Rigid body trajectories in different 6D spaces

—The objective of this paper is to show that the group SE(3) with an imposed Lie-Poisson structure can be used to determine the trajectory in a spatial frame of a rigid body in Euclidean space. Identical results for the trajectory are obtained in spherical and hyperbolic space by scaling the linear displacements appropriately, since the inﬂuence of the moments of inertia on the trajectories tend to zero as the scaling factor increases. The semi-direct product of the linear and rotational motions gives the trajectory from a body frame perspective. It is shown that this cannot be used to determine the trajectory in the spatial frame. The body frame trajectory is thus independent of the velocity coupling. In addition, it is shown that the analysis can be greatly simpliﬁed by aligning the axes of the spatial frame with the axis of symmetry which is unchanging for a natural system with no forces and rotation about an axis of symmetry.


I. INTRODUCTION
The motion of a rigid body in Euclidean space E 3 consists of 3 translations and 3 rotations about the centre of mass.The ways in which these are combined determine the calculated trajectory of the body.This paper considers several methods of determining the corresponding velocities and the resultant trajectories, and investigates the consequences of each method.
Previous work on Lie groups has used the special Euclidean group SE (3) with no imposed structure so that it is the semi-direct product of the translations in R 3 and rotations in the group SO (3).The mapping between the groups used by Holmes [6] and Marsden [9] is expressed in two formats in this paper: as the mapping itself and as the integration using SE (3).This method gives the trajectory in the body frame, which cannot be used to determine the trajectory in the spatial frame needed for many applications.The body frame trajectory is the independent of the velocity coupling.There is no natural way of weighting the rotations and translations to measure the distance (in 6 dimensions) along a trajectory using the semidirect product.
There is no bi-invariant Riemannian metric on SE (3).There are natural metrics on SO (4) and SO (1, 3) -the trace form, which can be inherited by SE (3) with the appropriate scaling.Etzet and McCarthy [3] used a metric on SO (4) as a model for a metric on SE (3).Larochelle et al. [8] projected SE (3) onto SO (4) to obtain a metric.In this paper, the linear displacements are scaled so that they are small C. Linton is with the School of Systems Engineering, University of Reading, Reading.UK (e-mail: c.l.linton@pgr.reading.ac.uk)W.Holderbaum is with the School of Systems Engineering, University of Reading, Reading.UK (e-mail: w.holderbaum@reading.ac.uk)J.Biggs is with the Department of Mechanical Engineering, University of Strathclyde, Glasgow, UK (e-mail: james.biggs@strath.ac.uk) compared with a unit hypersphere.This enables SE(3) to be projected vertically onto SO(4) and SO (1,3).This projection is extended by imposing the Lie-Poisson structure on SE (3) as mentioned in [2].
The 6D Lie groups SO(4), SO (1,3) and SE(3) with imposed Lie-Poisson structure are compared and shown to result in related trajectories, which approximate to the same values for small linear displacements.The trajectories are in a fixed frame which is the requirement for planning and controlling the motions.
Four methods of combining the translations and rotations are investigated: • A semi-direct product -where the rotation changes the body frame, but not the linear velocity itself.That is: there is no coupling of the angular and translational velocities.• The special Euclidean group SE (3) with an imposed Lie-Poisson structure, where the rotation induces a change in the linear velocity to conserve angular momentum.
• The rotation group SO (4) which can approximate the previous results if the linear displacements are scaled down.
• The Lorentz group SO (1,3).The latter is included as a continuation from the other groups, and not as a practical alternative.Finding the trajectory of a rigid body has 2 steps; 1) Determining how the velocity changes over time as the velocity components couple together .2) Integrating the velocity function down to the base manifold to give the trajectory.Both these depend on the group -for example, is the space that the group represents flat, convex or concave?The formulas for the general case are derived, but the examples are based on the simplest case with no external forces and rotation about an axis of symmetry.
Any combinations of rotations can be represented by rotation about a single axis of rotation.If this axis is also an axis of symmetry, the rotational axis does not change and there is no precession.By choosing the axes of the fixed frame appropriately, the initial motion of any body can be defined as an initial rotation about one axis, with some linear motion along and some motion perpendicular to that axis.Details are provided in the Appendix.In a natural system with no forces, the angular momentum remains constant and the resulting trajectories determined using the various methods can be compared.The linear displacement of the centre of mass is used.The initial velocities used in the examples are where f is a phasing constant.Using the rotation identified in the Appendix, this can be generalized to any initial conditions in a system with no forces so long as the rotation is about an axis of symmetry.
Section II introduces the basic ideas of Geometric Control theory and applies them to the rotation group SO (3).The linear motion is incorporated using the semi-direct product.Section III discusses the Lie-Poisson structure which is implicit for the rotation group SO (4) and the Lorentz group SO (1, 3), and can be imposed for SE (3).With that structure, the differential equations of motion can be derived from the Hamiltonian.The trajectories are found in Section IV.The results for each group are compared.A comparison of the results from SO (4) and the structured SE (3) in Section V show that they are interchangeable for small displacements.The strength of the coupling between the angular and translational velocities tend to the same value as the accuracy of the scaling is improved.The final Section VI compares the different methods of combining linear and rotational motions, and when the methods fail.

