《力学》课程教学资源（参考资料）惯量椭球 Stereo 3D Simulation of Rigid Body Inertia Ellipsoid for The Purpose of Unmanned Helicopter Autopilot Tuning.pdf_大学文库

International Journal of Engineering Science Invention ISSN (Online):2319-6734.ISSN (Print):2319-6726 www.ijesi.org Volume 3 Issue 8I August 2014I PP.28-35 Stereo 3D Simulation of Rigid Body Inertia Ellipsoid for The Purpose of Unmanned Helicopter Autopilot Tuning 'Petar Getsov,2Svetoslav Zabunov,Maya Gaydarova .Space Research and Technology Institute at the Bulgarian Academy of Sciences.Bulgaria Sofia University,Bulgaria ABSTRACT:The current paper aims at presenting the capabilities and benefits of an online stereo 3D simulation for the purpose of unmanned helicopter autopilot tuning.The parameters of the helicopter airframe are important for tuning the gains in the autopilot.The airframe is modelled as a rigid body whose inertial properties are fully described by the inertia ellipsoid.The inertia ellipsoid is another form of presenting the moment of inertia tensor of rigid bodies but instead of using a numerical approach the described method implements 3D graphical visualization.The current paper focuses on the benefits from stereoscopic graphical 3D presentation of the inertia ellipsoid and how such a method helps designers and researchers analyse, synthesise and tune unmanned helicopter autopilot algorithms.The simulation,subject to the current material, may be observed at the following web address:http://ialms.net/sim/. KEYWORDS:Inertia ellipsoid in stereo 3D simulation.Rigid body inertia ellipsoid.Unmanned helicopter autopilot tuning. I.INTRODUCTION The current article shows a new method of presenting the inertia ellipsoid of rigid bodies and its properties using a visual online environment.A simulation that has unrestricted access on the Internet demonstrates the inertia tensor of various rigid bodies to researchers in universities and institutes around the world.The major features of the advised simulation are the stereo 3D environment,in which the inertia ellipsoid of various rigid bodies is demonstrated along with translations of the ellipsoid and its moment of inertia tensor to any point in the body reference frame.Along with the inertia ellipsoid,the principal axes of inertia are also displayed.The simulation prints the diagonalized translated moment of inertia tensor and the diagonalizing rotation matrix(Figure 1).The simulation,described in the current paper,may be observed on web address: http://ialms.net/sim/. CAORANLRUT0R01 a50 Coeynight osect Sham al mnterfarr rementw ■emmtnf例a0■01n Bodv mass101.100kgl 10 wih amf (v2)fm] a山■5 01 IntrmaEtchim cemfhiiret [1/s](etkrat-ft) a001 ody drath [u.85.1m] 5 Shes angular mementum I shem anulr velecity lody hegtt [nn5 Iml 24■ 0U087 ■m■4则 nestid eflprstid Cmnter of themen ef erta trr CY 050250四■ Srtn0 anguarvelcy o5ou时s 07应 097■125n0 3251行5 000251,7 040■.31国014 0a8■19n98 03739g■125 Figure 1.Translated moment of inertia tensor and inertia ellipsoid to the vertex of a rectangular parallelepiped- shaped homogeneous rigid body and 3D graphical visualization www.ijesi.org 28 |Page

International Journal of Engineering Science Invention ISSN (Online): 2319 – 6734, ISSN (Print): 2319 – 6726 www.ijesi.org Volume 3 Issue 8 ǁ August 2014 ǁ PP.28-35 www.ijesi.org 28 | Page Stereo 3D Simulation of Rigid Body Inertia Ellipsoid for The Purpose of Unmanned Helicopter Autopilot Tuning 1 Petar Getsov, 2 Svetoslav Zabunov, 3Maya Gaydarova 1, 2 Space Research and Technology Institute at the Bulgarian Academy of Sciences, Bulgaria 3 Sofia University, Bulgaria ABSTRACT : The current paper aims at presenting the capabilities and benefits of an online stereo 3D simulation for the purpose of unmanned helicopter autopilot tuning. The parameters of the helicopter airframe are important for tuning the gains in the autopilot. The airframe is modelled as a rigid body whose inertial properties are fully described by the inertia ellipsoid. The inertia ellipsoid is another form of presenting the moment of inertia tensor of rigid bodies but instead of using a numerical approach the described method implements 3D graphical visualization. The