Sierpinski (<1K) Multivector Kinematics and Dynamics

    Kinematics is the study and representation of motion; dynamics envolves the notions of mass (inertia) and charge  addresses the reasons for motion. Dynamics includes the study of forces such as friction and gravity. The forces exerted on a body typically depend on the position, shape, and motion of the body, and the kinematical responce of the body to the applied forces depends on particular inertial properties of the body goverened by its mass distribution. The inertial properties determine how easily the body can be made to spin about, or accelerate along, a given axis. If the mass distribution of the body is fixed, its inertial properties remain constant when measured relative to the body.

    We can computationally model the motion of a body by determining the forces acting instantaneously on it at time t and assuming these forces to act uniformly on the body over the period [t,t+d]. The forces acting on the body may strive to fracture or deform the body and typically must be explicitly computed when determining kinematical responces so as to compare them with breaking or deformation thresholds. Assuming that the forces do not deform the body and that it remains rigid, the change in motion of the body due to them is independent of time and position and often also of body motion.
    We can think of the forces as imparting momentum to the body, and think of a body or a fluid possessing momentum by virtue of its motion. Momentum is ultimately a flow of mass. More generally, assigning any scalar property such as temperature or electric charge or baryon number to a body or to fluid particles generates a "current" of that property.

    We calculate the instantaeous forces acting on the body at t and consider them to apply uniformly over time interval [t,t+d] and model them by applying d times those forces to the body instanataneously at time t in the form of impulses individually effecting the body's motion so as to provide the desired change in momentum. We are, in effect, directly modelling the conservation of momentum.
    Gravitational forces are forces proportionate to the scalar inertial mass to which they are applied so that particulate accelerations are independant of particle mass.


    Suppose we have a 3D object H whose "centre" is located at a point c(t)ÎÂ3 and having orientation A(t)=(i,j,k) where i,j and k are t-dependant column vectors representing the "global coordinates" of the "orientation vectors" of the object. If i,j and k are always orthormal we have AAT=1 but more general A allow for "distorted" objects. Regardless of whether A is orthonormal, the "global coordinates" of a point having coordinates pL in the "local frame" of the object are given by p = c + ApL. Vector b and matrix A being expressed in global coordinates.
    The combination (c,A) representing a located (distorted) orientation is sometimes referred to as the frame of H but, as we use "frame" to describe an unpositioned set of N linearly independant ND 1-vectors , we will use the term stance for a located orientation. We will refer to a stance with an orthonormal orientation matrix as a rigid stance. Note that c is here the kinematical centre of motion which may be distinct from the dynamical centre of mass
     The conventional matrix method for representing 3D stances is to add a fourth "scaling" dimension to 3D point vectors with the understanding that r=(x,y,z,s)T=(x/s,y/s,z/s,1)T .
    A typical Â4×4 stance matrix would then be:
S= æ c1 ö
        ç A c2 ÷
        ç c3 ÷
        è 000l ø

where A is a conventional 3x3 transformation matrix. This acts on a 4D extended vector thus  
Sr= æ (Ar+c ö = æ (Ar+c)/l ö
è l ø è 1 ø

giving a translation by c and a scaling by l-1 in addition to the rotation and shearing effects of A.
These 4×4 transformation matrices combine according to the usual rules for matrix multiplication.

    In GHC we represent an ND stance with GHC multivector S = (½e¥c)R(-½le¥0) = (1+½e¥c)Rcoshl)- sinhl)e¥0) with rotor R Î e¥0* = i even and satisfying RR§=1 for a rigid stance . In the absence of a l dilation factor (-½le¥0) or an e¥0 negation factor,   stance S resides entirely within e¥* [ Recall that e¥* = -e¥* includes e¥ while excluding e0 ]
    When N=3, the R for a rigid stance is the standard quaternian representation of a 3D rotation.
    The GHC embedded point representaton of p is given by pG=SpLS-1 equivalent to S§(pL) º SpLS§ for a rigid stance. Typically Se¥S§=e¥ (or le¥ if S embodies some dilation) while Se0S§ = c = c+e0c2e¥.
    A direction aL in the local frame becomes scaled direction aG = lRaLR§ in the global frame . In GHC we can represent this as aGe¥ = SaLe¥S§ .
    The position and orientation factors of a stance rotor do not commute unless R§cR=c.

    S is nonunitary because since e¥=-2e0 we have (½e¥c)e¥c) = (½e¥c)e¥c) = (½e¥c) (e0c)    = (1+½e¥c)(1+e0c) . For Euclidean c this is 1+(½e¥+e0)c + ½c2 - ½e¥0c2 = 1+emc + ½c2 - ½e¥0c2 .
    This means that there is no antiHermitian logarithm for S as there is for R in a Euclidean space.

Composing Stances
    Programmers will typically represent the stance of an object relative to the stance of another "master" object. As an example consider a rotating gun turrent mounted in a bomber flying over a city at night. The stance of the gun turrent SQ¬T º SQ¬TQ would typically be implimented with respect to the "base frame" of the bomber. Assuming the bomber is orientated in the base frame with e1 as the forward direction, e2 as left, and e3 as upward, the turret might be considered as having a centre displaced by cTQ=(X,0,Z) from the centre of the bomber and an orientation RTQ=(-½fe12)qe13) corresponding to a roll-less turet. The notation aQ indicates that a is expressed relative to Q ( ie. "in Q coordinates"). Thus cTQ denotes the position of the centre of T in Q coordinate system. For a more general mobile turret, cTQ might be variable but confined to a particular k-curve fixed within the aircraft asuch as a straight rail or a hoop.
    The stance of the bomber S,¬Q might be expressed with respect to the stance of the Earth ,, perhaps with cQ, = [q,f,R,+h+a] where q and f are latitude and longitude coordinates, R, is the sealevel radius of an assumed spherical Earth, h is a ground height above or below sea level, and a is an altitude above ground level. The stance of the Earth S!¬, would typically be expressed with regard to the stance SSG of Sol which in turn would be expressed with respect to the galactic centre. All stances being rigid.
    Typically we neglect galactic rotation and consider the Sun to be at rest at 0G with identity orientation 1G so that SG¬!=1G=1 where G denotes the "global" or "galactic" rest frame of the "fixed stars" of the rendered night sky. We will refer to such "base frame" coordinates as the global coordinates with "global" being used here in the sense of "universal" rather than in relation to the "Globe" of the Earth.

    The global stance of the turrent is then given by the rotor
    SG¬T   =   SG¬!S!¬,S,¬QSQ¬T   =   S!¬,S,¬QSQ¬T     since SG¬!=1.
    Note that we compose or "cascade" relative stances from right to left rather than left to right with SA¬C = SA¬BSB¬C motivating the SB¬A º SB¬AB notation.
    Since SA¬B=SB¬A-1 is equal to SB¬A§ for a rigid stance, the stance of the aircraft __Acraft with respect to the city 4 is given by
    S4¬Q = S4¬,S,¬Q = S,¬4-1S,¬Q = S,¬4§S,¬Q which, when acting on a point in bomber coordinates, returns a point in city coordinates.

    Suppose we wish to emulate gunner Gus sitting in the turrent moving his head at least, and hypothesise a camera C positioned inside Gus' head having stance ST¬C camera. The stance of the city with respect to the camera, which we would use to render the city as seen Gus, is given by
    SC¬4 = SC¬TST¬QSQ¬4 = SC¬TSQ¬T§SQ¬4 converting city coordinates to camera coordinates.

