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.
Kinematics
Stances
Suppose we have a 3D object H whose "centre" is located at a point c(t)ÎÂ^{3} 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 AA^{T}=1
but more general A allow for "distorted" objects. Regardless of whether A is orthonormal,
the "global coordinates" of a point having coordinates p_{L} in the "local frame" of the object are given by
p = c + Ap_{L}. 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 Â_{4×4} stance matrix would then be:
S | = | æ | c_{1} | ö | |||
ç | A | c_{2} | ÷ | ||||
ç | c_{3} | ÷ | |||||
è | 0 | 0 | 0 | l | ø |
Sr | = | æ | (Ar+c | ö | = | æ | (Ar+c)/l | ö |
è | l | ø | è | 1 | ø |
In GHC we represent an ND stance with GHC multivector
S = (½e_{¥}c)^{↑}R(-½le_{¥0})^{↑} =
(1+½e_{¥}c)R( cosh(½l)- sinh(½l)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 e_{0}
]
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 p_{G}=Sp_{L}S^{-1}
equivalent to S_{§}(p_{L}) º Sp_{L}S^{§} for a rigid stance.
Typically Se_{¥}S^{§}=e_{¥} (or le_{¥} if S embodies some dilation) while
Se_{0}S^{§} = c = c+e_{0}+½c^{2}e_{¥}.
A direction a_{L} in the local frame becomes scaled direction
a_{G} = l^{↓}Ra_{L}R^{§} in the global frame .
In GHC we can represent this as a_{G}e_{¥} = Sa_{L}e_{¥}S^{§} .
The position and orientation factors of a stance rotor do not commute unless R^{§}cR=c.
S is nonunitary because since e_{¥}^{†}=-2e_{0} we have
(½e_{¥}c)^{↑} (½e_{¥}c)^{↑}^{†}
= (½e_{¥}c)^{↑} (½e_{¥}c)^{†}^{↑}
= (½e_{¥}c)^{↑} (e_{0}c)^{†}^{↑}
= (1+½e_{¥}c)(1+e_{0}c^{†}) .
For Euclidean c this is
1+(½e_{¥}+e_{0})c + ½c^{2} - ½e_{¥0}c^{2}
= 1+e_{m}c + ½c^{2} - ½e_{¥0}c^{2} .
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
S_{Q¬T} º S_{Q¬T}_{Q} would typically be implimented with respect
to the "base frame" of the bomber. Assuming the bomber is orientated in the base frame with e_{1} as the
forward direction, e_{2} as left, and e_{3} as upward, the turret might be considered as having a centre displaced by
c_{TQ}=(X,0,Z) from the centre of the bomber and an orientation
R_{TQ}=(-½fe_{12})^{↑}(½qe_{13})^{↑} corresponding to a roll-less turet.
The notation a_{Q} indicates that a is expressed
relative to Q ( ie. "in Q coordinates").
Thus c_{TQ} denotes the position of the centre of T in Q coordinate system.
For a more general mobile turret, c_{TQ} 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
c_{Q,} = [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 S_{S}_{G} 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
0_{G} with identity orientation 1_{G} so that S_{G¬!}=1_{G}=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
S_{G¬T}
= S_{G¬!}S_{!¬,}S_{,¬Q}S_{Q¬T}
= S_{!¬,}S_{,¬Q}S_{Q¬T}
since S_{G¬!}=1.
Note that we compose or "cascade" relative stances from right to left rather than left to right
with S_{A¬C} = S_{A¬B}S_{B¬C} motivating the S_{B¬A} º S_{B¬A}_{B} notation.
Since S_{A¬B}=S_{B¬A}^{-1} is equal to S_{B¬A}^{§} for a rigid stance, the
stance of the aircraft __Acraft with respect to the city 4 is given by
S_{4¬Q} = S_{4¬,}S_{,¬Q} = S_{,¬4}^{-1}S_{,¬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 S_{T¬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
S_{C¬4} = S_{C¬T}S_{T¬Q}S_{Q¬4} = S_{C¬T}S_{Q¬T}^{§}S_{Q¬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 Â_{3}^{%} @ Â_{4,1}. 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^{§} - Se_{0}S^{§} = S(a+½a^{2}e_{¥})S^{§} .
The former approach extends to evaluating RxR^{§} for xÎU_{N} as (Sxe_{¥}S^{§})¿e^{¥}.
However, it is usually preferable to simply reclaim R from S ,
ideally logically as the unexteneded U_{N} 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.
Kinematics
Let H be an N-D object moving through U^{N} 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(a_{1},a_{2},..)=(ba_{1},ba_{2},..) ;
(a_{1},a_{2},..)b=(a_{1}b,a_{2}b,.. ) .
Other multivector products (¿, Ù, ×, ...) are treated likewise ]
For brevity we will omitt the explicit (t) and write _{0} for (0)
so that A = RA_{0}R^{§}
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^{·} - ½w^{2})R
.
Writing w_{L} = R^{§}wR
= -2R^{§}R^{·}R^{§}R
= -2R^{§}R^{·}
for the local spin expressed in the frame of the object, we have
R^{·} = -½Rw_{L} .
[ 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^{·}
= (Ra_{0}R^{§})^{·}
= (R^{·}a_{0}R^{§} + Ra_{0}(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^{*}=we_{123}^{-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
W^{2}
= w^{2} + 2b^{·}Ùw generally nonscalar even for N=3 .
[ Proof : S^{·} = ½e_{¥}b^{·}R + (1+½e_{¥}b)R^{·}
= (½e_{¥}b^{·} - (1+½e_{¥}b)½w)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 p_{G}^{·} = p_{G}^{·} + (p_{G}¿p_{G}^{·})e_{¥}
perpendicular to null p_{G} is given by
p_{G}^{·} = p_{G}¿W + S_{§}(p_{L}^{·})
.
[ Proof :
p_{G}^{·}
= (Sp_{L}S^{§})^{·}
= S^{·}p_{L}S^{§} + Sp_{L}^{·}S^{§} + Sp_{L}(S^{·})^{§}
= -½WSp_{L}S^{§} + Sp_{L}^{·}S^{§} - ½Sp_{L}S^{§}W^{§}
= -½Wp_{G} + Sp_{L}^{·}S^{§} - ½p_{G}W^{§}
= -½(Wp_{G}-p_{G}W) + Sp_{L}^{·}S^{§}
= p_{G}¿W + S_{§}(p_{L}^{·})
.]
Now (p-b+e_{0})¿W
= b^{·} + (p-b)¿w + ((p-b)¿b^{·})e_{¥}
= v + ((p-b)¿b^{·})e_{¥}
is the instantaneous 1-velocity of p º p_{G} 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+e_{0})Ùe_{¥})×W
= ((p-b+e_{0})¿W)e_{¥}
= (b^{·} + (p-b)¿w)e_{¥}
providing the instamtaneous 2-blade velocity of body point p.
