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# BIODYNAMICS EGM 4590

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This 12 page Class Notes was uploaded by Rollin Mann DVM on Saturday September 19, 2015. The Class Notes belongs to EGM 4590 at University of Florida taught by Benjamin Fregly in Fall. Since its upload, it has received 15 views. For similar materials see /class/207067/egm-4590-university-of-florida in Engineering & Applied Science at University of Florida.

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Date Created: 09/19/15

THREEDIMENSIONAL ROTATIONS While working for a biodynamics company you are given the opportunity to demonstrate your knowledge of Autolev Matlab motion analysis human anatomy intermediate reference frames and threedimensional 3D rotations An orthopedic surgeon with whom you work wants to know a patient s left hip exion extension internalexternal rotation and abductionadduction during gait movement recorded with a markerbased motion analysis system A schematic of the le hip region is shown below along with the motion capture marker locations and recommended body segment coordinate systems pelvisy sacral Lasis I asis I g pe VSZ 1 3 r i u pelvisix I 4amp7 r Pelvis Femur lhighz Y superlor mth 7 lhighifranl Ihlghrear Z lateral 4 Lab la X anterior For this assignment your goal is to create the pelvis and thigh segment coordinate systems de ne a 3D rotation matrix between the pelvis and thigh and use inverse trigonometric calculations to compute two distinct sets of three Euler angles from the rotation matrix Before proceeding to the following pages you will need to download and save the 3D rotations homework les from the course web page To complete the assignment you need to perform the series of steps indicated below Body 3 1 2 3 Rotation Matrix 1 Using Autolev or by hand derive the rotation matrix resulting from a series of three simple rotations q1 q2 and q3 about body xed axes using the rotation sequence 123 2 Record the rotation matrix below feel free to use shorthand such as cosq1 C or sinq2 32 R 2 Body 3 17273 Body 3 3 1 2 Rotation Matrix 1 Using Autolev or by hand derive the rotation matrix resulting from a series of three simple rotations q1 q2 and q3 about body xed axes using the rotation sequence 312 2 Record the rotation matrix below feel free to use shorthand such as cosq1 C or sinq2 32 R 2 Body 3 37172 Matlab and Data Files Place the Matlab file Rotation573Dm and the six input data files riasis txt liasis txt sacral txt thighiupper txt thighifront txt and thighirear txt in your Matlab working directory Pelvis Segment Coordinate System In the appropriate section of ROtatiOl lSiBD m define the pelvis segment coordinate system for the current time frame f 1 Define a vector pe lvi sivl from the 17a 5 i s marker to the ria s i s marker 2 Normalize the vector to define the pelvis zaxis unit vector pelvi siz BJ Fregly EGM 4590 E 8094 9 50 Compute the vector cross product pelvisiv2 of one vector from the sacral marker to the rias i s marker and a second vector from the s acral marker to the 17a 5 i s marker Normalize the vector to de ne the pelvis yaxis unit vector pelvi 87y Compute the vector cross product pe lvi s7v3 of pe lvi siy and pe lvi s72 Normalize the vector to de ne the pelvis xaxis unit vector pelvi six De ne the coordinate system origin pelvi siorigin as the midpoint between the ria s i s marker and the 17a 5 i s marker De ne the rotation matrix rotimatifromil abitoipe lvi s describing the orientation of the pelvis segment relative to the laboratory coordinate system Compute the transposed rotation matrix rotimatifromipe lvi sitoil ab describing the orientation of the laboratory relative to the pelvis segment coordinate system Thigh Segment Coordinate System In the appropriate section of ROtatiOl lSiBD m de ne the thigh segment coordinate system for the current time frame f l 9 8094 De ne a vector thighivl from the thighirear marker to the thighifront marker Normalize the vector to de ne the thigh xaxis unit vector thi ghix Compute the vector cross product thi gh7v2 of one vector from the thi ghiuppe r marker to the thi ghire a r marker and a second vector from the thighiupper marker to the thighifront marker Normalize the vector to de ne the thigh zaxis unit vector thighiz Compute the vector cross product thi gh7v3 of thi ghiz and thi ghix Normalize the vector to de ne the thigh yaxis unit vector thighiy De ne the coordinate system origin thighiorigin as the midpoint between the thighifront marker and the thighirear marker De ne the rotation matrix rotimatifromil abitoithi gh describing the