    For N=3, the product of any number of <0;2>-grade normalised quaternian R is another such and the product of any number of <0;2;4>-grade stance rotors is likewise just another <0;2;4>-grade rotor within Â3% @ Â4,1. For N³4 however, combined stances accrue higher even grades.

    Stance rotors are applied to embedded points. If we wish to use a stance rotor S to rotate a scaled direction a, ie. to compute RaR§ rather than SaS§ options include evaluating (Sae¥S§)¿e¥ or SaS§ - Se0S§ = S(aa2e¥)S§ . The former approach extends to evaluating RxR§ for xÎUN as (Sxe¥S§)¿e¥. However, it is usually preferable to simply reclaim R from S , ideally logically as the unexteneded UN component of S rather than by evaluating (Se¥)¿e¥ (which assumes S has no e¥0 dilation component), and then calculating RxR§ within the unextended algebra.

    Let H be an N-D object moving through UN space and rigidly rotating about a motion centre which traverses a 1-curve path b(t). Let A(t) = R(t)§(A(0)) º R(t)(A(0))R(t)§ be the orientation matrix of the object at time t where unit even rotor R(t) satisfies R(t)R(t)§ = 1) . We can construct GHC stance rotor S(t)=(1+½b(t)e¥)R(t)). [  We define matrix-multivector products columnwise, treating matrix columns as seperate 1-vectors thus b(a1,a2,..)=(ba1,ba2,..) ; (a1,a2,..)b=(a1b,a2b,.. ) . Other multivector products (¿, Ù, ×, ...) are treated likewise ]

    For brevity we will omitt the explicit (t) and  write 0 for (0) so that A = RA0R§ and letting · denote differentiation with respect to t we have
    A· = -w×A = -w.A = A¿w where w º -2R·R§ is a pure bivector known as the classical spin or spin (aka. rotational bivelocity or angular velocity bivector) with R· = -½wR and R·· = -½(w· - ½w2)R .
    Writing wL = R§wR = -2R§R·R§R = -2R§R· for the local spin expressed in the frame of the object, we have R· = -½RwL .
[ Proof :   RR§=1 Þ R·R§ + R(R§)· = 0 Þ R·R§ = -R(R§)· = -(R·R§)§ so w=-w§ and is even for even R, so must have grade <2;6;10;14;...>.
    a· = (Ra0R§)· = (R·a0R§ + Ra0(R§)·) = (R·R§)a - a(R·R§) = 2(R·R§)×a = -a¿2(R·R§) = a×w = a¿w is a 1-vector for all 1-vector a(t) so w× must either preserve the grade of or annihilate every 1-vector. Hence w has zero <6;10;...>-grade components.  .]

    For N=3 we have the conventional rotational velocity 1-vector w=w*=we123-1 with a· = -w.a = -(w*).a = -(wÙa)i = -w×a = a×w.

    In GHC we have stance spin W º -2S·S§   =   (1+½e¥b)W(1-½e¥b)   =   (1+½e¥b)w(1-½e¥b) + b·e¥ where bivelocity W º w + b·e¥ embodies both the velocity b· and the spin w with W2 = w2 + 2b·Ùw generally nonscalar even for N=3 .
[ Proof :  S· = ½e¥b·R + (1+½e¥b)R·   =   (½e¥b· - (1+½e¥bw)R   =   ½(1+½e¥b)((1-½e¥b)e¥b· - w)R   =   -½(1+½e¥b)(b·e¥ + w)R
    Þ W º -2S·S§     =   (1+½e¥c)W(1-½e¥c)  .]

Reclaiming Velocities from Spins

    The nonnull 1-velocity pG· = pG· + (pG¿pG·)e¥ perpendicular to null pG is given by pG·   =   pG¿W   + S§(pL·) .
[ Proof : pG·   =   (SpLS§)·   =   S·pLS§   + SpL·S§ + SpL(S·)§   =   -½WSpLS§   + SpL·S§ - ½SpLS§W§
      =   -½WpG   + SpL·S§ - ½pGW§   =   -½(WpG-pGW)   + SpL·S§   =   pG¿W   + S§(pL·)  .]

    Now (p-b+e0)¿W = b· + (p-b)¿w + ((p-b)¿b·)e¥ = v + ((p-b)¿b·)e¥ is the instantaneous 1-velocity of p º pG but with an additional ((p-b)¿b·)e¥ component . We can eliminate this by representing velocity v with 2-blade v=ve¥ with
    ((p-b+e0)Ùe¥)×W = ((p-b+e0)¿W)e¥ = (b· + (p-b)¿w)e¥   providing the instamtaneous 2-blade velocity of body point p.
[ Proof :  ((p-b+e0)Ùe¥)×W = ((p-b)e¥-e¥0)×W = ((p-b+e0)Ùe¥)×W = ((p-b+e0)e¥)×W = ((p-b+e0)×W)e¥ = ((p-b+e0)¿W)e¥  .]
     Note that (p-b+e0)Ùe¥ = (p-b+e0+½(p-b)2e¥)Ùe¥ can be regarded as pcÙe¥ where pc is the GHC embedding of p-b interpreted as a point.

Iterating Stances

    Suppose we have S and W at a particular time t and we wish to evaluate S(t+d) for a small timestep d.
    We know S· = -½WS but setting S(t+d) = S(t)-½dW(t)S(t) = (1-½dW(t))S(t) results in S(t+d)S(t+d)§ = 1-¼d2W(t)2 intead of unity.
    The correct iteration is S(t+d) = (-½dW) S , or S(t+d) = S(-½dWL) for WL expressed in stance local (source) coordinates. This preserves the unit roticity of S(t+d) but since W2 is nonscalar it is computationally profligate and the simpler approach is to impliment velocity and spin seperately as S(t+d) = (1+½(c+dc·)e¥)(-½w(t))R(t) with w(t)2 scalar for N£3.

    To form W(t+d) given W·(t) we can take W(t+d)=W(t) + dW·(t) using bivector addition without subsequently violating the roticity of S but we may instead wish for acceleration control to be a precession and amplification of spin rather than simple addition of spin.
    Typically W·(t) will be some bivector "accceleration" function Q(c,c·,R,R·,t) = Q(W,W·,t) modelling the forces and torques acting upon the object at time t such as gravitational weight, air resistance, lift, engine thrust, and so forth.
    However we compute W(t+d), better O(d2) agreement can be obtained by evaluating w(td), forming S(t+d) as (1+½(cdc·)e¥)(-½w(td))R(t) , and then computing W(t+d).

    Thus we must iterate L timesteps multiplicatively with R(Ld) = (-½w((L-1)d) (-½w((L-2)d)...(-½w(d))(-½w(0))R(0) approximating R(Ld) by a particular sequence of L discrete rotations of intial state R(0). This generally differs from
     RS º (-½(w((L-1)d)+w((L-2)d)+..+w(d)+w(0)))R(0) approximating (-½ ò0Ld dtw(t))R(0)(L-1)d)+w((L-2)d)+..+w(d)+w(0)))R(0) because the w(ld) do not commute.
    RS can be regarded as a superposition of all R reachable by any combination of the particular L subrotations, with repeated uses allowed.

