第十五屆華為杯中國研究生數(shù)學(xué)建模競(jìng)題a題參考文獻(xiàn)2

1、c 2016 Society for Industrial and Applied MathematicsSIAM J. APPLIED DYNAMICAL SYSTEMS Vol. 15, No. 4, pp. 18061822Twisting SomersaultHolger R. Dullin and William TongAbstract.We give a dynamical system analysis of the twisting somersaults using a reduction to a time-dependent Euler equation for non

2、rigid body dynamics.The central idea is that after reductionthe twisting motion is apparent in a body frame, while the somersaulting (rotation about the fixed angular momentum vector in space) is recovered by a combination of dynamic phase and geometricphase. In the simplest “kick-model” the number

3、of somersaults m and the number of twists n are obtained through a rational rotation number W = m/n of a (rigid) Euler top. Using the full model with shape changes that take a realistic time we then derive the master twisting-somersault formula: an exact formula that relates the airborne time of the

4、 diver, the time spent in various stages of the dive, the numbers m and n, the energy in the stages, and the angular momentum by extending a geometric phase formula due to Cabrera J. Geom. Phys., 57 (2007), pp. 14051420. Numerical simulations for various dives agree perfectly with this formula where

5、 realistic parameters are taken from actual observations.Key words. nonrigid body dynamics, geometric phase, biomechanics, rotation numberAMS subject classifications. 70E55, 70E17, 37J35, 92C10DOI. 10.1137/15M10550971. Introduction. One of the most beautiful Olympic sports is springboard and platfor

6、m diving, where a typical dive consists of a number of somersaults and twists performed in a variety of forms. The athlete generates angular momentum at take-off and achieves the desired dive by executing shape changes while airborne. From a mathematical point of view the simpler class of dives are

7、those for which the rotation axis and hence the direction of angular velocity remain constant and only the values of the principal moments of inertia are changed by the shape change, but not the principal axis. This is typical in dives with somersaults in a tight tuck position with minimal moments o

8、f inertia. The mathematically much more interesting dives include a shape change that moves the principal axis and hence generates a motion in which the rotation axis is not constant. This is typical in twisting somersaults, the object of this paper. We are using modern tools from dynamical systems,

9、 in particular from geometric mechanics, to understand this situation. Our findings apply to coupled rigid body dynamics in general, e.g., in other sports like aerial skiing, or to spacecraft attitude control.The first correct description of the physics of the twisting somersault was given by Frohli

10、ch 5. Frohlich writes with regards to some publication from the 60s and 70s that “several booksReceived by the editors January 4, 2016; accepted for publication (in revised form) by C. Wulff July 27, 2016; published electronically October 4, 2016./journals/siads/15-4/M105509.htmlFu

11、nding: This research was supported by ARC Linkage grant LP100200245 and the New South Wales Institute of Sports.School of Mathematics and Statistics, The University of Sydney, Sydney, NSW 2006, Australia (holger.dullin .au, .au).1806Copyright by SIAM. Unauthorized rep

12、roduction of this article is prohibited.Downloaded 10/22/16 to 8. Redistribution subject to SIAM license or copyright; see /journals/ojsa.phpTWISTING SOMERSAULT1807written by or for coaches discuss somersaulting and twisting, and exhibit varying degrees of insight and/or

13、 confusion about the physics processes that occur.” A full fledged analysis has been developed by Yeadon in a series of classical papers 17, 18, 19, 20. Frohlich was the first to point out the importance of shape change for generating rotations even in the absence of angular momentum. Our analysis r

14、eveals how exactly a shape change generates a change in rotation in the presence of angular momentum. From a modern point of view this is a question raised in the seminal papers by Shapere and Wilczek 13, 14: “What is the most efficient way for a body to change its orientation?” Our answer involves

15、the generalization of geometric phase in rigid body dynamics 10 to shape-changing bodies recently obtained in 3.To be able to apply these ideas in our context we first derive a version of the Euler equation for a shape-changing body.Such equations have been obtained in principle in, e.g., 11, 6, 7,

