Modeling The DeviceFirst created: Sun Mar 26 2000
Last update: Sun Mar 26 2000
The Equilibrium Position
Under 3 different configurations of the magnet stacks, the minimum of the black curve indicates the equilibrium position of the float. (Click image for full picture).
The two stack design is the one with the broadest, flatest, equilibrium for the float. This ensures that small changes in gravity will make the float oscillate with an unusually large amplitude.
The potential energy (V) of the float has contributions from 3 sources: the bottom stack, the top stack and gravity.
V = k/r - k/(h - r) + mgrwhere
k is the magnetic strength of a stack h the distance between the stacks r the distance of the float above the bottom stack m the mass of the float g the acceleration from gravityWhether the magnetic potential varies as 1/r (monopole) or 1/r2 (dipole), or something in between, is not critical for the arguments that follow.
Equilibrium points are reached when the potential energy is a minimum (stable) or maximum (unstable),
d/dr V = 0 k/r2 + k/(h - r)2 - mg = 0 (1)
Provided h is big enough, there are 2 equilibrium points, a stable one at r0 = h(1 - q)/2 and an unstable one at r1 = h(1 + q)/2, where q is some small calculable quantity. The float at the stable position will oscillate under small perturbations. At the unstable position, it will invariably end up falling onto the top stack. A stopper just above h/2 float is needed to prevent this from occuring during adjustments.
The Oscillation Period
The oscillation period at the stable point is of great interest because oscillations are easily produced and their period is easily measured. The period T at the stable point is
T = 2 * pi * sqrt(Q) (2)and it does not depend on g. The quantity Q
m 1 Q = -- * ------------------- (3) 2k 1/r03 - 1/(h - r0)3is an important characteristic of the probe. It has dimensions of a time squared.
This result can be obtained by a formal expansion of the potential around the stable position r0. Assume r = r0 + dr, then
V(r) = V(r0 + dr) = C + k(1/r03 - 1/(h - r0)3)(dr)2 - k(1/r04 + 1/(h - r0)4)(dr)3 + k(1/r05 - 1/(h - r0)5)(dr)4 - ...C is some constant value. The term linear in dr vanishes at the stable point. For small dr, the cube an higher powers can be neglected and the potential is that of an harmonic oscillator. The expression for the period follows from there.
To achieve a long period one must try, according to (3), to get the stable and unstable points close together, on each side of h/2. This is tricky because one also needs to limit the oscillation amplitude to stay at all times on the stable side of the unstable point. The longer the period, the smaller the amplitude allowed and the more care is needed in designing the stopping mechanism. A heavier float and weaker magnets will also help in extending the period. The longest period I have been able to achieve is between 1 and 1.5 seconds.
Sensitivity To Changes In Gravity
Under a small change g -> g + dg the equilibrium position changes from r0 to r0 + dr, with dr satisfying
k/(r0 + dr)2 + k/(h - r0 - dr)2 - m(g + dg) = 0to first order
2k*(1/r03 - 1/(h - r0)3)*dr + mdg = 0 dr = - Q * dg (4)
If we can achieve a long period (2), then we also achieve high sensitivity to changes in g (4). Specifically, if the period is 1 sec, then in response to dg = 10-7g, we will see a displacement dr
dr = 1/(2*pi)22 9.8*10-7 (m) dr = 2.5*10-8 (m)This is exceedingly small.
Sensitivity To Changes In Temperature
The temperature has a significant effect on the magnetic strength k. Under a change k -> k + dk, relation (1) implies
(k + dk)/(r0 + dr)2 + (k + dk)/(h - r0 - dr)2 - mg = 0to first order
dk*mg/k + 2k*(1/r03 - 1/(h - r0)3)*dr = 0 dr = Q * g * dk/k (5)dr and dg have a similar depency on dk, proprortional to Q. This is because (1) really only depends on the combination k/(mg). For the dg effect to dominate, we would need dg >> g*dk/k, but ... this is not the case.
dg/g is about 10-7 for the Moon tides and dk/k is about 10-3 per degree F for a ceramic magnet. Therefore drTides/drF is about 10-4. The temperature will have to be kept constant to 10-4 F for tidal and thermal effects to have comparable magnitudes.
Sensitivity To Changes In Air Pressure
Changes in atmosperic pressure will also appear as a small change in g. Air is mostly N2, atomic weight 2*14, so its density is about 28/22.4 = 1.25 g/liter. The float displaces about 10 cm3 of air, 10-2 liter, so the buoyant force it experiences is about 0.0125 grams. Atmosperic pressure can vary by about 10 % so the buoyancy force can change the apparent weight of the float by about 0.00125 grams The float weights about 80 grams so the relative change in weight is 0.00125/80 = 1.6*10-5
One way to minimize this effect is to seal the experiment in a pressure tight container.
Prospects for measuring Moon and Sun tidal forces with this device are grim. The first challenge is to be sensitive enough to measure 10-7 to 10-8 m. If that is met, there remains the task of extracting the signal for the thermal and pressure backgrounds.