UN Spinor Representation
    We have here represented UN% position c as (½e¥c)§(e0) but provided c2 has the same sign as e12  we also have c = Ce1C§ where C = (c2)¼q(cÙe1)~) is a nonunit <0;2>-grade  rotor in UN satisfying (CC§)2 = e1-2c2 with C-1 = (e12)½ c-1C§ . Here q is the angle subtended betweem c and e1. We  can recover c=CC§ ; q = 2 cos-1(C<0>) ; (cÙe1)~ = C<2>~ .
    C is not uniquely determined by c because CA where A is any unit rotor with Ae1A§=e1 also serves. For N=3, A=(fe1i-1) incorporates a single arbitary phase angle but for N>3 we have more general guage rotations "about" e1.
    C· = ½c-1 c·Ce1
[ Proof :   c· = C·e1C§ + Ce1C§· = C·e1C§ + (C·e1C§)§ = 2(C·e1C§)<0;1;4;5;..> but c· is a 1-vector so c· = 2C·e1C§ = 2Ce1C·§    Þ c·Ce1 = 2C·  .]

    Since c = |c2|½ = CC§ we have c· = 2C·*C§ .
    In GHC we have S = (1+½e¥Ce1C§)R = C(c-1+e¥1)C§R |e1C§)R = c-1C(1+ce¥1)C§R   =   c-1C(ce¥1)C§R   =   C~(ce¥1)C~§R

    We can define monotonically increasing spinor time s = ò0t dt r-1 regardable as an "temporally accummulated closeness measure" of c to 0 with ds/dt=c-1 so that dC/ds º C'   =   C· dt/ds = cC· = ½c·Ce1 .
    Whence C'' = ½(c··c + ½c·2)C .
[ Proof : C'' = cC'· = ½c(c·Ce1)·   =   ½c(c·Ce1)· = ½c(c··Ce1 + c·C·e1) = ½c(c··Ce1 + ½c-1c·2Ce12) = ½(cc··Ce1C-1 + ½c·2e12)C = ½(cc··c~ + ½c·2e12)C = ½(c··c + ½c·2)C .  .]
    Now suppose that the acceleration has the form c·· = a - m'c-2c~ for an attractive inverse square force plus peturbation a.
    We then have a geometric form of the Kustaanheimo-Stiefel equation
    C'' = ½(ac - mc-1 + ½c·2)C which for a=0 has harmonic form C'' = ½EC where energy E= ½c·2 - mc-1 is conserved .

    The key advantage to working with C(s) rather than c(t) is that solutions to the unpeturbed harmonic equation C'' = ½EC linearly combine whareas solutions to Newtonian equation c·· = -mc-3c do not.
    We can reparameterise the orientation rotor R(t) as a function of spinor time R(s) with R'(s) = cR· .

    For E<0 corresponding to |c·| less than escape speed (2mc-1)½ we have oscillatory solution C(0) cos((-½E)½s) + C·(0)(-½E)) sin((-½E)½) of frequency (-½E)½ and spinor time period 2p(-½E) = 8p(-E) . Furthermore C(t) = cos( (-½E)½ s) C0 + sin( (-½E)½ s) C1 provides a solution with C(0)=C0 and Cp(-½E)) = C1 which is the natural constant energy trajectory from c(0) to c(T), having Cds(0) = (-½E)½C1 and Cds(½p(-½E)) = -(-½E)½C0.
    The arrival time t= ò0½p(-½E) ds r(s) is approximately ½p(-½E) r(0) for slow movement at large r and differs from the desired arrival time T.
    This suggests an alternte form of "initial conditions" in which we specify C0=C(0), E, and one of C±    specifying a "detemporalised" position passed through.
    A more general solution is  cosh(sbC0 + b-2 b  sinhsb (b) where b2 = ½E . For a single N=3 particle, the angular momentum h0 defines the orbit 2-plane so the motion is effectively 2D and it is natural to tahe b=¼|E|½h0-1h0 where h0=|h02|½ .

    Writing S(s) = (1+½e¥Ce1C§)R = (1+½Ce¥1C§)R = R + ½Ce¥1C§R and noting that R'' = (cR·)' = c(cR·)· = cc·R· + c2R··
    R' = cR· = -½cwR.
    R'' = cR'· = -½cc·wRccw·Rcw2R = -½c(c·w+cw·w2)

GHC spinor form

    A GHC variant of spinorising position in this way is to express a given UN% 1-vector y=ls  dual to a hypersphere of positive squared radius s2=r2>0 as as a dilated rotatation of the unit hypersphere dual as Be+B§ where B=(lr)½q(ye+e+)~) , q being the angle subtended by s and e+ , with cos(q) = r-1s¿e+ = ½r-1(-1+c2-r2) and sÙe+ = ce+ - ½(1+c2-r2)e¥0.
    Our spinor time is now s= ò0t dt(lr)-1 with B'' = ½(y··y + ½y·2)B .
    Now y· = l(c· + e¥(cc· - rr·)) + l·s     with y·2 = lc·2 + l·2r2 + l2(c·¿c) , and
    y·· = l(c·· + e¥(cc·· + c·2 - rr·· - r·2) + 2l·(c· + e¥(cc· - rr·)) + l··s     has e0 coordinate l·· and e¥ coordinate l(cc·· + c·2 - rr·· - r·2) + 2l·(cc· - rr·)) + ½l··(c2-r2) .

    Taking l=1 to keep y within e0¿y=1 we have
    y··y = s··s = (c·· + e¥(cc·· + c·2 - rr·· - r·2)(c+e0+ee½(c2-r2)) = c··c + (e¥c-1+e¥0)(cc·· + c·2 - rr·· - r·2) + c··e0 + + c··e¥(½(c2-r2)

    y·· = M(y)y requires l··=M(y) and hence lc·· + 2l·c· + ½l··c = M(y)lc Þ c··   =   ½M(y)c + 2l-1l·c· .
    If we want mc·· = m'(c)c~ = c-1m'(c)c we thus need M(y) = 2c-1m'(c)  and hence l· = ò0t dt   2c-1m'(c)  = ò0c dc   (c·)-1 2c-1m'(c)  =

    Setting l=r-1 for unit y gives l·=-r-2r· ; l·· = 2r-3r· - r-2r·· = r-2(2r-1r· - r··)

s··s = c··c + c··e0 + ½c··e¥(c2-r2) +   (cc·· + c·2 - rr·· - r·2)e¥(c+e0).
    If r(t) is constant then s·s = s·Ùs is a pure 2-blade with (s·s)2 = -s·2s2 = c·2r2 .
    The radial acceleration assumption

Subbody motion


    Returning to our aircraft gun turret example, and assuming for generality that the turret can traverse within the aircraft as well as rotate we have W4¬T4   =   W4¬Q4 + (1+½e¥cQ4)WQ¬T4 (1-½e¥cQ4)   =   W4¬Q4 + (1+½e¥cT4)WQ¬T4 (1-½e¥cT4)
      =   (1+½e¥cT4)§( W4¬Q4 + WQ¬T ) .
[ Proof :  W4¬T   =   -2S4¬T·S4¬T§   =   -2(S4¬QSQ¬T)·(S4¬QSQ¬T)§   =   -2(S4¬Q·SQ¬T + S4¬QSQ¬T·) SQ¬T§S4¬Q§
      =   W4¬Q + S4¬QWQ¬TQS4¬Q§   =   W4¬Q + (1+½e¥cQ4)R4¬Q ((1+½e¥cTQ)WQ¬TQ (1-½e¥cTQ) R4¬Q§ (1-½e¥cQ4)
      =   W4¬Q + (1+½e¥cQ4) (1+½e¥(cT4-cQ4))R4¬Q WQ¬TQR4¬Q§ (1-½e¥(cT4-cQ4)) (1-½e¥cQ4)
      =   W4¬Q + (1+½e¥cT4)R4¬QWQ¬TQR4¬Q§ (1-½e¥cT4)   =   W4¬Q + (1+½e¥cT4)WQ¬T4 (1-½e¥cT4)  .]