[ Proof : ((p-b+e_{0})Ùe_{¥})×W
= ((p-b)e_{¥}-e_{¥0})×W
= ((p-b+e_{0})Ùe_{¥})×W
= ((p-b+e_{0})e_{¥})×W
= ((p-b+e_{0})×W)e_{¥}
= ((p-b+e_{0})¿W)e_{¥}
.]
Note that
(p-b+e_{0})Ùe_{¥} =
(p-b+e_{0}+½(p-b)^{2}e_{¥})Ùe_{¥} can be regarded as p_{c}Ùe_{¥} where
p_{c} 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)^{§}
= 1-¼d^{2}W(t)^{2} intead of unity.
The correct iteration is
S(t+d) = (-½dW)^{↑} S ,
or S(t+d) = S(-½dW_{L})^{↑}
for W_{L} expressed in stance local (source) coordinates.
This preserves the unit roticity of S(t+d)
but since W^{2} 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}(d^{2}) agreement can be obtained by evaluating w(t+½d), forming
S(t+d) as
(1+½(c+½dc^{·})e_{¥})(-½w(t+½d))^{↑}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
R_{S} º
(-½(w((L-1)d)+w((L-2)d)+..+w(d)+w(0)))^{↑}R(0)
approximating
(-½ ò_{0}^{Ld} dtw(t))^{↑}R(0)(L-1)d)+w((L-2)d)+..+w(d)+w(0)))^{↑}R(0)
because the w(ld) do not commute.
R_{S} can be regarded as a superposition of all R reachable by any combination of the particular L subrotations, with
repeated uses allowed.
U_{N} Spinor Representation
We have here represented U_{N}^{%} position c as (½e_{¥}c)^{↑}_{§}(e_{0}) but
provided c^{2} has the same sign as e_{1}^{2} we also have
c = Ce_{1}C^{§}
where
C
= (c^{2})^{¼}(½q(cÙe_{1})^{~})^{↑}
is a nonunit <0;2>-grade rotor in U_{N} satisfying
(CC^{§})^{2} = e_{1}^{-2}c^{2} with
C^{-1} = (e_{1}^{2})^{½} c^{-1}C^{§} .
Here q is the angle subtended betweem c and e_{1}.
We can recover c=CC^{§} ; q = 2 cos^{-1}(C_{<0>}) ; (cÙe_{1})^{~} =
C_{<2>}^{~} .
C is not uniquely determined by c because CA where A
is any unit rotor with Ae_{1}A^{§}=e_{1} also serves.
For N=3, A=(fe_{1}i^{-1}) incorporates a single arbitary phase angle but for N>3 we have more general
guage rotations "about" e_{1}.
C^{·} = ½c^{-1} c^{·}Ce_{1}
[ Proof :
c^{·} = C^{·}e_{1}C^{§} + Ce_{1}C^{§}^{·}
= C^{·}e_{1}C^{§} + (C^{·}e_{1}C^{§})^{§}
= 2(C^{·}e_{1}C^{§})_{<0;1;4;5;..>}
but c^{·} is a 1-vector so
c^{·}
= 2C^{·}e_{1}C^{§}
= 2Ce_{1}C^{·}^{§}
Þ c^{·}Ce_{1} = 2C^{·}
.]
Since c = |c^{2}|^{½} = CC^{§}
we have c^{·} = 2C^{·}_{*}C^{§} .
In GHC we have S = (1+½e_{¥}Ce_{1}C^{§})R = C(c^{-1}+e_{¥1})C^{§}R
|e_{1}C^{§})R = c^{-1}C(1+ce_{¥1})C^{§}R
= c^{-1}C(ce_{¥1})^{↑}C^{§}R
= C^{~}(ce_{¥1})^{↑}C^{~}^{§}R
We can define monotonically increasing spinor time s = ò_{0}^{t} 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^{·}Ce_{1} .
Whence
C^{''}
= ½(c^{··}c + ½c^{·}^{2})C .
[ Proof :
C^{''}
= cC^{'}^{·} =
½c(c^{·}Ce_{1})^{·}
= ½c(c^{·}Ce_{1})^{·}
= ½c(c^{··}Ce_{1} + c^{·}C^{·}e_{1})
= ½c(c^{··}Ce_{1} + ½c^{-1}c^{·}^{2}Ce_{1}^{2})
= ½(cc^{··}Ce_{1}C^{-1} + ½c^{·}^{2}e_{1}^{2})C
= ½(cc^{··}c^{~} + ½c^{·}^{2}e_{1}^{2})C
= ½(c^{··}c + ½c^{·}^{2})C .
.]
Now suppose that the acceleration has the form
c^{··} = a - m'c^{-2}c^{~}
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^{-3}c 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)^{½}s) + C^{·}(0)(-½E))^{-½} sin((-½E)^{½})
of frequency (-½E)^{½} and spinor time period
2p(-½E)^{-½}
= 8p(-E)^{-½} . Furthermore
C(t) =
cos( (-½E)^{½} s) C_{0}
+ sin( (-½E)^{½} s) C_{1} provides a solution
with C(0)=C_{0} and
C(½p(-½E)^{-½}) = C_{1} which is the natural constant energy
trajectory from c(0) to c(T), having
Cds(0) = (-½E)^{½}C_{1}
and
Cds(½p(-½E)^{-½}) = -(-½E)^{½}C_{0}.
The arrival time t= ò_{0}^{½p(-½E)-½} ds r(s) is approximately
½p(-½E)^{-½} 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 C_{0}=C(0), E, and one of
C_{±}
specifying a "detemporalised" position passed through.
A more general solution is
cosh(sbC_{0} + b^{-2}
b
sinhsb
(b)
where b^{2} = ½E . For a single N=3 particle, the angular momentum h_{0} defines
the orbit 2-plane so the motion is effectively 2D and it is natural to tahe
b=¼|E|^{½}h_{0}^{-1}h_{0}
where h_{0}=|h_{0}^{2}|^{½} .
Writing S(s) = (1+½e_{¥}Ce_{1}C^{§})R
= (1+½Ce_{¥1}C^{§})R
= R + ½Ce_{¥1}C^{§}R
and noting that R^{''} = (cR^{·})^{'}
= c(cR^{·})^{·}
= cc^{·}R^{·} + c^{2}R^{··}
R^{'} = cR^{·} = -½cwR.
