Hi Chris,
 
  For a damped pendulum, we are dealing with the equation
 
  F = m(d2x/dt2) + c(dx/dt) + kx, where F is some forcing  function.
 
 m is mass, k is stiffness, and c is drag. Feynman defines  k=m*Omega0^2, 
c=m*Gamma, and Q=Omega0/Gamma, where Omega0 is the natural  frequency.
 
 In order to calculate the effect of a shunt load on the  pickup coil for 
damping, recall that the sensor output sensitivity is given by S  = B*L, where S 
is in units of volts per meter per second, B is field strength in  Teslas, and 
L is active wire length in meters.
 
  If we pass current through the coil, the resultant motor force is f  = S*I, 
where f is in newtons, and I is amperes.
 
  If we load the moving coil with the resistance of the coil in series  with 
an external resistance, the generated voltage is E = S * v, and the current  
is I = E/R, so the resulting drag is c*v = S*I = S*E/R = S*(S*v)/R.
 
  So, c = S*S/R, where c is drag in newtons per meter per second. From  the 
Feynman definitions, c = m * Gamma = m * Omega0 / Q. Since c from shunt  
damping is S^2 / R, we have the equality S *S / R = m * Omega0 / Q.
 
  Solving for Q, we obtain:
 
Q = R * m * Omega0 / S^2
 
  For a Q of unity, 
 
R = (Ps * S^2) / (2 * Pi * m), where Ps is sensor natural period in  seconds.
 
  When I plug in my assumed values of Ps = 12 seconds, S = 85 v-s/m,  and m = 
0.1 kg, the value for R is 138 kOhms.
 
  In reality, the value I need to get a Q of 1.0 is less  than half the 
above. I have no clue as to why. I calibrated the sensor output  several ways, all 
yielding the same result. The mass was determined by weighing  the solder wire 
mass and the coil assembly. To get better agreement, I would  have to 
considerably decrease my value for S, which incidentally would also  bring my 
calibrated seismic amplitudes into closer agreement with those recorded  by station 
PAL. My present calibration gives me only 40% of the amplitude  recorded at PAL.
 
  I hope someone out there can find a flaw in the mathematics. I am  not 
happy with the apparent disagreement between theory and experiment.
 
Cheers,
 
Bob
 
 





Hi Chris,
 
  For a damped pendulum, we are dealing with the equation
 
  F =3D m(d2x/dt2) + c(dx/dt) + kx, where F is some forcing=20
function.
 
 m is mass, k is stiffness, and c is drag. Feynman defines=20
k=3Dm*Omega0^2, c=3Dm*Gamma, and Q=3DOmega0/Gamma, where Omega0 is the natur=
al=20
frequency.
 
 In order to calculate the effect of a shunt load on the=
=20
pickup coil for damping, recall that the sensor output sensitivity is given=20=
by S=20
=3D B*L, where S is in units of volts per meter per second, B is field stren=
gth in=20
Teslas, and L is active wire length in meters.
 
  If we pass current through the coil, the resultant motor force i=
s f=20
=3D S*I, where f is in newtons, and I is amperes.
 
  If we load the moving coil with the resistance of the coil in se=
ries=20
with an external resistance, the generated voltage is E =3D S * v, and the c=
urrent=20
is I =3D E/R, so the resulting drag is c*v =3D S*I =3D S*E/R =3D S*(S*v)/R.<=
/DIV>
 
  So, c =3D S*S/R, where c is drag in newtons per meter per second=
.. From=20
the Feynman definitions, c =3D m * Gamma =3D m * Omega0 / Q. Since c from sh=
unt=20
damping is S^2 / R, we have the equality S *S / R =3D m * Omega0 / Q.
 
  Solving for Q, we obtain:
 
Q =3D R * m * Omega0 / S^2
 
  For a Q of unity, 
 
R =3D (Ps * S^2) / (2 * Pi * m), where Ps is sensor natural period in=20
seconds.
 
  When I plug in my assumed values of Ps =3D 12 seconds, S =3D 85=20=
v-s/m,=20
and m =3D 0.1 kg, the value for R is 138 kOhms.
 
  In reality, the value I need to get a Q of 1.0 is less=
=20
than half the above. I have no clue as to why. I calibrated the sensor outpu=
t=20
several ways, all yielding the same result. The mass was determined by weigh=
ing=20
the solder wire mass and the coil assembly. To get better agreement, I would=
=20
have to considerably decrease my value for S, which incidentally would also=20
bring my calibrated seismic amplitudes into closer agreement with those reco=
rded=20
by station PAL. My present calibration gives me only 40% of the amplitude=20
recorded at PAL.
 
  I hope someone out there can find a flaw in the mathematics. I a=
m=20
not happy with the apparent disagreement between theory and experiment.
 
Cheers,
 
Bob