II. GEOMETRIC CONTROL THEORY AND NOTATION
This section provides the basic ideas of Geometric Control theory.A fuller explanation is available in many texts such as [4].The rotations and translations are considered separately.This is used as an opportunity to introduce the necessary notation for combining them in the following sections.

A. Lie theory
A trajectory is represented by a matrix g (t) ∈ G, where G is the matrix Lie group which reflects the structure or shape of the space on which the trajectory lies.In Lie theory, the trajectory is pulled back to the origin by the action of g −1 (t).
The tangent field at the origin X ∈ g (where g is the Lie algebra) determines the trajectory through the expression One finds that If X = X (t), time dependent, an analytic solution is difficult to find in most cases, so the forward Euler method is used in this paper to demonstrate the general form of the solutions.
where s is the step length and n is the step number so that X n = X (t n ) is the tangent field at time t n = n s.The trajectory started at g (0) = I.
Fact.The forward Euler method uses Alternatively, assume X n is constant between times t n and t n+1 so that the incremental motion is given by δg n = exp (X n s).This increment is applied to the previous configuration using g n+1 = g n δg n which gives equation (3).

B. Rotations
Rotation in 3 dimensions is used to demonstrate the ideas of the previous section.A body rotating in space about its fixed centre of mass can have angular velocity {w i } for i ∈ {1, 2, 3} about three orthogonal axis {e i }.The resultant velocity X can be written in both coordinate and matrix forms as From this notation, the base matrices {e i } of the Lie algebra so (3) can be extracted.Any rotation in Euclidean space can be represented by a rotation about a single axis, so, by choosing the axes of the body frame appropriately, all initial rotation can be represented about the e 1 axis.The Appendix gives the required rotation of the body frame to achieve this.The initial rotational velocity can then be written as In a natural motion of a body rotating about an axis of symmetry, there are no forces and no coupling with other motions.(The discussion in Section III can be applied to SO (3)) to confirm this, with o 2 = o 3 ).The angular velocity is unchanging.The axis of rotation does not change (no precession).The attitude of the body Φ (t) ∈ SO (3) is the solution of The resulting attitude at time t is To represent a rigid body rotating about its centre of mass and moving through Euclidean space, the rotational and translational motions need to be combined.The semi-direct product is used to represent the motion from the body frame prospective.The rotational motion has already been determine.The linear velocity can be written in matrix and coordinate representations of where here {e i } is the orthogonal basis in R 3 .The semi-direct product enables the rotation to influence the linear motion.The two elements act on each other through an action • defined by as used by Marsden (p22 of [9]) and Holmes (p110 of [6]).This rotates the body frame.It can also be written in matrix form as which is the format of the Lie group SE (3) with no structure imposed.The attitude Φ (t) is found using equation (2).The displacement x at time t is found by integration as where Φ (s) is the rotation achieved at time s (since integration is the method of adding incremental changes in R 3 ).Alternatively the 2 velocity functions can be combined into one 4 × 4 matrix to give the equation which has the solution SE (3) .
The same result is obtained from the integration in equation ( 4).
In the simple example defined in expression (1), the initial translation velocity is T .In a natural motion of a body moving in Euclidean space, there are no forces and no coupling with other motions.The translational velocity is unchanging.The combined velocity function is and the configuration at time t is Thus the semi-direct product produces a trajectory as perceived in the body frame.Although the translational velocity is fixed at v 0 , the perceived trajectory is curved.In the body frame, the velocity in the y − z plane is v y so, at time t, the origin O is seen at an angle wt behind the body having traveled a distance v y t.The perceived radius of travel r is given by rwt = v y t so the perceived radius is v y /w as seen in Figure 1 and equation ( 5).In a spatial frame, a rigid body rotating about its centre of mass and moving through Euclidean space does not move in a straight line but moves by spirally about an axis.If the rotation is about a fixed axis with constant translational speed, then the trajectory is a screw motion with constant pitch as proved by Chasles in his Screw Theory of motion.The angular momentum induces a change in the translational velocity.
In the semi-direct product, the two velocity matrices were combined as 0 0 v φ to calculate the trajectory, where v is the 3 × 1 column matrix of linear velocities and φ is the 3 × 3 angular velocity matrix given above.In Section IV, the velocity matrix is taken as with ∈ {1, 0, −1} for so (4) , se (3) , so (1, 3) respectively.Other variations in the order of columns and rows are possible but produce equivalent results.Alternative values of ε create non closed groups and are not considered here.
In the next section, a structure is imposed on these groups which creates a relationship between the rotational and linear elements.With the Lie-Poisson structure, the velocities interact, with rotations inducing changes in the linear motion.The motions are described in the spatial frame, rather than the body frame.