R^{''} = cR^{'}^{·}
= -½cc^{·}wR
-½ccw^{·}R
+¼cw^{2}R
= -½c(c^{·}w+cw^{·}-½w^{2})
GHC spinor form
A GHC variant of spinorising position in this way is to express a given U^{N}^{%}
1-vector y=ls dual to a hypersphere of positive squared radius s^{2}=r^{2}>0 as
as a dilated rotatation of the unit hypersphere dual as Be_{+}B^{§} where
B=(lr)^{½}(½q(ye_{+}e_{+})^{~})^{↑} ,
q being the angle subtended by
s and e_{+} , with cos(q) = r^{-1}s¿e_{+} = ½r^{-1}(-1+c^{2}-r^{2})
and
sÙe_{+}
= ce_{+} - ½(1+c^{2}-r^{2})e_{¥0}.
Our spinor time is now s= ò_{0}^{t} 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^{·}^{2}r^{2}
+ l^{2}(c^{·}¿c) , and
y^{··} = l(c^{··} + e_{¥}(cc^{··} + c^{·}^{2} - rr^{··} - r^{·}^{2})
+ 2l^{·}(c^{·} + e_{¥}(cc^{·} - rr^{·})) + l^{··}s
has e_{0} coordinate l^{··} and e_{¥} coordinate
l(cc^{··} + c^{·}^{2} - rr^{··} - r^{·}^{2})
+ 2l^{·}(cc^{·} - rr^{·})) + ½l^{··}(c^{2}-r^{2})
.
Taking l=1 to keep y within e^{0}¿y=1 we have
y^{··}y =
s^{··}s =
(c^{··} + e_{¥}(cc^{··} + c^{·}^{2} - rr^{··} - r^{·}^{2})(c+e_{0}+ee½(c^{2}-r^{2}))
= c^{··}c + (e_{¥}c-1+e_{¥0})(cc^{··} + c^{·}^{2} - rr^{··} - r^{·}^{2})
+ c^{··}e_{0} +
+ c^{··}e_{¥}(½(c^{2}-r^{2})
y^{··} = M(y)y requires l^{··}=M(y) and hence
lc^{··} + 2l^{·}c^{·} + ½l^{··}c
= M(y)lc
Þ
c^{··}
= ½M(y)c
+ 2l^{-1}l^{·}c^{·} .
If we want mc^{··} = m'(c)c^{~} = c^{-1}m'(c)c we thus need
M(y) = 2c^{-1}m'(c) and hence
l^{·} = ò_{0}^{t} dt 2c^{-1}m'(c) =
ò_{0}^{c} dc (c^{·})^{-1} 2c^{-1}m'(c) =
Setting l=r^{-1} for unit y
gives l^{·}=-r^{-2}r^{·} ;
l^{··}
= 2r^{-3}r^{·} - r^{-2}r^{··}
= r^{-2}(2r^{-1}r^{·} - r^{··})
s^{··}s = c^{··}c + c^{··}e_{0}
+ ½c^{··}e_{¥}(c^{2}-r^{2}) +
(cc^{··} + c^{·}^{2} - rr^{··} - r^{·}^{2})e_{¥}(c+e_{0}).
If r(t) is constant then s^{·}s =
s^{·}Ùs is a pure 2-blade with
(s^{·}s)^{2} = -s^{·}^{2}s^{2} = c^{·}^{2}r^{2} .
The radial acceleration assumption
Subbody motion
Gunturret
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
W_{4¬T}_{4}
= W_{4¬Q}_{4}
+ (1+½e_{¥}c_{Q4})W_{Q¬T}_{4}
(1-½e_{¥}c_{Q4})
= W_{4¬Q}_{4}
+ (1+½e_{¥}c_{T4})W_{Q¬T}_{4}
(1-½e_{¥}c_{T4})
=
(1+½e_{¥}c_{T4})_{§}(
W_{4¬Q}_{4} + W_{Q¬T} )
.
[ Proof : W_{4¬T} = -2S_{4¬T}^{·}S_{4¬T}^{§}
= -2(S_{4¬Q}S_{Q¬T})^{·}(S_{4¬Q}S_{Q¬T})^{§}
= -2(S_{4¬Q}^{·}S_{Q¬T} + S_{4¬Q}S_{Q¬T}^{·})
S_{Q¬T}^{§}S_{4¬Q}^{§}
= W_{4¬Q}
+ S_{4¬Q}W_{Q¬T}_{Q}S_{4¬Q}^{§}
= W_{4¬Q}
+ (1+½e_{¥}c_{Q4})R_{4¬Q}
((1+½e_{¥}c_{TQ})W_{Q¬T}_{Q}
(1-½e_{¥}c_{TQ})
R_{4¬Q}^{§}
(1-½e_{¥}c_{Q4})
= W_{4¬Q}
+ (1+½e_{¥}c_{Q4})
(1+½e_{¥}(c_{T4}-c_{Q4}))R_{4¬Q}
W_{Q¬T}_{Q}R_{4¬Q}^{§}
(1-½e_{¥}(c_{T4}-c_{Q4}))
(1-½e_{¥}c_{Q4})
= W_{4¬Q} +
(1+½e_{¥}c_{T4})R_{4¬Q}W_{Q¬T}_{Q}R_{4¬Q}^{§}
(1-½e_{¥}c_{T4})
= W_{4¬Q}
+
(1+½e_{¥}c_{T4})W_{Q¬T}_{4}
(1-½e_{¥}c_{T4})
.]
The velocity of the turret is
v_{T4} =
v_{Q4} + R_{4¬Q}_{§}(
v_{TQ} + c_{TQ}¿w_{QQ} )
while its spin is
w_{T4}=R_{4¬Q}_{§}(w_{QQ}+w_{TQ})
giving bivelocity
W_{T4} =
(v_{Q4} +
R_{4¬Q}_{§}( v_{TQ} + c_{TQ}¿w_{QQ}) )e_{¥}
+
R_{4¬Q}_{§}( w_{QQ} + w_{TQ} )
= W_{Q4}
+ R_{4¬Q}_{§}(
(v_{TQ} + c_{TQ}¿w_{QQ} )e_{¥}
+ w_{TQ} ) .
Here c_{T} 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 c_{Q} is typically a fixed reference "centre point" of the aircraft rather than its dynamical mass centre which may move with respect to c_{Q} due to fuel slosh, turret movement, and so forth.
The instantaneous apparent bivelocity of the city 4 as percieved by the turret T is
-W_{TT} =
-R_{T4}^{§}_{§}(v_{Q4})e_{¥}
- R_{TQ}^{§}(
(v_{TQ}+c_{TQ}¿w_{QQ} )e_{¥}
+ w_{QQ} + w_{TQ}
)R_{TQ} .
