I've been improving my calculations, and the latest edits to my posts contain those improvements, but here's another summary that I think is my best yet:
1. Torque-free precession is when there is practically no external torque (i.e. the tiny torques due to Luna & the Sun are negligible) . Because there is no external torque, Earth's spin angular momentum vector is constant. But that doesn't mean that the same piece of turf has to be under the pole all the time. Earth can move so that different pieces of turf are under the pole. The "pole" (i.e. latitude & longitude of the piece of turf) can move (i.e. "precess") a lot, relative to Earth coordinates, even though the angular momentum vector is unchanged in "absolute" coordinates. (When this happens there is also a slight daily wiggling of Earth's spin vector in absolute coordinates, due to Earth's slight asphericity and Earth's being somewhat askew, which causes the spin vector to have a direction slightly different from the angular momentum vector, but that's pretty negligible.)
2. Earth has almost perfect symmetry about the z-axis of its spheroid. So, two of Earth's moments of inertia are equal, both slightly smaller than the z-axis moment of inertia. The equality of those two x & y-axis moments, simplifies "Euler's equations" (Euler's equations amount to a statement of conservation of angular momentum, expressed in rotating Earth coordinates by way of an expression for "Coriolis force" applied to the angular momentum vector itself). Then it's pretty simple to rearrange Euler's equations to show that Earth can "precess" so that the pole moves slowly along a line of Earth latitude. Even without Euler's equations, just considering conservation of energy and angular momentum, it is evident that two variables, (1) the latitude of this precession line, and (2) the period of Earth's rotation, are determined by two equations involving angular momentum and kinetic energy of rotation and Earth's moments of inertia. Angular momentum is a known constant, the same as it is today, but I need to know the kinetic energy (see #4) and the moments of inertia (i.e. the flattening) (see #5).
3. As long as the changes in kinetic energy and angular momentum are small, the instantaneous period of Earth's sidereal rotation changes only by a few seconds: e.g. 56 sec shorter, in the 18.90deg precession solution discussed below. But the latitude of the precession line, i.e. angle of precession, can become large with only small changes in kinetic energy and no change in angular momentum. Basically this is because Earth's flattening is small. When Earth's kinetic energy is exactly what it is now, all Earth can do is rotate with nearly perfect symmetry about its z-axis, as it does now; but if Earth somehow acquired more kinetic energy of rotation, it could precess at a large angle.
4. Earth's gravitational self-energy is large; about 1000x Earth's kinetic energy of rotation. The energy released by, say, a major decrease in Earth's flattening (like energy released by a falling weight or a skater pulling in his arms) is of the order of magnitude to enable large-angle torque-free precession. My best effort to calculate this energy of flattening, is that it is
8/5 * (k/3)^2 * Earth's total gravitational self-energy
"k" is the flattening ratio, 1/297. (In practice I make small adjustments to account for my belief that the core does not participate in this torque-free precession.) Half of the above energy, a "4/5" is simply due to generally greater distances between mass particles, when Earth is flattened. Half of the energy, i.e. a "4/5" is a "monatomic adiabatic perfect gas response" to the other half, assuming Earth isn't truly "incompressible": when Earth is less flattened, forces are greater, pressures are greater, and Earth shrinks slightly to a new equilibrium, releasing yet more potential gravitational energy, namely an "8/5" but due to "the virial theorem", half of that energy goes into heat, and half, another "4/5" is available for mechanical work, so 4/5 + 1/2 * 8/5 = 8/5 is the answer.
5. I also need to know how much Earth flattens. I simply averaged all the spheroids one would get with the poles at different points along the line of latitude. The answer is that the flattening is lessened by the factor
1 - 3/2 * sin^2 (alpha)
where alpha is the colatitude of the precession, i.e. the central half-angle of the precession cone.
6. Using my estimates of flattening and gravitational energy release, I have all that I need to solve those two equations in two unknowns, for Earth's spin rate and precession angle. I find that there are two solutions:
I. What is happening today.
II. Torque-free precession along the 71st (lat 71.10) parallel, with less than a minute difference in day length.
Between these two solutions are "energy trough" states, i.e., states that are easily traversible. I & II are like the two extreme swings of the pendulum.
7. Now here are the two reasons to believe that I am right in #4 and #5:
a) There is another equation, obtained from Euler's equations, that gives the period of the torque-free precession. The period turns out to be a year; if my estimate of the gravitational potential energy release, is increased by only 2% (believable, considering the crudity of my approximations) then the precession period is exactly one tropical year.
b) The torque-free precession is at such a latitude that it would explain Hapgood's estimates of the poles. With a precession period of one year, having the pole at Hudson's Bay in winter has about the same effect on ice cap position, as having the pole at Hudson's Bay year-round. Hapgood estimated the last three Ice Age pole latitudes (in today's Earth coordinates) as, most recent to least recent, 60, 72, & 63deg latitude.
