## PSN-L Email List Message

Subject: Re: Integrating in WinQuake
From: Barry Lotz barry_lotz@.............
Date: Mon, 23 Nov 2009 19:58:26 -0800 (PST)

```All
This seems to be a good one. I think we are having issues with semantics. Y=
ou have the movement of the sensor and movement of the ground. One would li=
ke the movement of the sensor ( which we can measure) to relate to movement=
of the earth. One can only measure displacement (distortion of a material =
- capacitive,LVDT, etc) or velocity(magnet/coil)=A0 of the sensor relative =
to it's base (excluding GPS). Assuming the proper damping ,=A0 for short pe=
riod sensors, the displacement of the sensor is linear with acceleration of=
the earth. With long period sensors, the movement of the sensor relates to=
the movement of the earth. If one is measuring displacement of the sensor =
with say a Lehman, the output would be related to the displacement of the e=
arth. The velocity of the sensor (magnet/coil) would relate to the velocity=
of the earth. With feedback systems I'm not so sure. With STM's feedback h=
e claimed that the displacement measured of the sensor related to the
velocity of the earth.
Regards
Barry

--- On Mon, 11/23/09, ChrisAtUpw@.......  wrote:

From: ChrisAtUpw@.......
Subject: Re: Integrating in WinQuake
To: psn-l@..............
Date: Monday, November 23, 2009, 6:09 PM

=0A=0A =0A=0AIn a message dated 23/11/2009, rsparks@.......... writes:=0ATh=
ere =0A  seems to be an ongoing difference of opinion of what constitutes a=
=0A =20
velocity detector or an acceleration detector.=A0 To my mind, if a =0A =20
seismometer automatically returns to a zero position, then it must be =0A =
=20
recording acceleration.=A0 All vertical seismometers do this because =0A  t=
hey=20
all are measuring against gravity, so they would all be acceleration =0A =
=20
devices.=A0 Excepting the sensitivity to tilt, all horizontal =0A  seismome=
ters=20
also return to a zero position so they could also be called =0A  accelerati=
on=20
sensitive devices.=A0 But maybe this description =0A  oversimplifies the si=
tuation.=0AHi Roger,=0A=A0=0A=A0=A0=A0=A0Sorry, but no. You seem to be=A0co=
nfusing =0Adevices which have AC and DC characteristics.=0ASome =0A  sensor=
s (capacitive and optical for example) clearly record=20
displacement =0A  (neither acceleration or velocity).=A0 =0A=A0=A0=A0=A0Agr=
eed.=0AOn the =0A  other hand, with magnetic/coil devices, velocity is alwa=
ys observed when =0A  electrical=20
output is observed so magnetic/coil devices are velocity =0A  detectors.=0A=
=A0=A0=A0=A0No. They may be either velocity or acceleration =0Adetectors. I=
f the coil is attached to a ~freely suspended, but damped, mass, you =0Aget=
a velocity detector - eg a Lehman. If it is attached to a mass suspended o=
n =0Aa spring, you get an acceleration detector. The length of the spring i=
s constant =0Aif the velocity is constant. It only changes in length and ca=
uses the coil to =0Amove if there is an acceleration.=0AA =0A  displacement=
sensor would record the relative distance from a some zero =0A =20
point at the instant of data read, the velocity sensor would record the =0A=
=20
relative velocity at the instant of data read,=A0 and acceleration =0A  sen=
sor would =0A=A0=A0=A0=A0detect the acceleration! eg a MEMs =0Aacclerometer=
..=0Abe a =0A  calculated number found by using the data from any two veloci=
ty=20
data =0A  points and any three displacement data points.=0A=A0=A0=A0=A0Agre=
ed.=0AFinally, =0A  if two seismometers, identical except for detectors, we=
re=20
placed side by =0A  side, they would both plot the identical earthquake wav=
e=20
form, assuming =0A  that the frequency characteristics were the same.=A0=0A=
=A0=A0=A0=A0No, they wouldn't. A velocity output is =0Aproportional to the =
differential (slope) of a position output plot with =0Atime.=0A=A0However, =
when it is recognized that displacement position is =0A  not time=20
sensitive but velocity is, the builder can expect dramatic =0A  frequency=
=20
response differences between displacement detectors and velocity =0A =20
detectors .=A0 Simply put, distance is distance, but velocity is the =0A =
=20
distance divided by the time needed to travel between two =0A  points.=A0=
=A0 As a=20
result, for velocity detectors, the longer the =0A  wave length, the less=
=20
energy for each instant resulting in decreased =0A  voltage (and current)=
=20
