# Focault's compass

1. Mar 12, 2004

### Charlls

Hi there,

the Focault compass its supposed to be a fast spinning disk that keeps pointing to the earth north

the spinning axis is mounted on top of a rotating base, angle which is supposed to oscillate when slighty displaced from the north direction

is this oscilatory behaviour occuring 'only' on the earth surface, or its true also in inertial far-away-from-earth frames?

i wrote the lagrangian for this thing using omega . I omega

lets assume first the problem in far-awar-from-earth-inertial frames, then i got a fixed basis Ex, Ey and Ez

the gyroscope has its principal axis like Ez, En and Ew, where En = cos(a)Ex + sin(a)Ey

and Ew = -sin(a)Ex + cos(a)Ey

the spinning axis of the gyroscope is En, which is also the symmetry axis

the gyroscope can also rotate in the Ez axis, with an angle a

so i wrote the I tensor in dyadic rep. like I = Iz EzEz + Ir EnEn

but EnEn = cos(a)^2 ExEx + sin(a)cos(a) [ ExEy + EyEx ] + sin(a)^2 EyEy

the rotation of the gyroscope can be represented as:
omega = a' Ez + omega En = a' Ez + omega cos(a) Ex + omega sin(a) Ey

so when you plug this rotation into the inertia tensor to get the kinetic energy, you get at the end:

Lagrangian = a'^2 Iz + omega^2 Ir

(remember that the dyads act with vectors like EiEj * Ek = (Ej . Ek) Ei, where . is the dot product between vectors)

so as you see, my lagrangian does NOT depend on a, so i cant get an oscillatory motion in this system

Im doing something blatantly wrong here?

any insights are welcome

Cheers

2. Mar 14, 2004

### pallidin

Not sure about the "Focault Compass" but what you are describing appears to me to be an effect governed by gyroscopic principles. As such, a regenerative spinning "disk" on a frictionless gymbal will maintain its spacial orientation regardless if you are on earth or in deep space.
In other words, if you set this device to spin with the disks orientation to be "level" with a particular point on earth, than carefully moving it a quadrillion miles away would still indicate that level plane from earth.

3. Mar 14, 2004

### Charlls

ok, but then that means my derivation is wrong at some point. where and how? i've been dealing with this problem the last 2 days only getting frustation from it

4. Mar 14, 2004

### Charlls

To avoid any confusions about what we are talking here, think of this apparatus as exactly the front wheel of a bike put upside down; the 'a' angle is the maneuvering angle; the 'omega' rotation itself is the rotation of the wheel.

The idea here is that if the axis of rotation of the wheel is aligned with the north direction, small deviations from that direction will oscillate, meaning that the wheel tends to keep aligned with the north direction; hence its being called 'focault compass'

Now, when you explicitly include the rotation of earth, you write first the angular momenta of the compass relative to the center of earth (ill be consistent with the previous choice of vector basis)

L = { I:cm: + Mcos(theta)^2 (R^2 - ErEr ) } (Qcos(theta)Er + Qsin(theta)En )

Here we used the steiner tranlated inertia tensor to the center of earth; where I:cm: is the inertia tensor about the center of mass of the compass, R is the radius of earth, theta is the lattitude angle, Q is the angular speed of earth's rotation (2*Pi / (24 hours))

To this angular momenta you add the momenta of the compass 'internal' degrees of freedom

L = I:cm: ( a' Er + omega cos(a) En + omega sin(a) Ew )

Where En and Ew are respectively the north and west direction in the surface of earth, omega is the angular speed of the wheel and a is the maneuvering angle; a' is its velocity

So, when you process all this algebraically, i got this kinetic energy:

Ir (a' + Qcos(theta))^2 + Iw (omega + Qcos(a)sin(theta))^2 + M R^2 Q^2 cos(theta)^2 sin(theta)^2

Where Iw is the symetrical axis inertia (the wheel inertia) and Ir is the inertia along the wheel plane

which indeed has small oscillations behaviour (apply lagrange equation dL/da = d(dL/da')/dt )

i got finally (asumming the earth rotation velocity and the lattitude angle staying constant, and approximating sin(a) = a :)

a'' = - (omega*Iw /(2*Ir)) a

Putting this all aside, just like you Pallidin, i think that this behaviour is also present when the wheel also stays in an inertial frame. If this is true, my first derivation must be wrong somewhere. The question is where?

.oO( help! im clueless ! )