PSN-L Email List Message
Subject: Re: Geophone Damping Mass Slew rate
From: Pete Rowe ptrowe@.........
Date: Thu, 16 Dec 2010 15:29:38 -0800 (PST)
Hi Ed
If you are very careful , you can push the pendulum off center a small amou=
nt and watch the results of lower and lower shunt resistance. This is how I=
originally set mine up. Don't be afraid to experiment with this. It is pre=
tty cool to watch the damping while you adjust the pot.
Thanks ot Bob McClure=A0 you can calculate the damping resistor=A0 and not =
need to goof around. (I guess I'm just a goof around kind of guy)
Let us know how you do.
Pete
--- On Thu, 12/16/10, Ed Ianni wrote:
From: Ed Ianni
Subject: Re: Geophone Damping Mass Slew rate
To: psnlist@..............
Date: Thursday, December 16, 2010, 2:08 PM
=0A=0A =0A=0A =0A =0A Hi Pete;
=0A =20
=0A Concerning your damping advice, I have my "screen" in a different=0A=
room from my seismograph, therefore I can't see it as I perform your=0A=
procedure. Would there be a VISIBLE dampening of the=0A oscillations=
as the coil travels=A0 through the magnetic field at=0A the seismogra=
ph.=A0=A0 Thanks, Ed.=20
=0A =20
=0A =20
=0A On 12/16/2010 8:59 PM, Pete Rowe wrote:=0A =0A =0A =
=0A =0A =20
=0A Hi Ted
=0A With no damping, if you slightly push the arm off center=
=0A and let go of it you will see on the Winsdr trace on the=
=0A screen oscillate back and forth across zero for quite some=
=0A time. With a 1 k pot across the coil, adjust the pot=A0 su=
ch=0A that the oscillation goes across center and back to the=
=0A center with a tiny amount of overshoot. I can send you a=
=0A picture of this if it doesn't make sense.
=0A The pot value is correct for just slightly underdamped.=0A=
You can measure the pot and replace it with a fixed=0A =
resistor.
=0A =20
=0A I will get back to you tomorrow with answers to the=0A =
questions in your e mail about parts=A0 availability.
=0A =20
=0A Pete
=0A --- On Thu, 12/16/10, Ted Channel =
=0A wrote:
=0A =20
=0A From: Ted Channel
=0A Subject: Re: Geophone Damping Mass Slew rate
=0A To: psnlist@..............
=0A Date: Thursday, December 16, 2010, 12:46 PM
=0A =20
=0A =0A Thanks again=0A =
Pete...............If I have a coil which measures=0A =
1.2K ohms,=A0 and another at 2.4K=0A ohms...........=
...what approx. value should I=0A try?=A0=A0=A0=A0 Shou=
ld I use a Pot and adjust until I get=0A the effect I =
need and then measure the resistor=0A value?=0A =
=A0=0A Ted=0A =0A =
----- Original=0A Message ----- =0A =
From:=0A Pete Rowe=0A =0A =
To: psnlist@................. =0A =
Sent:=0A Thursday, December 16, 2010 1:=
18 PM=0A Subject: Re:=0A Geophone D=
amping Mass Slew rate=0A =20
=0A =0A =0A =0A =
=0A =20
=0A =20
=0A --- On Thu, 12/16/10, Ted Channel =0A wrote:
=0A =20
=0A From: Ted Channel
=0A Subject: Re: Geophone Damping Mass Slew=0A=
rate
=0A To: psnlist@..............
=0A Date: Thursday, December 16, 2010, 12:11=
=0A PM
=0A =20
=0A =0A =0A =
This is=0A k=
ind of the same=0A subject..............=
.....I=0A understand you can dampen a coi=
l=0A with a resistor,=A0 =0A =
=A0=0A Does=0A =
anyone use this method in place of a=0A =
Lehman style magnet plate? I=0A =
have for 20 years and
=0A =0A Do =
you=0A give up anything (voltage) by usi=
ng=0A a resistor on the coil?My=0A =
coil measures about 10 k ohms and=0A =
I use about 1 k for critical=0A =
damping. I don't lose any=0A =
sensitivity
=0A =0A If =
it=0A works, why do we still use magnet=
=0A plates to damp? It works so=0A =
well, I don't know why it isn't=0A =
more popular.