[ Proof :
W_{TT}
= R_{T4}^{§} W_{T4} R_{T4}
= (R_{Q4}R_{TQ})^{§} W_{T4}
(R_{Q4}R_{TQ})
= R_{TQ}^{§}
R_{Q4}^{§}
W_{T4}
R_{Q4}R_{TQ}
=
R_{TQ}^{§}
R_{Q4}^{§}
(
v_{Q4}e_{¥} + R_{Q4}_{§}(
v_{TQ}+c_{TQ}¿w_{QQ} )e_{¥}
+ w_{QQ} + w_{TQ})
R_{Q4}R_{TQ}
=
R_{TQ}^{§}R_{Q4}^{§}
v_{Q4}
R_{Q4}R_{TQ}e_{¥}
+
R_{TQ}^{§}
(v_{TQ}+c_{TQ}¿w_{QQ} )e_{¥}
+ w_{QQ} + w_{TQ})
R_{TQ}
.]
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_{,}_{!}De_{1}R_{,}_{!}^{§})R_{,}_{!}
= R_{,}_{!}(1+½De_{¥}e_{1})
where D(t) is the distance from ! to , and
R_{,}_{!}=(½w_{,}te_{12})^{↑} where
w_{,}=2p year^{-1} .
Thus the , frame has e_{3} constant normal to the earth orbit plane in the solar north direction and e_{1} always pointing directly away from the sun
while S_{,¬,} = R_{,¬,}_{,} is a Â_{3} 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_{!¬,}e_{123}^{-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 e_{3}_{!})
, 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 Â^{N},
and suppose at time t the i^{th} particle has mass m_{i}, position r_{i}, and velocity v_{i} where
i ranges through some indexing set I.
Let m º å_{i}m_{i} denote the total mass and c º m^{-1} å_{i}m_{i}r_{i}
denote the instantaneous centre of mass at time t.
The condition for these particles to form a rigid body is that
v_{i} º r_{i}^{·} = v + (r_{i}-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
(r_{i}-b+e_{0})¿W
= (r_{i}-b+e_{0})×W
= r_{bi}¿W
= v_{i} + ((r_{i}-b)¿v)e_{¥}
where stance bivelocity W=ve_{¥}+w
and r_{bi} º r_{i}-b + e_{0} + ½(r_{i}-b)^{2}e_{¥} .
Hence å_{i}m_{i}(r_{i}-b+e_{0})¿W = å_{i}m_{i}v_{i} + m((c-b)¿v)e_{¥}
and we can eliminate the e_{¥} compoenent by considering
2-blade velocity
v_{i} º v_{i}e_{¥} = v_{i}Ùe_{¥} = v_{i}×e_{¥} = ((r_{i}-b+e_{0})×W)×e_{¥}
.
Kinetic Energy
The classical kinetic energy of a particle having velocity v_{i} and scalar mass m_{i}³0 is the scalar
E_{i} = ½m_{i}v_{i}^{2} , though this
is actually a |v|<<c approximation to the relativistic
E = m_{i}c^{2}(1+(c^{-1}v_{i})^{2})^{½} - m_{i}c^{2}
= ½m_{i}(v_{i}^{2} - ¼c^{-2}v_{i}^{4} + _{O}(c^{-4}v_{i}^{6}))
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
E_{w,v} = å_{i}½m_{i}(v + (r_{i}-b)¿w)^{2}
= ½mv^{2}
+ å_{i}½m_{i}((r_{i}-b)¿w)^{2}
+ å_{i}m_{i}v¿((r_{i}-b)¿w)
= ½mv^{2}
+ å_{i}½m_{i}((r_{i}-b)¿w)^{2}
+ mv¿((c-b)¿w) .
If w is a 2-blade
with w^{2}=-w^{2}
then the middle term is
å_{i}½m_{i}(¯_{w}(r_{i}-b)w)^{2}
= -å_{i}½m_{i}(¯_{w}(r_{i}-b))^{2}w^{2}
= ½I_{w,b}w^{2}
where scalar
I_{w,b} º å_{i}m_{i}(¯_{w}(r_{i}-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 = ½mv^{2} - ½mI_{w}w^{2}
= ½m(v^{2} ± (k_{w}w)^{2})
according as w^{2} = -/+w^{2} ;
where I_{w} º I_{w,c}
and scalar radius of gyration
k_{w,b} º (m^{-1}I_{w,b})^{½}
is defined so that
I_{w,b} = mk_{w,b}^{2} .
For N>3 and a general 2-vector spin w the
å_{i}½m_{i}((r_{i}-b)¿w)^{2} energy term is more complicated.
BiMomentum
The linear momentum of a particle having position r_{i} Î Â^{N} and scalar mass m_{i}³0
moving with velocity v_{i}ÎÂ^{N} at time t is just m_{i}v_{i}.
Its 2-blade angular momentum about a point a, or line
L={a+ln : lÎÂ}
is h_{a} = (r_{i}-a)Ù(m_{i}v_{i}), or
((r_{i}-a)-(r_{i}-a)¿n)Ùm_{i}v_{i}
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:
In GHC we have
(r_{ai}Ùe_{0})×(v_{i}e_{¥}) = ((r_{i}-a)e_{0}+½(r_{i}-a)^{2}e_{¥0})×(v_{i}e_{¥})
= (r_{i}-a)Ùv_{i} + ((r_{i}-a)¿v_{i})e_{¥0} - ½(r_{i}-a)^{2}v_{i}e_{¥}
= (r_{i}-a)Ùv_{i} + (((r_{i}-a)^{2})^{·} + (r_{i}-a)cnta^{·})e_{¥0} - ½(r_{i}-a)^{2}v_{i}e_{¥}
where
r_{ai}ºr_{i}-a+e_{0}+½(r_{i}-a)^{2}e_{¥}
and r_{ai}Ùe_{0} can be regarded as the bipoint {0,r_{i}-a}.
We can eliminate the unwanted terms by forming 3-blade
rotational momentum
m_{i}((r_{ai}Ùe_{0})×(v_{i}e_{¥}))Ùe_{¥} = m_{i}(r_{i}-a)Ùv_{i}Ùe_{¥}
but it is more natural to form 3-blade bimomentum
m_{i}(r_{i}-a+e_{0})Ùv_{i}
= m_{i}((r_{i}-a+e_{0})Ùv_{i})Ùe_{¥}
= m_{i}((r_{i}-a)Ùv_{i})e_{¥} + m_{i}v_{i}e_{¥0} containing both the rotational momentum
m_{i}((r_{i}-a)Ùv_{i})e_{¥}
about a
and the linear momentum m_{i}v_{i}e_{¥0} as 3-blades.