current paper focuses on the benefits from stereoscopic graphical 3D presentation of the inertia ellipsoid and how such a method helps designers and researchers analyse, synthesise and tune unmanned helicopter autopilot algorithms. The simulation, subject to the current material, may be observed at the following web address: http://ialms.net/sim/. KEYWORDS: Inertia ellipsoid in stereo 3D simulation, Rigid body inertia ellipsoid, Unmanned helicopter autopilot tuning. I. INTRODUCTION The current article shows a new method of presenting the inertia ellipsoid of rigid bodies and its properties using a visual online environment. A simulation that has unrestricted access on the Internet demonstrates the inertia tensor of various rigid bodies to researchers in universities and institutes around the world. The major features of the advised simulation are the stereo 3D environment, in which the inertia ellipsoid of various rigid bodies is demonstrated along with translations of the ellipsoid and its moment of inertia tensor to any point in the body reference frame. Along with the inertia ellipsoid, the principal axes of inertia are also displayed. The simulation prints the diagonalized translated moment of inertia tensor and the diagonalizing rotation matrix (Figure 1). The simulation, described in the current paper, may be observed on web address: http://ialms.net/sim/. Figure 1. Translated moment of inertia tensor and inertia ellipsoid to the vertex of a rectangular parallelepipedshaped homogeneous rigid body and 3D graphical visualization

Stereo 3d Simulation Of Rigid Body Inertia... II.MATHEMATICAL FOUNDATIONS OF THE DESCRIBED 3D SIMULATION To prove the consistency and fidelity of the presented simulation,a concise introduction to the inertial properties of rigid bodies follows.Moment of inertia of a rigid body about a given axis describes quantitatively the body inertia behaviour during a rotation about that axis.If a rigid body has volume V then its moment of inertia of about axis OO'is: where p is the body density at the location of elementary mass dm=pdl and r is the perpendicular radius- vector from the axis of rotation OO'to the elementary mass.About any given axis the rigid body has a certain and generally different moment of inertia.Each moment of inertia could be calculated using the equation above. An alternative way of calculating the moment of inertia is by using the relation between the angular momentum and angular velocity: i=, Here vector L is the angular momentum,and vector is the angular velocity of the rotational motion. Although useful,this equation is applicable only to rotations that are realized about a principal axis of inertia. Only then L and are parallel to each other and the relation between them is presented using a product with a scalar value,as in the example above.Such a rotation is the rotation of an axially symmetric homogenous rigid body about its axis of symmetry. Generally,vectors L and are not parallel,but a relation connecting them still exists and it is a tensor of second rang called moment of inertia tensor: 1o= y 1 1 1 This tensor is in respect to a point O,having that axis of rotation passes through this point.To define the relation between L and these two vectors should be presented in a matrix form(one-row-matrixes): L=@lo It follows that components of vector L are: Lx=@xlxx+oylxy+0I Ly=@xlxy+@ylw+ly L:=@xlx+yly+01- In this general case,the momentary axis of rotation coincides with The inertia about this axis of rotation creates an angular momentum about this same axis that is the projection of the angular momentum vector L along vector In matrix form we have: LO Lo=- -(0 is the transposed matrix of c Substituting the angular momentum with the product of angular velocity and the moment of inertia tensor yields: olo Lo= -=noIn@=1o0, 0 or the projection of the L along the axis of rotation is derived from the magnitude of and the scalar quantity I.It is called a reduced moment of inertia from the moment of inertia tensor about a given axis of rotation www.ijesi.org 29 Page