    The velocity of the turret is vT4   =   vQ4 + R4¬Q§( vTQ + cTQ¿wQQ ) while its spin is wT4=R4¬Q§(wQQ+wTQ) giving bivelocity
    WT4 = (vQ4 + R4¬Q§( vTQ + cTQ¿wQQ) )e¥ + R4¬Q§( wQQ + wTQ )   =   WQ4 + R4¬Q§( (vTQ + cTQ¿wQQ )e¥ + wTQ ) .

    Here cT is the position of the kinematical motion centre or attachment point of the turret rather than the centre of mass of the turret  considered in isolation. Similarly cQ is typically a fixed reference "centre point" of the aircraft rather than its dynamical mass centre which may move with respect to cQ due to fuel slosh, turret movement, and so forth.

    The instantaneous apparent bivelocity of the city 4 as percieved by the turret T is
    -WTT   =   -RT4§§(vQ4)e¥ - RTQ§( (vTQ+cTQ¿wQQ )e¥ + wQQ + wTQ )RTQ .
[ Proof : WTT   =   RT4§ WT4 RT4   =   (RQ4RTQ)§ WT4 (RQ4RTQ)   =   RTQ§ RQ4§ WT4 RQ4RTQ
      =   RTQ§ RQ4§ ( vQ4e¥ + RQ4§( vTQ+cTQ¿wQQ )e¥ + wQQ + wTQ) RQ4RTQ
      =   RTQ§RQ4§ vQ4 RQ4RTQe¥ + RTQ§ (vTQ+cTQ¿wQQ )e¥ + wQQ + wTQ) RTQ  .]

Rotating Earth Frame

    Consider S!¬,, the stance of the spinning Earth (,) with respect to the Sun (!). We will neglect the other planets and moons and assume the sun to be at rest at the centre of the solar system with the earth to orbitting it, rather than the more correct orbit of the sun-earth mass centre.
    To construct S!¬, it is useful to form an intermediate heliocentric stance which is centered on the Earth as it orbits the sun but does not include the Earth's own daily spin. This is
    S!¬,   =   (1+½e¥R,!De1R,!§)R,!   =   R,!(1+½De¥e1) where D(t) is the distance from ! to ,   and R,!=(½w,te12) where w,=2p year-1 . Thus the , frame has e3 constant normal to the earth orbit plane in the solar north direction and e1 always pointing directly away from the sun while S,¬, = R,¬,, is a Â3 quaternian rotor having period one solar day.
    S!¬,! = S!¬,! S,¬,, has positional velocity component as for S!¬, and rotational velocity component R!¬,!  =  R!¬,!R,¬,, with
    R!¬,!·   =   R!¬,!·R,¬,, + R!¬,!R,¬,,·   =   ½(w!¬,!R!¬,!R,¬,, + R!¬,!w,¬,,R,¬,,)
      =   ½(w!¬,!R!¬,! + R!¬,! (R!¬,!§ w,¬,! R!¬,!) R,¬,,)   =   ½(w!¬,! + w,¬,!)R!¬,!
    Hence w!¬,!   =   w!¬,! + w,¬,! as expected and
    w!¬,,   =   w!¬,, + w,¬,,   =   R!¬,!§w!¬,!R!¬,! + w,¬,, since w,¬,, = w,¬,, .

    w!¬, º w!¬,! is approximately constant providing the Earth's combined spin axis w,! = w!¬,e123-1 in the assumed inertial Sun frame. The Earth rotates about this axis with period one sidereal day. The sidereal day is shorter than the solar day (by approx 3.6 min) because the Earth spins in the same sense as it orbits the sun (clockwise as seen from below the solar south pole, looking toward the solar north pole along e3!) , albeit about a spin axis inclined at approx 24 o to the orbit axis.  

Rigid Body Dynamics

Rigid Bodies
    Consider a set H of point-masses moving in ÂN, and suppose at time t the ith particle has mass mi, position ri, and velocity vi where i ranges through some indexing set I. Let m º åimi denote the total mass and c º m-1 åimiri denote the instantaneous centre of mass at time t.
    The condition for these particles to form a rigid body is that vi º ri· = v + (ri-b)¿w     where 1-vector drift velocity v = b· , 2-vector spin w, and spin centre point b may vary with t but are the same for each particle.
    In GHC we have (ri-b+e0)¿W = (ri-b+e0)×W = rbi¿W = vi + ((ri-b)¿v)e¥ where stance bivelocity W=ve¥+w and rbi º ri-b + e0 + ½(ri-b)2e¥ .
    Hence åimi(ri-b+e0)¿W = åimivi + m((c-b)¿v)e¥ and we can eliminate the e¥ compoenent by considering 2-blade velocity vi º vie¥ = viÙe¥ = vi×e¥ = ((ri-b+e0)×W)×e¥ .

Kinetic Energy

    The classical kinetic energy of a particle having velocity vi and scalar mass mi³0 is the scalar Ei = ½mivi2 , though this is actually a |v|<<c approximation to the relativistic E = mic2(1+(c-1vi)2)½ - mic2 = ½mi(vi2 - ¼c-2vi4 + O(c-4vi6)) where c denotes the speed of light.

    The net classical kinetic energy for a rigid body with drift velocity v rotating with spin w about b is thus
    Ew,v   =   åi½mi(v + (ri-b)¿w)2   =   ½mv2 + åi½mi((ri-b)¿w)2 + åimiv¿((ri-b)¿w)   =   ½mv2 + åi½mi((ri-b)¿w)2 + mv¿((c-b)¿w) .
    If w is a 2-blade with w2=-w2 then the middle term is åi½mi(¯w(ri-b)w)2   =   -åi½mi(¯w(ri-b))2w2   =   ½Iw,bw2 where scalar
    Iw,b º åimi(¯w(ri-b))2 is the moment of inertia for the body about b "within" 2-blade w, regardable as spining about an "axis" of "direction" w passing through b . As usual, ¯w(x) º (x¿w)w-1 denotes projection into w.

    Thus an ND body spinning about its centre of mass (b=c) with 2-blade spin w has kinetic energy E = ½mv2 - ½mIww2 = ½m(v2 ± (kww)2) according as w2 = -/+w2 ; where Iw º Iw,c and scalar radius of gyration kw,b º (m-1Iw,b)½ is defined so that Iw,b = mkw,b2 .
    For N>3 and a general 2-vector spin w the åi½mi((ri-b)¿w)2 energy term is more complicated.