orientation of the thigh segment relative to the laboratory coordinate system Rotation Matrix In the appropriate section of Rotations73Dm compute the rotation matrix rotimatifromipelvisitoithigh describing the orientation of the thigh segment relative to the pelvis segment coordinate system for the current time frame f Angles from Rotation Matrix BJ Fregly EGM 4590 1 In the appropriate section of Rotations73Dm assuming a Body 3 123 rotation sequence determine the hip joint angles from the rotation matrix using inverse trigonometric calculations for the current time frame f E Fquot 0 Compute the hip abductionadduction angle abiadductionil 2 3 or q about the 1aXis Compute the hip intemaleXtemal rotation angle i niext e rna 171 2 3 or qg about the 2 aXis Compute the hip eXionextension angle fl exiext ens i onil 2 3 or q3 about the 3aXis 2 In the appropriate section of Rotations73Dm assuming a Body 3 312 rotation sequence determine the hip joint angles from the rotation matrix using inverse trigonometric calculations for the current time frame f E Fquot 0 Compute the hip eXionextension angle fl exiext ens i on73 l 2 or q about the 3aXis Compute the hip abductionadduction angle abiadductioni3l 2 or qg about the 1aXis Compute the hip intemaleXtemal rotation angle i niexternali3l 2 or q3 about the 2 aXis Generating Results 1 After saving your work run the Matlab file Rotations73Dm by entering the file name at the Matlab command prompt 2 Following the animation of the markers within the laboratory coordinate system Matlab will automatically generate three plots that you must print and include with your assignment Discussion Considering your goal of quantifying a patient s hip joint movement which rotation sequence ie 123 or 312 would you choose Why BJ Fregly EGM 4590 BIODYNAMICS PRIMER DEFINITIONS Kinematics The study of motion without regard to the forces that produce it Thus knowledge of mass and inertia is not required Dynamics The study of motion with regard to the forces that produce it Thus knowledge of mass and inertia is essential NEWTON39S LAWS 1 2 F 0 A body will remain in a state of rest or uniform motion unless a force is applied to change it 2 2 F ma Force is required to produce a change in acceleration not velocity where the acceleration must always be calculated with respect to a xed ie inertial or Newtonian reference frame The Earth provides such a reference frame for objects moving close to the surface of the Earth 3 FA 8 FBA Law of action and reaction interaction forces are equal and opposite TYPES OF MODELS Lumped Mass Models all mass is lumped at the mass center of each body and no mass distribution is considered in the analysis ie all moments and products of inertia are assumed to be zero Distributed Mass Models in addition to a lumped mass located at the mass center the distribution of mass throughout the body is accounted for by moments and products of inertia ROTATION MATRICES These matrices define the relationship between unit vectors xed in one reference frame and those xed in another To make things as simple as possible unit vectors xed in each reference frame should be defined such that all unit vectors from every reference frame align when all of the rotations are zero For planar problems the relationships between the various unit vectors can then be determined by taking dot products or by trigonometry AN GULAR VELOCITY Definition A A A db2 db3 db1 b b b b b dt 3 2 dt 1 3 dt 2 BJ Fregly March 25 2001 where Adbl dl i 1 2 3 can be determined by using the rotation matrix that relates the b1 unit vectors to the a unit vectors This method of calculating A DB is rarely necessary in practice The Million Dollar Formula A B ds ds A DB gtlts dt dt This equation turns a vector derivative into a cross product It allows you to take the rst time derivative of a vector s in a reference frame other than the one in which the vector is expressed Specifically it is useful when you want to calculate Adsd1 but you have s and A DB expressed in terms of b rather than a unit vectors In essence this relationship allows you to avoid reexpressing s in terms of a unit vectors in order to calculate the desired derivative in reference frame A Simple Angular Velocity A033 aal abl When rigid bodies A and B share a common unit vector during some specified time period then A033 can be found immediately from the above relationship without resorting to the de nition of angular velocity given above This situation is commonly encountered when solving dynamics problems Addition Theorem for Angular Velocities A mBA 011 41 DA Aw71 031 A39n DE The individual angular velocities A39 00A