Stereo 3d Simulation Of Rigid Body Inertia... www.ijesi.org 29 | Page II. MATHEMATICAL FOUNDATIONS OF THE DESCRIBED 3D SIMULATION To prove the consistency and fidelity of the presented simulation, a concise introduction to the inertial properties of rigid bodies follows. Moment of inertia of a rigid body about a given axis describes quantitatively the body inertia behaviour during a rotation about that axis. If a rigid body has volume V then its moment of inertia of about axis OO is:      V V IOO r dm r dV 2 2 , where  is the body density at the location of elementary mass dm  dV and r is the perpendicular radiusvector from the axis of rotation OO to the elementary mass. About any given axis the rigid body has a certain and generally different moment of inertia. Each moment of inertia could be calculated using the equation above. An alternative way of calculating the moment of inertia is by using the relation between the angular momentum and angular velocity: L I    , Here vector L  is the angular momentum, and vector   is the angular velocity of the rotational motion. Although useful, this equation is applicable only to rotations that are realized about a principal axis of inertia. Only then L  and   are parallel to each other and the relation between them is presented using a product with a scalar value, as in the example above. Such a rotation is the rotation of an axially symmetric homogenous rigid body about its axis of symmetry. Generally, vectors L  and   are not parallel, but a relation connecting them still exists and it is a tensor of second rang called moment of inertia tensor:            xz yz zz xy yy yz xx xy xz O I I I I I I I I I I This tensor is in respect to a point O , having that axis of rotation passes through this point. To define the relation between L  and   , these two vectors should be presented in a matrix form (one-row-matrixes): L  ωI O It follows that components of vector L  are: x x xx y xy z xz L  I  I  I y x xy y yy z yz L  I  I  I z x xz y yz z zz L  I  I  I In this general case, the momentary axis of rotation coincides with   . The inertia about this axis of rotation creates an angular momentum about this same axis that is the projection of the angular momentum vector L  along vector   . In matrix form we have:  ~ Lω Lω  ( ~ ω is the transposed matrix of ω ) Substituting the angular momentum with the product of angular velocity and the moment of inertia tensor yields:    ω nωInω ω ωIω L    I ~ ~ , or the projection of the L  along the axis of rotation is derived from the magnitude of   and the scalar quantity Iω . It is called a reduced moment of inertia from the moment of inertia tensor about a given axis of rotation

Stereo 3d Simulation Of Rigid Body Inertia... 0 defined by its direction unit vector n or n= in vector form.Thus the reduced moment ofieria may be expressed using the components of the unit vector n and the moment of inertia tensor: Io=nolno=nglxx +nlwy+nI+2nxny lgy +2ngn-le+2nyn-Iy Because vector n has unit length,its components could be substituted with the direction cosines defining the axis of rotation: 1=Ix cos a+lw cos B+I=cosy+21 cosacosB+21cosacosy+21=cosBcosy At the same time,the reduced moment of inertia may be expressed through L and as follows: lo=Lo=Lo- 002 The inertial ellipsoid enables the researcher to observe in stereoscopic 3D graphical scene the moment of inertia tensor properties,which is essential for the following autopilot synthesis labour.The ellipsoid is defined in such a manner that its three axes are the reciprocals of the square roots of the principal moments of inertia(Figure 1,2,3 and 4): y2 22 +存+ Q =1←1xx2+1wy2+1_22=1 Lamera:la成udc[010u】 18.90 Full Screen Copyright ialms.net Camera:longtude [1m1阙】 43.40 start smulation ■O5 tereo mode Panse Show all interface elements Orthogonal parallelepped Body mass [0.1.100kg] with arm rl (x.Y'2)[m] 00■0.0■05 10 0.0t 05 External friction coefficient [Ns7/rad](Mf=kf) Body width [o.05..m] 1.0 Internal friction coefficient [/s](dEkrot-kfi.d) 0.001 Body depth [] 0.5 ■Show angudar velocity Body height [005-Im] 0.2 Moment of inertiatencor [kgmz] MShow body coordinate axes 0.24 0.10 0.0 Show rigid body 000■0870.0 Inertial ellipsoid Starting orientation (rotation by X,Y,2)[] Center