A GHC body bimomentum is thus a 3-vector formed by the taking the outter product of an e_{0}^{*} 2-vector
with e_{¥},
H_{ia,b}(W) º
m_{i}(r_{i}-a+e_{0})Ù((r_{i}-b+e_{0})¿W)Ùe_{¥}
=
m_{i}(
(r_{i}-a+e_{0})Ù((r_{i}-b)¿w)
+ (r_{i}-a+e_{0})Ùb^{·}
)Ùe_{¥}
=
(h_{ia,b}(b^{·},w)
+ m_{i}e_{0}Ù((r_{i}-b)¿w+b^{·})
)Ùe_{¥}
embodying both angular momentum about a and linear momentum
m_{i}((r_{i}-b)¿w+b^{·}).
=
m_{i}( (r_{i}-c+e_{0} + c-a)Ù((r_{i}-c+e_{0} + c-b)¿W)Ùe_{¥}
=
m_{i}( (r_{i}-c+e_{0})Ù((r_{i}-c+e_{0})¿W)
+ (c-a)Ù((r_{i}-c+e_{0})¿W)
+ (r_{i}-c+e_{0})Ù((c-b)¿W)
+ (c-a)Ù((c-b)¿W)
)Ùe_{¥}
=
m_{i}( (r_{i}-c+e_{0})Ù((r_{i}-c)¿w)
+ (c-a)Ù((r_{i}-c)¿w)
+ (r_{i}-c+e_{0})Ù((c-b)¿w)
+ (c-a)Ù((c-b)¿w)
+ (r_{i}-c+e_{0})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
h_{a,b}(v,w)
= å_{i}h_{ia,b}(v,w)
= å_{i}h_{ia,b}m_{i}(r_{i}-a)Ù((r_{i}-b)¿w)
= å_{i}(m_{i}(r_{i}-c)^{2}w - m_{i}(r_{i}-c)((r_{i}-c)¿w))
+ m(c-a)Ùv
+ m(c-a)Ù((b-c)¿w)
= h_{c}(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
h_{c}(w) º
å_{i}m_{i}(r_{i}-c)Ù((r_{i}-c)¿w)
= å_{i}m_{i}(r_{i}-c)×((r_{i}-c)×w)
= å_{i}m_{i}((r_{i}-c)^{2}w - (r_{i}-c)((r_{i}-c)¿w)) .
This third expression can be used to extend h_{c} over scalars with
h_{c}(l) º lå_{i}m_{i}((r_{i}-c)^{2} .
For 2-blade w we have
h_{c}(w)
= å_{i}m_{i}(r_{i}-c)¯_{w}(r_{i}-c)w
= (I_{w} + å_{i}m_{i}^_{w}(r_{i}-c)¯_{w}(r_{i}-c))w .
For nondegenerate solid bodies with m_{i}>0 that extend into N_{0} dimensions,
linear h_{c}(w) is nonzero for every nonzero w
and so invertible.
h_{c}(w) is symmetric (self adjoint) in that
c.h_{c}(w)
= w.h_{c}(c)
= h_{c}(c).w
and has postive real eigenvalues for postive mass m_{i}>0.
We naturally decompose w(t) = å_{k=1}^{½N(N-1)} f^{k}(t)f_{k}(t)
where f_{k}(t) = R_{§}(¦_{k}(0)) are the unit principle spins.
Letting D_{0}(c) = å_{k}I_{k}(c¿f^{k}(0))f_{k}
we have h_{c}(c) = R_{§}(D_{0}(R_{§}^{-1}(c)))
º R_{§}D_{0}R_{§}^{-1}(c) and
h_{c}^{-1}(c) = R_{§}^{-1}(D_{0}^{-1}(R_{§}(c)))
Because h_{c} is symmetric, the squared angular momentum
h_{c}(w)^{2} =
w¿h_{c}^{2}(w)
= å_{k=1}^{½N(N-1)}
±_{k}
I_{k}^{2} f^{k}^{2}
while
w¿h_{c}(w)
= å_{k=1}^{½N(N-1)}
±_{k}
I_{k} f^{k}^{2}
= -å_{i}m_{i}((r_{i}-c)¿w)^{2}
= -2E_{w,0}
provides the rotational kinetic energy. Here ±_{k} denotes the sign of
f_{k}¿f_{k} .
[ Proof : Symmetry follows from
c.((r_{i}-c)Ù((r_{i}-c).w) =
((r_{i}-c).w).(c.(r_{i}-c)) =
(c.(r_{i}-c)) . ((r_{i}-c).w) =
((r_{i}-c).c).(w.(r_{i}-c)) =
w.((r_{i}-c)Ù((r_{i}-c).c)
.]
The time derivative
h_{c}^{·}(c) =
h_{c}(w×c)
- w×h_{c}(c) .
[ Proof :
å_{i}m_{i}((r_{i}-c)×((r_{i}-c)×c))^{·}
= å_{i}m_{i}((r_{i}-c)^{·}×((r_{i}-c)×c)
+ (r_{i}-c)×((r_{i}-c)^{·}×c)
= å_{i}m_{i}(
((r_{i}-c)×w)×((r_{i}-c)×c)
- (r_{i}-c)×(c×((r_{i}-c)×w)))
= å_{i}m_{i}(
(r_{i}-c)×(w×(c×(r_{i}-c)))
+ ((r_{i}-c)×w)×((r_{i}-c)×c)
- (r_{i}-c)×(
w×(c×(r_{i}-c))
+ c×((r_{i}-c)×w)) )
= å_{i}m_{i}(
(r_{i}-c)×(
((r_{i}-c)×c)×w
+ ((r_{i}-c)×c)×(w×(r_{i}-c))
- (r_{i}-c)×(
w×(c×(r_{i}-c))
+ c×((r_{i}-c)×w)) )
= å_{i}m_{i}(-w×((r_{i}-c)×((r_{i}-c)×c))
+ (r_{i}-c)×((r_{i}-c)×(w×c)))
= -w×h_{c}(c) +
h_{c}(w×c)
.]
In the absence of external torques w thus satisfies Euler equation
h_{c}^{·}(w) + h_{c}(w^{·}) = 0
Þ
w^{·} = h_{c}^{-1}( h_{c}^{·}(w))
Now w×h_{c}(w)
= å_{i}å_{j}
I_{j}
f^{i}
f^{j}
f_{i}×
f_{j}
=
= å_{i<j}
(I_{j}-I_{i})
f^{i}
f^{j}
f_{i}×
f_{j}
but it is more useful to exploit the conservation of h_{c}(w) with
w^{·} =
h_{c}^{-1}( -w×h_{0})
= h_{c}^{-1}( -å_{i<j}
(I_{j}-I_{i})
f^{i}(t)f^{j}(0)
f_{i}(t)×f_{j}(0) )
= -å_{k} I_{k}^{-1}
h_{c}^{-1}( -å_{i<j}
(I_{j}-I_{i})
f^{i}(t)f^{j}(0)
(f^{k}(t)¿
(f_{i}(t)×f_{j}(0) )) f_{k}
Now
(f^{k}(t)¿
(f_{i}(t)×f_{j}(0) ))
= R_{§}(f^{k}(0))¿
(R_{§}(f_{i}(0))×f_{j}(0) ))
= f^{k}(0))¿
R_{§}^{-1}(R_{§}(f_{i}(0))×f_{j}(0) )))
= f^{k}(0))¿
R_{§}^{-1}(R_{§}(f_{i}(0)))×
R_{§}^{-1}(f_{j}(0) )))
= f^{k}(0))¿
(f_{i}(0)×
R_{§}^{-1}(f_{j}(0) )))
But w^{·} = -2R^{··}R^{§}
-2R^{·}R^{·}^{§} so
w^{·}R = -2R^{··}
-2R^{·}R^{·}^{§}R so .... ????
