Charles,

I did some work on this question a couple of years back and got the same
answers Chris is proposing that as the angle increases the restoring force
increases.  Consider the pivot point on a rolling contact will move
laterally on the boom end.  The horizontal pressure of the mass support
geometry will now fall on an extended line making a right triangle with the
boom axis and which is longer than the axis and thus the axis is the stable
position with no moment about the cg.  Use the mass as the origin of your
support force system and sum moments and forces since a stable mass is the
premise of the operation.  My experience is that support flexure or lose
threads in supporting legs are the barriers to settup stability and the
floor of the basement is the controlling factor over days or weeks.  I am
also finding that with the stronger magnets I have obtained that I am
fighting magnetic effects not seen before where even my damping plate may be
causing instability.

Randy
----- Original Message ----- 
From: 
To: 
Sent: Tuesday, July 08, 2008 02:00
Subject: Digest from 07/07/2008 00:01:28


>
> .------ ------ ------ ------ ------ ------ ------ ------ ------ ------.
> | Message 1                                                           |
> '------ ------ ------ ------ ------ ------ ------ ------ ------ ------'
> Subject: Re: Is a Lehman geometry rolling pivot inherently unstable?
> From:    Charles Patton 
> Date:    Mon, 07 Jul 2008 00:02:39 -0700
>
> Brett, Chris,
> Thanks to Chris for tossing the wrench in the works -- :-) -- I have now
> spent the day trying to resurrect trig math I haven't done seriously for
> 35 years.  I think I have the formula for Chris's bottom pivot, but I'm
> still psyching out the upper pivot, so I haven't started sims yet.  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 - the bob going lower as it swings.  But this seems to fly
> in the face of Chris's success in long period Lehman. The answer may lie
> in the combination in that the gate is twisting as it swings so the
> vertical position of the bob would play an important part of the
> stability.  There has to be some point that the beam twists about, and
> if the bob is mounted above or below this point the twist could
> compensate or increase the trajectory error of the bob.
>
> The morning is spoken for, so I'll try to get back on the problem in the
> afternoon.
>
> Now I'll take a big leap 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 = C to P.  All angles are in radians.
> Then:
> B = T(R/K)
> G = (R+K)/cosB
> x & y are referenced from C
> x =  G(sin(T-B))
> y =  G(cos(T-B))
>
> If there's interest, I can try to do a standard proof deriving the
> above.  I didn't get to the formulas above with a step-by-step written
> proof, so it very well could be flawed.  Good for discussion though.
>
> Anyway, later.
> Chas.
>
>
> .------ ------ ------ ------ ------ ------ ------ ------ ------ ------.
> | Message 2                                                           |
> '------ ------ ------ ------ ------ ------ ------ ------ ------ ------'
> Subject: Re: Is a Lehman geometry rolling pivot inherently unstable?
> From:    ChrisAtUpw@.......
> Date:    Mon, 7 Jul 2008 20:36:46 EDT
>
>
> --part1_cf4.316dc149.35a4109e_boundary
> Content-Type: text/plain; charset="UTF-8"
> Content-Transfer-Encoding: quoted-printable
> Content-Language: en
>
> 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
>
> --part1_cf4.316dc149.35a4109e_boundary
> Content-Type: text/html; charset="UTF-8"
> Content-Transfer-Encoding: quoted-printable
> Content-Language: en
>
> In a
me=
> ssage dated 2008/07/07, charles.r.patton@........ writes:

> 

> 
 : 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px">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.  COLOR=3D"#000000" BACK=3D"#ffffff" style=3D"BACKGROUND-COLOR: #ffffff"
SIZE=
> =3D2 PTSIZE=3D10 FAMILY=3D"SANSSERIF" FACE=3D"Arial"
LANG=3D"0"> >

> 

>  #ffffff" SIZE=3D2 PTSIZE=3D10 FAMILY=3D"SANSSERIF" FACE=3D"Arial"
LANG=3D"0"=
> >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.

>  

> 
 : 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px">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
 R>
> 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 R>
> 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   
>
> --part1_cf4.316dc149.35a4109e_boundary--
>
>
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