III. STRUCTURE OF THE LIE ALGEBRAS
In this section, the structure of the Lie groups are identified in terms of structure constants.This structure (or shape of the space) determines how motion in one direction influences motion in another, and how the motions add together to arrive at a configuration in space.
• The rotation group SO (4) has an obvious structure in that it represents rotations.The space has positive curvature which is related to the fixed Casimir C = 6 i=1 p 2 i where the {p i } are the momentum components.The group has a bi-linear map and is a Poisson manifold.
• SO (1, 3) has some similar characteristics.It is a space with negative curvature in some directions since one of the fixed Casimirs is There is thus no automatic relative weighting between the rotational and linear elements.
More comparative data is provided by Jurdjevic [7].In all cases, there is another fixed Casimir C = 3 i=1 p i p i+3 which pairs the momentum types but adds no information about the relative weighting of the two types.
The base matrices for the 6-dimensional Lie algebras being considered can be seen from the equation with ∈ {1, 0, −1} for so (4) , se (3) , so (1, 3) respectively.The orthogonal basis for the dual Lie algebra g * is e i given by e i = Ie i where I = I 4×4 , the unit matrix, is a bi-linear form.
From these base matrices, the structure of the group is quantified in terms of the structure constants in sub-section (III-A).For a group with a Lie Poisson structure, these same constants provide an interaction between the functions on the algebra in sub-section (III-B).If those functions are the Hamiltonian and the momentum components, the equations of motion can be expressed in terms of the structure constants in sub-section (III-C).Finally in sub-section (III-D), the velocity matrix is found for the three Lie algebras.

A. Structure Constants
In many situations, the addition of two motions depend on the order in which they occur -rotate then move, or move then rotate.This is reflected in the non-associative matrix multiplication: AB = BA in most cases.Structure constants c k ij are used to describe this non-associative action in a Lie algebra g.They are defined using the Lie bracket by (see [4] p56) The value of the structure constants are easily shown by matrix multiplication to be, by using ∈ {1, 0, −1} for so (4), se (3) and so(1, 3),

B. Poisson bracket
In order to develop the coordinate equations for these Lie groups, it is necessary to introduce the Poisson bracket and show that it describes the structure in the same way as the Lie bracket.
The canonical form of the Poisson bracket is (see [5] p20 onwards) with F (p, q) and E (p, q) being functions on the cotangent space with canonical coordinates (q, p), representing position and momentum.To express this in other coordinates, write q = q (z i ) and p = p (z j ) and get, using partial differentiation, If these functions are pulled back to the origin, there is no positional dependence and so For a Lie algebra with a Poisson structure, it can be shown that {p i , p j } = −c k ij p k (see [5] p50) and the Poisson bracket describes the structure in a similar way to the Lie Bracket.Hence the Poisson relationship between any functions E and