are usually but need not be simple angular velocities This equation is particularly useful when we know absolute angular velocities ie measured with respect to the inertial reference frame and we want to calculate relative angular velocities ie the angular velocity of one rigid body measured with respect to another AN GULAR ACCELERATION Definition AaB AdAmB BdAmB dt dt The nice thing about calculating AaB is that the time derivative can be performed in either reference frame A or B the proof involves the use of the Million Dollar Formula No addition theorem for angular accelerations exists similar to that for angular velocities BJ Fregly April 2 2003 VELOCITY AND ACCELERATION Definition A PAdeP v dt A A P dv AaP dt where P is the point whose velocity and acceleration are desired A is the reference frame in which the velocity and acceleration of P are to be calculated and O is any point fixed in reference frame A Iudicious selection of the point O can greatly simplify the amount of work required to calculate A VP To carry out DP and AVP must rst be reexpressed in terms of the indicated vector derivatives in reference frame A p unit vectors xed in reference frame A which may require the use of one or more rotation matrices Also if the acceleration of point P is to be used in the formulation of dynamics equations then it must be calculated with respect to a fixed ie inertial or Newtonian reference frame N Two Points Fixed on a Rigid Body AVPAVQAmB XPQP AaPAaQAmB XADB XPQPAOCB XPQP where Q and P are points xed on a rigid body B P is the point whose velocity and acceleration are desired B moves in a references frame A and pQP AvQ A219 A038 and A068 are known These relationships are often easier to use than the definitions of velocity and acceleration given above since it is not necessary to reexpress pop in terms of unit vectors xed in reference frame A lN39ERTIA General Considerations While mass defines a rigid body s resistance to translational motion inertia de nes its resistance to rotational motion To determine the inertia of a rigid body we must specify three key ingredients 0 Point position vectors 0 Mass bodies andor particles 0 Directions between one and three We therefore speak of the moment or product of inertia for a body S rigid or set or particles relative to some point O for mutually perpendicular directions a and a 139 j 123 The various moments of inertia ij or ondiagonal elements and products of inertia 139 j or offdiagonal elements can be collected into a single inertia matrix 130 called the inertia matrix of S relative to O for a i 1 2 3 as follow s 13 Fregly April 2 2003 30 30 30 111 112 113 30 30 30 30 I 112 122 123 30 30 30 113 123 133 where the entries of this matrix are symmetric ie 150 fl0 Every inertia matrix is intimately related to a set of three mutually perpendicular unit vectors and each entry in this matrix is associated with a particular pair of unit vectors as indicated by the subscripts on the moment or product of inertia For example the moment of inertia 1131 O is associated with unit vectors a1 and 211 while the product of inertia 11320 is associated with unit vectors a1 and a2 Since ISO can be expressed in terms of an infinite number of sets of mutually perpendicular unit vectors there are an in nite number of inertia matrices of S relative to O with each matrix corresponding to a unique set of unit vectors Furthermore changing the point O with respect to which the inertia matrix is calculated will change the individual entries in the matrix Physical Interpretation In physical terms the entries of the inertia matrix 13 0 tell us whether or not rotation about one direction a of the inertia matrix would produce motion about another direction a if all axes passed through the point O For example if 11320 0 then rotation about the 211 direction would result in motion about the 212 direction as well This phenomenon is called dynamic coupling If the directions a i 1 2 3 are determined such that all the products of inertia equal zero ie 150 0 for 139 j then we call a i 1 2 3 the principal directions of S relative to point O Furthermore when the principal directions are relative to the mass center S of S then we call them central principal directions and the corresponding principal moments of inertia are the smallest possible moments of inertia for S Note that different points O will usually possess different principal directions Practical Calculation Inertia matrices are needed to calculate angular momentum for use in the Angular Momentum Principle see below Since this principle can involve calculating angular momentum about various points