## PSN-L Email List Message

Subject: Re: Article on gravimeters
From: John Hernlund hernlund@.......
Date: Mon, 08 May 2000 01:11:29 -0700 (MST)

```Okay, I went through and did the math for the problem of two accelerometers (I
know, I am definitely a nerd) in a rotating frame of reference.  The two
accelerometers are at opposite ends of the spinning frame, both at a radius of
"r."  The angular velocity of the frame is "w."

To say that a gravity gradient exists is to say that there is a curvature in
the gravitational potential "U" in the vicinity of the spinning disk.  There
are many ways to deal with this, but I let the gravitational potential take
the form:

U = K + 0.5*ax^2 + b*xy + 0.5*cy^2 + d*x + e*y

Since this problem is being done in a plane, only x and y coordinates are
needed.  x and y could be east and north, or down and east, or which ever
combination you want.  The acceleration due to gravity is the gradient of the
potential, which is (in vector form):

g = [a*x + b*y + d,c*y + b*x +e]

The gravity gradient is then (in matrix form):

Dg =	[ a   b	]
[ b   c	]

So we seek to find the values for a, b, and c only in the gradient
determination.  Saving you all the math, the signals from the two
accelerometers (s1 and s2), when figured in terms of rotation, noise, and
gravity can be added together to yield a very simple equation:

S = s1 + s2 = 2r[w^2 - a*cos(wt)cos(wt) - 2b*sin(wt)cos(wt) -c*sin(wt)sin(wt)]

The signals take this form if you adjust your time so that they lie on the
x-axis at t = 0.  Notice that the values for d and e, the values for the
gravitational field itself, have cancelled out along with the noise.

Lets modify this record further and let T = -S/(2r) + w^2 and the above
equation becomes simply:

T = a*cos(wt)cos(wt) + 2b*sin(wt)cos(wt) + c*sin(wt)sin(wt)

Because we know what r, w, and t are in the data set we can fit the data to
this formula and obtain values for a, b, and c.  Viola!  There is the gradient
in that plane.  (This assumes a negligible phase shift in the response for the
instrument, but that could also be dealt with)

The unanswered question, of course, is whether we can mount two accelerometers
that are sensitive enough at the frequency w to make out exactly what a, b,
and c are.  The values for a, b, and c are very small I am sure.  We can
estimate what they would be in the vertical direction on the Earth by
considering two points one meter apart at the surface.  The gravity is given
as g = -GM/r^2 (G = gravitational const., M is mass of Earth, and r is radius
from Earth's center).  The vertical gradient is the derivative with respect to
km, its mass 5.98E24 kg, and G is 6.67E-11 (SI units) the vertical gradient is

Lets say we want to detect a mountain that is 1 kilometer away (to its
middle) has a density of about 3000 kg/m^3 (or 3 g/cc, about that of many
rocks) and is one cubic kilometer in volume.  We get about 2E-7 s^-2.  That
value is about one-tenth of the value of the vertical gradient, and is larger
than I expected.  This is because of the 1/r^3 form of the equation for the
gradient.  This type of instrument will be extremely sensitive to things that
are close.  The gradient direction will also tend to point towards high
density bodies and away from low density objects.  This would be great for
finding salt domes and such.

Anyways, I think that this is definitely possible.  I am almost itching to try
it myself.  The key does seem to be finding a couple of accelerometers that
are sensitive enough, and also rugged enough to take the beating they are sure
to get.  High performance piezo-electric accelerometers would be perfect for
the task but also very expensive or impossible to find for the sensitivity
required.  One pair on a single spinning frame is all that is needed I think,
because it could be made to tilt in any general direction.  Surely though, the
bear is probably lurking in the details.

John Hernlund
E-mail: hernlund@.......
WWW: http://www.public.asu.edu/~hernlund/

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