In a message dated 2008/07/07, charles.r.patton@........ writes:

> Thanks to Chris for tossing the wrench in the works -- :-) -- I have now=20
> spent the day trying to resurrect trig math I haven=E2=80=99t done serious=
ly for=20
> 35 years.  I think I have the formula for Chris=E2=80=99s bottom pivot, bu=
t I=E2=80=99m=20
> still psyching out the upper pivot, so I haven=E2=80=99t started sims yet.=
  What=20
> I think at this point is that as the gate swings the lower point=20
> trajectory tightens up (radius decreases) with swing while the upper=20
> pivot flattens out (radius increases) which I would assume leads to=20
> unstability =E2=80=93 the bob going lower as it swings.  But this seems to=
 fly=20
> in the face of Chris=E2=80=99s success in long period Lehman.=20

Hi Charles,

       I am not entirely sure how you are analysing it, but you seem to be=20
having problems.

       Consider a sphere of radius R with the flat end of the arm of length=20=
L=20
resting up against it. If the arm swings through a small angle d, the=20
distance between the centre of mass and the axis of rotation increases by ~=20=
L x=20
(Rxd/L)^2 / 2. The separation must increase, since the contact point on the=20=
face=20
has rolled sideways by Rxd from it's minimum central position. Whether the m=
ass=20
rises or not depends on the behavoir of the top hinge. For a wire top=20
suspension the centre of curvature should be ~constant and the mass should t=
end rise=20
slightly, but the wire clamp must have well defined edges. If it does not, t=
he=20
top centre of rotation will move sightly away from the vertical, increasing=20
the stability, but decreasing the response linearity.
=20
> Now I=E2=80=99ll take a big leap of faith and put forth the formulas I thi=
nk=20
> describe the lower pivot.  Assume a cylinder of radius R with a beam of=20
> length K having a flat face resting on the cylinder.  Angle T is the=20
> angle of the contact point of cylinder/face (beam angle).  The angle of=20
> a line thru the center C of the cylinder and plumb bob P is equal to=20
> angle T minus angle B.  Line G =3D C to P.  All angles are in radians.
> Then:
> B =3D T(R/K)
> G =3D (R+K)/cosB
>         x & y are referenced from C
> x =3D  G(sin(T-B))
> y =3D  G(cos(T-B))
>=20
> If there=E2=80=99s interest, I can try to do a standard proof deriving the=
=20
> above.  I didn=E2=80=99t get to the formulas above with a step-by-step wri=
tten=20
> proof, so it very well could be flawed.

       Out of interest, what types of system were giving trouble in ''so man=
y=20
anecdotal stories about the difficulty of adjusting Lehmans in the long=20
period realm''?? You seem to be implying that there must be some inherent pr=
oblem,=20
when something inadequate in the construction seems the more likely=20
explanation.=20
       I would expect to get mechanical problems with some of the amateur=20
designs previously described.

       Regards,

       Chris Chapman  =20
In a me=
ssage dated 2008/07/07, charles.r.patton@........ writes:



Thanks to Chris for tossing the=
 wrench in the works -- :-) -- I have now 

spent the day trying to resurrect trig math I haven=E2=80=99t done seriously=
 for 

35 years.  I think I have the formula for Chris=E2=80=99s bottom pivot,=
 but I=E2=80=99m 

still psyching out the upper pivot, so I haven=E2=80=99t started sims yet.&n=
bsp; What 

I think at this point is that as the gate swings the lower point 

trajectory tightens up (radius decreases) with swing while the upper 

pivot flattens out (radius increases) which I would assume leads to 

unstability =E2=80=93 the bob going lower as it swings.  But this seems=
 to fly 

in the face of Chris=E2=80=99s success in long period Lehman. 



Hi Charles,



       I am not entirely sure how you are anal=
ysing it, but you seem to be having problems.



       Consider a sphere of radius R with the=20=
flat end of the arm of length L resting up against it. If the arm swings thr=
ough a small angle d, the distance between the centre of mass and the axis o=
f rotation increases by ~ L x (Rxd/L)^2 / 2. The separation must increase, s=
ince the contact point on the face has rolled sideways by Rxd from it's mini=
mum central position. Whether the mass rises or not depends on the behavoir=20=
of the top hinge. For a wire top suspension the centre of curvature should b=
e ~constant and the mass should tend rise slightly, but the wire clamp must=20=
have well defined edges. If it does not, the top centre of rotation will mov=
e sightly away from the vertical, increasing the stability, but decreasing t=
he response linearity.

 

Now I=E2=80=99ll take a big lea=
p of faith and put forth the formulas I think 

describe the lower pivot.  Assume a cylinder of radius R with a beam of=
 

length K having a flat face resting on the cylinder.  Angle T is the 
angle of the contact point of cylinder/face (beam angle).  The angle of=
 

a line thru the center C of the cylinder and plumb bob P is equal to 

angle T minus angle B.  Line G =3D C to P.  All angles are in radi=
ans.

Then:

B =3D T(R/K)

G =3D (R+K)/cosB

        x & y are referenced from C
x =3D  G(sin(T-B))

y =3D  G(cos(T-B))



If there=E2=80=99s interest, I can try to do a standard proof deriving the <=
BR>
above.  I didn=E2=80=99t get to the formulas above with a step-by-step=20=
written 

proof, so it very well could be flawed.




       Out of interest, what types of system w=
ere giving trouble in ''so many anecdotal stories about the difficulty of ad=
justing Lehmans in the long period realm''?? You seem to be implying that th=
ere must be some inherent problem, when something inadequate in the construc=
tion seems the more likely explanation. 

       I would expect to get mechanical proble=
ms with some of the amateur designs previously described.



       Regards,



       Chris Chapman