=0A Pete Rowe
=0A =0A =A0=
=0A Thanks,=0A =
Ted=0A =0A =
-----=0A Original Message ----- =
=0A From: Christopher=0A =
Chapman =0A To:=0A =
psnlist@................. =
=0A Sent:=0A =
Thursday, December 16, 2010 10:00 AM=0A =
Subject:=0A Re: Geop=
hone Damping Mass Slew rate=0A =20
=0A =0A =
=0A =0A =
=0A =0A =
=0A =
=0A =0A =
=0A =
=0A =
I have=0A =
been playing=0A =
around with a=0A =
dismantled=0A =
SPZ geophone=0A =
damping and find=A0if=0A =
you short the=0A =
leads to the=0A =
coil and=0A =
then drop the=0A =
mass=A0
=0A the mass will=
=0A descend=0A =
slowly, I=0A =
guess from=0A =
meeting=A0the=0A =
generator=0A =
current being=0A =
dissipated=0A =
by the=A0internal=0A =
resistance.=A0
=0A =A0
=0A Questions=0A =
arise from=0A =
this.=A0
=0A =A0
=0A 1. if there=0A=
were no=0A =
internal=0A =
resistance=0A =
(superconductor)=A0=A0=0A =
then when the=0A =
leads are=0A =
shorted the=0A =
mass would=0A =
fall=A0
=0A =A0 a bit then=
=0A stay forever=
=0A hovering ?=A0
=0A =A0
=0A =A0=A0=A0=0A =
Correct=0A =
=A0=0A =
2.the=0A =
geophone is=0A =
acting just=0A =
like a=0A =
rotating=0A =
electrical generator=A0
=0A armature devel=
oping=0A torque un=
der=0A electrical=
=0A loads ?=A0
=0A =A0
=0A =A0=A0=A0=0A =
Correct=0A =
=A0=0A =
3. It=0A =
seems to me=0A =
that seismic=0A =
noise rarely=0A =
hits the resonant=0A =
frequency and=A0
=0A =A0 you=0A =
might do=0A =
better to not=0A =
increase the=0A =
damping over=0A =
what already=A0
=0A =A0 is=0A =
there in the=0A =
mechanical and=0A =
physical sense=0A =
?=0A =
=A0=0A =
=A0=A0=A0=0A =
No. The=0A =
critical=0A =
damping allows=0A =
you to get=A0an=0A =
output voltage=0A =
flat with=0A =
frequency=0A =
above the=0A reson=
ant=0A frequency.=
=A0
=0A =A0
=0A 4. Does the=0A=
rate at which=0A =
the geophone=0A =
mass drop under=0A =
heavy damping=A0rep=
resent=0A some new=
=0A fundamental=0A=
Eigen frequency?=
=0A =20
=0A =A0=A0=A0=A0No=
..=0A The resonance=
=0A is determined=
=0A by the mass of=
=0A the armature=
=0A and the spring=
=0A constant. =A0=
=0A =A0=0A =
5. Can=0A =
anyone=0A =
provide me=0A =
with high=0A =
school math=A0models=0A =
which=0A =
represent the=0A =
mechanical=0A =
and=A0electrical=0A =
behaviors of=0A =
the geophone?=A0
=0A =A0 High schoo=
l=0A math being=0A=
trig and=0A =
algebra=A0minus=0A =
the calculus?=0A =
=A0=0A =
=A0=A0=A0=0A =
The theory is=0A =
freely=0A =
available on=0A =
the Internet.=0A =
=A0=0A =
an=0A =
Eigen freq not=0A =
contained in EQ=0A =
signals then=0A =
do no=0A damping=
=0A at all?=A0
=0A This=0A =
[little=0A =
damping]=0A =
should work=0A =
for weak EQ signals=0A =
and=0A =
not close=0A =
strong ones ?=A0=0A =
=0A =
=A0=0A =
=A0=A0=A0=A0=A0=0A =
An undamped=0A g=
eophone has a=0A s=
ingle=0A frequency=
=0A peaked=0A =
response -=0A =
definitely NOT=0A =
what you=0A =
want!=A0 =0A =
=20
=0A =A0=A0=A0=A0=
=0A Regards,=0A =
=0A =
=A0=0A =
=A0=A0=A0=A0=0A =
Chris Chapman=0A =
=0A =
=0A =0A =
=0A =
=0A =0A =
=0A =0A =
=0A =0A =
=0A =0A =0A =
=0A =0A =20
=0A =0A =0A =0A =0A=
=0A =0A =0A =20
=0A =0A =20
Hi Ed
If you are very careful , you can push the =
pendulum off center a small amount and watch the results of lower and lower=
shunt resistance. This is how I originally set mine up. Don't be afraid to=
experiment with this. It is pretty cool to watch the damping while you adj=
ust the pot.