C. Hamiltonian Flow
For a Poisson manifold, Bloch [1] p121 defines X H , the Hamiltonian vector field of H, as the unique vector field such that df dt = X H (f ) = {f, H} for all f From equation ( 8) above, the Hamiltonian vector field is and, for all functions f , With a Poisson structure naturally or imposed, the coordinate equation for Lie algebras is found by setting f = p i for i ∈ {1, 2, 3, 4, 5, 6} where {p i } are the components of momentum.The relationship with the velocities is p i = m i v i for i ∈ {1, 2, 3} and p i+3 = o i w i , where the linear velocities are {v 1 , v 2 , v 3 }, the angular velocities are {w 1 , w 2 , w 3 }, the moments of inertia are {o 1 , o 2 , o 3 } and the added masses are {m 1 , m 2 , m 3 } allowing for a different mass in each direction.The Hamiltonian for a natural motion, with no forces, is the kinetic energy so Velocity functions for the groups The velocity functions change with time in each direction as determined in this section and are shown in Table I.The velocities for E 3 SO (3) arise from the separate sub-groups and were found earlier.For the Lie groups, the differential equations for the momentum components are found using equation (10), the Hamiltonian given above and the structure constants found by matrix multiplication of the base matrices.First the Hamiltonian is written in terms of momentum components where {k i } = {o i w i } = {p i+3 } are the angular momentum components.The required differential equations for the momentum are, from equation (10) and the structure constants, In order to demonstrate the properties of the various groups, the following simplifying assumptions are made: m i = m for all i.The rotation is assumed to be about the axis of symmetry so o 3 = o 2 .The momentum equations become Continuing with the simple example from expression (1), the initial momentum components are The angular velocities are easily found to be constants The linear velocity components can be determined from These velocities are shown in Table I.In summary, the initial velocities remain unchanged in the semi-direct product.There is no interaction between the components.In all three Lie groups with the Lie-Poisson structure, the rotation induces a change in the linear velocities perpendicular to the axis of rotation.In SE(3) (with imposed structure), the frequency of the change is equal to the initial rotation w.In SO(4) and SO(1, 3), the frequency of the change is influenced by the ratio of the moment of inertia and the mass of the body.

IV. TRAJECTORIES DETERMINED FOR THE LIE GROUPS
Having found the velocity functions, they are integrated in this section to determine the trajectory.The resulting trajectories are analyzed in the next section.
From Table I using the base matrices shown in equation ( 6), the velocity matrix X is given by X (t) = v x e 1 +v y cos (w t + f ) e 2 +v y sin (w t + f ) e 3 +we 4 with w = 1 − o1 m w and ∈ {1, 0, −1} for so (4) , se (3) , so (1, 3) respectively.The trajectory is found by equation ( 2).Because the velocity matrix is time dependent, the trajectory is calculated by the forward Euler method of numerical iteration using the simple equation ( 3) where s is the step length and n is the step number so that X n = X (t n ) and t n = n s.
The figures show the displacements x, y and the rotation θ for the three Lie groups, with initial velocities {v x , v y , w} = {0.014,0.01, 0.09}, f = 0, o1 m = 0.6 and step length s = 0.1.The displacement along the axis of rotation as shown in Figure 2   The rotation of the body is fixed at the original angular velocity, as shown in Figure 3, the plot of sin (θ) where θ is the orientation at time t.
The trajectories in all 3 spaces approximate a screw motion about the original axis of rotation.The main difference in the trajectories is determined by the ratio of the moment of inertia and the mass: o 1 /m.With o 1 /m = 0.6, the difference in angular frequency and amplitude can be seen in Figure 4.The angular frequency is w y w = w (1 − o 1 /m) and the amplitude is y m v y /w y .
For all four combinations of linear and rotations, the linear and rotational displacement functions shown in Table II indicate the form of the solutions.There are additional small magnitude terms which add lower frequency variations and  curve the trajectory away from the main axis of the screw.The largest difference is in SO(4) and is shown in Figure 5.
To conclude this section, the trajectory functions shown in Table II are summarized.The motion determined in the semi-direct product R 3 SO (3) is that experienced in the body frame.In SE(3), with the imposed Poisson structure, the motion is in the spatial frame -see Figure 6.The actual and perceived distances traveled are both v y t.The perceived angle of rotation is the actual rotation of the trajectory plus the rotation of the body.The actual radius of travel (v y /w 5 ) is half the perceived radius since w 5 = 2w.In order to conserve angular momentum about the origin, the body performs a screw motion and the linear velocity changes in the spatial frame.In SO(4) and SO(1, 3), the radius of the screw motion is influenced by the ratio of the moment of inertia and the mass of the body.
The perceived trajectory (calculated using the semi-direct product) is independent of the actual trajectory (whether calculated using the Lie-Poisson structure of SE (3) or working with the uncoupled velocities).Further details are given in the Appendix.
Over longer time frames, the structure of the group causes the trajectories to diverge from this simple model.