16、4. Our form of the equations is particularly simple, and we derive the explicit form of the important time-dependent terms for a system of coupled rigid bodies. By writing the equations in a particular frame we find beauty and simplicity in the equations of motion (see Theorems 1 and 2). We then tak

17、e a simple particular system of just two coupled rigid bodies (the “one-armed diver”) and show how a twisting somersault can be achieved with this model. An even simpler model is the diver with a rotor analyzed in 2, in which all the stages of the dive can be analytically solved for. In the present

18、paper we use an analytically solvable approximation in which the shape change is instantaneous. In this kick-model the dynamics is described by a piecewise smooth dynamical system and we use an extension of Montgomerys geometric phase formula 10 for the reconstruction; see Theorem 6. A similar exten

19、sion for a different rotation angle was first obtained by Bates, Cushman, and Savev 1. Throughout the paper we emphasize the geometric mechanics point of view. Hence the translational and rotational symmetry of the problem is reduced, and thus Euler-type equa- tions are found in a co-moving frame. I

20、n this reduced description the amount of somersault (i.e., the amount of rotation about the fixed angular momentum vector in space) is not present. Reconstruction allows us to recover this angle by solving an additional differential equation driven by the solution of the reduced equations. It turns

21、out that for a closed loop in shape space the somersault angle can be recovered by a geometric phase formula due to 3; seeTheorem 9.The structure of the paper is as follows. In section 2 we derive the equations of motion for a system of coupled rigid bodies that is changing shape. The resulting Eule

22、r-type equations are the basis for the following analysis. In section 3 we discuss a simplified kick-model, in which the shape change is impulsive. The kick changes the trajectory and the energy, but not the total angular momentum. In section 4 the full model is analyzed, without the kick assumption

23、. Unlike the previous section, here we have to resort to numerics to compute some of the terms. But we show that using the generalized geometric phase formula due to Cabrera3 gives an exact description and a beautiful geometric interpretation of the mechanics behind the twisting somersault.2. Euler

24、equations for coupled rigid bodies. Let l be the constant angular momentum vector in a space fixed frame. Rigid body dynamics usually use a body-frame because in that frame the tensor of inertia is constant. The change from one coordinate system to the other isCopyright by SIAM. Unauthorized reprodu

25、ction of this article is prohibited.Downloaded 10/22/16 to 8. Redistribution subject to SIAM license or copyright; see /journals/ojsa.php1808HOLGER R. DULLIN AND WILLIAM TONGgiven by a rotation matrix R = R(t) SO(3) such that l = RL. In the body frame the vector L is des

26、cribed as a moving vector and only its length remains constant since R SO(3). The angular velocity in the body frame is the vector such that v = RtR v for any vector v R3. Even though for a system of coupled rigid bodies the tensor of inertia is generallynot a constant, a body frame still gives the

27、simplest equations of motion.Theorem 1. The equations of motion for a shape-changing body with angular momentum vector L R3 in a body frame areL(1)= L ,where the angular velocity R3 is given by(2) = I1(L A),I = I(t) is the tensor of inertia, and A = A(t) is a “momentum shift” (or “shape momen- tum”)

28、 generated by the shape change.Proof. The basic assumption is that the shape change is such that the angular momentum is constant. Let l be the vector of angular momentum in the space fixed frame; then l = RL.Taking the time derivative gives 0 = R L + RL and hence L = RtR L = L = L . The interesting

29、 dynamics is all hidden in the relation between and L.Let q = RQ, where Q is the position of a point in the body B in the body frame, and q is the corresponding point in the space fixed frame. The relation between L and is obtainedfrom the definition of angular momentum which is q q integrated over

30、the body B. The relation q = RQ is for a rigid body; for a deforming body we label each point by Q in the body frame but allow for an additional shape change S, so that q = RSQ. We assume thatS : R3 R3 is volume preserving, which means that the determinant of the Jacobian matrix of S is 1. The defor