When N=2, setting w=w^{*}=we_{12}^{-1} we have body linear momentum m(v+(c-b)w) = m(v+w(c-b)e_{12}) and pseudoscalar angular momentum h_{c}(w)=h_{c}(w)e_{12} with dual scalar angular momentum h_{c}(w) º h_{c}(w)e_{12}^{-1} = wå_{i}m_{i}(r_{i}-c)^{2} ; although a third spacial dimension e_{3} is often invoked to "hold" 1-vector angular momentum and spin h_{c}(w) = wh_{c}(e_{3})e_{3} .
When N=3, setting w=w^{*}=we_{123}^{-1} we have body
linear momentum m(v+(c-b)¿w)
= m(v+((c-b)Ùw)^{-*})
= m(v-(c-b)×w)
= m(v+w×(c-b))
and 2-blade body angular momentum h_{c}(w) with dual angular momentum 1-tensor
h_{c}(w) º h_{c}(w_idual)^{*}
= å_{i}m_{i}((r_{i}-c)^{2}w - (r_{i}-c)((r_{i}-c)¿w)
= å_{i}m_{i}((r_{i}-c)×(w×(r_{i}-c)))
= B_{c}w
where 1-vector spin w º we_{123}^{-1}
and symmetric Â_{3×3} matrix
B_{c} º S_{i}m_{i}((r_{i}-c)^{2}1-(r_{i}-c)(r_{i}-c)^{T})
= S_{i}m_{i}((r_{i}-c)Ä(r_{i}-c))
is the matrix representation of the instantaneous inertia tensor of H about its mass centre.
Once again we can extend h_{c}() over scalars with
h_{c}(l) = lå_{i}m_{i}((r_{i}-c)^{2} .
Note that a¿h_{c}(a) =
= å_{i}m_{i}((r_{i}-c)^{2}a^{2} - ((r_{i}-c)¿a)^{2})
provides
å_{i}m_{i}((r_{i}-c)^{2}a^{2} =
a^{2}å_{i}m_{i}(r_{i}-c)^{2} - a¿h_{c}(a) .
Time derivative
h_{c}^{·}(c^{*}) = w×h_{c}(c)
- h_{c}(c)×w
[ Proof :
h_{c}^{·}(c)^{*}
= h_{c}(c×w)^{*} - (w×h_{c}(c)^{*}
= (((h_{c}(c)^{*})×(w^{*}))^{*}
- ((w^{*})×(h_{c}(c)^{*}))^{*}
= ((h_{c}(c)×w)^{*}
- (w×h_{c}(c))^{*}
= -h_{c}(c)×w
+ w×h_{c}(c)
.]
More generally for any a we have symmetric matrix (with non-negative elements on the leading diagonal) which we may write explicitly as
B_{a} | =S_{i}m_{i} | æ | (y_{i}-a_{y})^{2}+(z_{i}-a_{z})^{2} | -(x_{i}-a_{x})(y_{i}-a_{y}) | -(x_{i}-a_{x})(z_{i}-a_{z}) | ö |
ç | -(x_{i}-a_{x})(y_{i}-a_{y}) | (x_{i}-a_{x})^{2}+(z_{i}-a_{z})^{2} | -(x_{i}-a_{x})(y_{i}-a_{y}) | ÷ | ||
è | -(x_{i}-a_{x})(z_{i}-a_{z}) | -(y_{i}-a_{y})(z_{i}-a_{z}) | (x_{i}-a_{x})^{2}+(y_{i}-a_{y})^{2} | ø |
B_{a} | = | æ | I_{x} | 0 | 0 | ö |
ç | 0 | I_{y} | 0 | ÷ | ||
è | 0 | 0 | I_{z} | ø |
Let
f_{1}(0),
f_{2}(0),..f_{½N(N-1})(0)
be the ½N(N-1) principle spins for the at t=0
so that
f_{k})(t) = R(t)f_{k}(0)R(t)^{§} is the _kth principle spinat time t.
The GHC 3-vector bimomentum is å_{j}I_{j} R(t))f_{j}(0)R(t)^{§} e_{¥}
+ mc^{·}e_{¥0} =
= m(å_{j}k_{j}^{2} R)f_{k}(0)R^{§}
+ e_{0}c^{·})Ùe_{¥}
.
If 3D body H has a rotational symmetry axis k then
h_{c}(w) = h_{1}w
+ (h_{3}-h_{1})(k^{-1}¿w)k .
If k^{2}=1 then B=h_{1}1 + (h_{3}-h_{1})kk^{T} .
For general N the inertia tensor has a symmetric Â_{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}(f_{k}(0)¿f_{k}(0))^{-1} h_{c}(f_{k})¿(f_{k}))^{½} where f_{k} is the _kth eigenbivector. The gyration radii can be regarded as eigenvalues for m^{-½}h_{c}^{½}(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
h_{c}^{·}(w) =
h_{c}(w)× |w .
so the rate of change of angular momentum
(h_{c}(w))^{·} =
h_{c}^{·}(w) + h_{c}(w^{·})
is
h_{c}(w)× |w for constant spin w^{·}=0
and since
h_{c}(w)Ùw=0
this
means w must be an eigen 2-vector of h_{c}.
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
h_{c}(w) = j constant
implying w¿h_{c}^{2}(w) = j¿j.
The rotational energy w¿h_{c}(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¿h_{c}^{2}(w) = j^{2}
and the energy ellipse w¿h_{c}(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 w_{L} .
When two gyration radii are equal, The projections of the energy and squared momentum ellipses
into w_{1}Ùw_{2} are cocentred circles, intersecting
everywhere provided I_{1}^{2} = j^{2} and
E = ½I_{1} = ½(j^{2})^{½}, and nowhere else, corresponding
to spin of quantised magnitude 2I_{1}^{-2}E possible about any axis in
w_{1}Ùw_{2}.