V. COMPARING THE DIFFERENT TRAJECTORIES
In this section, the trajectories identified previously are compared in more detail.The configuration in SO (4) is shown Table II APPROXIMATE DISPLACEMENT FUNCTIONS FOR SMALL DISPLACEMENTS to approximate the configuration in SE (3) for small linear displacements.Thus SE (3) can be projected onto SO (4) (and vice versa).The induced Lie-Poisson structure has coupling of the velocities for SE (3).The variation in the frequency of the spiral motion across the groups is analyzed and found to be eliminated for small displacements.

A. Projecting SO(4) vertically onto SE (3)
Intuitively, the 2 linear displacements and one rotation of a planar body can be represented on the surface of a large sphere.Equivalently, the environment can be scaled down to a unit sphere, so that the linear displacement are small compared to 1, using some linear unit (such as kilometer or light-year).Similarly 3D spatial displacements can be approximated on the surface of a unit hypersphere.Three axes are the rotation axes of the body frame, and there is no constraints on the size of those rotations (θ 1 , θ 2 , θ 3 ).The small angular displacements (x, y, z) from the other axes are the spacial displacements, after scaling down by a factor R.
Any element g ∈ SO (4) can be written as are the two sub-elements (and a = 1 − x ).
The corresponding group element in SE (3) is The difference between the two groups is found from the exponential series expansion to be Ignoring the values in the first row, the difference is of order A similar projection can be done from SO (1, 3) onto SE (3).

B. Scaling the linear displacements
Since the projection from SO (4) to SE (3) is accurate to second order in x, a scaling factor S ∈ R (as suggested in [8]) can be chosen so that the plotted displacement x p of the rigid body in R 3 is x p = Sx and x is the calculated displacement which is kept within the required condition |x| < √ δ where the required accuracy is δ (a proportion).
This scaling has a consequence on the inertia terms.The moment of inertia is a function of the mass m and the square of the size of the body r.So scaling the linear dimensions to be small compared with the unit hypersphere reduces the moment of inertia to order of S 2 .
The angular frequency of the linear velocity terms tend to the angular velocity, as can be seen from Table I.
for all three groups as the accuracy of the projection increases.For small displacements, the three groups produce the same results.for {v x t, v y t} 1.This gives the displacement in the spatial frame.
The scaling factor S also has the effect of changing the relative weighting of the rotational and linear displacements.The linear displacements are shrunk and have less weighting in any comparison, such as distance (in 6 dimensions) traveled along the trajectory.This was demonstrated above -equation ( 12) -where the angular frequency tends to the same value irrespective of the linear size of the body.There is an innate difference between rotations and translations which cannot be overcome by some arbitrary metric.There is no bi-invariant Riemannian metric on SE (3).The requirements of each system will determine what (if any) a metric can be defined for that situation.

VI. DISCUSSION
Four methods of combining linear motion in 3 dimensions with a rotation have been considered.A single rotation is used since any 3 rotations about orthonormal axes can be replaced by a single rotation about one rotated axis, and if that axis is also an axis of symmetry, precession is avoided.The rotation of the axes necessary to achieve this is given on the Appendix.The linear motion results in linear displacements and these can be scaled by a factor S so that they are small compared to a unit hypersphere.Small linear displacements on a sphere are generated by small angular displacements.
The four methods considered are 1) semi-direct product, where the motion is described in the body frame.There is no coupling of the angular and translational velocities.This is often referred to as SE (3) because the integration of the velocities can be done using that group.2) SE (3) with induced Lie-Poisson structure.The angular momentum induces a change in the linear velocity.The resulting trajectory is in a fixed frame and is a spiral motion about an axis parallel to the original axis of rotation.The frequency of the screw motion is twice the original angular rotation.Whatever the frequency of the screw motion, the perceived motion in the body frame is always that produced by the semi-direct product.Higher frequencies reduce the radius of the screw maintaining the angular momentum.A frequency of w generates a radius of v y /w and maintains a velocity of v y around the screw -see Table II.3) SO (4) where the linear motions are interpreted as small angular displacements from three of the six axes.The frequency of the spiral motion is reduced as the motion of inertia about the original axis of rotation increases.When considering small linear dimensions, this reduction in frequency is eliminated -see equation ( 12).The trajectories for small linear displacements are the same as those produced using SE (3) with induced Lie-Poisson structure.The difference arises from the term x (t) = sin (v x t) compared with x (t) = v x t, which feeds back into the displacements in other directions.As the rate of change of x (t) decreases, the momentum is transferred to the other directions.The rate of change of y(t) increases faster.This is not shown in the approximate functions in Table II.4) SO (1,3).The same considerations apply to this group with the following differences.The frequency of the spiral motion is increased as the motion of inertia about the original axis of rotation increases.As the displacement x (t) = sinh (v x t) increases exponentially, the displacement in the other directions decrease expo-nentially.The displacement y(t) decreases by an extra term with the form of cosh function.

