⎤⎢⎣⎡−+−+−=4122312212se u u L 41u u L 1u u k 21W()()()(12312221212se u u u u L 21u u L 23u u k W δ−δ⎥⎦⎤⎢⎣⎡−+−+−=δ) and 2211a u F u F W δ+δ=δ.The principle of virtual work gives()()()⎥⎦⎤⎢⎣⎡−+−+−−=3122212121u u L 21u u L 23u u k F and()()(⎥⎦⎤⎢⎣⎡−+−+−=3122212122u u L 21u u L 23u u k F ), represented in matrix form as()()()()()⎟⎟⎠⎞⎜⎜⎝⎛⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−+−−−−−−−+−+⎥⎦⎤⎢⎣⎡−−=⎟⎟⎠⎞⎜⎜⎝⎛21121212121221u u L 3u u 1L 3u u 1L 3u u 1L 3u u 1L 2u u k 3k k k k F Fwhich is of the form[]u F G L K K += where is called the geometrical stiffness matrix and is the usual linear stiffnessmatrix. G K L KA common approximation used, depending on the magnitude of L /u u 12−, is⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡−−+⎥⎦⎤⎢⎣⎡−−=⎟⎟⎠⎞⎜⎜⎝⎛2121u u 1111L 2P 3k k k k F F where ()12u u k P −=.The fact that is non-linear (even in its approximate form) means that iterative solution procedures are required to be employed to determine the unknown displacements. G KNote that the approximate form is arrived at using the following strain energy expression()()⎥⎦⎤⎢⎣⎡−+−=312212se u u L 1u u k 21WExample:The strain energies for the springs in the above system (fixed at node 1) are k 1 F 2u 23k 2u3F321⎥⎦⎤⎢⎣⎡+=1322211seL u u k 21W and ()()⎥⎦⎤⎢⎣⎡−+−=323222322se u u L 1u u k 21WUse the principle of virtual work to obtain the assembled linear and geometrical stiffness matrices.()()()3322a 2322322322122212se1sese u F u F W u u u u L 23u u k u L 2u 3u k W W W δ+δ=δ=δ−δ⎥⎦⎤⎢⎣⎡−+−+δ⎥⎦⎤⎢⎣⎡+=δ+δ=δThus ()(⎥⎦⎤⎢⎣⎡−+−−⎥⎦⎤⎢⎣⎡+=2232232122212u u L 23u u k L 2u 3u k F ) and ()()⎥⎦⎤⎢⎣⎡−+−=22322323u u L 23u u k F⎟⎟⎠⎞⎜⎜⎝⎛⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡αα−α−α+α+⎥⎦⎤⎢⎣⎡−−+=⎟⎟⎠⎞⎜⎜⎝⎛32222212222132u u k k k k k F F where 1211L 2u k 3=α and ()23222u u L 2k 3−=α.Note that the element stiffness matrices are[][]111k K α+= and ⎥⎦⎤⎢⎣⎡αα−α−α+⎥⎦⎤⎢⎣⎡−−=222222222k k k k Kand it is evident how these should be assembled to form the assembled linear and geometrical stiffness matrices.2v21u 1v1u 2kxBar subject to longitudinal and lateral deflectionConsider a bar of length L and cross sectional area A represented by a spring element and subject to longitudinal and lateral displacements u and v, respectively.The normal strain is 2211x v 21x u 21x u E ⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂+∂∂= and the associated strain energy∫∫∫∫⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂+∂∂==Ω=Ω=212121x x 22x x 211x x V se dx x v 21x u 21x u EA 21dx E EA 21dx A dV W ∫⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛∂∂⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂≈21x x 232se dx x v x u x u x u EA 21WConsider further a linear displacement field of the form ()21u L x u L x L x u ⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛−= and()21v L x v L x L x v ⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛−=, and note thatL u u x u 12−=∂∂ and L v v x v 12−=∂∂. ()()()()⎥⎦⎤⎢⎣⎡−−+−+−=L v v u u L u u u u L EA 21W 21212312212se()()()()()()()1212121221221212se v v L v v u u k u u L 2v v L 2u u 3u u k W δ−δ⎦⎤⎢⎣⎡−−+δ−δ⎥⎦⎤⎢⎣⎡−+−+−=δ2v 22h 21v 11h 1a v F u F v F u F W δ+δ+δ+δ=δ and the principle of virtual work gives()()()⎥⎦⎤⎢⎣⎡−+−+−−=L 2v v L 2u u 3u u k F 21221212h1and ()()⎥⎦⎤⎢⎣⎡−−−=L v v u u k F 1212v1 ()()()⎥⎦⎤⎢⎣⎡−+−+−=L 2v v L 2u u 3u u k F 21221212h2and ()()⎥⎦⎤⎢⎣⎡−−=L v v u u k F 1212v2()()⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−−+⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−+⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−=⎟⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎜⎝⎛22111212v 2h 2v 1h 1v u v u 101005.105.1101005.105.1Lu u k 1010000010100000L2v v k 0000010100000101k F F F FExam Standard Question:The spring system depicted in the Figure consists of two massless springs of equal length , which are attached to fixed boundaries by means of pin-joints at nodes 1 and 2. The springs are connected to a slider atnode 3. The slider is constrained to move in a frictionless channel whose axis is 45 to the horizontal. Each spring has the same stiffness . The slider is subjected to an external force F 1L =0L /EA k =3 whose direction is along the axis of the frictionless channel.FigureAssume the springs have strain density ⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛∂∂⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂=Ω232x v x u x u x u E 21.(i) Write expressions for the longitudinal and lateral displacements for each spring at node 3 in terms of thedisplacement along the channel at node 3.(ii) In terms of displacement along the channel at node 3, write expressions for the total strain energy W of thespring-mass system. In addition, specify the variation in work done se a W δ resulting from the application of the force .3F (iii) Use the principle of virtual work to find a relationship between the magnitude of and the displacementalong the channel at node 3. 3FSolution:(i) Directional vectors for springs and channel are: ()2113e e 321e +=and ()2123e e 321e +−= and (21c 3e e 21e +=). Perpendicular vectors are: ()2113e 3e 21e +−=⊥and ()2123e 3e 21e +=⊥Deflection c 333e U u =, so 2U v u 333==.Longitudinal displacement: ()3122U u e 331313+=⋅=δ, ()3122U u e 332323−=⋅=δ.Lateral displacement: ()3122U u e 331313+−=⋅=δ⊥⊥, ()3122U u e 332323+=⋅=δ⊥⊥(ii) The strain energy density for element 1 is ⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛δ⎟⎠⎞⎜⎝⎛δ+⎟⎠⎞⎜⎝⎛δ+⎟⎠⎞⎜⎝⎛δ=Ω⊥21313313213L L L L E 21 The strain energy density for element 2 is ⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛δ⎟⎠⎞⎜⎝⎛δ+⎟⎠⎞⎜⎝⎛δ+⎟⎠⎞⎜⎝⎛δ=Ω⊥22323323223L L L L E 21 The total strain energy with substitution of 1L = gives()()()()[]()()()()[][]3322312232332322321313313213se U Uk 21k 21k 21W α+α=δδ+δ+δ+δδ+δ+δ=⊥⊥where and are constants determined on collecting up terms on substitution of and .1α2α231313,,δδδ⊥⊥δ2333a U F W δ=δ.(iii) The principle of virtual work gives⎥⎦⎤⎢⎣⎡α+α=⇒δ=δ=δ⎥⎦⎤⎢⎣⎡α+α=δ32133332323231se U 23kU F U F W U U 23U k WPin-jointed structuresThe example above is a pin-jointed structure. A reasonable good approximation reported in the literature for strain energy density, commonly used with pin-jointed structures, is⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛∂∂⎟⎠⎞⎜⎝⎛∂∂+⎟⎠⎞⎜⎝⎛∂∂=Ω22x v x u x u E 21This arises from strain-energy approximation 211x v 21x u E ⎟⎠⎞⎜⎝⎛∂∂+∂∂=. Can be used when 22x v x u ⎟⎠⎞⎜⎝⎛∂∂<<⎟⎠⎞⎜⎝⎛∂∂.。
量子时间晶体
a r X i v :1202.2539v 2 [q u a n t -p h ] 11 J u l 2012MIT-CTP /4348Quantum Time CrystalsFrank WilczekCenter for Theoretical PhysicsDepartment of Physics,Massachusetts Institute of Technology Cambridge Massachusetts 02139USASome subtleties and apparent difficulties associated with the notion of spontaneous breaking of time translation symmetry in quantum mechanics are identified and resolved.A model exhibit-ing that phenomenon is displayed.The possibility and significance of breaking of imaginary time translation symmetry is discussed.PACS numbers:11.30-j,5.45.Xt,3.75.LmSymmetry and its spontaneous breaking is a central theme in modern physics.Perhaps no symmetry is more fundamental than time translation symmetry,since time translation symmetry underlies both the reproducibility of experience and,within the standard dynamical frame-works,the conservation of energy.So it is natural to consider the question,whether time translation symme-try might be spontaneously broken in a closed quantum-mechanical system.That is the question we will consider,and answer affirmatively,here.Here we are considering the possibility of time crystals,analogous to ordinary crystals in space.They represent spontaneous emergence of a clock within a time-invariant dynamical system.Classical time crystals are considered in a companion paper [1];here the primary emphasis is on quantum theory.Several considerations might seem to make the possi-bility of quantum time crystals implausible.The Heisen-berg equation of motion for an operator with no intrinsic time dependence readsΨ|˙O|Ψ =i Ψ|[H,O ]|Ψ →Ψ=ΨE0,(1)where the last step applies to any eigenstate ΨE of H .This seems to preclude the possibility of an order pa-rameter that could indicate the spontaneous breaking of infinitesimal time translation symmetry.Also,the very concept of “ground state”implies state of lowest energy;but in any state of definite energy (it seems)the Hamiltonian must act trivially.Finally,a system with spontaneous breaking of time translation symmetry in its ground state must have some sort of motion in its ground state,and is therefore perilously close to fitting the definition of a perpetual motion machine.Ring Particle Model :And yet there is a familiar phys-ical phenomenon that almost does the job.A supercon-ductor,in the right circumstances,can support a stable current-carrying ground state.Specifically,this occurs if we have a superconducting ring threaded by a flux that is a fraction of the flux quantum.If the current is con-stant then nothing changes in time,so time translation symmetry is not broken;but clearly there is a sense in which something is moving.We can display the essence of this situation in a simple model,that displays its formal structure clearly.Con-sider a particle with charge q and unit mass,confined to a ring of unit radius that is threaded by flux 2πα/q .The Lagrangian,canonical (angular)momentum,and Hamil-tonian for this system are respectivelyL =12(πφ−α)2(2)πφ,through its role as generator of (angular)translations,and in view of the Heisenberg commutation relations,is realized as −i ∂2(l −α)2.(3)The lowest energy state will occur for the integer l 0that makes l −αsmallest.If αis not an integer,we will havel 0|˙φ|l 0 =l 0−α=0.(4)The case when αis half an odd integer requires spe-cial consideration.In that case we will have two distinctstates |α±12Λ=kφ/q.Note that the totalflux is not invariant un-der this topologically non-trivial gauge transformation,which cannot be extended smoothly offthe ring,so L is modified.Second,that the operation of time-reversal T, implemented by complex conjugation of wave-functions, takes|l →|−l and leaves the dynamics invariant if simultaneouslyα→−α.Putting these observations to-gether,we see that the combined operation˜T=G2αT(5) leaves the Lagrangian invariant;it is a symmetry of the dynamics and maps|l →|2α−l .˜T interchanges |l±α →|l∓α .Thus the degeneracy between those states is a consequence of a modified time-reversal sym-metry.We can choose combinations|α+12 that simultaneously diagonalize H and˜T;for these com-binations the expectation value of˙φvanishes. Returning to the generic case:Forαthat are not half-integral time-reversal symmetry is not merely modified, but simply broken,and there is no degeneracy.How do we reconcile l0|˙φ|l0 =0with Eqn.(1)?The point is that˙φ,despite appearances,is neither the time-derivative of a legitimate operator nor the commutator of the Hamiltonian with one,sinceφ,acting on wave func-tions in Hilbert space,is multivalued.By way of contrast operators corresponding to single-valued functions ofφ, spanned by trigonometric functions O k=e ikφ,do satisfy Eqn.(1)for the eigenstates|Ψ =|l .Wave functions of the quantized ring particle model correspond to the(classical)wave functions that ap-pear in the Landau-Ginzburg theory of superconductiv-ity.Those wave functions,in turn,heuristically describe the wave function for macroscopic occupation of the single-particle quantum state appropriate to a Cooper pair,regarded as a particle.Under this correspondence, the non-vanishing expectation value of˙φfor the ground state of the ring particle subject to fractionalflux maps onto the persistent current in a superconducting ring. Symmetry Breaking and Observability:As mentioned previously,the choice of a ground state that violates time translation symmetryτmust be based on some criterion other than energy minimization.But what might seem to be a special difficulty with breakingτ,because of its connection to the Hamiltonian,actually arises in only slightly different form for all cases of spontaneous sym-metry breaking.Consider for example the breaking of number(or dually,phase)symmetry.We characterize such breaking through a complex order parameter,Φ, that acquires a non-zero expectation value,which we can take to be real:0|Φ|0 =v=0.(6) We also have states|σ related to|0 by the symme-try operation.These are all energetically degenerate and mutually orthogonal(see below),and satisfyσ|Φ|σ =ve iσ.(7)The superposition|Ω =13 form a lump and,in view of Eqn.(4),that they will wantto move.So we can expect that the physical ground statefeatures a moving lump,which manifestly breaksτ.To make contact with the argument of the previoussection,we need an appropriate notion of locality.Forsimplicity we assume that the particles have an addi-tional integer label,besides the common angleφ,andthat the physical observables are offinite range in theadditional label.