Earth powered gyroscope?

 There has been much speculation about the possibility of exploiting the precession torque of a gyroscope in such a way that it would be self-sustaining.The general observations which lead to this speculation are:1.  A fixed gyroscope with its axis perpendicular to the axis of the earth resists the earth's rotation.2.  This resisting torque, multiplied by 2PI radians, is the amount of rotational kinetic energy drawn from the earth in one day.The question is, can this kinetic energy loss of the earth be fed back as an energy gain for the gyroscope?If so, could the energy feedback rate be sufficient to overcome frictional losses?And, of course, the bigger question: Is it possible to build a gyroscope which would not only be self-sustaining, but have useful energy left over?It turns out that this last question is easier to answer than the first two.  Let's assume that the answer is affirmative for all three questions and that we have designed a frictionless device which is 100% efficient in converting the braking energy applied to the earth into electrical power.As a proof of concept, we set about building a demonstration device which will light a 100watt light bulb.  Since the braking energy rate is the dot product of the resisting torque and the angular velocity of the earth, the required braking torque is easily determined.$T\cdot \Omega =100 watts$One watt is one newton-meter per second, and the rotation rate of the earth is 0.00007272 radians per second.  The braking torque which would reduce the earth's rotational energy at the rate of 100watts is:$T =\frac{100}{0.00007272}=1,375,137 Nm =1,014,249 ftlb$It would be less than impressive if it required a machine larger than a desk to light one light bulb.  Assuming a gyroscope shaft three feet long, the side force on the bearings at each end would be approximately 330,000lb.  To put three hundred and thirty thousand pounds in perspective, it is equivalent to six 747-400 engines at maximum thrust.  (Extreme side force vastly increases the  difficulty of maintaining low friction in the bearings.)While the braking torque required to bleed kinetic energy from the earth at the rate of 100watts is extreme, the angular momentum required to generate this amount of precession torque should be discouraging to all but the most credulous.  Precession torque is equal to the time derivative of angular momentum.  The time derivative of any vector due to rotation is the cross product of the angular velocity and the vector.$\dot{L}=\Omega \times L=T$$\therefore L=\frac{T}{\Omega }$$L =\frac{1,014,249}{0.00007272}=13,947,318,481 ft\cdot lb\cdot sec$While it appears to be impractical to generate power on a large scale from rotational energy of the earth, the answers to the first two questions might still be of interest.  The one factor that makes rotational energy of the earth such a poor power source is the extremely low rotation rate.  If the energy source had a larger rotation rate, affirmative answers to the first to questions might lead to efficient mechanisms for power conversion from high torque, low RPM power sources to low torque, high RPM electrical generators. Most of the information concerning this hypothesis is anecdotal and distinctively lacking any experimental data or critical mathematical analysis.  This web page will attempt to provide the latter.The following diagram represents a gyroscope located at the Equator with its axis fixed relative to the surface of the earth.Upper case Omega is the rotation vector of the earth and lower case omega is the rotation vector of the gyroscope.The local earth-fixed orthogonal coordinates are east, north, and up.  North and east are in the horizontal plane, and up is perpendicular to the horizontal plane.  (We're at the Equator, so the axis of the earth is parallel to the north axis.)In the configuration to be analyzed, the gyroscope's axis is tilted to the west by the angle theta, while keeping it in the up-east plane.  (Theta is a rotation about the north axis, which remains in the plane of the gyroscope.)The local reference frame is not only rotating around the north axis, it also has an easterly inertial velocity of Omega times the radius of the earth.  This easterly trajectory has a radius of curvature equal to the radius of the earth, giving the local coordinate system a centripetal acceleration in the down (minus up) direction.  For this analysis, the gyroscope consists of a thin circular ring of mass, m, and moment of inertia r2m.  A small differential mass can be expressed by a small differential angle.$dm=\frac{m}{2\pi }d\phi$ 11/20/2012*********  to be continued ************