Modeling The Device
First created:
Sat Jun 2 14:22:33 EDT 2001
Last update:
Sat Jun 2 14:22:33 EDT 2001

The Equilibrium Position
The equilibrium position of the 2 pendulums is achieved when
they are both aligned vertically with the local gravity.
In this configuration the pendulums will oscillate around a stable
equilibrium under small perturbations.
The force that restores the pendulums to the equilibrium position
comes from the coupling magnets. The magnet dipoles on the two strips
attract each other.
The potential energy (V) of the float has contributions from 3
sources: the top pendulum, the bottom pendulum and the magnets.
V(a,b,r) = mgh(1  cos(a)) + mgh*cos(b)  k/r

(1)

where
m measures the mass of a pendulum
g is the acceleration from gravity
h is the length of a pendulum
a is the angular displacement of the top pendulum, in radians
b is the angular displacement of the bottom pendulum, in radians
k is the magnetic strength of the magnets
r the distance between the magnets


Whether the magnetic potential varies as 1/r (monopole) or 1/r^{2} (dipole),
or something in between,
is not critical for the arguments that follow.
The potential energy has a minimum at
where r_{0} is the minimum distance between the magnets.
Under the small change
a new equilibrium will be reached at
r = r_{0} + dr
a = b = da


where both dr and da are small and
2h*cos(da) + r + dr = 2h + r
dr = 2h(1  cos(da)) = h(da)^{2}


We can use this to constrain dr and db in terms of da
and express the potential as an effective potential in da alone
V(da) = V_{0} + k/r_{0}^{2}*dr
V(da) = h*k/r_{0}^{2}*da^{2}


After dropping the constant V_{0}.
Since a = 0 + da,
V(a) = h*k/r_{0}^{2}*a^{2}

(2)

Around the equilibrium
the system behaves like an harmonic oscillator,
as most systems do.

The Oscillation Period
The period of oscillation for small perturbations around the equilibrium
can be read from (2).
The kinetic energy for the system is
E(d/dt a) = m/2*(h*(d/dt a))^{2}


The equation of motion is
md^{2}/dt^{2} (h^{2}*a) = 2h*k/r_{0}^{2}*a


A solution is
Where the period T is given by
mh^{2}(2*pi/T)^{2} = 2kh/r_{0}^{2}
T = 2*pi*r_{0}*sqrt(h*m/k)

(3)

The period is easy to observe and measure.
A period of 3 seconds is not difficult to achieve.

Sensitivity To Changes In Gravity
If we introduce a small horizontal shift in gravity
with g vertical and dg horizontal,
the restoring force from the top pendulum is cancelled
by the destabilizing force of the bottom pendulum and a dx will develop.
The force that will restore equilibrium comes from the dr that develops
as the coupling magnets separate. Equating the forces
h*m*dg = k/r_{0}^{2}*dr
dr = h*m*r_{0}^{2}/k*dg


The distinguishing feature of pendulums is the square root relation between the
horizontal (dx) and vertical displacements (dr).
dx = h*da
= sqrt(h*dr)
= h*r_{0}*sqrt(m/k*dg)


In terms of the period,
dx = T/(2*pi)*sqrt(h*dg)

(4)

And here is why this device in interesting:
dx is proportional to the square root of dg.
This greatly increases the instrument's sensitivity.
Specifically, with a T of 3 sec and an h of 25 cm,
the response to a dg of 10^{7}g
will be a displacement dx = h*da
dx = 3/(2*pi)sqrt(0.25*9.8*10^{7})
= 4.8*10^{4} meters


Between 1 and 0.1 millimeter.
This should be easily measurable.

Sensitivity To Changes In Temperature
A higher temperature will cause the coupling magnets to weaken and the device
to expand. However the frame expands at the same rate so that effect is muted.
So there will be a lower k and maybe a smaller r_{0}.
This will amplify an existing displacement, but it should not cause a
displacement to appear when none existed.

Sensitivity To Changes In Air Pressure
Buoyancy shifts may vary the weight of the components (mg)
and thus the amplitude of a displacement but they should not create
a displacement when none existed.

Conclusion
Prospects for measuring horizontal components of tidal forces with this
apparatus appear to be good.