31、mation S need not be linear, but we assume that we are in a frame inwhich the center of mass is fixed at the origin, so that S fixes that point. Nowq = R SQ + RSQ + RSQ = RRtR SQ + RSQ = R( SQ) + RSS1SQ= R( Q + SS1Q ),(3)where Q = SQ. Thus we haveq q = RQ R( Q + SS1Q )= R(|Q |21 Q Q t) + R(Q SS1Q )

32、.(4)B= (Q ), so thatNowR l is defined by integrating over the deformed bodywith density l =q q dQ and using l = RL givesZZL =(|Q |21 Q Q t) dQ +Q SS1Q dQ .BB(5)The first term is the tensor of inertia I of the shape changed body, and the constant term defines the shape momentum A so that(6)as claimed

33、.L = I + A,Copyright by SIAM. Unauthorized reproduction of this article is prohibited.Downloaded 10/22/16 to 8. Redistribution subject to SIAM license or copyright; see /journals/ojsa.phpTWISTING SOMERSAULT1809Remark 1.1. Explicit formulas for I and A in the case of a sy

34、stem of coupled rigid bodies are given in the next theorem. When I is constant and A = 0 the equations reduce to the classical Euler equations for a rigid body.Remark 1.2. For arbitrary time dependence of I and A the total angular momentum |L|is conserved; in fact, it is a Casimir of the Poisson str

35、ucture f, g = .Remark 1.3. The equations are Hamiltonian with respect to this Poisson structure with11Hamiltonian H = (L A)I(L A) such that = H/L.2For a system of coupled rigid bodies the shape change S is given by rotations of the individual segments relative to some reference segment, typically th

36、e trunk. The orientation of the reference segment is given by the rotation matrix R so that l = RL. The system of rigid bodies is described by a tree that describes the connectivity of the bodies; see the thesis of Tong 15 for the details.Denote by C the overall center of mass, and by Ci the positio

37、n of the center of mass of body Bi relative to C. Each bodys mass is denoted by mi, and its orientation by Ri , where i denotes the set of angles necessary to describe its relative orientation (e.g., a single angle for a pin joint, or three angles for a ball and socket joint). All orientations are m

38、easured relative to the reference segment, so that the orientation of Bi in the space fixed frame is given byRRi . The angular velocity i is the relative angular velocity corresponding to Ri , so that the angular velocity of Bi in the space fixed frame is Rt i + i . Finally, Ii is the tensor of iner

39、tia of Bi in a local frame with center at Ci and coordinate axes aligned with the principle axes of inertia. With this notation we have the following.Theorem 2. For a system of coupled rigid bodies we haveXI =R IiRt2t(7)(| 1 C+ mC )iCiiiiiandXA =(mC C + R I ),(8)i ii iiiwhere mi is the mass, Ci the

40、position of the center of mass, Ri the relative orientation, iRtiR i v= i v for all v R3, and Ii the tensor ofthe relative angular velocity such thatinertia of body Bi. The sum is over all bodies Bi including the reference segment, for which the rotation is simply given by 1.Proof. The basic transfo

41、rmation law for body Bi in the tree of coupled rigid bodies is qi = RRi (Ci + Qi). Repeating the calculation in the proof of Theorem 1 with this particular S and summing over the bodies gives the result. We will skip the derivation of Ci in terms of the shape change and the geometry of the model and

42、 refer the reader to 15 for the de- tails.Remark 2.1. In the formula for I the first term is the moment of inertia of the segment transformed to the frame of the reference segment, while the second term comes from the parallel axis theorem (see, e.g., 8), applied to the center of mass of the segment

43、 relative to the overall center of mass.Copyright by SIAM. Unauthorized reproduction of this article is prohibited.Downloaded 10/22/16 to 8. Redistribution subject to SIAM license or copyright; see /journals/ojsa.php1810HOLGER R. DULLIN AND WILLIAM TONGRemark 2.2. In the

