Re: Fw: rear suspention

[Dave Dahlgren <ddahlgren@snet.net>]

So how would you do this practically. With the arm 2 feet off center that is a
greater than 6 ft wide rear end if you allow 1 ft at each end for suspension
components and wheel offset.. The 6 ft long arm seems problematic as well. If
the arm is shorter I assume you would have to move it further right to gain
mechanical advantage as well.

>From your explanation it seems like the math goes as follows

torque in * RP ratio = torque out(pinion lift)
pinion lift / arm length = force to resist
arm length(ft) / pinion ratio = offset to counteract lift(ft)

so if we had a 1000 ft lb input and a 5:1 rear gear
1000 * 5 = 5000 pinion lift
5000 / 6ft arm = 833 lbs
6 ft arm / 5:1 pinion  = 1.2 ft offset

or if we had 100 ft lbs and a 1:1 RP ratio
100 * 1 = 100 # pinion lift
100/ 6ft arm=16.6 # to resist
6 ft arm / 1:1 rp ratio = 6 ft offset..
What am I missing here it seems very counter intuitive..
Dave

John Burk wrote:
> 
> 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) .