(Imagine an array of separate rings,displaced along an axis,so that the coordinates of parti-cle j are(φ,x=ja).)I will return to these conceptualissues below,after describing the construction.An appropriate Hamiltonian isH=Nj=11N−1Nj=k,1δ(φj−φk)(13)≡Nj=11∂t= N j=11∂t=1λφ,k2)E=−r2λ(1−k2λλ(20)in terms of the complete elliptic integrals E(k2),K(k2).We can solve E(k2)K(k2)=πλ2.Here dn(u,0)reduces to a con-stant,and E=−1/4.Asλincreases beyond that value krapidly approaches1,as does E(k2).dn(u,k2)→sech uand E→−λ2/8in that limit.Of course the constantsolution with E=−λ/2πexists for any value ofλ,butwhenλexceeds the critical value the inhomogeneous so-lution is more favorable energetically.These results havesimple qualitative interpretations.The hyperbolic secantis the famous soliton of the non-linear Schr¨o dinger equa-tion on a line.If that soliton is not too big it can be de-formed,without prohibitive energy cost,tofit on a unitcircle.The parameterβreflects spontaneous breaking of(ordinary)translation symmetry.Here that breaking isoccurring through a kind of phase separation.Our Hamiltonian is closely related,formally,to theLieb-Liniger model[3],but because we consider ultra-weak(∼1/N)attraction instead of repulsion,the groundstate physics is very different.Since our extremely inho-mogeneous approximate ground state does not supportlow-energy,long-wavelength modes(apart from overalltranslation),it has no serious infrared sensitivity.Now since non-zeroαcan be interpreted as magneticflux through the ring,we might anticipate,from Fara-day’s law,that as we turn it on,starting fromα=0,ourlump of charge will feel a simple torque.(Note that sinceFaraday’s law is a formal consequence of the mathemat-ics of gauge potentials,its use does not require additionalhypotheses.)We can also apply“gauge transformations”,as in the discussion around Eqn.(5).These observationsare reflected mathematically in the following construc-tion:For any l,we solvei∂ψl2(−i∂φ−α)2ψl−λ|ψl|2ψl,(21)withψl(φ,t)=e−ilφ˜ψ(φ+(l+α)t,t)i∂˜ψ2(−i∂φ)2˜ψ−λ|˜ψ|2˜ψ+(l+α)24energy trajectory.The parameterβ,which parameterizes an orbit of(ordinary)translation symmetry,changes at a constant rate;bothτand translation symmetry are broken,but a combination remains intact.Now let us return to address the conceptual issues alluded to earlier.Our model Hamiltonian was non-local,but we required observables to be local.That schizophrenic distinction can be appropriate,since the Hamiltonian might be–and,for our rather artificial dynamics,would have to be–carefully engineered,as opposed to being constructed from easily implemented, natural observables.Moreover it is not unlikely that the assumption of all-to-all coupling could be relaxed,in par-ticular by locating the rings at the nodes of a multidimen-sional lattice and limiting the couplings to afinite range. Were we literally considering charged particles con-fined to a common ring,and treating the electromag-neticfield dynamically,our moving lump of charge would radiate.The electromagneticfield provides modes that couple to all the particles,and in effect provide observers who manifestly violate the framework of Eqn.(12).That permits,and enforces,relaxation to a|k state.Simple variations can ameliorate this issue,e of multi-poles in place of single charges,embedding the system in a cavity,or simply arranging that the motion is slow.A more radical variation,that also addresses the unrealistic assumption of attraction among the charges,while still obtaining spatial non-uniformity,would be to consider charged particles on a ring that form–through repul-sion!–a Wigner lattice.Imaginary Time Crystals:In the standard treatment offinite temperature quantum systems using path inte-gral techniques,one considers configurations whose ar-guments involve imaginary values of the time,and im-poses imaginary-time periodicity in the inverse temper-atureβ=1/T.In this set-up the whole action is con-verted,in effect,into a potential energy:time derivatives map onto gradients in imaginary time,which is treated on the same footing as the spatial variables.At the level of the action,there is symmetry under translations in imaginary time(iTime).But since iTime appears,in this formulation,on the same footing as the spatial variables,it is natural to consider the possibility that for appropriate systems the dominant configurations in the path integral are iTime crystals.Let the iTime crystal have preferred periodλ.Whenβis an integer multiple ofλthe crystal willfit without distortion,but otherwise it must be squeezed or stretched,or incorporate defects.Periodic behavior of thermodynamics quantities in1/T,with periodλ,arise,and provide an experimental diagnostic.Integration over the collective coordinate for the broken symmetry contributes to the entropy,even at zero temperature.Inspired by the spatial crystal-iTime crystal analogy,one might also consider the possibility of iTime glasses(iGlasses),which would likewise have residual entropy,but no simple order,or iQuasicrystals.Comments: 1.It is interesting to speculate that a (considerably)more elaborate quantum-mechanical sys-tem,whose states could be interpreted as collections of qubits,might be engineered to traverse,in its ground con-figuration,a programmed landscape of structured states in Hilbert space over time.2.Fields or particles in the presence of a time crystal background will be subject to energy-changing processes, analogous to crystalline Umklapp processes.In either case the apparent non-conservation is in reality a trans-fer to the background.(In our earlier model,O(1/N) corrections to the background motion arise.)3.Many questions that arise in connection with any spontaneous ordering,including the nature of transitions into or out of the order atfinite temperature,critical dimensionality,defects and solitons,and low-energy phe-nomenology,likewise pose themselves for time crystal-lization.There are also interesting issues around the classification of space-time periodic orderings(roughly speaking,four dimensional crystals[4]).4.The a.c.Josephson effect is a semi-macroscopic oscillatory phenomenon related in spirit to time crystal-lization.It requires,however,a voltage difference that must be sustained externally.5.Quantum time crystals based on the classical time crystals of[1],which use singular Hamiltonians,can be constructed by combining the ideas of this paper with those of[5],[6].The appearance of swallowtail band structures in[7],and emergence of complicated frequency dependence in modelingfinite response times[1],as in[8], suggest possible areas of application. Acknowledgements I thank B.Halperin,Hong Liu,J. Maldacena,and especially Al Shapere for helpful com-ments.This work is supported in part by DOE grant DE-FG02-05ER41360.References[1]A.Shapere and F.Wilczek,Classical Time CrystalsarXiv:1202.2537(2012).[2]Compare F.Strocchi,Symmetry Breaking(Springer,sec-ond edition2008).[3]E.Lieb and W.Liniger,Phys.Rev.1301605(1963).[4]H.Brown,R.B¨u low,J.Neub¨u ser,H.Wondratschekand H.Zassenhaus Crystallographic Groups of Four-Dimensional Space(Wiley,1978).[5]M.Henneaux,C.Teitelboim,J.Zanelli Phys.Rev.A364417(1987).[6]A.Shapere,F.Wilczek Branched Quantization(paper inpreparation).[7]B.T.Seaman,L.D.Carr,M.J.Holland Phys.Rev.A72033602(2005).[8]G.Georges,G.Kotliar,W.Krauth,M.Rozenberg Rev.Mod.Phys.681(1996).。
协方差
1
18.05 class 7, Covariance and Correlation, Spring 2014
2
2.2
Sums and integrals for computing covariance
Since covariance is defined as an expected value we compute it in the usual way as a sum or integral. Discrete case: If X and Y have joint pmf p(xi , yj ) then
n m
n
m
p(xi , yj )xi yj − µX µY .CovFra bibliotekX, Y ) =
i=1 j =1
p(xi , yj )(xi − µX )(yj − µY ) =
i=1 j =1
Continuous case: If X and Y have joint pdf f (x, y ) over range [a, b] × [c, d] then
18.05 class 7, Covariance and Correlation, Spring 2014
3
So Cov(XY ) = E (XY ) − µX µY =
5 1 −1= . 4 4 Next we redo the computation of Cov(X, Y ) using the properties of covariance. As usual, let Xi be the result of the ith flip, so Xi ∼ Bernoulli(0.5). We have X = X1 + X2 and Y = X2 + X3 . We know E (Xi ) = 1/2 and Var(Xi ) = 1/4. Therefore using Property 2 of covariance, we have Cov(X, Y ) = Cov(X1 +X2 , X2 +X3 ) = Cov(X1 , X2 )+Cov(X1 , X3 )+Cov(X2 , X2 )+Cov(X2 , X3 ). Since the different tosses are independent we know Cov(X1 , X2 ) = Cov(X1 , X3 ) = Cov(X2 , X3 ) = 0. Looking at the expression for Cov(X, Y ) there is only one non-zero term Cov(X, Y ) = Cov(X2 , X2 ) = Var(X2 ) = 1 . 4
机械英语考试试题及答案
机械英语考试试题及答案一、选择题(每题2分,共20分)1. The term "mechanical engineering" refers to:A. The study of machinesB. The design and manufacture of mechanical systemsC. The operation of machineryD. The maintenance of mechanical equipment答案:B2. What is the function of a bearing in a mechanical system?A. To reduce frictionB. To increase efficiencyC. To provide powerD. To transmit motion答案:A3. The process of converting thermal energy into mechanical energy is known as:A. ElectrificationB. CombustionC. ThermodynamicsD. Hydrodynamics答案:C4. In mechanical design, the principle of "KISS" stands for:A. Keep It Simple, StupidB. Keep It Short and SimpleC. Keep It Simple and SafeD. Keep It Simple, Smart答案:A5. A gear train is used to:A. Change the direction of motionB. Increase the speed of rotationC. Decrease the speed of rotationD. All of the above答案:D6. What does CAD stand for in mechanical engineering?A. Computer-Aided DesignB. Computer-Aided DraftingC. Computer-Aided DevelopmentD. Computer-Aided Diagnostics答案:A7. The SI unit for force is:A. NewtonB. JouleC. PascalD. Watt答案:A8. What is the purpose of a flywheel in a mechanical system?A. To store energyB. To increase speedC. To reduce noiseD. To dissipate heat答案:A9. The term "hydraulics" is associated with the study of:A. Fluid dynamicsB. Solid mechanicsC. Structural analysisD. Thermal engineering答案:A10. The process of cutting a material to a specific shape is known as:A. MachiningB. CastingC. ForgingD. Extrusion答案:A二、填空题(每空1分,共10分)11. The formula for calculating the moment of a force is \( F \times d \), where \( F \) is the force and \( d \) is the_______.答案:distance from the pivot12. A _______ is a device that converts linear motion into rotational motion.答案:crank13. In a four-stroke internal combustion engine, the four strokes are intake, compression, _______, and exhaust.答案:power14. The _______ of a material is its ability to resist deformation under load.答案:stiffness15. The term "overhaul" in mechanical maintenance refers to a thorough inspection and _______ of a machine or its parts.答案:repair16. The _______ of a machine is the study of how forces act on and within a body.答案: statics17. A _______ is a type of machine that uses a screw to convert rotational motion into linear motion.答案:screw jack18. The _______ of a system is the point around which the system rotates.答案:pivot19. The _______ of a lever is the ratio of the effort arm to the load arm.答案:mechanical advantage20. The _______ is a type of bearing that allows for rotation with minimal friction.答案:ball bearing三、简答题(每题5分,共30分)21. Explain the difference between static and dynamic equilibrium in mechanical systems.答案:Static equilibrium refers to a state where the net force and net moment acting on a body are zero, resulting in no acceleration. Dynamic equilibrium occurs when the net force is zero, but the body is in motion with constant velocity.22. What is the purpose of a clutch in a vehicle?答案:A clutch is used to engage and disengage the power transmission from the engine to the transmission system, allowing the vehicle to start, stop, and change gears smoothly.23. Describe the function of a governor in an engine.答案:A governor is a device that automatically controls the speed of an engine by regulating the fuel supply or the valve settings, ensuring the engine operates within safespeed limits.24. What are the three primary types of joints in structural engineering?答案:The three primary types of joints are pinned joints, fixed joints, and sliding joints, each serving different purposes in connecting and supporting structural elements.25. Explain the。
积分用英语怎么说
积分用英语怎么说商家为了刺激消费者消费,是一种变相营销的手段。
比如,满多少积分可换购某样商品。
积分获取的途径:购物,做任务,参加某种活动。
那么你知道积分用英语怎么说吗?下面来学习一下吧。
积分英语说法:integral积分的相关短语:不定积分indefinite integral ; antiderivative ; Primitive function ; indefinite integration数值积分 numerical integration ; Integration point ; Integral approximation ; numerical integral积分表lists of integrals ; Integration formulae ; Standard integral ; table of integral指数积分Exponential integral ; Expint ; Ei maple ; Well function黎曼积分 Riemann integral ; Riemann integrability ; riemann calculus ; Riemann Sums积分符号 integral symbol ; Integral sign ; Signs For Integrals ; Integralzeichen曲线积分line integral ; curvilinear integral ; Curved Line Integral积分时间Integral Time ; Integration time ; reset time ; Integrationszeit积分赛 Points Race ; tournament ; Score-O ; Ranked Match 积分的英语例句:1. He graduated with a GPA of 3.8.他毕业时各科成绩的平均积分点为3.8。