of the moment ot inertia tensor 000 00■000.0 Starting angular velocity about [rad/s] Transloted moment of mertid tensor 000.7■02 0.240.00■000■ 0.000.87000 0.000.00104 Diagonalizyng rotation 1.000.00000 0.001.0000■ a.000.001.00 Diagonalized tenso (elements of themain di) 0.24087104 Figure 2.The inertia ellipsoid of the non-displaced moment of inertia tensor What the simulation offers to the researcher The simulation was developed in order to facilitate researchers and scientists while studying the mechanical properties of unmanned helicopter airframes and their input in autopilot algorithms.Further the simulation helps engineers in unmanned helicopter autopilot analysis and synthesis.The simulation helps the construction of the inertia ellipsoid of different rigid bodies and the observation of the principal moments of inertia.To make clear the effectiveness of the simulation while solving mechanics problems,several tasks are disclosed as examples.The solutions could be observed in 3D stereo mode.Each step of a solution has its www.ijesi.org 30|Page

Stereo 3d Simulation Of Rigid Body Inertia... www.ijesi.org 30 | Page defined by its direction unit vector nω or       n   in vector form. Thus the reduced moment of inertia may be expressed using the components of the unit vector    n and the moment of inertia tensor: x xx y yy z zz x y xy x z xz y z yz I n I n I n I 2n n I 2n n I 2n n I ~ 2 2 2 ω  nωInω       , Because vector    n has unit length, its components could be substituted with the direction cosines defining the axis of rotation: cos  cos  cos  2 coscos 2 coscos 2 cos cos 2 2 2 xx yy z z xy xz yz Iω  I  I  I  I  I  I At the same time, the reduced moment of inertia may be expressed through L  and   as follows: 2 ~   ω Lω ω   L I The inertial ellipsoid enables the researcher to observe in stereoscopic 3D graphical scene the moment of inertia tensor properties, which is essential for the following autopilot synthesis labour. The ellipsoid is defined in such a manner that its three axes are the reciprocals of the square roots of the principal moments of inertia (Figure 1, 2, 3 and 4): 1 1 2 2 2 2 2 2 2 2 2     I x  I y  I z  c z b y a x x x y y zz Figure 2. The inertia ellipsoid of the non-displaced moment of inertia tensor What the simulation offers to the researcher The simulation was developed in order to facilitate researchers and scientists while studying the mechanical properties of unmanned helicopter airframes and their input in autopilot algorithms. Further the simulation helps engineers in unmanned helicopter autopilot analysis and synthesis. The simulation helps the construction of the inertia ellipsoid of different rigid bodies and the observation of the principal moments of inertia. To make clear the effectiveness of the simulation while solving mechanics problems, several tasks are disclosed as examples. The solutions could be observed in 3D stereo mode. Each step of a solution has its

Stereo 3d Simulation Of Rigid Body Inertia... graphical representation,which clarifies the notion of the used mathematical formulae and presents the acquired results and solutions in graphical manner.Hence the simulation is useful in drawing practical understanding among researchers. Task 1 Let's have a homogenous rigid body.Its shape is a rectangular parallelepiped and its mass is 10 kg (Figure 2).Body reference frame origin coincides with the centre of mass and its axes are parallel to the body edges.The body dimensions are 1.0,0.5,0.2 m along the Ox,Ov,O=axes respectively.The task is to calculate the moment of inertia tensor for the centre of mass I. Note:Without applying similarity transformation,the moment of inertia tensor is diagonal [ 0 0 I=0 1 0 and correspond to the principle moments of inertia of the centre of mass 0 along the body reference frame coordinate axes.This follows from the body homogenous and symmetrical properties. 1660 Camerac longitude【-180180】 54.90 start simulation ■口5 tereo mode Show all interfoce elements Orthogonal Parallelepiped wth arm rf (.