44、 formula for A the first term is the internal angular momentum gen- erated by the change of the relative center of mass, while the second term originates from therelative angular velocity.Remark 2.3. When there is no shape change, then C i = 0 and i = 0, and hence A = 0.Remark 2.4. The vectors Ci, C

45、 i, and i are determined by the set of time-dependent matrices Ri (the time-dependent “shape”) and the joint positions of the coupled rigid bodiePs (the time-independent “geometry” of the model); see 15 for the details. In particular,also mi Ci = 0.In order to describe and numerically compute the ro

46、tation matrix R that determines the position of the body in space, we use quaternions. This is convenient because unlike Euler angles the description in quaternions is free of singularities. Specifically, we write Rx = qxq, where the quaternion q is q = q0 + q1i + q2j + q3k and the vector x on the l

47、eft-hand side and the pure quaternion x = x1i + x2j + x3k on the right-hand side; see, e.g., 15 for more details.3. A simple model for twisting somersault. Instead of the full complexity of a realistic coupled rigid body model for the human body, e.g., with 11 segments 16 or more, here we are going

48、to show that even when all but one arm is kept fixed it is still possible to do a twisting somersault. The formulas we derive are completely general, so that more complicated shape changes can be studied in the same framework. But in order to explain the essential ingredients of the twisting somersa

49、ult we choose to discuss a simple example. A typical dive consists of a number of phases or stages in which the body shape is either fixed or not. Again, it is not necessary to make this distinction; the equations of motion are general, and one could study dives where the shape is changing throughou

50、t. However, the assumption of rigid body motions for certain times is satisfied to a good approximation in reality and makes the analysis simpler and more explicit. The stages where shape change occurs are relatively short, and considerable time is spent in rotation with a fixed shape. This observat

51、ion motivates our first approximate model, in which the shape changes are assumed to be impulsive. Hence we have instantaneous transitions between solution curves of rigid bodies with different tensors of inertia and different energy, but the same angular momentum. A simple twisting somersault hence

52、 looks like this: The motion starts out as a steady rotation about a principal axis resulting in pure somersault (stage 1), and typically this is about the axis of the middle principle moment of inertia which has unstable equilibrium. After some time a shape change occurs; in our case one arm goes d

53、own (stage 2). This makes the body asymmetric and generates some tilt between the new principal axis and the constant angular momentum vector. As a result the body starts twisting with constant shape (stage 3) until another shape change (stage 4) stops the twist, for which the body then resumes pure

54、 somersaulting motion (stage 5) until head first entry in the water. The amount of time spent in each of the five stages is denoted by i, where i = 1, . . . , 5. We also use the subscripts s for the somersaulting stages 1 and 5, and the subscript t for the twisting phase 3.11 l2Is,yy .1The energy fo

55、r pure somersault in stages 1 and 5 is E= L IL =In the kick-ssss22model, stages 2 and 4 do not take up any time, so 2 = 4 = 0, but they do change the energyand the tensor of inertia. On the momentum sphere |L|2 = l2 the dive thus appears like this; see Figure 1: For some time the trajectory of L on

56、the L-sphere remains at the equilibriumCopyright by SIAM. Unauthorized reproduction of this article is prohibited.Downloaded 10/22/16 to 8. Redistribution subject to SIAM license or copyright; see /journals/ojsa.phpTWISTING SOMERSAULT1811Figure 1. Twisting somersault on

57、the sphere |L| = l where the shape change is a kick. The region A boundedbythestage 3 orbit of L and the equator (dashed) is shaded in dark blue.point Lx = Lz = 0; then it is kicked into a periodic orbit with larger energy describing the twisting motion. Depending on the total available time a number of full (or half) revolutions are done on this orbit, until another kick brings the solution back to the unstable equilibrium point where it started (or on the opposite end with negative Ly). Finally, some time is sp