分析力学第一次作业解答
1.1 Spherical coordinates. Many of the problems which we will solve will possess spherical symmetry, so that the most natural coordinates are spherical coordinates. We'll need to know how to express the position, velocity, acceleration, and kinetic energy of a particle in this coordinate system.The spherical coordinates (,,)r θφare related to the rectangular coordinates (x, y, z ) through sin cos x r θφ=,sin sin y r θφ=, cos z r θ=. The unit vectors are(a) Draw a careful figure which illustrates the meaning of the spherical coordinates and their relation to the rectangular coordinates.(b) Show that this is indeed on orthogonal coordinate system. In other words, show that 0,0,0r r e e e e e e θφθφ⋅=⋅=⋅=. Also show that each of the vectors,,r e e e θφhas a length of one. Finally, show that the coordinate system is “right-handed”, so that ,,r r r e e e e e e e e e θφθφφθ×=×=×=(c)Use these relations to show that(d) In spherical coordinates the position vector is r =r re . Use your results above to calculate the velocity v = r i, the kinetic energy 2/2T mv =, and the acceleration a = r iof a particle in spherical coordinates.Solution: (b) 22(sin cos sin sin cos )(cos cos cos sin sin )sin cos cos sin cos sin sin cos 0r x y z x y z e e e e e e e e θθφθφθθφθφθθθφθθφθθ=+++−=+−= i isin cos sin sin cos sin cos cos cos cos sin sin x y zr x y e e e e e e e e θφθφθφθφφθφθφθ×==−+=−The others are straightforward. (c)(cos cos sin sin )(cos sin sin cos )sin (cos cos cos sin sin )sin (sin cos )sin r x y zx y z x y e e e e e e e e e e e θφθθθθφφθθφθφφθθθθθθφθθφφφθθφ=−++−=+−+−+=+i i i i i i i i i iThe others are straightforward.(d) This closely follows the derivation in class for cylindrical coordinates. The key point to remember is that the unit vectors are time dependent, so you need to use Eqs.above and the product rule when carrying out the di_erentiations. The results are1.2 Central forces. A central force is a force directed along the line connecting the two interacting particles which only depends on the separation between the particles. Assuming that the center of the force is at the origin (r = 0), then the central force can be written as F (r ) = f (r ) e r , with r the distance from the origin. Important examples of central forces are the gravitational force, the electrostatic force (Coulomb's law), and the force exerted by a spring on a mass.(a) Assume that a particle of mass m is acted upon by a central force. The natural coordinates to describe the motion of the particle are spherical coordinates (,,)r θφ. Separate Newton's Second Law in spherical coordinates, and write down the three equations of motion (one for each component).(b) Show that for any f (r ), a solution of the θand φ components of these equationshas /2θπ=and 2mr l φ=i, with l a constant which is independent of time. What is the physical significance of these results?(c) Use the conditions above to simplify the equation for the radial coordinate r .Show thatSolution:(a) Use the results from Problem 1; using m a = Fin spherical coordinates, we have(b)Taking(independent of time), then; substituting theseresults into Eqs above, we findMultiply both sides by r; we can then write this asSo that , with l a constant (independent of time). If we recognize as the tangential component of the particle momentum, then is the angular momentum about the center of force. Therefore, the condition is simply a statement of conservation of angular momentum about the center of force. The factthat means that the motion occurs in a plane(c)Substitute , , and into, we can get the result1.4 Spherical pendulum. A mass m is attached to the ceiling by an inextensible (i.e. fixed length), massless string of length b, as shown in the figure.(a) Draw a free-body diagram for the mass, showing all forces acting upon it.(b) Use F = m a in spherical coordinates to derive the equations of motion forθand φ(c) Use your equations of motion to show that the quantity(d) Use your result from part (c) to eliminate the dependence upon φin your equation of motion for θ. You should find thatSolution:(a)The figure is below(b)Resolving the forces into components along the unit vectors, we haveUse F = m a and the expression for acceleration in spherical coordinates along with theconstraint that r = b and to obtain the equations of motionAndAs a check, note that when the equation forθiswhich is the equation of motion for the plane pendulum.(c)DifferentiateFactoring sinθfrom this expression, we see that it is indeed zero. The quantity is the angular momentum about the z-axis, which is conserved.(d)The angular momentum is a constant whose value is determined from initialconditions. Usingwe immediately obtain3.1 Lagrangian vs. Newtonian mechanics . Construct the Lagrangians and derive the equations of motion for (a) the spherical pendulum and (b) the bead on the hoop. In both cases compare your results with the results which you obtained in previous assignments using Newton's Second Law. Solution:(a) Let's begin with the spherical pendulum. In spherical coordinates the kinetic energy isThe pendulum string has a constant length R , so r = R and 0r =i. There are twodegrees of freedom, corresponding to the two generalized coordinates θand φ. The kinetic energy is thenSetting the zero of the potential energy to be when the string is horizontal, we haveThe Lagrangian is thenLagrange's equation for θiswhich, after a little rearranging, yieldsFor φ, we see that L is independent of φ, so that φis a cyclic coordinate. Thuswhereis a constant.(b) For the bead on the hoop, we'll use the angle θ(the angle that the line connecting the bead to the center of the hoop makes with the vertical) as our generalized coordinate. The constraint is that r = R so that 0r =i, and that the hoop rotates with a constant angular velocity, so that. There is now only one degree of freedom, so there is one generalized coordinate, which we will choose to be θ. From Equation above, we see that the kinetic energy isIf we choose the potential energy to be zero at , thenThe Lagrangian isLagrange's equation isTaking the derivatives, and rearranging a bit, we obtain the equation of motionNote that in both cases the equations of motion are the same as we obtained using Newton's Second Law. However, using the Lagrangian method we do not obtain the forces of constraint.3.2 Moving pendulum. A plane pendulum of length l has a bob of mass m , which is attached to a cart of mass M which can move on a frictionless, horizontal table.(a) Express the rectangular coordinatesof the pendulum bob in terms of thecoordinates. (b) Using X and θas your generalized coordinates, construct the Lagrangian for this system.(c) Using Lagrange's equations, derive the equations of motion for the coupled pen- dulum and cart system.(d) Find all first integrals of the motion (i.e.,find the conserved quantities)Solution(a) The coordinates of the pendulum bob are and cos b y θ=−the components of the velocity are then(b)The kinetic energy of the bob is thenand the kinetic energy of the cart isThe potential energy of the bob isPulling all of this together, we have for the Lagrangian(c)The equations of motion are obtained from Lagrange's equationsForθwe havewhich after some algebra becomesFor X we haveThe Lagrangian is independent of X (X is a cyclic coordinate), so that is a constant. This is a statement of the conservation of the x-component of the momentum of the system:The value of P x is determined from the initial conditions. We can differentiate this equation again to obtainWe could use this equation to eliminate the fromto get a single,nonlinear second order di_erential equation for θ(d) In addition to the momentum P x, the total energy of the system is conserved:3.3 Pendulum A pendulum consisting of a rigid massless rod of length l with a mass m at one end moves in a uniform gravitational field g. The point of support is notstationary but moves in the vertical direction with a displacement y(t) which is a givenfunction of time.(a) Find the Lagrangian and the equations of motion.(b) Show that the equation is the same as that for pendulum with fixed support, with g replaced by ()eff g t . What is ()eff g t ? Solution:(a) The Lagrangian is given by:221()2m m m L T V m x y mgy =−=+−i iWhere()sin ()()()cos ()m m x t l t y t y t l t θθ==−⇒cos sin m m x l y y l θθθθ==+i iiii⇒2221(2sin )cos 2L m l y l y mgy mgl θθθθ=++−+i i i iNote that there is a single generaliged coordinate, ()t θ(()m x t and ()m y t are notindependent). y(t) is NOT an independent coordinate, The Lagrangian and the equations of motion are20sin cos sin 0d L L d ml ml y ml y mgl dt dt θθθθθθθ∂∂⎛⎞⎡⎤−=⇒+−+=⎜⎟⎢⎥∂∂⎣⎦⎝⎠i i i i sin sin 0l y g θθθ⇒++=iiii(b) Rewrite as: ()sin 0eff g t lθθ+=iiWhere()(1)eff yg t g g=+ii。
1PPC控制器故障代码查询重要
Rexroth IndraControl VCP 20Industrial Hydraulics Electric Drivesand ControlsLinear Motion andAssembly Technologies PneumaticsServiceAutomationMobileHydraulicsRexroth VisualMotion 10 Multi-Axis Machine Control R911306327 Edition 01Troubleshooting GuideAbout this Documentation Rexroth VisualMotion 10 Troubleshooting Guide DOK-VISMOT-VM*-10VRS**-WA01-EN-PRexroth VisualMotion 10Multi-Axis Machine ControlTroubleshooting Guide DOK-VISMOT-VM*-10VRS**-WA01-EN-P Document Number, 120-2300-B323-01/ENPart of Box Set, 20-10V-EN (MN R911306370)This documentation describes …•the use of VisualMotion Toolkit for assitance in diagnostics •the proper steps for indentifing diagnostic faults • and the suggested remedies for clearing faults Description ReleaseDateNotes DOK-VISMOT-VM*-10VRS**-WA01-EN-P 08/2004Initial release© 2004 Bosch Rexroth AGCopying this document, giving it to others and the use or communicationof the contents thereof without express authority, are forbidden. Offendersare liable for the payment of damages. All rights are reserved in the eventof the grant of a patent or the registration of a utility model or design(DIN 34-1).The specified data is for product description purposes only and may notbe deemed to be guaranteed unless expressly confirmed in the contract.All rights are reserved with respect to the content of this documentationand the availability of the product.Bosch Rexroth AGBgm.-Dr.-Nebel-Str. 2 • D-97816 Lohr a. MainTel.: +49 (0)93 52/40-0 • Fax: +49 (0)93 52/40-48 85 • Telex: 68 94 21Bosch Rexroth Corporation • Electric Drives and Controls5150 Prairie Stone Parkway • Hoffman Estates, IL 60192 • USATel.: 847-645-3600 • Fax: 847-645-6201/Dept. ESG4 (DPJ)This document has been printed on chlorine-free bleached paper.Title Type of DocumentationDocument TypecodeInternal File Reference Purpose of Documentation Record of Revisions Copyright Validity Published byNoteRexroth VisualMotion 10 Troubleshooting Guide Table of Contents I Table of Contents1VisualMotion Tools for Diagnosing1-1 The Diagnostics Menu.............................................................................................................1-1System Diagnostics.................................................................................................................1-1Tasks Diagnostics...................................................................................................................1-3Drive Overview….....................................................................................................................1-42Monitoring and Diagnostics2-12.1System Diagnostics - Codes and Message...................................................................................2-1Parameters..............................................................................................................................2-2DriveTop..................................................................................................................................2-32.2Control Startup Messages.............................................................................................................2-4PPC Boot-Up Sequence..........................................................................................................2-4Control Firmware Sequence....................................................................................................2-42.3Status Messages (001-199)...........................................................................................................2-5001 Initializing System.............................................................................................................2-5002 Parameter Mode...............................................................................................................2-5003 Initializing Drives...............................................................................................................2-5004 System is Ready...............................................................................................................2-5005 Manual