inertia matrices must also be calculated relative to various points In practice if a rigid body S possesses uniform mass density then the inertia matrix of S relative to any point O can be calculated by following a simple threestep process The rst step is to calculate the inertia matrix ISAquot for S relative to its mass center S for the principal directions 51 i 1 2 3 using tabulated information eg as in the back of the course textbook 1513quot 0 0 ISSquot 0 IiiSquot 0 0 0 1313quot 13 Fregly April 2 2003 The reason the offdiagonal entries of ISAquot are zero is because the associated directions are the principal directions The second step is to calculate the inertia matrix ISHOabout point O for the principal directions 51 i 1 2 3 ofa fictitious particle possessing the same mass as S and located at S 2 2 x2 x3 x1x2 x1x3 SO 2 2 I m x1x2 x1 x3 x2x3 2 2 x1x3 x2x3 x1 x2 where 703 xls1 xzs2 x3s3 The third and nal step is to sum the results ISO ISSquot ISO where it is important to note that both ISAquot and ISHO are defined in terms of the same set of mutually perpendicular unit vectors This result is simply a more general way of writing the parallel axis theorem If 130 for example S consists of two bodies A and B then can be found from ISO IAO IBO IAAquot IAO IBBquot IBO where all of the inertia matrices on the right hand side can be found from either tabulated data or the inertia matrix for a single particle MOMENTUM PRINCIPLES Linear Momentum Principle The linear momentum N LS of a particle body or system of particles and or bodies S in a Newtonian reference frame N is de ned as N LS ms N VS where ms is the mass of S and S is the mass center of S If S is a system composed of a combination of particles and rigid bodies A B Z then NLS can be found from NLSNLANLB NLZ which is often advantageous since it eliminates the need to calculate the position and velocity of the mass center S of S which changes during movement Once N LS has been found the linear momentum principle states that NdNLS FS 2 dt which equals ms N 213 from Newton s Second Law This principle can be applied to a single particle a single rigid body or any desired grouping of particles and rigid bodies 13 Fregly April 2 2003 Angular Momentum Principle The angular momentum NHSP of a particle or group of particles S relative to any point P in a Newtonian reference frame N is de ned as NHSP PSXNLS P This expression for NHSP sometimes referred to as the moment of momentum about point P is used for lumped mass models Note that if point P is chosen as the mass center 5 of S then NHSS will be zero for a group of particles If the angular momentum NHSO N HS P N HS 0 of S relative to any other point O is desired and is known then can be found from NHSONHSP pOPXNLS This equation is advantageous when S is composed of multiple particles since it eliminates the need to recalculate an angular momentum for each of the individual particles If on the other hand S is a rigid body then the angular momentum NHSS ofS relative to its mass center S in a Newtonian reference frame N is given by NHSSquot ISSINDS NHSS is called the central angular momentum of S in N and is particularly easy to calculate since ISAquot is generally available from a table To perform the indicated dot multiplication ISAquot and N as must be expressed in terms of the same unit vectors Since ISAquot will typically be expressed in terms of unit vectors xed in rigid body S it is often convenient to express N 003 in terms of these unit vectors If the angular momentum NHSO NHSS NHSO of rigid body S relative to some other point O is desired and is known then can be found from NHSONHSS pOSXNLS which is similar to the equation above for a group of particles If S is a system composed of a combination of particles and rigid bodies A B Z then NHSO can be found from NHSONHAONHBO IHNHZO If S is a system of particles this equation is advantageous since it eliminates the need to calculate the position and velocity of the mass center S of S just as with linear momentum Once N HS 0 has been found the an ular momentum rinci 1e states that g P P 13 Fregly April 2 2003 NdNHSO M0 2 dt which holds true ifund only ifpoint O is either 1 the mass center S of S or 2 any point fixed in the inertial reference frame N The sum of moments acting on S about point O includes only contact and distance forces and torques The inertia forces and torques are accounted for by the first time derivative of the angular momentum of S about O in the Newtonian reference frame N As with the linear momentum principle this principle can be applied to a single particle a single rigid body or any desired