PSN-L Email List Message

Subject: Re: Pivot paper discussion
From: ChrisAtUpw@.......
Date: Wed, 22 Oct 2008 18:13:41 EDT


In a message dated 2008/10/22, charles.r.patton@........ writes:

> Randy Pratt asked:
> > I see references to twist in the revision history so is
> > this twist resulting from upper and lower pivot travel in opposite
> > direction?  If so would not the support structure height play a role=20
> as well
> > as angle of the support wire to the boom?

Hi Randy,

       It does. The axis tilts as the spheres roll sideways, or the plane=20
rolls on the sphere. If the mass is lower than the centre of the rolling axi=
s,=20
this adds to the stability. If it is above this, it can detract from the=20
stability.

> That revision was early in the discussion with Brett and Chris.  I had=20
> taken a simplistic approach to the path based on the contact point.=20
> Latter I agreed with Brett on the approach to finding the locus center=20
> as is embodied in the current XLS calculations.  That If a locus center=20
> point exists that makes the locus a circular arc, then the twist doesn=E2=
=80=99t=20
> exist.  But even if twist exists, you choose your bob center-of-gravity=20
> roughly half way in the Z axis between the upper and lower pivot points,=20
> and its effect  should drop out of the considerations.  Either way, I=20
> stopped considering it.
>=20
> > Why was the locus chosen as a point behind the plate rather than the=20
> contact
> > point? Reason - In a Lehman, translation of a rolling pivot will=20
> result in unsymetric forces on the boom. =20

    Sorry but I don't understand what you are trying to say?

       The compression in the boom and the > tension
> > in the support wire will move out of the vertical plane.  A horizontal
> > friction force must be present to prevent slippage of the ball and this
> > force is also a moment around the mass.  These forces would have to be
> > analysed from the contact point. Larger angles and larger balls would
> > increase this effect.
> Please see my answer to Bob McClure on 10/16, posted to the PSN list.=20
> The =E2=80=9Ccenter point=E2=80=9D is chosen to give the =E2=80=9C best fi=
t=E2=80=9D to the calculated=20
> locus.  It is not chosen for any other purpose than to answer=20
> mathematically whether there is a close fit circular arc to the locus=20
> that has been calculated.  If there is, then it can be said that the bob=20
> with regards to the vertical beam travels a circular arc therefore a=20
> pendulum fit can be realized.  You may be right about secondary effects=20
> of the offset mass.=20
> >
> > An upper pivot does not travel a semicircle around the circumference in
> > relation to a near horizontal swing. The semicircle is tilted by the=20
> support
> > wire angle in relation to the near vertical swing axis.  Do you agree?

       Yes, but it is the swing axis itself which rotates.

> No, because a ball always has circular cross section and tilt only=20
> matters to the extent that it changes the length of CP. =20

    I don't agree.

So the > hypothetical center of the arc may not be where the physical contac=
t=20
>=20
> point is, but as you adjust the period of the swing, you will actually=20
> be adjusting these centers in space to be almost in a vertical line,=20
> with the upper just a bit forward to give the bob a pendulum arc that is=20
> lowest in the center of the swing.
> This was the whole point of doing all this analysis =E2=80=93 trying to=20
> understand just what the weight/bob was experiencing and what=20
> arrangements of pivots had the best circular fit.  Additionally what=20
> shows up is the ball on plate inferior to plate on ball with regard to=20
> the side slipping you touch on.
>=20
> > Construction could change this I guess with rigid upper boom structure.


> But even if twist exists, you choose your bob center-of-gravity roughly=20
> half
> way in the Z axis between the upper and lower pivot points, and its effect=
=20
> should drop out of the considerations.

       It doesn't, since the swing axis is rotating about it's centre in the=
=20
near vertical plane. The gradient is still non linear.

> --This point seems to be a large deviation from the traditional horizontal
> boom with double support wires and or underslung mass to prevent rocking o=
n
> the axis. It may be the improvement point we need to consider and construc=
t.