Mode....................................................................................................................2-5006 Automatic Mode: ABCD....................................................................................................2-5007 Program Running: ABCD.................................................................................................2-6008 Single-Stepping: ABCD....................................................................................................2-6009 Select Parameter Mode to Continue................................................................................2-6010 Breakpoint Reached: ABCD.............................................................................................2-6018 Please cycle power to continue........................................................................................2-6019 Executing User Initialization Task....................................................................................2-62.4Warning Messages (201-399).......................................................................................................2-7201 Invalid jog type or axis selected........................................................................................2-7202 Drive %d is not ready.......................................................................................................2-7203 Power Fail detected..........................................................................................................2-7204 Sercos ring was disconnected..........................................................................................2-8205 Parameter transfer warning in Task %c...........................................................................2-8207 Axis %d position limit reached..........................................................................................2-8208 Lost Fieldbus Connection.................................................................................................2-9209 Fieldbus Mapping Timeout...............................................................................................2-9210 File System Defrag: %d completed................................................................................2-10211 Program- & Data memory cleared..................................................................................2-10212 Option Card PLS Warning, see ext. diag.......................................................................2-10213 Sercos cycle time changed.............................................................................................2-11214 PCI Bus Cyclic Mapping Timeout...................................................................................2-11 DOK-VISMOT-VM*-10VRS**-WA01-EN-PII Table of Contents Rexroth VisualMotion 10 Troubleshooting Guide215 RECO I/O Failure, see ext. diag.....................................................................................2-11216 Control PLS %d warning, see ext. diag..........................................................................2-12217 PCI Bus Communication, see ext. diag..........................................................................2-12218 PCI Bus Register Mapping Timeout...............................................................................2-13219 PCI Bus Lifecounter Timeout.........................................................................................2-13220 Excessive deviation in PMG%d, see ext. diag...............................................................2-13221 Excessive Master Position Slip Deviation......................................................................2-13222 ELS Config. Warning, see ext. diag...............................................................................2-14223 PCI Bus reset occurred, cyclic data are invalid..............................................................2-14225 System booted................................................................................................................2-14226 RS485 Serial Communication Error (port X1%d)...........................................................2-15227 Control Over-temperature Warning................................................................................2-15228 Control - SYSTEM WARNING.......................................................................................2-152.5Shutdown Messages (400 - 599).................................................................................................2-16400 EMERGENCY STOP......................................................................................................2-16401 Sercos Controller Error: %02d........................................................................................2-16402 Sercos Config. Error: see ext. diag................................................................................2-16403 System Error see ext. diag.............................................................................................2-17405 Phase %d: Drive did not respond...................................................................................2-17407 Drive %d Phase 3 Switch Error......................................................................................2-17409 Sercos Disconnect Error.................................................................................................2-18411 Drive %d Phase 4 Switch Error......................................................................................2-18412 No drives were found on ring..........................................................................................2-18414 Parameters were lost......................................................................................................2-19415 Drive %d was not found..................................................................................................2-19416 Invalid Instruction at %04x..............................................................................................2-19417 SYSTEM ERROR: pSOS #%04x...................................................................................2-19418 No program is active.......................................................................................................2-20419 Invalid Program File: code = %d....................................................................................2-20420 Drive %d Shutdown Error...............................................................................................2-20421 User Program Stack Overflow........................................................................................2-20422 Parameter transfer error in Task %c..............................................................................2-21423 Unimplemented Instruction.............................................................................................2-21425 Instruction error: see Task %c diag................................................................................2-21426 Drive %d is not ready.....................................................................................................2-22427 Calc: invalid table index %d............................................................................................2-22428 Calc: division by zero......................................................................................................2-22429 Calc: too many operands................................................................................................2-22430 Calc: invalid operator......................................................................................................2-23431 Calc error: see Task %c diag.........................................................................................2-23432 Calc: too many nested expressions...............................................................................2-23433 Setup instruction outside of a task.................................................................................2-23434 Axis %d configured more than once...............................................................................2-23435 Axis %d is not assigned to a task...................................................................................2-24436 General Compiler Error: %04x.......................................................................................2-24438 Invalid Axis Selected: %d...............................................................................................2-24DOK-VISMOT-VM*-10VRS**-WA01-EN-PRexroth VisualMotion 10 Troubleshooting Guide Table of Contents III439 Axis %d: Invalid Motion Type.........................................................................................2-24440 I/O Transfer Error: see task diag....................................................................................2-25450 Event %d: invalid event type..........................................................................................2-25451 Invalid event number ‘%d’..............................................................................................2-25452 More than %d event timers armed.................................................................................2-25453 Homing param. transfer error: %d..................................................................................2-25454 Axis %d homing not complete........................................................................................2-26459 Axis %d target position out of bounds............................................................................2-26460 Invalid program %d from binary inputs...........................................................................2-26463 Ratio command: invalid ratio..........................................................................................2-26464 Can't activate while program running.............................................................................2-27465 Drive %d config. error, see ext. diag..............................................................................2-27467 Invalid ELS Master Option..............................................................................................2-27468 ELS adjustment out of bounds.......................................................................................2-27470 Axis %d velocity > maximum..........................................................................................2-28474 Drive %d cyclic data size too large.................................................................................2-28477 Axis D: probe edge not configured.................................................................................2-28478 Calc: operand out of range.............................................................................................2-28483 Parameter Init. Error: see Task %c diag........................................................................2-29484 Control SYSTEM ERROR..............................................................................................2-29486 Sercos Device %d is not a drive.....................................................................................2-29487 CAM %d is invalid or not stored.....................................................................................2-29488 CAM Error: See Task %c diag........................................................................................2-30489 More than %d CAM axes selected.................................................................................2-30490 System Memory Allocation Error....................................................................................2-30492 Programs were lost, see ext. diag..................................................................................2-30496 Can't execute this instruction from an event..................................................................2-31497 Limit switch config. error, see ext. diag.........................................................................2-31498 Drive %d Shutdown Warning..........................................................................................2-32499 Axis number %d not supported in this version...............................................................2-32500 Axis %d is not referenced...............................................................................................2-32501 Drive %d comm. error, see ext. diag..............................................................................2-33502 ELS and cams not supported in this version..................................................................2-33504 Communication Timeout.................................................................................................2-33505 Axis %d is not configured...............................................................................................2-33508 User Watchdog