Thanks ot Bob McClure you can calculate the dampi=
ng resistor and not need to goof around. (I guess I'm just a goof aro=
und kind of guy)
Let us know how you do.
Pete
--- On Thu, 1=
2/16/10, Ed Ianni
<edwianni1@...........> wrote:
=0A=
=0A =0A=0A =0A =0A Hi Pete;
=0A
=0A Con=
cerning your damping advice, I have my "screen" in a different=0A room f=
rom my seismograph, therefore I can't see it as I perform your=0A proced=
ure. Would there be a VISIBLE dampening of the=0A oscillations as=
the coil travels through the magnetic field at=0A the seismo=
graph. Thanks, Ed.
=0A
=0A
=0A On 12/1=
6/2010 8:59 PM, Pete Rowe wrote:=0A =0A <=
table border=3D"0" cellpadding=3D"0" cellspacing=3D"0">=0A =
=0A =0A
=0A Hi Ted
=0A With no damping, if you sl=
ightly push the arm off center=0A and let go of it you will se=
e on the Winsdr trace on the=0A screen oscillate back and fort=
h across zero for quite some=0A time. With a 1 k pot across th=
e coil, adjust the pot such=0A that the oscillation goes=
across center and back to the=0A center with a tiny amount of=
overshoot. I can send you a=0A picture of this if it doesn't =
make sense.
=0A The pot value is correct for just slightly =
underdamped.=0A You can measure the pot and replace it with a =
fixed=0A resistor.
=0A
=0A I =
will get back to you tomorrow with answers to the=0A questions=
in your e mail about parts availability.
=0A
=0A=
Pete
=0A --- On Thu, 12/16/10, Ted Channel=
<tchannel@............>=0A wrote:
=
=0A
=0A From: Ted =
Channel <tchannel@............>
=0A Subject: =
Re: Geophone Damping Mass Slew rate
=0A To: psnlist@=
webtronics.com
=0A Date: Thursday, December 16, 2010,=
12:46 PM
=0A
=0A =0A Thanks again=0A Pete...............If I =
have a coil which measures=0A 1.2K ohms, and ano=
ther at 2.4K=0A ohms.............what approx. value sh=
ould I=0A try? Should I use a =
Pot and adjust until I get=0A the effect I need and th=
en measure the resistor=0A value? =0A =
=0A Ted =0A =0A ----- Original=0A Message ----- =
=0A =0A =0A Sent:=0A Thursday, December=
16, 2010 1:18 PM =0A Subject: Re:=0A Geophone Damping Mass Slew =
rate =0A
=0A
=0A =
=
=0A =0A =0A =
=0A
=
=0A --- On Thu, 12/16/10, Ted Channel <=
tchannel@............>=0A =
wrote:
=0A
=0A From: Ted Channel <tchannel@............>
=0A S=
ubject: Re: Geophone Damping Mass Slew=0A rate=
=0A To: psnlist@webtro=
nics.com
=0A Date: Thursday, December 1=
6, 2010, 12:11=0A PM
=0A =
=0A =0A =0A This is=0A kind of the same=0A =
subject..................I=0A =
understand you can dampen a coil=0A =
with a resistor, =0A =
=0A Does=0A any=
one use this method in place of a=0A Leh=
man style magnet plate? I=0A =
have for 20 years and
=0A =
=0A <=
div>Do you=0A =
give up anything (voltage) by using=0A =
a resistor on the coil?My=0A =
coil measures about 10 k ohms and=0A =
I use about 1 k for critical=0A =
damping. I don't lose any=0A =
sensitivity
=0A =
If it=0A works, why do we=
still use magnet=0A plates to damp? It works so=0A =
well, I don't know why it isn't=0A =
more popular.