       You still need to prevent the boom from rotating about it's long axis=
..=20
A V wire upper support is most desirable. The problem is that the damping=20
force is not usually precisely on the line joining the centre of mass to the=
=20
lower point of rotation. This is a common cause of unwanted resonances, sinc=
e such=20
motions may be inadequately damped.

       Regards,

       Chris Chapman
In a me=
ssage dated 2008/10/22, charles.r.patton@........ writes:

Randy Pratt asked:
> I see references to twist in the revision history so is
> this twist resulting from upper and lower pivot travel in opposite
> direction?  If so would not the support structure height play a ro= le
as well
> as angle of the support wire to the boom?


Hi Randy,

       It does. The axis tilts as the spheres=20= roll sideways, or the plane rolls on the sphere. If the mass is lower than t= he centre of the rolling axis, this adds to the stability. If it is above th= is, it can detract from the stability.


That revision was early in the=20= discussion with Brett and Chris.  I had
taken a simplistic approach to the path based on the contact point.
Latter I agreed with Brett on the approach to finding the locus center
as is embodied in the current XLS calculations.  That If a locus center=
point exists that makes the locus a circular arc, then the twist doesn=E2= =80=99t
exist.  But even if twist exists, you choose your bob center-of-gravity=
roughly half way in the Z axis between the upper and lower pivot points, and its effect  should drop out of the considerations.  Either way= , I
stopped considering it.

> Why was the locus chosen as a point behind the plate rather than the contact
> point? Reason - In a Lehman, translation of a rolling pivot will
result in unsymetric forces on the boom. 


    Sorry but I don't understand what you are trying to say?<= BR>
       The compression in the boom and the tension
> in the support wire will move out of the vertical plane.  A horizo= ntal
> friction force must be present to prevent slippage of the ball and this=
> force is also a moment around the mass.  These forces would have t= o be
> analysed from the contact point. Larger angles and larger balls would > increase this effect.
Please see my answer to Bob McClure on 10/16, posted to the PSN list.
The =E2=80=9Ccenter point=E2=80=9D is chosen to give the =E2=80=9C best fit= =E2=80=9D to the calculated
locus.  It is not chosen for any other purpose than to answer
mathematically whether there is a close fit circular arc to the locus
that has been calculated.  If there is, then it can be said that the bo= b
with regards to the vertical beam travels a circular arc therefore a
pendulum fit can be realized.  You may be right about secondary effects=
of the offset mass.
>
> An upper pivot does not travel a semicircle around the circumference in=
> relation to a near horizontal swing. The semicircle is tilted by the su= pport
> wire angle in relation to the near vertical swing axis.  Do you ag= ree?


       Yes, but it is the swing axis itself w= hich rotates.

No, because a ball always has c= ircular cross section and tilt only
matters to the extent that it changes the length of CP. 


    I don't agree.

So the
hypothetical center of t= he arc may not be where the physical contact
point is, but as you adjust the period of the swing, you will actually
be adjusting these centers in space to be almost in a vertical line,
with the upper just a bit forward to give the bob a pendulum arc that is lowest in the center of the swing.
This was the whole point of doing all this analysis =E2=80=93 trying to
understand just what the weight/bob was experiencing and what
arrangements of pivots had the best circular fit.  Additionally what shows up is the ball on plate inferior to plate on ball with regard to
the side slipping you touch on.

> Construction could change this I guess with rigid upper boom structure.=



But even if twist exists, you c= hoose your bob center-of-gravity roughly half
way in the Z axis between the upper and lower pivot points, and its effect s= hould drop out of the considerations.


       It doesn't, since the swing axis is ro= tating about it's centre in the near vertical plane. The gradient is still n= on linear.

--This point seems to be a larg= e deviation from the traditional horizontal
boom with double support wires and or underslung mass to prevent rocking on<= BR> the axis. It may be the improvement point we need to consider and construct.=


       You still need to prevent the boom from= rotating about it's long axis. A V wire upper support is most desirable. Th= e problem is that the damping force is not usually precisely on the line joi= ning the centre of mass to the lower point of rotation. This is a common cau= se of unwanted resonances, since such motions may be inadequately damped.
       Regards,

       Chris Chapman

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