Timeout.................................................................................................2-33509 Control System Timing Error (%d).................................................................................2-34515 PLC Communications Error............................................................................................2-34516 More than %d registration functions enabled.................................................................2-34519 Lost Fieldbus/PLC Connection.......................................................................................2-35520 Fieldbus Mapping Timeout.............................................................................................2-35521 Invalid Virtual Master ID: %d..........................................................................................2-36522 Invalid ELS Master ID: %d..............................................................................................2-36523 IFS status, facility = 0x%x..............................................................................................2-36524 Hardware Watchdog timeout..........................................................................................2-36525 I/O Configuration error, see ext. diag.............................................................................2-36 DOK-VISMOT-VM*-10VRS**-WA01-EN-PIV Table of Contents Rexroth VisualMotion 10 Troubleshooting Guide526 Sercos Multiplex Channel Config, see ext. diag.............................................................2-37527 Control Initialization Error, see ext. diag.........................................................................2-38528 System Event %d Occurred...........................................................................................2-38529 Invalid ELS Group ID: %d...............................................................................................2-38530 CAM %d is active, can't overwrite..................................................................................2-39531 Invalid variable for Fieldbus/PCI Bus Mapping...............................................................2-39532 Power fail brown out condition detected.........................................................................2-39533 Multiple instances of index CAM: %d found...................................................................2-39534 Hardware Version Not Supported..................................................................................2-40539 Invalid Parameter Number..............................................................................................2-40540 Option Card PLS error....................................................................................................2-40541 Link Ring Error, see ext. diag.........................................................................................2-41542 PLC Cyclic Mapping Timeout.........................................................................................2-42543 PCI Bus Runtime Error...................................................................................................2-42544 RECO I/O Failure, see ext. diag.....................................................................................2-42545 Invalid Coordinated Articulation Function ID: %d...........................................................2-43546 Multiple Instance of Coordinated Articulation Function with ID: %d...............................2-43547 Task %c Coordinated Articulation Error, see ext. diag...................................................2-43548 Invalid Kinematic Number: %d.......................................................................................2-43549 Fieldbus Initialization Error.............................................................................................2-43550 User Initialization Task Timeout.....................................................................................2-44551 Master Slip Config. Error, see ext. diag..........................................................................2-44552 Excessive Master Position Slip Deviation......................................................................2-44553 Invalid Parameter Detected, see C-0-2002....................................................................2-44554 Excessive Deviation in PMG%d, see ext. diag...............................................................2-45555 PCI Bus Register Mapping Timeout...............................................................................2-45556 PCI Bus Lifecounter Timeout.........................................................................................2-45557 PMG%d Maximum allowed deviation window is Zero....................................................2-45558 PMG%d Only 1 axis parameterized...............................................................................2-46559 PMG%d Number of offsets does not match number of Axis..........................................2-46560 PMG%d Max. allowed dev. window is larger than 25% of Modulo................................2-46561 PMG%d Offset is larger than Modulo.............................................................................2-46562 PMG%d Parameterized Axis is not in system................................................................2-46563 Invalid Task Specified, Must be A-D..............................................................................2-46564 PMG%d Invalid configuration, see ext. diag...................................................................2-46565 Axis %d: Configuration error, see ext. diag....................................................................2-47566 Filter sample rate and cutoff frequency mismatch.........................................................2-47567 ELS Config. Error, see ext. diag.....................................................................................2-47568 Axis %d: Assigned Task is Not Defined.........................................................................2-48570 ELS Max. Vel. Exceeded, see ext. diag.........................................................................2-48571 No Program Found.........................................................................................................2-49572 PCI Bus reset occurred, cyclic data is invalid.................................................................2-49573 CAM %d is being built....................................................................................................2-49575 ELS Master for ELS Group %d is invalid........................................................................2-49576 Event for input I%d is already armed, cannot arm again...............................................2-50577 Restored non volatile memory from compact flash........................................................2-50DOK-VISMOT-VM*-10VRS**-WA01-EN-PRexroth VisualMotion 10 Troubleshooting Guide Table of Contents V578 Virtual Master %d Exceeded Its Max. Vel., see ext. diag...............................................2-50579 Group %d Exceeded Its Jog Velocity, see ext. diag......................................................2-51580 pROBE Error Occurred in Task:0x%04X.......................................................................2-51581 Probe Function for Axis # is locked by the PLC.............................................................2-51582 Integrated PLC: PLC Stopped in Operation Mode.........................................................2-51583 Integrated PLC: Internal System Error...........................................................................2-51584 ELS System Master %d is invalid, see ext. diag............................................................2-51585 Drive %d separate deceleration not supported..............................................................2-52586 Master Encoder Card Error, see ext diag.......................................................................2-522.6Integrated PLC Status Messages................................................................................................2-536001 Integrated PLC: Running..............................................................................................2-536002 Integrated PLC: Stopped..............................................................................................2-536002 Integrated PLC: Stopped at Breakpoint........................................................................2-532.7Integrated PLC Error Codes........................................................................................................2-530016 Integrated PLC: Software Watchdog Error...................................................................2-530019 Integrated PLC: Program Checksum Error..................................................................2-530020 Integrated PLC: Fieldbus Master Error.........................................................................2-540021 Integrated PLC: I/O Update Error.................................................................................2-542000 Integrated PLC: Internal SIS System Error..................................................................2-542001 Integrated PLC: Internal Acyclic Access Error.............................................................2-542002 Integrated PLC: Internal Acyclic Memory Error............................................................2-542003 Integrated PLC: PLC Configuration Error.....................................................................2-552004 Integrated PLC: File Access Error................................................................................2-552005 Integrated PLC: Internal Fatal Task Error....................................................................2-556011 Integrated PLC: PLC Program Stopped in Operation Mode........................................2-556012 Integrated PLC: General Error.....................................................................................2-552.8Communication Error Codes and Messages...............................................................................2-56!01 Sercos Error Code # xxxx...............................................................................................2-56!02 Invalid Parameter Number..............................................................................................2-57!03 Data is Read Only...........................................................................................................2-57!04 Write Protected in this mode/phase...............................................................................2-57!05 Greater than maximum value.........................................................................................2-57!06 Less than minimum value...............................................................................................2-57!07 Data is Invalid.................................................................................................................2-57!08 Drive was not found........................................................................................................2-57!09 Drive not ready for communication.................................................................................2-57!10 Drive is not responding...................................................................................................2-57!11 Service channel is not open...........................................................................................2-57!12 Invalid Command Class..................................................................................................2-57!13 Checksum Error: xx (xx= checksum that control calculated).........................................2-58!14 Invalid Command Subclass............................................................................................2-58!15 Invalid Parameter Set.....................................................................................................2-58!16 List already in progress..................................................................................................2-58!17 Invalid Sequence Number..............................................................................................2-58!18 List has not started.........................................................................................................2-58!19 List is finished.................................................................................................................2-58 DOK-VISMOT-VM*-10VRS**-WA01-EN-P。
化工原理英文教材流体输送阻力系数Flow past immersed bodies
For low Reynolds numbers the drag force for a sphere conforms to a theoretical equation called Stokes´law, which may be written
FD 3ud p
(7-3)
From Eq.(7-3), the drag coefficient predicted by Stokes´ law is, using Eq(7-1)
The total integrated drag from wall shear is called
wall drag.