grouping of particles and rigid bodies DYNAMICS METHODS NewtonEuler Method The NewtonEuler method of deriving dynamics equations involving treating every body as a separate free body and solving for ALL reaction forces and torques Thus for a 2D problem you will need to form 3 equations 2 translations and 1 rotational for every body while for a 3D problem you will need to form 6 equations 3 translational and 3 rotational for each body If nbodies is the number of bodies this means that the total number of unknowns will always be 3nbodies for a 2D problem and 6nbodies for a 3D problem These unknowns will be combinations of ij s and reaction forces and torques For example a 2D problem with a single rigid segment connected to ground by a pin joint will have 3 unknowns that must be solved one ij and two reaction forces horizontal and vertical The translational equations come directly from the linear momentum principle while the rotational equations come directly from the angular momentum principle where the mass center must be chosen as the point about which to sum moments The Motion Law The Motion Law often attributed incorrectly to D Alambert replaces the linear and angular momentum principles for a free body S S can be a single body or group of bodies with the related equations Fgontact SP SP SP MContact Msttancz MInzma 0 s s Fsttancz FInema 0 S DIEM are all distance where Fgmm are all contact forces acting on S eg ground contact forces F forces acting on S eg gravity forces and FISWm are all inertia forces acting on S For any individual body B FIB quotMm can be easily calculate as F3 mBNaB Inert a which is equivalent to N dNLBdl for body B 13 Fregly April 2 2003 As indicated above these contact distance and inertia forces also have moment counterparts However unlike the angular momentum principle the moments may be taken about ANY point P This is the primary advantage of the Motion Law over the NewtonEuler method This advantage eliminates the need to solve large systems of equations for large numbers of unwanted reaction forces and torques Furthermore it makes the formulation and solution of dynamics problems identical conceptually to the solution of statics problems The contact moment M2452 is simply the moment about point P of all contact forces acting on free body S Similarly the distance moment MSP BMW is the moment about point P of all distance forces acting on S The inertia moment Min213210 is the quantity that is different as it is composed of two contributions the moment of all inertia forces about point P plus a new quantity called an inertia torque For a rigid body B the inertia torque is defined as TBBquot NaB IIBBNw8 XIBBINO N which is equivalent to dNHB Bdt for body B based on the Million Dollar Formula Thus the inertia moment M Z m for body B about point P becomes BP PBquot B BBquot MInema P XFInzrtm T Note that if point P is taken as B the mass center of body B then the cross product above goes to zero and the inertia moment becomes identical to the inertia torque SOLUTION PROCESS FOR DYNAMICS PROBLEMS 1 De ne reference frames unit vectors xed in each reference frame and unknown translations and rotations corresponding to the degrees of freedom N V Draw free body diagrams realizing that groups of individual bodies can be treated as a quotfree bodyquot 9 V Apply the NewtonEuler or Motion Law method choosing directions in which to sum forces and points about which to sum moments in such as way as to minimize the need to calculate unwanted reaction forces and moments Create a table listing which body or group of bodies will be treated as the quotfree bodyquot which equation is to be used eg F to represent sum of forces and MP to represent the moment about some point P in which direction a dot product will be taken and finally what the resulting unknowns will be ie second time derivatives of the unknown translations and rotations and possibly reaction forces and torques a V Calculate important kinema c quantities ie the acceleration of every point with mass the angular velocity and acceleration of every body with inertia with respect to the inertial reference frame and express them in terms of time derivatives of the unknown translations and rotations II V Solve the resulting linear system of equations for the second time derivatives of the unknown translations and rotations and for the unknown reaction forces and torques if present 0 V If possible integrate the second time derivative equations analytically to determine the unknown translations and rotations as a function of time 13 Fregly April 2 2003

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