[m] 10 Sndy radius【.5.,1m] .01■ 10 ■Interal frictinn coeficent D/s】(dEkrot=-k.d) 0.001 Body depth [0.05..Im] 0.5 Body height [o.05.1m] 02 ■5 how coordnate ate Moment of inertia tensor [kg.m2] 0.240 ☑show rio时body 0nm0.700 Wiewt Starting orientatson (rotatinn by X.Ye) Center of the moment of inertia tensor (XY2) 00■0 0.502500 anguiar velocity about[rad/s] of ine 00■07■02 0B7-125000 1z53370.00 0.000006,17 0000001.00■ 092030■0.00 0380920.00 ts of the main daganal) .170353.88■ Figure 3.Displacement of the moment of inertia tensor along the Ox and Oy axes Solution to task 1: Homogenous rectangular parallelepiped moment of inertia tensor at its centre of mass in a reference frame oriented along the body edges is derived by the following formula: b2+c2 0 0 1 Ic= 0 a2+c2 0 0 0 a2+b2 Here m=10kg is the body mass,and a=1m,b=0.5m and c=0.2m are the body dimensions.Hence, the answer follows to be: 「0.24 0 0 1= 0 0.87 0 0 0 1.04 www.ijesi.org 31 |Page

Stereo 3d Simulation Of Rigid Body Inertia... Task 2 Calculate the translated moment of inertia tensor from task I at the vertex of the body that has only positive coordinates in the body reference frame. Solution to task 2: The parallel axis theorem(Huygens-Steiner theorem)solves this problem. (1) L,=Ie+mr1-r⑧r→ 1+m+2） -mrxry -mrxr: 0.97 -1.25-0.507 2 mrxry 1y+m2+2 -1.253.47 -0.25 -mrxr: -mryr: 1+m2+3 -0.50 -0.254.17 In the above equation matrix 1 is the 3x3 identity matrix,vector r is the translation vector.The body is homogenous and its centre of mass and geometrical centre coincide.It follows that the translation vector pointing to the vertex of the body with only positive coordinates is equal to 0.5 0.25 0.1(see Figure 1). Figure 3 shows a translation of the moment of inertia tensor along both the Ox and Oy axes.Note that in the case of translation along a single coordinate axis,a translated tensor is diagonal.In the case of moment of inertia tensor translation along two axes,four of the translated tensor's non-diagonal components are zeroes. Task 3 Diagonalize the translated moment of inertia tensor from task 2. Solution to task 3:Using the theory of symmetric matrixes one could always diagonalize a 3x3 symmetric matrix or symmetric rang 2 tensor.The diagonalized tensor will have only three non-zero values and they will be found in the main diagonal.The other three values(products of inertia)found in the non-dagonalized tensor will be represented by the transformation used to diagonalize the tensor.This transformation is a similarity transformation realized through applying a rotation matrix R.We can prove this assumption,taking into account that under a certain rotation R the principle axes of the moment of inertia tensor coincide with the axes of the body reference frame and the rotated tensor becomes diagonalized: 0 0 (2) ID=RIR= 0 Ipyy 0 0 0 ID:> From the eigenvectors p and eigenvalues of the pursued diagonalized tensor we have: (3) nDlD=nDi→nDlD-nDA=0 This matrix equation shows a homogenous system of three linear equations having a non-trivial solution only if the determinant of its coefficients is zero: (4) D-1,l=0→(0p-0Dw-AuDe=-)=0 Formula(4)has three real roots:=IDx,=IDyy and=ID=,hence the diagonalized tensor is: [2 0.0 (5) 1D= 0 2 0 0 03 After we apply the similarity transformation(2)to equation(3)we get: nDiID=npRR=nDi→nDRI=nDRA→ (6) nl=n;→nl-n4=0 www.ijesi.org 32 Page

Stereo 3d Simulation Of Rigid Body Inertia... www.ijesi.org 32 | Page Task 2 Calculate the translated moment of inertia tensor from task 1 at the vertex of the body that has only positive coordinates in the body reference frame. Solution to task 2: The parallel axis theorem (Huygens-Steiner theorem) solves this problem. (1) It  Ic  mrr ~ 1  r  r                                           - 0.50 - 0.25 4.17 -1.25 3.47 - 0.25 0.97 -1.25 - 0.50 2 2 2 2 2 2 x z y z z z x y x y yy x z y z xx y z x y x z t mr r mr r I m r r mr r I m r r mr r I m r r mr r mr r I In the above equation matrix 1 is the 33 identity matrix, vector r is the translation vector. The body is homogenous and its centre of mass and geometrical centre coincide. It follows that the translation vector pointing to the vertex of the body with only positive coordinates is equal to 0.5 0.25 0.1 (see Figure 1). Figure 3 shows a translation of the moment of