    The linear momentum of a particle having position ri Î ÂN and scalar mass mi³0 moving with velocity viÎÂN at time t is just mivi.
    Its 2-blade angular momentum about a point a, or line L={a+ln : lÎÂ} is ha = (ri-a)Ù(mivi), or ((ri-a)-(ri-a)¿n)Ùmivi respectively.
    The angular momentum of a system about a point or line is the sum of the angular momenta of its constituent particles about that point or line while the linear momentum of a system is the sum of the linear momenta of its constituent particles.
    We have the following two principles of Newtonian mechanics:

  1. The rate of change of linear momentum of a system is equal to the sum of the external forces.
  2. The rate of change of angular momentum of a system about: (a) a fixed point or line, or (b) a point or line moving with the centre of mass of the system, is equal to the total moment of external forces about that point or line.
    A particle of mass mi at position ri with instantaneous velocity vi=v+(ri-b)¿w has 2-blade angular momentum about point a given by
    hia,b(v,w) º mi(ri-a)×vi   =   mi(ri-a)Ùvi   =   mi(ri-a)Ù(v+(ri-b)¿w)   =   mi(ri-a)Ù(v+(ri-c)¿w) + mi(ri-a)Ù((b-c)¿w)
      =   mi( (ri-c)Ù((ri-c)¿w) + (ri-c)Ùv + (c-a)Ù(v+(ri-c)¿w) + (ri-c)Ù((b-c)¿w) + (c-a)Ù((b-c)¿w) ) so that
    hic(v,w) º hic,c(v,w)   =   mi( (ri-c)Ù((ri-c)¿w) + (ri-c)Ùv )
    Since a×(a×b) = aÙ(a¿b) = aÙ(ab) = a2b - a¿(ab) = a2b - a¿(aÙb) = a2b - a(aÙb) has dual (aÙ(a¿b))* = a2b* - a(a¿(b*)) , we have
    ((ri-c)Ù((ri-c)¿w))*   =   (ri-c)2w* - (ri-c)((ri-c)¿(w*)) so
    hic(v,w)   =   mi( (ri-c)2w - i2 (ri-c)((ri-c)¿(w*))* + (ri-c)Ùv ) .

    In GHC we have (raiÙe0)×(vie¥) = ((ri-a)e0+½(ri-a)2e¥0)×(vie¥) = (ri-a)Ùvi + ((ri-a)¿vi)e¥0 - ½(ri-a)2vie¥ = (ri-a)Ùvi + (((ri-a)2)· + (ri-a)cnta·)e¥0 - ½(ri-a)2vie¥ where raiºri-a+e0+½(ri-a)2e¥ and raiÙe0 can be regarded as the bipoint {0,ri-a}. We can eliminate the unwanted terms by forming 3-blade   rotational momentum mi((raiÙe0)×(vie¥))Ùe¥ = mi(ri-a)ÙviÙe¥ but it is more natural to form 3-blade bimomentum mi(ri-a+e0)Ùvi = mi((ri-a+e0)Ùvi)Ùe¥ = mi((ri-a)Ùvi)e¥ + mivie¥0 containing both the rotational momentum mi((ri-a)Ùvi)e¥ about a and the linear momentum mivie¥0 as 3-blades.
    A GHC body bimomentum is thus a 3-vector formed by the taking the outter product of an e0* 2-vector with e¥,
    Hia,b(W) º mi(ri-a+e0)Ù((ri-b+e0)¿W)Ùe¥   =   mi( (ri-a+e0)Ù((ri-b)¿w) + (ri-a+e0)Ùb· )Ùe¥   =   (hia,b(b·,w)   + mie0Ù((ri-b)¿w+b·) )Ùe¥ embodying both angular momentum about a and linear momentum mi((ri-b)¿w+b·).

  =   mi( (ri-c+e0 + c-a)Ù((ri-c+e0 + c-b)¿W)Ùe¥
      =   mi( (ri-c+e0)Ù((ri-c+e0)¿W) + (c-a)Ù((ri-c+e0)¿W) + (ri-c+e0)Ù((c-b)¿W) + (c-a)Ù((c-b)¿W) )Ùe¥
      =   mi( (ri-c+e0)Ù((ri-c)¿w) + (c-a)Ù((ri-c)¿w) + (ri-c+e0)Ù((c-b)¿w) + (c-a)Ù((c-b)¿w)
    + (ri-c+e0)Ùb· + (c-a)Ùb· )Ùe¥

Inertia Tensors

N-D Inertia 2-Multiform
    The net linear momentum of a rigid body of mass centre c spinning about b is m(v + (c-b)¿w) while its net angular momentum about a is
    ha,b(v,w)   =   åihia,b(v,w)   =   åihia,bmi(ri-a)Ù((ri-b)¿w)   =   åi(mi(ri-c)2w - mi(ri-c)((ri-c)¿w)) + m(c-a)Ùv + m(c-a)Ù((b-c)¿w)
      =   hc(w) + m(c-a)Ù(v + (b-c)¿w) .
    The Inertia 2-multiform returning the angular momentum about the mass centre for the body due to a 2-vector spin w about its mass centre (a=b=c) is the linear function of w defined by
    hc(w) º åimi(ri-c)Ù((ri-c)¿w)   =   åimi(ri-c)×((ri-c)×w)   =   åimi((ri-c)2w - (ri-c)((ri-c)¿w)) .
    This third expression can be used to extend hc over scalars with hc(l) º låimi((ri-c)2 .
    For 2-blade w we have hc(w)   =   åimi(ri-c)¯w(ri-c)w   =   (Iw +  åimi^w(ri-c)¯w(ri-c))w .
    For nondegenerate solid bodies with mi>0 that extend into N0 dimensions, linear hc(w) is nonzero for every nonzero w and so invertible.
    hc(w) is symmetric (self adjoint) in that c.hc(w) = w.hc(c) = hc(c).w and has postive real eigenvalues for postive mass mi>0.
    We naturally decompose w(t) = åk=1½N(N-1) fk(t)fk(t) where fk(t) = R§(¦k(0)) are the unit principle spins.
    Letting D0(c) = åkIk(c¿fk(0))fk we have hc(c) = R§(D0(R§-1(c))) º R§D0R§-1(c) and hc-1(c) = R§-1(D0-1(R§(c)))

    Because hc is symmetric, the squared angular momentum hc(w)2 = w¿hc2(w)   =   åk=1½N(N-1)  ±k Ik2 fk2 while
    w¿hc(w)   =   åk=1½N(N-1)  ±k Ik fk2   =   -åimi((ri-c)¿w)2   =   -2Ew,0 provides the rotational kinetic energy. Here ±k denotes the sign of fk¿fk .
[ Proof :  Symmetry follows from c.((ri-c)Ù((ri-c).w) = ((ri-c).w).(c.(ri-c)) = (c.(ri-c)) . ((ri-c).w) = ((ri-c).c).(w.(ri-c)) = w.((ri-c)Ù((ri-c).c)  .]

    The time derivative hc·(c)   =   hc(w×c) - w×hc(c) .
[ Proof : åimi((ri-c)×((ri-c)×c))·   =   åimi((ri-c)·×((ri-c)×c) + (ri-c)×((ri-c)·×c)   =   åimi( ((ri-c)×w)×((ri-c)×c) - (ri-c)×(c×((ri-c)×w)))
      =   åimi( (ri-c)×(w×(c×(ri-c))) + ((ri-c)×w)×((ri-c)×c) - (ri-c)×( w×(c×(ri-c)) + c×((ri-c)×w)) )
      =   åimi( (ri-c)×( ((ri-c)×c)×w + ((ri-c)×c)×(w×(ri-c)) - (ri-c)×( w×(c×(ri-c)) + c×((ri-c)×w)) )
      =   åimi(-w×((ri-c)×((ri-c)×c)) + (ri-c)×((ri-c)×(w×c)))   =   -w×hc(c) + hc(w×c)  .]