In GHC we have 3-vector body bimomentum about a due to a stance spin W about b
H_{a,b}(W)
= å_{i}H_{ia,b}(W)
= å_{i}m_{i}
( (r_{i}-c+e_{0})Ù((r_{i}-c+e_{0})¿W)
+ (c-a)Ù((r_{i}-c+e_{0})¿W)
+ (r_{i}-c+e_{0})Ù((c-b)¿W)
+ (c-a)Ù((c-b)¿W)
)Ùe_{¥}
= (å_{i}m_{i}(r_{i}-c+e_{0})Ù((r_{i}-c+e_{0})¿W)
+ m(c-a)Ù(e_{0}¿W)
+ me_{0}Ù((c-b)¿W)
+ m(c-a)Ù((c-b)¿W)
)Ùe_{¥}
= H_{c}(W)Ùe_{¥}
+ m((c-a)Ùv
+ (c-a+e_{0})Ù((c-b)¿W)
)Ùe_{¥}
where 2-vector
H_{c}(W)
= å_{i}m_{i}(r_{i}-c+e_{0})Ù((r_{i}-c+e_{0})¿W)
is linear and symmetric in W ;
has vanishing "expansion" component
å_{i}m_{i}((r_{i}-c)¿v)e_{¥}
= å_{i}m_{i}((r_{i}-c)^{2})^{·}e_{¥}
;
and has H_{a}(w) = bv(h)_{a}(w).
.
For spin about the mass centre b=c we have
H_{a,c}(W)
= H_{c}(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
H_{c}(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
I_{e12,a} = å_{i}m_{i}(r_{i}-a)^{2}.
We define the moment of inertia about the line
L=a+ln as
I_{L}=å_{i}m_{i}s_{i}^{2}
where s_{i}=(r_{i}-a)-(r_{i}-a).
n is the perpendicular from r_{i} 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 S_{i} 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 ma^{2}/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
I_{AB} = ò_{0,c}p((c-x)b/c)^{3}/3dx
= [-p(c-x)^{4}(b/c)^{3}/12]_{0,c} = pcb^{3}/12.
Now p = (mass of H)/(area of H) = 2m/(bc) and therefore I_{AB} = mb^{2}/6. Similarly I_{AC} = mc^{2}/6.
We now make use of the theorem of perpendicular axies to obtain I_{A}.
We have I_{A} =
òp(x^{2}+y^{2})dS where the integration is over the surface of H.
But this seperates to I_{A} = òpx^{2}dS +
òpy^{2}dS = I_{AC} + I_{AB}
= m(b^{2}+c^{2})/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 I_{G} from
I_{A.} We know from the last chapter that
I_{A} = I_{G} + mAG^{2}.
Now AG = AN/3 where N is the perpendicular projection of A to BC. But AN=bc/a so we have I_{G} = m(b^{2}/6+c^{2}/6+b^{2}c^{2}/(9a^{2})). 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
Body | Principle Axes (at mass centre) | Principle moments of inertia | |
Rod of length a | Perp. to rod at centre | ma^{2}/12 | |
Perp. to rod at end | ma^{2}/3 | ||
Right triangle ABC (A=p/2) | Perp. to plane at A | m(b^{2}+c^{2})/6 | |
Perp. to plane at B | m(b^{2}/6+c^{2}/2) | ||
Perp. to plane at C | m(b^{2}/2+c^{2}/6) | ||
AB | mb^{2}/6 | ||
BC | mb^{2}c^{2}/6a^{2} | ||
Perp. to plane at G | m(b^{2}/6+c^{2}/6+b^{2}c^{2}/9a^{2}) | ||
Isosceles triangle base 2w,hieght h | Perp. to plane at apex | m(w^{2}/6+h^{2}/2) | |
Perp. to plane at G | m(w^{2}/6+h^{2}/18) | ||
Line of symmetry | mb^{2}/6 | ||
Rectangle with edges b,c | Parallel to edge b | mc^{2}/12 | |
Edge b | mc^{2}/3 | ||
Perp. to plane at G | m(b^{2}+c^{2})/12 | ||
Diagonal | mb^{2}c^{2}/6(b^{2}+c^{2}) | ||
Disk radius a | Diameter | ma^{2}/4 | |
Perp. to plane at G | ma^{2}/2 | ||
Hoop radius a | Diameter | ma^{2}/2 | |
Perp. to plane at G | ma^{2} | ||
Arc radius a angle 2a | Perp. to plane at centre of arc | ma^{2} | |
Perp. to plane at G | ma^{2} (1-(sin^{2}a) /a^{2}) | ||
Regular n-sided polygon side a | Edge if n even | ma^{2}(1/3+5( tan(n^{-1}p))^{-2})/16 | |
Perp. to plane at G | ma^{2}(1/3+( tan(n^{-1}p))^{-2})/8 |
If we cohere a
a body having of mass m_{1} and central inertia tensor B_{1}_{c}
with a body having mass m_{2} , and central inertia tensor B_{2}_{c+d}
then we create a composite body having centre c'=c + m_{2}(m_{1}+m_{2})^{-1} d
and central inertia tensor
B'_{c'} =
B_{1}_{c}
+ m_{1} (m_{2}(m_{1}+m_{2})^{-1} d)Äm_{2}(m_{1}+m_{2})^{-1} d)
+ B_{2}_{c+d} + m_{2} (m_{1}(m_{1}+m_{2})^{-1} d)Äm_{1}(m_{1}+m_{2})^{-1} d)
= B_{1}_{c} + B_{2}_{c+d}
+ m_{1}m_{2}(m_{1}+m_{2})^{-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=A^{T}r' in the universal frame.
If we write B_{a} for the inertia tensor of H in the universal frame then
B_{a} = A^{T}B_{a}A .
[ Proof :
Since AA^{T}=1 we have
B_{a} = S_{i}m_{i}((r_{i}'-a')^{2}1 -(r_{i}'-a')(r_{i}'-a')^{T})
= S_{i}m_{i}((r_{i}-a)^{2})1 - (A^{T}(r_{i}-a))(A^{T}(r_{i}-a))^{T}
= S_{i}m_{i}((r_{i}-a)^{2}1 - A^{T}(r_{i}-a)(r_{i}-a)^{T}A)
= S_{i}m_{i}(A^{T}((r_{i}-a)^{2}-(r_{i}-a)(r_{i}-a)^{T})A)
= A^{T}(S_{i}m_{i}((r_{i}-a)^{2}-(r_{i}-a)(r_{i}-a)^{T}))A
= A^{T}B_{a}A.
.]
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 B_{a}w, while in the universal frame it is B_{a}'w' = A^{T}B_{a}AA^{T}w = A^{T}B_{a}w.
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.).