detected at each instant (for any defined magnetic =0A  field).=0A=A0=A0=A0=
=A0Regards,=0A=A0=0A=A0=A0=A0=A0Chris Chapman
AllThis seems to be a good one. I think w=
e are having issues with semantics. You have the movement of the sensor and movement of the ground. One would like the move=
ment of the sensor ( which we can measure) to relate to movement of the ear=
th. One can only measure displacement (distortion of a material - capacitiv=
e,LVDT, etc) or velocity(magnet/coil)  of the sensor relative to it's =
base (excluding GPS). Assuming the proper damping ,  for short period =
sensors, the displacement of the earth. With long period sensors, the movement of=
the sensor relates to the movement of the earth. If one is measuring displ=
acement of
the sensor with say a Lehman, the output would be related to the displacem=
ent of the earth. The velocity of the sensor (magnet/coil) would relate to =
the velocity of the earth. With feedback systems I'm not so sure. With STM'=
s feedback he claimed that the displacement measured of the sensor related to the velocity of the <=
span style=3D"text-decoration: underline;">earth.RegardsBarr=
y--- On Mon, 11/23/09, ChrisAtUpw@....... <ChrisAtUpw@=
aol.com> wrote:From: ChrisAtUpw=
@....... <ChrisAtUpw@.......>Subject: Re: Integrating in WinQuake=
To: psn-l@..............Date: Monday, November 23, 2009, 6:09 PM=0A=0A =0A=0AIn a message dated 23/11/2009, =
rsparks@.......... writes:=0AThere =0A  =
seems to be an ongoing difference of opinion of what constitutes a =0A  velocity detector or an acceleration detector.  To my mind, if a =0A =
seismometer automatically returns to a zero position, then it must be =
=0A  recording acceleration.  All vertical seismometers do this be=
cause =0A  they all are measuring against gravity, so they would all be=
acceleration =0A  devices.  Excepting the sensitivity to tilt, al=
l horizontal =0A  seismometers also return to a zero position so they c=
ould also be called =0A  acceleration sensitive devices.  But mayb=
e this description =0A  oversimplifies the situation.=
=0AHi Roger,=0A =0A    =
Sorry, but no. You seem to be confusing =0Adevices which have AC and D=
C characteristics.=0ASome =0A  sensors (=
capacitive and optical for example) clearly record displacement =0A  (n=
either acceleration or velocity).  =0A &=
nbsp;  Agreed.=0AOn the =0A  o=
ther hand, with magnetic/coil devices, velocity is always observed when =0A=
electrical output is observed so magnetic/coil devices are velocity =
=0A  detectors.=0A    No. They=
may be either velocity or acceleration =0Adetectors. If the coil is attach=
ed to a ~freely suspended, but damped, mass, you =0Aget a velocity detector=
- eg a Lehman. If it is attached to a mass suspended on =0Aa spring, you g=
et an acceleration detector. The length of the spring is constant =0Aif the=
velocity is constant. It only changes in length and causes the coil to =0A=
move if there is an acceleration.=0AA =
=0A  displacement sensor would record the relative distance from a some zer=
o =0A  point at the instant of data read, the velocity sensor would rec=
ord the =0A  relative velocity at the instant of data read,  and a=
cceleration =0A  sensor would =0A   =
; detect the acceleration! eg a MEMs =0Aacclerometer.=0Abe a =0A  calculated number found by using the data fr=
om any two velocity data =0A  points and any three displacement data po=
ints.=0A    Agreed.=0AFinally, =0A  if two seismometers, identical ex=
cept for detectors, were placed side by =0A  side, they would both plot=
the identical earthquake wave form, assuming =0A  that the frequency c=
haracteristics were the same. =0A  =
No, they wouldn't. A velocity output is =0Aproportional to the =
differential (slope) of a position output plot with =0Atime.=0A However, when it is recognized that displaceme=
nt position is =0A  not time sensitive but velocity is, the builder can=
expect dramatic =0A  frequency response differences between displaceme=
nt detectors and velocity =0A  detectors .  Simply put, distance i=
s distance, but velocity is the =0A  distance divided by the time neede=
d to travel between two =0A  points.   As a result, for veloc=
ity detectors, the longer the =0A  wave length, the less energy for eac=
h instant resulting in decreased =0A  voltage (and current) detected at=
each instant (for any defined magnetic =0A  field).=0A=
Regards,=0A =0A&nbs=
p;   Chris Chapman=0A<=

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