    In the absence of external torques w thus satisfies Euler equation hc·(w) + hc(w·) = 0 Þ w· = hc-1( hc·(w))
    Now w×hc(w)   =   åiåj Ij fi fj fi× fj   =     =   åi<j (Ij-Ii) fi fj fi× fj but it is more useful to exploit the conservation of hc(w) with
    w· = hc-1( -w×h0)   =   hc-1( -åi<j (Ij-Ii) fi(t)fj(0) fi(t)×fj(0) )
      =   -åk Ik-1 hc-1( -åi<j (Ij-Ii) fi(t)fj(0) (fk(t)¿ (fi(t)×fj(0) )) fk
    Now (fk(t)¿ (fi(t)×fj(0) ))   =   R§(fk(0))¿ (R§(fi(0))×fj(0) ))   =   fk(0))¿ R§-1(R§(fi(0))×fj(0) )))   =   fk(0))¿ R§-1(R§(fi(0)))× R§-1(fj(0) )))   =   fk(0))¿ (fi(0)× R§-1(fj(0) )))
    But w· = -2R··R§ -2R·R·§ so w·R = -2R·· -2R·R·§R so .... ????

    When N=2, setting w=w*=we12-1 we have body linear momentum m(v+(c-b)w) = m(v+w(c-b)e12) and pseudoscalar angular momentum hc(w)=hc(w)e12 with dual scalar angular momentum hc(w) º hc(w)e12-1 = wåimi(ri-c)2 ; although a third spacial dimension e3 is often invoked to "hold" 1-vector angular momentum and spin hc(w) = whc(e3)e3 .

    When N=3, setting w=w*=we123-1 we have body linear momentum m(v+(c-b)¿w)   =   m(v+((c-b)Ùw)-*)   =   m(v-(c-bw)   =   m(v+w×(c-b)) and 2-blade body angular momentum hc(w) with dual angular momentum 1-tensor
    hc(w) º hc(w_idual)*   =   åimi((ri-c)2w - (ri-c)((ri-c)¿w)   =   åimi((ri-c)×(w×(ri-c)))   =   Bcw    where 1-vector spin w º we123-1 and symmetric Â3×3 matrix  
    Bc º Simi((ri-c)21-(ri-c)(ri-c)T)   =   Simi((ri-c)Ä(ri-c)) is the matrix representation of the instantaneous inertia tensor of H about its mass centre.
    Once again we can extend hc() over scalars with hc(l) = låimi((ri-c)2 .
    Note that a¿hc(a) =   =   åimi((ri-c)2a2 - ((ri-c)¿a)2) provides åimi((ri-c)2a2 = a2åimi(ri-c)2 - a¿hc(a) .
    Time derivative hc·(c*) = w×hc(c) - hc(cw
[ Proof : hc·(c)*   =   hc(c×w)* - (w×hc(c)*   =   (((hc(c)*)×(w*))* - ((w*)×(hc(c)*))*   =   ((hc(c)×w)* - (w×hc(c))*   =   -hc(cw + w×hc(c)  .]
    More generally for any a we have symmetric matrix (with non-negative elements on the leading diagonal) which we may write explicitly as
Ba=Simi æ (yi-ay)2+(zi-az)2   -(xi-ax)(yi-ay)-(xi-ax)(zi-az) ö
ç -(xi-ax)(yi-ay)(xi-ax)2+(zi-az)2-(xi-ax)(yi-ay) ÷
è -(xi-ax)(zi-az) -(yi-ay)(zi-az) (xi-ax)2+(yi-ay)2 ø

    It follows from a mathematical result concerning the diagonalizability of symmetric matrices, that for any body H there will be a set of principle axies OXYZ in which Ba has the diagonal form
Ba= æ Ix00 ö
ç 0Iy0 ÷
è 00Iz ø
greatly simplifying our matrix multiplications. Thus in 3D every rigid body is inertially equivalent to an ellipsoid and a 3D rotational inertia tensor is representable by a unit quaternian rotor and three positive scalars _Var(I1),_Var(I2),_Var(I3).
    The principle axies of H are the eigenvectors of both H and H-1 and finding them is not always easy (although axies of symmetry are always principle axies, as are normals to planes of symmetry for points in the plane of symmetry).

    Let f1(0), f2(0),..f½N(N-1)(0) be the ½N(N-1) principle spins for the at t=0 so that fk)(t) = R(t)fk(0)R(t)§ is the _kth principle spinat time t. The GHC 3-vector bimomentum is åjIj R(t))fj(0)R(t)§ e¥ + mc·e¥0 = = m(åjkj2 R)fk(0)R§ + e0c·)Ùe¥ .
    If 3D body H has a rotational symmetry axis k then hc(w) = h1w + (h3-h1)(k-1¿w)k . If k2=1 then B=h11 + (h3-h1)kkT .

    For general N the inertia tensor has a symmetric ½N(N=1)×½N(N-1) matrix form which we can diagnonalise with ½N(N-1) orthogonal (w¿c=0) eigen 2-vector principle spins forming a principle biframe and define ½N(N-1) gyration radii (m-1(fk(0)¿fk(0))-1   hc(fk)¿(fk))½ where fk is the _kth eigenbivector. The gyration radii can be regarded as eigenvalues for mhc½(c). . We say the body is assymetric if the ½N(N-1) diagonal eigenvalues are distinct and centrocentric if they are all the same.

    For N=3 the principle biframe is dual to a principle frame comprising 3 orthogonal 1-vectors. For N=4 or 5 the principle biframe emodies a principle frame of N 1-vectors but carries more information. For N=6 there is no natural way to construct 1-vectors from 2-vectors and we have only the principle biframe.
    For N>3 the principle biframe basis of orthogonal 2-vectors is more complicated than an N-D rotation of an orthogonal 2-blades basis but for N=3 it is dual to a rotated 1-vector basis so any body is inertially equivalent to a rotatated ellipsoid shell.
    Now hc·(w)   =   hc(w)× |w . so the rate of change of angular momentum (hc(w))·   =   hc·(w) + hc(w·) is hc(w)× |w for constant spin w·=0 and since hc(w)Ùw=0 this means w must be an eigen 2-vector of hc.
    Thus, perhspas surprisingly, in the absence of external torques, for an assymmetric N-D body constant spin is only possible for multiples of the ½N(N-1) principle spins so that for a 3D assymetric body constant spin is only possible about the three principle axies. More generally, in the absence of applied torques, spin w(t) varies in such a way as to keep angular momentum hc(w) = j constant implying w¿hc2(w) = j¿j. The rotational energy w¿hc(w) = w¿j = -2E is also constant.

    For N=3 this corresponds to w restricted to the 1-curve intersections of the two coaligned elliposids w¿hc2(w) = j2 and the energy ellipse w¿hc(w) = 2E tangent to the invariable plane w¿j = 2E .
    Such ellipsoids intersect in simple closed loops known as polhodes so precession tends to be periodic in that wL . When two gyration radii are equal, The projections of the energy and squared momentum ellipses into w1Ùw2 are cocentred circles, intersecting everywhere provided I12 = j2 and E = ½I1 = ½(j2)½, and nowhere else, corresponding to spin of quantised magnitude 2I1-2E possible about any axis in w1Ùw2.