The total integrated drag from pressure is called form drag.
In potential flow (ideal fluids) there is no wall drag.
R(7-e2)
Drag coefficients of typical shapes
In Fig, curves of CD(ζ) vs. Re are shown for spheres, long cylinders, and disks. The axis of the cylinder and the face of the disk are perpendicular to the direction of flow.
(7-25)
du dt
ae
p p
Ap
Vp p
u2 2
(7-30)
Terminal velocity
The drag always increases with velocity, the acceleration du/dt of a particle decreases with time and approaches to zero.
Thethree-bodyproblem
The ThreeBody ProblemJoakim Hirvonen Gr¨u tzeliusKarlstad University December26,2004 Department of Engineeringsciences,Physics and MathematicsAnalytical Mechanics5pExaminator:Prof J¨u rgen F¨u chsAbstractThe main topic of this project is to give a mathematical description of the three body problem.A direct application to this problem is a rotating two-body system such as Sun-Jupiter,this rotating system is going to be treated in detail.At the end of the project there is a short discussion about the Ascending nodes and the Lagrangian points.Contents1Historical Background21.1Origin of The Three Body Problem (2)1.2Introduction to The Three Body Problem (2)2Mathematical Description of The Three Body Problem32.1The differential equations of the problem (3)2.2Reduction to the6th Order (4)3The Restricted Three Body Problem6A Ascending Node8B Lagrangian Points91Historical Background1.1Origin of The Three Body ProblemSince bodies in the solar system are approximately spherical and their dimensions extremely small when compared with the distances between them,they can be considered as point masses.Hence the origin of the problem can be thought of as being synonymous with the foundation of modern dynamical astronomy.This part of celestial mechanics,which connects the mechanical and physical causes with the observed phe-nomena,began with the introduction of Newton’s theory of gravitation. From the time of the publication of the Principia in1687,it became important to verify whether Newton’s law alone was capable of render-ing a complete understanding of how celestial bodies move in space.In order to pursue this line of investigation,it was necessary to ascertain the relative motion of n bodies attracting one another according to the Newtonian law.Newton himself had geometrically solved the problem of the two bodies for two spheres moving under their mutual gravitational attrac-tion,and in1710Johann Bernoulli had proved that the motion of one particle with respect to the other is described by a conic section.In 1734Daniel Bernoulli won a French academy prize for his analytical treatment of the two body problem,and the problem was solved in detail by Euler1744.Meanwhile work was already in progress on the higher dimensional problem.Driven by the needs of navigation for knowledge about the motion of the moon,researchers scrutinized the system formed by the sun,the earth and the moon,and the lunar theory quickly dominated the early research into the problem.1.2Introduction to The Three Body ProblemThe three body problem,which was described by Whittaker as”the most celebrated of all dynamical problems”[1]and which fulfilled for Hilbert the necessary criteria for a good mathematical problem,can be simply stated:three particles move in space under their mutual gravitational attraction;given their initial conditions,determine their subsequent motion.Like many mathematical problems,the simplicity of its statement belies the complexity of its solution.For although the one and two body problems can be solved in closed form by means of elementary functions,the three body problem is a complicated linear problem,and no similar type of solution exists.Apart from its intrinsic appeal as a simple-to-state problem,the three body problem has a further attribute which has contributed to its attraction for potential solvers:its intimate link with the funda-mental question of the stability of the solar system.Over the years attempts tofind a solution spawned a wealth of research,and between 1750and the beginning of the twentieth century more than800papersrelating to the problem were published,invoking a roll call of many distinguished mathematicians and astronomers.And hence,as is often the case with such problems,its importance is now perceived as much in the mathematical advances generated by attempts at its solution as in the actual problem itself.These advances have come in many different fields,including,in recent times,the theory of dynamical problems.To clarify the mathematical difficulties associated with the problem we will begin with a mathematical description.2Mathematical Description of The Three Body Problem2.1The differential equations of the problemLet us suppose that the three bodies under consideration to be at the points P i,with masses m i and coordinates q ij in an inertial reference frame(i,j=1,2,3).The distance between them are large,so we can think of them as point particles.We denote the distance between them as P ij=r ij,where r ij=|q i−q j|.Due to Newton’s law of gravitation,the force of attraction between the i th and j th becomes Gm i m j/r2ij ,andthe corresponding term in the potential energy becomes−Gm i m j/r ij. Then the potential energy of the hole system isV=−Gm1m2r12+m1m3r13+m2m3r23(1)where G is the gravitational constant.With the help of Newtons equa-tion of motion and that F=−∂V∂r ,then choosing units so that G isequal to one,the equations of motion becomem1d2q1idt2=−∂V∂q1i(2)m2d2q2idt2=−∂V∂q2i(3)m3d2q3idt2=−∂V∂q3i(4)ord2q1i dt2=m2(q2i−q1i)r312+m3(q3i−q1i)r313(5)d2q2i dt2=m1(q1i−q2i)r321+m3(q3i−q2i)r323(6)d2q3i dt2=m1(q1i−q3i)r331+m2(q2i−q3i)r332(7)where i=1,2,3.The problem is therefore described by nine second-order differential equations or by18equations of thefirst orderdq1i dt =˙q1i,dq2idt=˙q2i,dq3idt=˙q3im1d˙q1idt=−∂V∂q1i,m2d˙q2idt=−∂V∂q2i,m3d˙q3idt=−∂V∂q3i(8)So for a closed solution to the problem,the system needs18independent integrals.However,it is only possible tofind12such integrals,and the system can therefore only be reduced to one of order six.As will be shown below,this is achived through the use of the so-called ten classic integrals,the six integrals of the motion of the centre of mass,the three integrals of angular momentum,and the energy integral,together with the elimination of the time and the elimination of what is called the ascending node.Illustration of the ascending node and other orbital parameters can be found in appendix A.2.2Reduction to the6th OrderWhen multiplying equation(5),(6)and(7)by m i a summation can be performed to give three equations3i=1m id2q ijdt2=0,(j=1,2,3),(9)if we integrate these equation twice we get the equations3i=1m i q ij=A j t+B j,(j=1,2,3),(10)in which the A j and B j are constants of integration.These equations show that the centre of mass of the three particles either remains at rest or moves uniformly in space in a straight line.This is expected since there are no forces acting except the mutual attractions of the particles.The six constants serve to describe the motion of the centre of mass in the original arbitrary inertial coordinate system and play no part in the motion of the bodies about the centre of mass.If thefirst equation of(5)multiplied by−q12,thefirst equation of (6)by−q22and thefirst equation of(7)by−q32,and in equation(5) the second equation is multiplied by q11,the second equation of(6)by q21,and the second equation of(7)by q31,and these two sets are added together,this will give us3i=1m i q i1d2q i2dt2−3i=1m i q i2d2q i1dt2=0,(11)and two similar equations can be obtained by a cyclic change of the variables(x,y,z).The three equations can then be integrated to give3i=1m iq i2dq i3dt−q i3dq i2dt=C1(12)3i=1m iq i3dq i1dt−q i1dq i3dt=C2(13)3i=1m iq i1dq i2dt−q i2dq i1dt=C3.(14)These equations represent the conservation of angular momentum for the system.That is,they show that the angular momentum of the three particles around each of the coordinate axes is constant throughout the motion.Equation(5),(6)and(7)can be written in the formm i d2q ijdt2=−∂V∂q ij.(15)Multiplying by dq ijdt and summing gives,since V is a function of thecoordinates only,3i,j=1p ijd2q ijdt2=−dVdt.(16)This equation can then be integrated to give3 i,j=1p2ij2m i=−V+C,(17)where C is a constant of integration.Furthermore,since the left-hand side of the equation represents the kinetic energy T of the system, the integral can be put in the form T+V=C,which expresses the conservation of energy.Twofinal reductions can then be made to the order of the system. First,the time can be eliminated by using one of the dependent vari-ables as an independent variable which is used in the section treating the restricted three body problem,and,second,a reduction can be made by the so called elimination of the nodes.Thus through use of the classical integrals and these last two inte-grals,the original system of order18can be reduced to a system of order six.Furthermore,this result can be generalised to the n body problem.In this case the differential equations constitute a system of order6n.By using the same integrals this system can be reduced to a system of order(6n−12).3The Restricted Three Body ProblemIf we consider a body of unit mass(e.g.,an asteroid)moving in the field of a heavy body of mass M(e.g.,the Sun)and a much lighter body(e.g.,Jupiter)of mass m.Also assume that the heavy bodies are in circular orbit around each other with angular frequencyΩ;the effect of the light body on them is negligible.All three bodies move in the same plane.This is the restricted three body problem.When ignoring the effect of the asteroid,we can solve the two body problem to getMm M+m RΩ2=GMmR2,⇒Ω2=G(M+m)R3.(18)When we choose the center of mass of the heavy bodies as the ori-gin of a polar coordinate system.Then the position of the Sun is at (νR,π−Ωt)and Jupiter is at((1−ν)R,Ωt),where R is the distancebetween them,andν=mM+m .The distance from the asteroid to theSun isρ1(t)=r2+ν2R2+2νrR cos(θ−Ωt)(19)and to Jupiter isρ2(t)=r2+(1−ν)2R2+2(1−ν)rR cos(θ−Ωt).