A question that should be asked during physical design of any product
incorporating sensors is: “Where should be sensors be located within
the product?” Like almost anything else in life, “It depends
… ” is the answer. This post discusses accelerometer placement.
We’ll touch on magnetic sensors next time around.
If your application is such that the sensor is statically fixed in space, or
at least not rotating, it doesn’t really matter where you mount the
accelerometer on your board. But if the device can rotate, things get a lot
more interesting.
The rate of change transport theorem, which can be found in
any rigid body dynamics textbook, relates the rate of change of a
vector r as observed in two different reference
frames. For our purposes, the first is the fixed (earth) frame and the
second (body frame) rotates and translates relative to the first.
In equation form:
(dr/dt)_{f }=(dr/dt)_{r} +
ω X r_{r} (Eqn. 1)
where the subscripts indicate the frame of reference.
The transport theorem is a fundamental relationship, based upon basic
principles of geometry. It applies to ANY vector observable from two
independent reference frames. You should note that both left and right
sides of the equation are vectors.
We are going to repeatedly apply the transport theorem to derive a
relationship defining the acceleration of an object in the fixed earth frame
based upon sensor measurements made in the body frame.
In graphical form, consider movable point “P”, observable
from two different frames of reference. “O_{f}” is
the origin of the fixed earth reference frame. “O_{r}”
is the origin of the rotating reference frame.
By basic geometry
r’ = R + r (Eqn. 2)
Note that
V = (dr/dt)_{f}
V is the rate of change of vector r as viewed in the fixed reference
frame. It can also be viewed as the velocity of P as viewed in the fixed
earth frame
V_{r} = (dr/dt)_{r}
V_{r }is the he rate of change of vector r as viewed in
the rotating reference frame. It can also be viewed as the velocity
of P as viewed within the rotating reference frame.
ω = the angular velocity of the rotating frame relative to the fixed
earth frame
by differentiating Eqn. 2, we get
(dr’/dt)_{f} = (dR/dt)_{f} +
(dr/dt)_{f} (Eqn. 3)
which we can also write as
V_{r’} = V + (dr/dt)_{f} (Eqn. 4)
The quantity V can be interpreted as the translational velocity of the
rotating frame relative to the fixed frame of reference. V_{r’}
is the velocity of point P as observed in the fixed earth frame.
Typically, this is the quantity we are trying to measure.
We can apply the transport theorem to the right most term in Eqn. 4 to
get
V_{r’} = V + (dr/dt)_{r} + ω X
r
(Eqn. 5)
V_{r’} = V + V_{r}
+ ω X r (Eqn. 6)
Now differentiate and take advantage of the math identity:
d/dt (A X B) = (dA/dt X B) + (A X dB/dt) (Eqn. 7)
to get
A_{r’} = (dV_{r’}/dt)_{f }= (dV/dt)_{f} + (dV_{r}/dt)_{f} +
(dω/dt)_{f} X r + ω X (dr/dt)_{f} (Eqn. 8)
Let
A = (dV/dt)_{f}
where A is the acceleration of the second frame of reference
relative to the fixed earth frame.
At the same time, apply the transport theorem again to substitute
(dV_{r}/dt)_{f }=(dV_{r}/dt)_{r}+ ω X
V_{r} (Eqn. 9)
to get
A_{r’} = A+
(dV_{r}/dt)_{r}+ ω X V_{r}+ (dω/dt)_{f} X r + ω X (dr/dt)_{f}
(Eqn. 10)
Let
α = (dω/dt)_{f }
where α is the rate of angular acceleration, and
A_{r} = (dV_{r}/dt)_{r}
where A_{r} is the rate of acceleration of P relative to
O_{r}. Then
A_{r’} = A+
A_{r}+ ω X V_{r}+ α X
r + ω X (dr/dt)_{f} (Eqn. 11)
and one last substitution using the transport theorem for the right most term
in Eqn. 11 to get
A_{r’} = A+ A_{r}+ ω X
V_{r}+ α X r + ω X
((dr/dt)_{r} + ω X r) (Eqn. 12)
A_{r’} = A+ A_{r}+ ω X
V_{r}+ α X r + ω X (V_{r} + ω X r) (Eqn. 13)
A_{r’} = A+ A_{r}+ ω X
V_{r}+ α X r +
ω X V_{r} + ω X (ω X r)) (Eqn. 14)
A_{r’ }=

A + A_{r} +

<– Inertial Acceleration



2ω X V_{r} +

<– Coriolis Acceleration



α X r + 
<– Euler Acceleration



ω X (ω X r))

<– Centripetal Acceleration

(Eqn. 15) 
Eqn. 15 is a classic result, and can give you a lot of insight into tradeoffs
associated with accelerometer placement. To see how, let’s modify
our previous figure by locating O_{r} at the center of mass of a
portable consumer device. Point P represents the location of our
accelerometer within the same device. Point O_{f} is
essentially the center of the earth.
Since points P and O_{r} are fixed relative to one another (they are
part of the same rigid body), A_{r} and
V_{r} are zero by definition and drop out of the
equation. If the rotational rate of the device is zero, then ω=0 and both Coriolis and Centripetal acceleration terms drop out of the
equation. Finally, if α (the rotational
acceleration) is zero, there will be no Euler acceleration. If all these
terms hold true, then the value measured by our accelerometer is
A_{r’} = A = the translational rate between the
two frames of reference
which is exactly what we would like. But what if ω≠0?
If we collapse point P back into the center of mass (O_{r}), then both
Euler and Centripetal accelerations still drop out.
What happens if you ignore these effects? Let’s work an example
assuming that angular acceleration is zero, but angular rotation is not:
A_{r’ }=

A + A_{r} + 
<– Inertial Acceleration 


ω X (ω X r)) 
<– Centripetal Acceleration 
(Eqn. 16) 
To bound the problem, I’m going to take advantage of the identity:
A X B = A B sin Θ n
where Θ is the angle between vectors A and B, and n is
a unit vector perpendicular to the plane containing both A and B.
Looking at just the extremes, we can see that the Centripetal acceleration
ranges between +/ ω^{2}r in magnitude, depending upon
the axis of rotation and accelerometer offset from center of mass.
Assume
 ω = 2π radians/sec
 r = 5 cm
Putting numbers to Eqn. 16, we get
Centripetal Acceleration = +/ 0.05 (2π)^{2} =+/ 1.97
m/s^{2}.
That’s better than 1/5 gravities! At a very modest rate of
rotation!
Basically, sensor fusion mathematics get much easier when the
accelerometer is at the center of mass (which is also the center of the
rotating frame of reference). If you can make the assumption that, in
your use case, the device is in a quasistatic state with no rotation, it
doesn’t matter. If not, you need to place the sensor at the
center of mass OR account for the effects mathematically. That involves
keeping continuous track of the axis and rate of rotation and then making the
adjustments required to separate out the Centripetal acceleration.
References:

Dynamics of Particles and Rigid Bodies: A Systematic Approach, Anil V. Rao,
Cambridge University Press, 2006

Motion in a NonInertial Frame, Alain J. Brizard, Saint Michael’s
College.