    In GHC we have 3-vector body bimomentum about a due to a stance spin W about b
    Ha,b(W)   =   åiHia,b(W)   =   åimi ( (ri-c+e0)Ù((ri-c+e0)¿W) + (c-a)Ù((ri-c+e0)¿W) + (ri-c+e0)Ù((c-b)¿W) + (c-a)Ù((c-b)¿W) )Ùe¥
      =   (åimi(ri-c+e0)Ù((ri-c+e0)¿W) + m(c-a)Ù(e0¿W) + me0Ù((c-b)¿W) + m(c-a)Ù((c-b)¿W) )Ùe¥
      =   Hc(W)Ùe¥ + m((c-a)Ùv + (c-a+e0)Ù((c-b)¿W) )Ùe¥ where 2-vector Hc(W)   =   åimi(ri-c+e0)Ù((ri-c+e0)¿W) is linear and symmetric in W ; has vanishing "expansion" component åimi((ri-c)¿v)e¥ = åimi((ri-c)2)·e¥ ; and has Ha(w) = bv(h)a(w). .
    For spin about the mass centre b=c we have Ha,c(W)   =   Hc(w)Ùe¥ + m(c-a)Ùc·Ùe¥ .

    We are interested in the inverse tensor H-1 taking a 3-vector biimpulse T back to the 2-vector bivelocity kinematic response DW it induces in the body. The kinematic response embodies a momental change   Hc(DW)   which should equal the applied biimpulse,

Inertia Tensors for Particular 2D Laminae
    The inertia tensor of a 2D lamina H is simply the scalar moment of inertia Ie12,a = åimi(ri-a)2. We define the moment of inertia about the line L=a+ln as IL=åimisi2 where si=(ri-a)-(ri-a). n is the perpendicular from ri to L.
    To calculate the moments of inertia of a lamina H we consider H as a large number of low-mass particles and replace the Si with a surface integral over H. There are various tricks to make this easier and these can be found in any good Mechanics textbook (eg. "Principles of Mechanics" by J.L.Synge & B.A.Griffith (McGraw-Hill 1959)) .

Example: Right Angled Triangle
    We construct the moment of inertia of a right-angled triangular lamina about its centre of mass G given that the moment of inertia of a rod of length a and mass m about one end is ma2/3.
    [ Add picture here]
    Let A be the right-angled corner, B and C the other two. Let BC=a,AB=c,CA=b. We first find the moment of inertia of H about the edge AB. To do this we consider H as being made up of seperate strips of width dq perpendicular to AB. The length of such a strip a distance x from AC is (c-x)b/c and its mass dqm is its area times p the mass/area density of H ie. dm = p(c-x)b dx/c. The moment of inertia of this strip about AB is the same as the moment of inertia a rod of identical length and mass about its endpoint, specifically dm((c-x)b/c)2/3 = p((c-x)b/c)3/3 dx. Summing the moments of inertia of these strips gives
    IAB = ò0,cp((c-x)b/c)3/3dx = [-p(c-x)4(b/c)3/12]0,c = pcb3/12.
    Now p = (mass of H)/(area of H) = 2m/(bc) and therefore IAB = mb2/6. Similarly IAC = mc2/6.
    We now make use of the theorem of perpendicular axies to obtain IA. We have IAòp(x2+y2)dS where the integration is over the surface of H. But this seperates to IA = òpx2dS + òpy2dS = IAC + IAB = m(b2+c2)/6.
    This is an example of the general result that the moment of inertia of a plane distribution of matter about a point P in the plane (and also about an axis through P perpendicular to the plane) is equal to the sum of the moments of inertia about any two pendicular axies in the plane passing through P.
    Finally we use the theorem of parallel axies to obtain IG from IA. We know from the last chapter that IA = IG + mAG2. Now AG = AN/3 where N is the perpendicular projection of A to BC.  But AN=bc/a  so we have IG = m(b2/6+c2/6+b2c2/(9a2)). This is an example of the general result that the moment of inertia of a body about a line L is equal to that about a parallel line through the mass centre of the body plus the mass of the body times the square of the perpendicular distance of its mass centre from L.
    The moments of inertia of various lamina are given below
BodyPrinciple Axes (at mass centre)Principle moments of inertia
Rod of length aPerp. to rod at centrema2/12
Perp. to rod at end ma2/3
Right triangle ABC (A=p/2) Perp. to plane at A m(b2+c2)/6
Perp. to plane at B m(b2/6+c2/2)
Perp. to plane at C m(b2/2+c2/6)
AB mb2/6
BC mb2c2/6a2
Perp. to plane at G m(b2/6+c2/6+b2c2/9a2)
Isosceles triangle base 2w,hieght h Perp. to plane at apex m(w2/6+h2/2)
Perp. to plane at G m(w2/6+h2/18)
Line of symmetry mb2/6
Rectangle with edges b,c        Parallel to edge b mc2/12
Edge b mc2/3
Perp. to plane at G m(b2+c2)/12
Diagonal mb2c2/6(b2+c2)
Disk radius a Diameter ma2/4
Perp. to plane at G ma2/2
Hoop radius a Diameter ma2/2
Perp. to plane at G ma2
Arc radius a angle 2a Perp. to plane at centre of arc ma2
Perp. to plane at G ma2 (1-(sin2a) /a2)
Regular n-sided   polygon side a Edge if n even ma2(1/3+5( tan(n-1p))-2)/16
Perp. to plane at G ma2(1/3+( tan(n-1p))-2)/8

Inertia Tensors For Particular 3D Bodies

    Unlike the 2D case, the inertia tensor of a body moving in 3D generally varies with the orientation of the body. It is pointless evaluating the correct inertia tensors for complex objects if anything reflecting the rough dimensions and symmetry of an object will give acceptable collision behaviour.
    If c is the mass centre of a rigid body then for arbitary displacment d we have Bc+d = Bc + mdÄd = Bc + m(d21-ddT) and hence the symmetry Bc+d = Bc-d.
[ Proof :  Bc+d = Simi((ri-c-d)Ä(ri-c-d)   =   Simi(ri-c-d)21-((ri-c-d)(ri-c-d)T)   =   Simi(ri-c)21+d2I-((ri-c)(ri-c)T-ddT))
    = Simi((ri-c)Ä(ri-c))+(dÄd)) = Bc + mdÄd   ; all the linear terms in ri-c vanishing due to mass centre definition of c  .]

    If we cohere a a body having of mass m1 and central inertia tensor B1c with a body having mass m2 , and central inertia tensor B2c+d then we create a composite body having centre c'=c + m2(m1+m2)-1 d and central inertia tensor
     B'c' = B1c + m1 (m2(m1+m2)-1 d)Äm2(m1+m2)-1 d) + B2c+d + m2 (m1(m1+m2)-1 d)Äm1(m1+m2)-1 d)
    = B1c + B2c+d + m1m2(m1+m2)-1 dÄd

    Assume that the inertia tensor of a body H about its mass centre is known to be B' with respect to axies OX'Y'Z' fixed in the body and suppose the body has orientation matrix A with respect to some universal coorodinate system OXYZ so that a point with coordinates r' with respect to the OX'Y'Z' frame has coordinates r=ATr' in the universal frame.
    If we write Ba for the inertia tensor of H in the universal frame then Ba = ATBaA .
[ Proof :   Since AAT=1 we have
    Ba = Simi((ri'-a')21 -(ri'-a')(ri'-a')T) = Simi((ri-a)2)1 - (AT(ri-a))(AT(ri-a))T = Simi((ri-a)21 - AT(ri-a)(ri-a)TA)
    = Simi(AT((ri-a)2-(ri-a)(ri-a)T)A) = AT(Simi((ri-a)2-(ri-a)(ri-a)T))A = ATBaA.  .]

