From: ChrisAtUpw@.......

Date: Mon, 7 Jul 2008 20:36:46 EDT

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 theangle 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 Cx =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