(20)The Lagrangian for the motion of the asteroid isL=12˙r2+12r2˙θ2+G(M+m)1−νρ1(t)+νρ2(t).(21)In this coordinate system the Lagrangian has an explicit time depen-dence:the Hamiltonian is not conserved.We change variables to χ=θ−Ωt to getL=12˙r2+12r2(˙χ+Ω)2+G(M+m)1−νr1+νr2(22)wherer1=r2+ν2R2+2νrR cosχ(23)andr2=r2+(1−ν)2R2−2(1−ν)rR cosχ(24)are now independent of time.Now the Hamiltonian in the rotating frame,H=˙r ∂L∂˙r+˙χ∂L∂˙χ−L=12˙r2+12r2˙χ2−G(M+m)r22R3+1−νr1+νr2(25)is a constant of the motion.This is called the Jacobi integral in classical literature.The Hamiltonian is of the form H=T+V where T is the kinetic energy and V is an effective potential energy:V(r,χ)=−G(M+m)r22R3+1−νr1+νr2.(26)It consist of the gravitational potential energy plus a term due to the centrifugal barrier,since we are in a rotating coordinate system.The effective potential V(r,χ)is conveniently expressed in terms of the distances to the massive bodies,V(r1,r2)=−GMr212R3+1r1+mr222R3+1r2(27)using the identity1νr21+11−νr22=1ν(1−ν)r22+R2.(28)(We have removed an irrelevant constant from the potential).Sometimes it is convenient to use cartesian coordinates,in which the lagrangian and hamiltonian areL=12˙x2+12˙y2+Ω(x˙y−y˙x)−V(x,y).(29) H=12˙x2+12˙y2+V(x,y).(30)It is obvious from the above formula for the potential as a function of r1and r2that r1=r2=R is an extremum of the potential.There are two ways this can happen:the asteroid can form an equilateral triangle with the Sun and Jupiter on either side of the line joining them.These are the Lagrange points L4and L5.These are actually maxima of the potential.In spite of this fact,they correspond to stable equilibrium points because of the effect of the velocity dependent forces.A discussion of the Lagrangian points can be found in appendix B.[2].A Ascending NodeThe ascending node is one of the orbital nodes,a point in the orbit of an object where it crosses the plane of the ecliptic from the south celestial hemisphere to the north celestial hemisphere in the direction of motion.Because of this,the ascending node of the orbit of the Earth’s moon is one of only two places where a lunar or solar eclipse can occur.The line of nodes is the intersection of the object’s orbital plane with the ecliptic,and runs between the ascending and descending nodes.Figure1:Illustration of orbital parameters.B Lagrangian PointsA location in space around a rotating two-body system(such as the Earth-Moon or Sun-Jupiter)where the pulls of the gravitating bodies combine to form a point at which a third body of negligible mass would be stationary relative to the two bodies.There arefive Lagrangian points in all,which can be seen infigure1below,three of which areFigure2:Illustration of the Lagrangian points.unstable because the slightest disturbance to any object located at one of them causes the object to drift away permanently.Until recently,this meant that the unstable Lagrangian points seemed to have no practical application for spaceflight.Now,however,they are known to have immense significance and have become the basis for chaotic control.In addition,growing numbers of spacecraft are being placed in halo orbits around the L1and L2points;station-keeping,in the form of regular thrusterfirings,are needed to maintain these orbits(which are around empty points in space!).The NASA Sun-observing probes SOHO and ACE currently orbit around L1,while future spacecraft to be placed in L2halo orbits include the Next Generation Space Telescope and the European Space Agency’s Herschel,GAIA,and Darwin spacecraft.In many ways these points are ideal for observing both near and far reaches of space since spacecraft can orbit around them far from disturbing influences,such as that of Earth’s magnetosphere.L1is well-suited to solar observations;L2offers uninterrupted observations of deep space, since the spacecraft can be oriented so that the Earth,Moon and Sun remain”behind”it at all times,and enables the entire celestial sphere to be observed over the course of one year.L3hasn’t been utilized for spaceflight because it lies on the opposide side of the Sun from Earth. The remaining two Lagrangian points,L4and L5,lie at the vertices of equilateral triangles formed with the two main gravitating masses and in their orbital plane.They are also referred to as libration points since if any objects located at them are disturbed,the objects simply wobble back and forth,or librate.References[1]A treatise on the analytical dynamics of particles and rigid bodies,4th edition,E.T Whittaker,Cambridge University Press(1937).[2]Methods of Celestial Mechanics,D.Brouwer and G.M.Clemence,Academic Press,NY(1961)[3]Poincar´e and the three body problem,June Barrow-Green,Amer-ican Mathematical Society,Printed in USA(1997).[4]/encyclopedia/L/Lagpoint.html[5]Dynamical systems,G.D.Birkhoff,American mathematical society1927.。
PDLAMMPS近场动力学
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芯片测试仪设备操作指南说明书
MechanicsOscillationsElliptical Oscillation of a String PendulumDESCRIPTION OF ELLIPTICAL OSCILLATIONS OF A STRING PENDULUM AS THE SU-PERIMPOSITION OF TWO COMPONENTS PERPENDICULAR TO ONE ANOTHER.UE1050121 06/15 MEC/UDFig. 1: Experiment set-upGENERAL PRINCIPLESDepending on the initial conditions, a suitable suspended string pendulum will oscillate in such a way that the bob’s motion describes an ellipse for small pendulum deflections. If the motion is resolved into two perpendicu-lar components, there will be a phase difference between those components.This experiment will investigate the relationship by measuring the oscillations with the help of two perpendicularly mounted dynamic force sensors. The amplitude of the components and their phase difference will then be evaluated. The phase shift between the oscillations will be shown directly by displaying the oscillations on a dual-channel oscilloscope.Three special cases shed light on the situation:a) If the pendulum swings along the line bisecting the two force sensors, the phase shift φ = 0°.b) If the pendulum swings along a line perpendicular to that bisecting the two force sensors, the phase shift φ = 180°.c) If the pendulum bob moves in a circle, the phase shift φ =oscillation directions of the string pendulum under in-vestigationLIST OF EQUIPMENT1 SW String Pendulum Set 1012854 (U61025)1 SW Stand Equipment Set 1012849 (U61022)1 SW Sensors Set @230 V 1012850 (U61023-230) or1 SW Sensors Set (@115 V 1012851 (U61023-115) 1 USB Oscilloscope 2x50 MHz 1017264 (U112491)SET-UP∙Screw the stand rods with both external and internal threads into the outer threaded sockets of the base plate. ∙Extend both rods by screwing rods with external thread only onto the ends of them.∙Attach double clamps near the top of both stand rods and turn them to point inwards so that the slots are vertical and facing one another.∙Attach both springs from the spring module to the lugs on the cross bar (angled side).∙Hang the large loop of string from the lug on the flat side.Fig. 3 Assembly of spring module∙Connect the springs and vector plate to the hook of a dynamic force sensor with a small loop of string and care-fully pull everything taut.∙Attach the force sensor with the screw tightened by hand. ∙Attach the second force sensor in the same way.Fig. 4 Attachment of dynamic force sensors to spring module∙Pull the string through the eyelet of the spring module (in the middle of the metal disc).∙Thread the end of the string through the two holes of the length adjustment slider.Fig. 5 Set up of string3B Scientific GmbH, Rudorffweg 8, 21031 Hamburg, Germany, ∙Clamp the cross bar into the slots of the two double clamps, suspend a weight from the end of the string and set up the height of the pendulum using the length ad-justment slider.Fig. 6 Attachment of cross bar in double clamp ∙ Connect the force sensors to the inputs for channels A and B of the MEC amplifier board.∙ Connect outputs A and B of the MEC control unit to channels CH1 and CH2 of the oscilloscope.EXPERIMENT PROCEDURE∙Set the oscilloscope time base time/div to 1 s, select a vertical deflection for channels CH1 and CH2 of 50 mV DC and set the trigger to “Edge” mode, “Normal” sweep, “Source CH1” and “Slope +”.∙Slightly deflect the string pendulum and allow it to oscil-late in a plane which bisects the alignment of the two force sensors (oscillation path a in Fig. 2). Observe the oscilloscope trace and save it.∙Slightly deflect the string pendulum and allow it to oscil-late in a plane which is perpendicular to the one which bi-sects the two force sensors (oscillation path b in Fig. 2). Observe the oscilloscope trace and save it.∙Slightly deflect the string pendulum and allow it to oscil-late in a circle (oscillation path c in Fig. 2). Observe the oscilloscope trace and save it.SAMPLE MEASUREMENT AND EVALUA-TIONWhen the pendulum is oscillating in the plane of the bisecting angle between the sensors, the two sensors will experience symmetric loading (oscillation path a in Fig. 2). The signals from the two force sensors will be in phase, i.e. the phase shift between them will be φ= 0° (Fig. 7).Fig. 7: Oscillation components for a string pendulum swingingalong the line bisecting the two force sensorsWhen the pendulum is oscillating in the plane perpendicular to the bisecting angle between the sensors, the two sensors will experience asymmetric loading (oscillation path b in Fig. 2). The signals from the two force sensors will be wholly out of phase, i.e. the phase shift between them will be φ= 180° (Fig. 8).Fig. 8: Oscillation components for a string pendulum swingingalong the line perpendicular to that bisecting the two force sensorsThe circular oscillation is a superimposition of the oscillations along the plane of the bisecting angle between the sensors and the angle perpendicular to it with a phase shift of φ = 90°(Fig. 9).Fig. 9: Oscillation components for a string pendulum describ-ing a circle。