=0A =
Pete Rowe
=0A =0A =
=0A =
Thanks,=0A =
Ted =0A =0A =
-----=0A =
Original Message ----- =0A =
=0A =0A Sent:=0A Thurs=
day, December 16, 2010 10:00 AM =0A <=
div style=3D"font: 10pt arial;">Subject:=0A =
Re: Geophone Damping Mass Slew rate=0A =
=0A
=0A =
=0A =0A =
=0A =0A =0A =
=0A =
=0A =
=0A =
=0A I have=0A =
been playing=0A =
around with a=0A =
dism=
antled=0A S=
PZ geophone=0A =
damping and find if=0A =
you short the=0A =
leads to the=0A =
coil a=
nd=0A then =
drop the=0A mass&n=
bsp;
=0A the ma=
ss will=0A descend=
=0A slowly, I=0A =
guess from=0A =
meeting the=0A =
generator=0A =
current being=0A =
dissipated=
=0A by the internal=0A =
resistance.
=0A =
=0A =
Questions=0A =
arise from=0A =
this.
=0A =
=0A =
1. if there=0A =
were no=0A =
internal=0A re=
sistance=0A (super=
conductor) =0A =
then when the=0A =
leads are=0A =
shorted the=0A m=
ass would=0A fall&=
nbsp;
=0A =
; a bit then=0A st=
ay forever=0A hove=
ring ?
=0A =
=0A =0A =
Correct =0A =
=0A =
2.the=0A =
geophone is=0A =
acting just=0A =
like a=0A =
rotating=0A electr=
ical generator
=0A =
armature developing=0A torque under=
=0A electrical=0A =
loads ?
=
=0A
=0A =
&=
nbsp; =0A =
Correct =0A =
=0A =
3. It=0A se=
ems to me=0A that =
seismic=0A noise r=
arely=0A hits the =
resonant=0A =
frequency and
=0A =
you=0A =
might do=0A =
better to not=0A =
increase the=0A =
damping over=0A =
what already
=0A =
is=0A there in the=
=0A mechanical and=
=0A physical sense=
=0A ? =0A =
=0A =
=
=0A No=
.. The=0A critical=
=0A damping allows=
=0A you to get&nbs=
p;an=0A output vol=
tage=0A flat with=
=0A frequency=0A =
above the=0A =
resonant=0A =
frequency.
=0A=
=0A =
4. Does the=0A =
rate at which=0A =
the geophone=0A =
mass drop =
under=0A he=
avy damping represent=0A =
some new=0A =
fundamental=0A =
Eigen frequency? =
=0A
=0A =
&=
nbsp; No.=0A =
The resonance=0A =
is determined=0A =
by the mass of=0A =
the armature=0A =
and the spring=0A =
constant.
=0A =
=0A =
5. Can=0A =
anyone=0A =
provide me=0A =
with high=0A =
school math models=0A =
which=0A =
represent the=
=0A mechanical=0A =
and electrical=0A =
behaviors of=0A =
the geophone?
=0A =
High school=0A =
math being=0A =
trig and=0A =
algebra minus=0A =
the calculus? =0A =
=0A =0A =
The theory is=0A =
freely=0A =
available on=0A =
the Internet. =0A =
=0A =
an=0A =
Eigen freq not=0A =
contained in EQ=0A =
signals then=0A =
do no=0A damping=0A =
at all?
=0A =
This=0A =
[little=0A =
damping]=0A =
should work=0A =
for weak EQ signals=0A=
=
and=0A not =
close=0A strong on=
es ? =0A =
=
; =0A =0A =
An undamped=0A =
geophone has a=0A =
single=0A =
frequency=0A =
peaked=0A =
response -=0A =
definitely NOT=0A =
what you=0A wa=
nt! =0A =
=0A <=
font size=3D"2"> =0A =
Regards,
=0A =
=0A =
=0A =
=0A =
Chris Chapman=0A =0A =0A =
=0A =
=0A =0A =0A =0A
| =0A =
=0A =0A =0A =
=0A =0A <=
/div>=0A =0A
| =0A =
=0A =0A |
=0A =0A =0A =
=0A =0A |
<=
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