    If w is measured with respect to H's frame rather than the universal frame, (as is often the case in flight simulators which fragment rotations into roll,dive/climb, and yaw) the angular momentum of H expressed in its own frame is Baw, while in the universal frame it is Ba'w' = ATBaAATw = ATBaw.

    The principle moments of inertia given below assume uniform mass density of the bodies, which is not true of the objects commonly found in computer games (tanks, spaceships etc.).
BodyPrinciple Axes (at mass centre) Principle moments of inertia
Rod of length aLine of rod0
Any perp. to rod ma2/12
Third Perp.ma2/12
Right triangle ABC (A=p}/2)Perp. to planem(b2/6+c2/6+b2c2/9a2)
Isosceles triangle base 2w,hieght h Line of symmetry mb2/6
Parallel to basemh2/18
Perp. to planem(w2/6+h2/18)
Rectangle with edges b,c Parallel to edge bmc2/12
Parallel to edge cmb2/12
Perp. to planem(b2+c2)/12
Disk radius a Any diameter ma2/4
Perp. diameter ma2/4
Perp. to plane ma2/2
Hoop radius a Any diameter ma2/2
Perp. diameter ma2/2
Perp. to plane ma2
Regular 2n-sided polygon side a Line of symmetryma2(1/3+5/tan2(p/2n))/16
Other line of symmetryma2(1/3+5/tan2(p/2n))/16
Perp. to planema2(1/3+1/tan2(p/2n))/8
Cylinder radius a , length l Any diameter m(3a2+l2)/12
Perp. diameter m(3a2+l2)/12
Axis of cylinder ma2/2                   
Cylindrical shell radius a length l Any diameter
Perp. diameter
Axis of cylinder                      
Sphere radius a  Any three perp.axies 2ma2/5
Spherical shell   radius a Any three perp.axies2ma2/3
Ellipsoid Semiaxis a 5-1m(b2+c2)
Semiaxis b m(a2+c2)/5
Semiaxis c m(a2+b2)/5
Ellipsoid shellSemiaxis a1/3m(b2+c2)
Semiaxis b1/3m(a2+c2)
Semiaxis c1/3m(a2+b2)

Multi-Particle Systems
    For a nonrigid body we realx the assumption ri·=b·+(ri-b)¿w for common but vaiable w,b, and b·. If the forces exerted by the particles on eachother are directed along the lines joining the particles and there are no external forces then the total linear momentum åimiri· is conserved as are the total rotational momentae åimiriÙri· about 0 and j=åimi(ri-c)Ùri· about mass centre cº m-1åimiri.
    Differentiating ri = c + R§(riL) provides ri· = c· + R·riLR§ + RriL·R§ + RriLR§ · Þ
    riL· = R§(ri·-c·)R - R§R·riL - riLR§·R = R§(ri·-c·)R + R·R§riL - riLR·R§ = R§(ri·-c·)R + 2(R·R§)×riL = R§(ri-c)·R - w×riL where w º -2R·R§ .
    Hence riLriL· = R§(ri-c)(ri·-c·)R + 2riL((R·R§)×riL)   =   R§(ri-c)(ri·-c·)R + riL(w×riL) Þ
    åimiriLÙriL·     =   R§(åimi(ri-c)Ùri·)R + åimiriLÙ(w×riL)   =   R§jR - hcL(w)   =   jL - hcL(w) so the net angular momentum about 0L=c in the bodyframe as measured in the rotating bodyframe is jL , the constant net angular momentum about c as measured in the changing bodyframe minus the "intrinsic" hcL(w).

    The _neterm(principle biframe) for a nonrigid mutliparticle system is the is the principle biframe for a rigid body having instantaneously identical mass distribution. It is determined solely by the instananeous values of ri and mi, independant of the ri·. A better choice of baseframe depending also on the ri· is provided by a choice of R yielding åimiriLÙriL·=0 so that jL = hcL(w) provides a bodyframe in which the angular momentum is equivalent to that of a rigid body roatating at w about 0L.
    We can construct such an R integratively via w =  -2R·R§ = hcL-1(jL) Þ R· = -½hcL-1(jL)R .

Multiparticle Potentials
    Suppose the particle at ri generates a Coulomb potential fi(x) = qi|x-ri|-1 whwre qi is the charge of the particle. For a gravitational potential we set qi=-Gmi. The total potential at x is expresible as a multipole expansion
    f(x) = åiqi|x-ri|-1   =   |x-c|-1 åiqiåk=0¥  (x-c)-2k Pk(x-c,ri-c)   =   |x-c|-1 åiqiåk=0¥  |x-c|-k Pk((x-c)~,ri-c)
    = |x-c|-1 ((åiqi) + (x-c)-4 P2(x-c,ri-c) + O(|x-c|-3)) since the åiqiP1(x-c,ri-c) = (x-c)¿åiri-c vanishes for charge center c=(åiqi)-1 åiqiri .
    A spherically symmetric matter distribution has f(x) =   (åiqi)|x-c|-1 equivalent to a single particle at c, so we can regard the |x-c|-1åiqiåk=2¥  |x-c|-k Pk((x-c)~,ri-c) terms as corrections for the asymmetry of the matter distribution.
    For N=3 we have åimiP2(x-c,ri-c)   =   ½(x-c)¿q(x-c) where q(a) º 2ahc(1) - 3hc(a) is the gravitational quadrapole 1-tensor.
[ Proof : åimiP2(x-c,ri-c)   =   åimi½(3((x-c)¿(ri-c))2 - (x-c)2(ri-c)2)   =       (3/2)( åimi(x-c)2(ri-c)2 - (x-c)¿hc(x-c)) - ½åiqi(x-c)2(ri-c)2 )
    = (x-c)2åimi(ri-c)2 - (3/2)(x-c)¿hc(x-c))   =   ½(x-c)¿(2(x-c)åimi(ri-c)2 - 3hc(x-c))  .]

    Hence f(x) = |x-c|-1(m + ½(x-c)-2 (x-c)¿q(x-c) + O(|x-c|-1)) and so the total gravitational force extered on a test mass m0 at x is
    f(x) = -m0Ñxf(x) = -(x-c)-2G(m(x-c)~ - |x-c|-3 ((x-c)¿q(x-c)((x-c)~  + (x-c)-2 q(x-c) + O(|x-c|-2)).
      =  = -(x-c)-2G((m - |x-c|-3 ((x-c)¿q(x-c))(x-c)~ + (x-c)-2 q(x-c) + O(|x-c|-?))

    The torque about c exerted on the system due to a pointsource of mass m0 at x  is åi(ri-c)Ù(-fi(x)) = -åi(ri-x)Ùfi(x) - åi(x-c)Ùfi(x) = -(x-c)Ùåifi(x) where fi(x) = -m0Ñx fi(x) parallel to ri-x is the force exerted on the pointsource by the particle at ri.
    Thus the total torque -(x-c)Ùåifi   =   3Gm0|x-c|-5 (x-c)Ùhc(x-c)   =   3Gm0 |x-c|-3) (x-c)~Ùhc((x-c)~) decreases with |x-x|-3 rather than |x-x|-2 and vanishes for a spherically symmetric body with hc(a)=Ia.

Note that a¿hc(a) =   =   åimi((ri-c)2a2 - ((ri-c)¿a)2) provides åimi((ri-c)2a2 = a2åimi(ri-c)2 - a¿hc(a) .

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