Stable Distributions
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Contents
I Univariate Stable Distributions
1 Basic Properties of Univariate Stable Distributions 1.1 Definition of stable . . . . . . . . . . . . . . . 1.2 Other definitions of stable . . . . . . . . . . . . 1.3 Parameterizations of stable laws . . . . . . . . 1.4 Densities and distribution functions . . . . . . 1.5 Tail probabilities, moments and quantiles . . . 1.6 Sums of stable random variables . . . . . . . . 1.7 Simulation . . . . . . . . . . . . . . . . . . . . 1.8 Generalized Central Limit Theorem . . . . . . 1.9 Problems . . . . . . . . . . . . . . . . . . . . 2 Modeling with Stable Distributions 2.1 Lighthouse problem . . . . . . . . . 2.2 Distribution of masses in space . . . 2.3 Random walks . . . . . . . . . . . 2.4 Hitting time for Brownian motion . 2.5 Fractional diffusions . . . . . . . . 2.6 Economic applications . . . . . . . 2.6.1 Stock prices . . . . . . . . . 2.6.2 Foreign exchange rates . . . 2.6.3 Value at risk . . . . . . . . . 2.6.4 Other economic applications 2.6.5 Multiple assets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
不要情感类的英语单词
不要情感类的英语单词Certainly! Here's a list of English words that are notrelated to emotions:1. Astronomy - The scientific study of celestial objects, space, and the physical universe as a whole.2. Biosphere - The global sum of all ecosystems.3. Calculus - A branch of mathematics that deals with limits, functions, derivatives, integrals, and infinite series.4. Dendrochronology - The scientific method of dating tree rings to date the age of a tree or the time at which it was felled.5. Ecosystem - A community of living organisms in conjunction with the non-living components of their environment,interacting as a system.6. Fjord - A long, narrow inlet with steep sides or cliffs, created by a glacier.7. Geology - The study of the solid Earth, the rocks of which it is composed, and the processes by which they change over time.8. Hydrosphere - The total amount of water on Earth's surface.9. Ionosphere - The upper region of the atmosphere, extending from about 60 kilometers to several hundred kilometers above the Earth, where ions and free electrons are present.10. Junction - A place where two or more things (such as roads, railroads, or rivers) meet or cross.11. Kilowatt - A unit of power equal to 1,000 watts.12. Lithosphere - The rigid outer layer of the Earth,consisting of the crust and the upper part of the mantle.13. Magnetosphere - The region of space surrounding andbonded to a celestial body, like a planet, through which its magnetic field lines are frozen into the plasma.14. Nanotechnology - The manipulation of matter on an atomic, molecular, and supramolecular scale.15. Ozone - A highly reactive form of oxygen with thechemical formula O₃.16. Photosynthesis - The process by which green plants and some other organisms use sunlight to synthesize foods withthe help of chlorophyll pigments.17. Quantum - A discrete amount or quantity, as of energy or charge, as prescribed by the laws governing a physical system.18. Relativity - A theory, primarily formulated by Albert Einstein, of the relations between space and time.19. Stratosphere - The second layer of the Earth's atmosphere, just above the troposphere and below the mesosphere.20. Thermodynamics - The study of the relationships between heat and other forms of energy.These words cover a range of scientific and technicalsubjects and are devoid of emotional connotations.。
meep简介
Meep(或MEEP)是一个免费的有限差分时域(FDTD)模拟软件包在麻省理工学院开发的模型电磁系统,伴随着我们MPB的本征模包。
Its features include:它的功能包括:Meep Meep发行说明Introduction简介Installation安装Tutorial教程Reference参考C++Tutorial C++教程C++Reference C++参考Acknowledgements致谢License and Copyright许可和版权•Free software under the GNU GPL.根据GNU GPL的免费软件。
•Simulation in1d,2d,3d,and cylindrical coordinates.模拟一维,二维,三维,圆柱坐标。
•Distributed memory parallelism on any system supporting the MPI standard.在任何支持的系统内存并行分布式的MPI标准。
Portable to any Unix-like system(GNU/Linux is fine).移植到任何类Unix系统(GNU/Linux的是罚款)。
•Arbitrary anisotropic electric permittivityεand magnetic permeabilityμ,along with dispersiveε(ω)andμ(ω)(including loss/gain)and nonlinear(Kerr&Pockels)dielectric and magnetic materials,and electric/magnetic conductivitiesσ.任意各向异性介电常数ε和磁导率μ,随着分散ε(ω)和μ(ω)(包括损耗/增益)和非线性(克尔电光)介电和磁材料,电/磁导率σ。
离散的情况凸多面体
1 离散的情况:凸多面体Minkowski 指出一个凸多面体完全可以用它的面的面积和方向来表示。
而我们可以通过球面上的质点来方便的表示面的面积和方向。
该球面称为高斯球面。
令每个点的质量等于相应面的面积,来获取一个凸多面体的EGI 。
如图所示:2.1 EGI 的性质EGI 的质量等于相应多面体的表面积和。
如果多面体是闭合的,任何一对相反的方向可以得到相同的投影面积。
这使得我们可以计算出EGI 质心的位置。
当我们从远处看一个多面体时,会感到比它的实际大小要小。
假设视觉方向的方向向量为v ,单位法向量si ,面的实际面积大小O i ,则有视觉面积为(si ·v )O i ,该视觉面以v 为法向量。
所有视觉面的表面积和为A(v )={|0}()i si v si v Oi ≥∑,当从相反的方向看该多面体时,视觉面的表面积和为A(-v )={|0}()i si v si v Oi ≥∑,显然A(v )=A(-v ),因此()0alli alli si v Oi siOi v ⎡⎤⋅=⋅=⎢⎥⎣⎦∑∑,故0a l i s i O i =∑。
所以EGI 的质心在原点( at the origin)上。
2.2 建立四面体具有公共边的面称为相邻面。
但是两个相邻面映射到高斯球面上的质点并不是相邻的关系。
这就很难从EGI 中重新建立一个四面体,因为无法确定面的相邻关系。
但可以找到每个面相对于质点的偏移量。
一个四面体的结构比较简单,每一个面都和其他面相邻。
四面体的形状可以由四个面的法向量表示,大小是个未知量。
设四面体的每个面的法向量分别为,,,a b c d ,面积分别为A ,B ,C ,D ,求质心到每个面的距离a,b,c,d ,进而可以求出四个顶点A,B,C,D 的位置,确定四面体的大小。
性质:一个三角形的中心到一条边的距离等于该边所对的顶点到这条边距离的三分之一(以证明过)。
类似的,在一个四面体中,质心到四个面的距离等于相应面的对顶点到该面的距离的四分之一。
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Keywords: discrete time optimal control, discrete time calculus of variations, discrete time mechanics, discrete time Pontryagin extremals, quasi-invariance up to difference gauge terms, discrete version of Noether’s theorem. Mathematics Subject Classification 2000: 49-99, 39A12.
arXiv:math/0301276v1 [math.OC] 24 Jan 2003
Integrals of Motion for Discrete-Time Optimal Control Problems∗
Delfim F. M. Torres delfim@mat.ua.pt Department of Mathematics University of Aveiro 3810-193 Aveiro, Portugal http://www.mat.ua.pt/delfim
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Discrete-Time Optimal Control
Without loss of generality (cf. [24, §2]), we consider the discrete optimal control problem in Lagrange form. The time k is a discrete variable: k ∈ Z. The horizon consists of N periods, k = M, M + 1, . . . , M + N − 1, where M and N are fixed
1
along the extremals, are obtained. The result is, however, much more than a theorem. It is an universal principle, which can be formalized in a precise statement, as a theorem, on very different contexts and, for each such context, under very different assumptions. Let us consider, for example, classical mechanics or, more generally, the calculus of variations. Typically, Noether transformations are considered to be point-transformations (they are considered to be functions of coordinates and time), but one can consider more general transformations depending also on velocities and higher derivatives [10] or within the broader context of dynamical symmetries [20]. For an example of an integral of motion which comes from an invariance transformation depending on velocities, see [23]. In most formulations of Noether’s principle, the Noether transformations keep the integral functional invariant (cf. e.g. [15, §1.5]). It is possible, however, to consider transformations of the problem up to an exact differential (cf. e.g. [31, p. 73]), called a gauge-term [33]. Once strictly-invariance of the integral functional is no more imposed, one can think considering additional terms in the variation of the Lagrangian – see the quasi-invariance and semi-invariance notions introduced by the author respectively in [38] and [39]. Formulations of Noether’s principle are possible for problems of the calculus of variations: on Euclidean spaces (cf. e.g. [18]) or on manifolds (cf. e.g. [19]); with single or multiple integrals (cf. e.g. [5]); with higher-order derivatives (cf. e.g. [1]); with holonomic or nonholonomic constraints (cf. e.g. [43, Ch. 7], [35]); and so on. Other contexts for which Noether’s theorems are available include supermechanics [8], control systems [42, 25], and optimal control (see e.g. [9, 6, 37, 40]). For a survey see [36, 41]. Here wermulation of the Noether’s principle in the discrete time setting. For a description of discrete time mechanics, discrete time calculus of variations, and discrete optimal control see, e.g., [13, 14, 28, 29], [7], and [12]. Illustrative examples of real-life problems which can be modeled in such framework can be found in [34, Ch. 8]. Versions of the Noether’s principle for the discrete calculus of variations, and applicable to discrete analogues of classical mechanics, appeared earlier in [16, 17, 21, 3, 11, 44], motivated by the advances of numerical and computational methods. There, the discrete analog of Noether’s theorem is obtained from the discrete analog of the Euler-Lagrange equations. To the best of our knowledge, no Noether type theorem is available for the discrete time optimal control setting. One such formulation is our concern here. The result is obtained from the discrete time version of the Pontryagin maximum principle. As corollaries, we obtain generalizations of the previous results for first-order and higher-order discrete problems of the calculus of variations which are quasi-invariant and not necessarily invariant.
∗ Accepted to be presented at the IFAC Workshop on Control Applications of Optimization – CAO’2003 – to be held in Visegr´ ad, Hungary, 30 June – 2 July 2003, and to appear in the respective conference proceedings. The date of this version is January 24, 2003.
Abstract We obtain a discrete time analog of E. Noether’s theorem in Optimal Control, asserting that integrals of motion associated to the discrete time Pontryagin Maximum Principle can be computed from the quasiinvariance properties of the discrete time Lagrangian and discrete time control system. As corollaries, results for first-order and higher-order discrete problems of the calculus of variations are obtained.