VII. CONCLUSION
Trajectories of a rigid body in Euclidean space are determined by the linear and rotational velocities, which can interact.This paper has separated the step of finding the velocities from integrating the velocities into a trajectory.The semi-direct product of the linear and rotational velocities consider the velocities separately without coupling them.The three 6D Lie groups use the Lie-Poisson structure to reflect the rotational influence on the linear velocity.The angular moments of inertia decrease the frequency of the spiral motion in the rotation group SO (4), have no impact in SE (3) and increase it in SO (1, 3).For small displacements, the impact on the linear velocity is the same for all three groups.
The velocities of the semi-direct product can be integrated using the matrix format of SE (3), and results in the usual identification between semi-direct product and Special Euclidean group.The resulting trajectory is in the body frame and this cannot be interpreted into the spatial frame, since the perceived trajectory is independent of the velocity coupling.The three Lie groups integrate the motion in a spherical, flat or hyperbolic space to give the trajectory in a spatial frame.For small displacements, these are the same (within any chosen degree of accuracy) and reflect the trajectory in a fixed frame.It differs from the body frame trajectory created by the semidirect product.
Any combinations of rotations can be represented by rotation about a single axis of rotation.If this axis is also an axis of symmetry, the rotational axis is fixed.In a natural system with no forces and a body symmetric about the axis of rotation, the analysis can be simplified by aligning one axis of the spatial frame with the rotation.
Due to the innate differences between rotational and translational motion, there cannot be a metric that applies to a rigid body motion in Euclidean space, without imposing some relationship between the relative motions which will depend on the system being measured.

A. ROTATION TO GENERALIZE THE MOTION
The simplified example of rotation about one axis, with initial linear motion along and perpendicular to that axis, can be generalized by rotating the coordinate frame.This is valid for SO (4), SE (3) and SO (1,3), and also SO (3) by dropping the first row and first column of the rotation matrix.
Assume that the generalized initial velocities are given by It follows that v 2 y + v 2 z is a constant and so the 2 linear components can be written as x cos (wt + f ) x sin (wt + f ) B. CONVERSION FROM BODY FRAME TO SPATIAL FRAME Selig [10] states that the body frame X and an alternative frame viewpoints X are related through the expression where u is the displacement vector between the two spaces and R is the relative rotation between them.The body frame trajectory (in the y-z plane) in the simple example of this paper is A trajectory with any value of n can be found by changing the displacement vector u.The radius of the trajectory in the spatial frame cannot be determined.This confirms the geometric diagrams provided in the paper.The body frame trajectory cannot be interpreted into a spatial frame without additional information.

Figure 1 .
Figure 1.Actual motion in determined using semi-direct product = c 6 21 = c 4 32 = − They are the same for the dual algebra, g * .
is a straight line for SE (3) (the dot-dash line).It has a sinusoidal function in SO (4) (the dotted line) with a

Figure 2 .Figure 3 .
Figure 2. Displacement along the axes of rotation

Figure 4 .
Figure 4. Induced screw motion about the axis of rotation

V = v 1 e 1 + v 2 e 2 + v 3 e 3 + w 1 e 4 + w 5 e 5 + w 6 e 6 3 i=1 3 i=1+ w 2 3 •linear momentum v 2 z +v 2 y = 3 i=1 v 2 i −v 2 x
By applying, a rotation R to this, the initial velocities are simplified toRV R −1 = v x e 1 + v y e 2 + v z e 3 + we 4 v i w i , the x component is fixed at v x = 1 w v i w i • v y = (v 2 w 3 − v 3 w 2 ) / w 2 2 Conserving