inertia tensor along both the Ox and Oy axes. Note that in the case of translation along a single coordinate axis, a translated tensor is diagonal. In the case of moment of inertia tensor translation along two axes, four of the translated tensor’s non-diagonal components are zeroes. Task 3 Diagonalize the translated moment of inertia tensor from task 2. Solution to task 3: Using the theory of symmetric matrixes one could always diagonalize a 33 symmetric matrix or symmetric rang 2 tensor. The diagonalized tensor will have only three non-zero values and they will be found in the main diagonal. The other three values (products of inertia) found in the non-dagonalized tensor will be represented by the transformation used to diagonalize the tensor. This transformation is a similarity transformation realized through applying a rotation matrix R . We can prove this assumption, taking into account that under a certain rotation R the principle axes of the moment of inertia tensor coincide with the axes of the body reference frame and the rotated tensor becomes diagonalized: (2)             Dzz Dyy Dxx D I I I 0 0 0 0 0 0 ~ I R IR From the eigenvectors nDi  xDi yDi zDi and eigenvalues i of the pursued diagonalized tensor we have: (3) nDiID  nDii nDiID nDii  0 This matrix equation shows a homogenous system of three linear equations having a non-trivial solution only if the determinant of its coefficients is zero: (4) ID 1i  0  IDxx  i IDyy  iIDzz  i   0 Formula (4) has three real roots: Dxx  I 1 , Dyy  I 2 and Dzz  I 3 , hence the diagonalized tensor is: (5)            3 2 1 0 0 0 0 0 0    ID After we apply the similarity transformation (2) to equation (3) we get: nD iID  nD iR ~ IR  nD ii  nD iR ~ I  nD iR ~ i  (6) ni I  nii ni I nii  0

Stereo 3d Simulation Of Rigid Body Inertia... www.ijesi.org 34 | Page (12)       1 0 0 0 2 2 2                i i i i xz i yz i zz i i xy i yy i i yz i xx i i xy i xz x y z x I y I z I x I y I z I x I y I z I    System (12) enables i y and i z to be expressed through i x , using the first three equations, and substituting in the last equation: (13)                 1 1 2 2 2                                              z z i xy xz yz xx i yz xy xz yy i xz xy yz xx i yz xy xz i z z i xy xz yz xx i yz xy xz i i yy i xz xy yz xx i yz xy xz i i I I I I I I I I I I I I I I I I x I I I I I I I I z x I I I I I I I I y x         From the last equation of system (13) i x is found as follows: (14)         2 2 1 1                             z z i xy xz yz xx i yz xy xz yy i xz xy yz xx i yz xy xz i I I I I I I I I I I I I I I I I x     We find the values of i y and i z by substituting i x in the first two equations of (13). Note that the components i i i x , y ,z of eigenvector ni do not have defined sign. There are two possibilities yielding an eigenvector ni along a defined line, but with two possible opposite directions. The correct direction will be found later. After finding all three eigenvectors n1 , n2 and n3 their components are substituted in (11) to generate the sought rotation matrix R (see Figure 1):               0.08 0.19 0.98 0.40 0.91 0.14 0.91 0.38 0.15 R However, if the incorrect directions (signs in (14)) were chosen, matrix R would yield an inverse (improper) rotation. This fact can be verified by calculating the determinant of R , which should be 1 : R 1 This means that matrix R is a proper rotation. In the case when the determinant is 1, matrix R should be corrected to a proper rotation by multiplying it with a negative identity (changing direction of all eigenvectors) or by changing the sign of one of its rows (changing direction of one of the eigenvectors). Visit http://ialms.net/sim/ to simulate other examples. III. CONCLUSION The described in this article simulation of rigid body properties in a stereo 3D virtual environment enables researchers and engineers to observe setups that are impossible to be created in laboratory conditions. Such an approach reveals important insights for the scientists working on unmanned helicopter autopilot analysis, synthesis and tuning. Authors of the presented material express their gratitude to Assoc. Professor Vesela Decheva