Body | Principle Axes (at mass centre) | Principle moments of inertia | |
Rod of length a | Line of rod | 0 | |
Any perp. to rod | ma^{2}/12 | ||
Third Perp. | ma^{2}/12 | ||
Right triangle ABC (A=p}/2) | Perp. to plane | m(b^{2}/6+c^{2}/6+b^{2}c^{2}/9a^{2}) | |
Isosceles triangle base 2w,hieght h | Line of symmetry | mb^{2}/6 | |
Parallel to base | mh^{2}/18 | ||
Perp. to plane | m(w^{2}/6+h^{2}/18) | ||
Rectangle with edges b,c | Parallel to edge b | mc^{2}/12 | |
Parallel to edge c | mb^{2}/12 | ||
Perp. to plane | m(b^{2}+c^{2})/12 | ||
Disk radius a | Any diameter | ma^{2}/4 | |
Perp. diameter | ma^{2}/4 | ||
Perp. to plane | ma^{2}/2 | ||
Hoop radius a | Any diameter | ma^{2}/2 | |
Perp. diameter | ma^{2}/2 | ||
Perp. to plane | ma^{2} | ||
Regular 2n-sided polygon side a | Line of symmetry | ma^{2}(1/3+5/tan^{2}(p/2n))/16 | |
Other line of symmetry | ma^{2}(1/3+5/tan^{2}(p/2n))/16 | ||
Perp. to plane | ma^{2}(1/3+1/tan^{2}(p/2n))/8 | ||
Cylinder radius a , length l | Any diameter | m(3a^{2}+l^{2})/12 | |
Perp. diameter | m(3a^{2}+l^{2})/12 | ||
Axis of cylinder | ma^{2}/2 | ||
Cylindrical shell radius a length l | Any diameter | ||
Perp. diameter | |||
Axis of cylinder | |||
Sphere radius a | Any three perp.axies | 2ma^{2}/5 | |
Spherical shell radius a | Any three perp.axies | 2ma^{2}/3 | |
Ellipsoid | Semiaxis a | 5^{-1}m(b^{2}+c^{2}) | |
Semiaxis b | m(a^{2}+c^{2})/5 | ||
Semiaxis c | m(a^{2}+b^{2})/5 | ||
Ellipsoid shell | Semiaxis a | 1/3m(b^{2}+c^{2}) | |
Semiaxis b | 1/3m(a^{2}+c^{2}) | ||
Semiaxis c | 1/3m(a^{2}+b^{2}) |
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 r_{i} and m_{i}, independant of the r_{i}^{·}. A better choice of baseframe depending also on the r_{i}^{·}
is provided by a choice of R yielding
å_{i}m_{i}r_{iL}Ùr_{iL}^{·}=0 so that
j_{L} = h_{cL}(w) provides a bodyframe in which the angular momentum
is equivalent to that of a rigid body roatating at w about 0_{L}.
We can construct such an R integratively via w = -2R^{·}R^{§} =
h_{cL}^{-1}(j_{L})
Þ
R^{·}
= -½h_{cL}^{-1}(j_{L})R .
Multiparticle Potentials
Suppose the particle at r_{i} generates a Coulomb potential f_{i}(x) = q_{i}|x-r_{i}|^{-1} whwre q_{i} is the charge
of the particle. For a gravitational potential we set q_{i}=-Gm_{i}. The total potential at x is
expresible as a multipole expansion
f(x) = å_{i}q_{i}|x-r_{i}|^{-1}
= |x-c|^{-1} å_{i}q_{i}å_{k=0}^{¥} (x-c)^{-2k} P_{k}(x-c,r_{i}-c)
= |x-c|^{-1} å_{i}q_{i}å_{k=0}^{¥} |x-c|^{-k} P_{k}((x-c)^{~},r_{i}-c)
= |x-c|^{-1}
((å_{i}q_{i}) + (x-c)^{-4} P_{2}(x-c,r_{i}-c) + _{O}(|x-c|^{-3}))
since the
å_{i}q_{i}P_{1}(x-c,r_{i}-c)
= (x-c)¿å_{i}r_{i}-c vanishes for charge center c=(å_{i}q_{i})^{-1} å_{i}q_{i}r_{i} .
A spherically symmetric matter distribution has f(x) = (å_{i}q_{i})|x-c|^{-1} equivalent to
a single particle at c, so we can regard the
|x-c|^{-1}å_{i}q_{i}å_{k=2}^{¥} |x-c|^{-k} P_{k}((x-c)^{~},r_{i}-c)
terms as corrections for the asymmetry of the matter distribution.
For N=3 we have
å_{i}m_{i}P_{2}(x-c,r_{i}-c)
= ½(x-c)¿q(x-c)
where q(a) º 2ah_{c}(1) - 3h_{c}(a) is the
gravitational quadrapole 1-tensor.
[ Proof :
å_{i}m_{i}P_{2}(x-c,r_{i}-c)
= å_{i}m_{i}½(3((x-c)¿(r_{i}-c))^{2} - (x-c)^{2}(r_{i}-c)^{2})
=
(3/2)(
å_{i}m_{i}(x-c)^{2}(r_{i}-c)^{2} - (x-c)¿h_{c}(x-c))
- ½å_{i}q_{i}(x-c)^{2}(r_{i}-c)^{2} )
= (x-c)^{2}å_{i}m_{i}(r_{i}-c)^{2} - (3/2)(x-c)¿h_{c}(x-c))
= ½(x-c)¿(2(x-c)å_{i}m_{i}(r_{i}-c)^{2} - 3h_{c}(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 m_{0} at x is
f(x) = -m_{0}Ñ_{x}f(x)
= -(x-c)^{-2}G(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)^{-2}G((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 m_{0} at x is
å_{i}(r_{i}-c)Ù(-f_{i}(x))
= -å_{i}(r_{i}-x)Ùf_{i}(x)
- å_{i}(x-c)Ùf_{i}(x)
= -(x-c)Ùå_{i}f_{i}(x)
where f_{i}(x) = -m_{0}Ñ_{x} f_{i}(x) parallel to r_{i}-x is the force exerted on the pointsource by the particle at r_{i}.
Thus the total torque
-(x-c)Ùå_{i}f_{i}
= 3Gm_{0}|x-c|^{-5} (x-c)Ùh_{c}(x-c)
= 3Gm_{0} |x-c|^{-3}) (x-c)^{~}Ùh_{c}((x-c)^{~})
decreases with |x-x|^{-3} rather than
|x-x|^{-2} and vanishes for a spherically symmetric body with h_{c}(a)=Ia.
Note that a¿h_{c}(a) =
= å_{i}m_{i}((r_{i}-c)^{2}a^{2} - ((r_{i}-c)¿a)^{2})
provides
å_{i}m_{i}((r_{i}-c)^{2}a^{2} =
a^{2}å_{i}m_{i}(r_{i}-c)^{2} - a¿h_{c}(a) .
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