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Fw: rear suspention

To: <land-speed@autox.team.net>
Subject: Fw: rear suspention
From: "John Burk" <joyseydevil@comcast.net>
Date: Mon, 13 Oct 2003 22:00:30 -0400
From: "John Burk" <joyseydevil@comcast.net>
To: <ddahlgren@snet.net>
Sent: Monday, October 13, 2003 6:51 PM
Subject: Re: rear suspention


> Hi Dave . My friend Bill came up with this suspension idea in the 60's .
> Later NASCAR "copied" it . If a hypothetical rear had an input torque of
500
> ' # and 3:1 ratio (1500 ' # output) it would take 250 # downward on a 6 '
> lever arm to resist pinion lift . If the lever was 2 ' off center (to the
> right) the 250# downward push times the 2 ' off set cancels the effect of
> drive shaft twist (equal tire loading without undesired roll stiffness) .
> ----- Original Message -----
> From: "Dave Dahlgren" <ddahlgren@snet.net>
> To: "John Burk" <joyseydevil@comcast.net>
> Cc: <land-speed@autox.team.net>
> Sent: Tuesday, October 14, 2003 4:34 AM
> Subject: Re: rear suspention
>
>
> > This sounds like a NASCAR modified rear suspension..
> > How does the math work for the offset what are the units used give an
> example
> > please.
> > Dave Dahlgren
> >
> >
> >
> > >
> > > A simpler
> > > fix
> > > ; use a single torque arm a little to the right of the drive shaft
.The
> > > offset
> > > needs to be the torque arm length divided by the ring & pinion ratio .
> With
> > > this design the torque arm exactly cancels the undesirable effects of
> drive
> > > shaft torque , wheel loading stays equal under acceleration or
> deceleration
> > > and roll stiffness can be as desired . The only drawback is  wheel
> loading
> > > is
> > > unequal under braking . Only the torque arm should resist axel torque
> > > (single
> > > link on each side) and only the single links should locate the rear
> front to
> > > rear . The front of the torque arm  needs to be free front to rear
(slot
> or
> > > vertical link) so there is no torque arm arc to fight with the side
link
> > > arcs
> > > . Hope my description is clear enough . John





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