In a message dated 06/12/2009, tchannel@............ writes:

Hi Folks,   This is the basic process I  use to adj. a Lehman.
1.  I assembly everything and level  everything.    Once this is completed 
the vertical part of the  frame is plumb.   The boom, and frame are parallel 
with  the earth and the boom is resting on the centreline.
2.  I time the period and start raising the  front adjusting leg, tweaking 
the side adj. legs as necessary to maintain the  boom on the centre line.  
When I get a stable 20 second period I am  finished.
 
These are my questions:     When I perform this set up, the vertical frame 
member is now tilting backward,  and the mass end of the boom has be raised. 
 Now the boom is no  longer parallel to the earth.
1.  "What would happen" if I extended the  wire support, lowering the boom 
back to parallel to the earth, but not  changing the vertical member.   To 
my knowledge I have never done  this, I just leave the boom with mass end 
raised via the front adj.  leg.
2.  Should I make this adjustment, returning  the boom to parallel, by 
lengthen the support wire?
I have tried it, but the adjustments are so small  I can't measure the 
results.



Hi Ted,
 
    The period is determined by 1) the angle 'A'  between the vertical and 
the true swing axis and by 2) the radius of  gyration 'k' of the pendulum. k 
depends on the weight and shape of the mass and  of the arm. It is usual to 
use a rigid light tubular arm. Lifting or  lowering the arm a bit will 
provide a Cosine Theta correction to k, so  you could see a few parts per 
thousand change. 
 
    Usual values of the swing angle are about 1/3  degree for a 20 second 
period and a 24" arm. T = 2xPixSqrt(k / gxsin(A)) Avoid  very short arms, say 
12". The value of A becomes very small and the pendulum  will experience 
tilt drift due to very small ground movements as the seasons,  rainfall and 
temperatures change. You need to design your sensor system so  that it's 
output is independent of any tilt drift.
    The layout of the sensor and the magnetic damping  with the frame 
should be your primary concern.
 
    Regards,
 
    Chris Chapman






In a message dated 06/12/2009, tchannel@............ writes:

  
Hi Folks,   This is the basic=
 process I=20
  use to adj. a Lehman.
  
1.  I assembly everything and leve=
l=20
  everything.    Once this is completed the vertical part=
 of the=20
  frame is plumb.   The boom, and frame are parallel=
 with=20
  the earth and the boom is resting on the centreline.
  
2.  I time the period and start ra=
ising the=20
  front adjusting leg, tweaking the side adj. legs as necessary to maintai=
n the=20
  boom on the centre line.  When I get a stable 20 second period I am=
=20
  finished.
  
 
  
These are my questions:  =
; =20
  When I perform this set up, the vertical frame member is now tilting bac=
kward,=20
  and the mass end of the boom has be raised.  Now the boom is=
 no=20
  longer parallel to the earth.
  
1.  "What would happen" if I exten=
ded the=20
  wire support, lowering the boom back to parallel to the earth, but not=
=20
  changing the vertical member.   To my knowledge I have never=
 done=20
  this, I just leave the boom with mass end raised via the front adj.=20
  leg.
  
2.  Should I make this adjustment,=
 returning=20
  the boom to parallel, by lengthen the support wire?
  
I have tried it, but the adjustments ar=
e so small=20
  I can't measure the results.

Hi Ted,
 
    The period is determined by 1) the angle 'A'=
=20
between the vertical and the true swing axis and by 2) the radius of=
=20
gyration 'k' of the pendulum. k depends on the weight and shape of the mas=
s and=20
of the arm. It is usual to use a rigid light tubular arm. Lifting or=
=20
lowering the arm a bit will provide a Cosine Theta correction to k, so=20
you could see a few parts per thousand change. 
 
    Usual values of the swing angle are about 1/3=
=20
degree for a 20 second period and a 24" arm. T =3D 2xPixSqrt(k / gxsin(A))=
 Avoid=20
very short arms, say 12". The value of A becomes very small and the pendul=
um=20
will experience tilt drift due to very small ground movements as the seaso=
ns,=20
rainfall and temperatures change. You need to design your sensor syst=
em so=20
that it's output is independent of any tilt drift.
    The layout of the sensor and the magnetic dam=
ping=20
with the frame should be your primary concern.
 
    Regards,
 
    Chris Chapman