Don't you just love it when one of them smart fellers comes along and tries to
describe things by combin'in writhmatic with the alphebet!!!!!!
Serously it was a good argument. I have no idea if its correct
You flip this switch and press that button. Now just drive it and see what
happens.
Mike B. TLS#1
Jeff Lloyd wrote:
> see what happens, you leave the darn door open and a mechanical engineer
> stumbles in, gosh, but lemme tell you the boy knows his stuff. but umm this
> is racin cars, not rocket sicence, right Vince?
>
> Jeff
>
> >From: TeamZ3@aol.com
> >Reply-To: TeamZ3@aol.com
> >To: autox@autox.team.net
> >CC: johncof@ibm.net
> >Subject: Swaybar end geometry, my take
> >Date: Wed, 4 Aug 1999 00:28:02 EDT
> >
> >Think about what a swaybar is and how it works. I simplify the argument by
> >considering only one axle end of the vehicle. In order for the body to
> >roll
> >with a swaybar attached it must carry some portion of the inner unsprung
> >weight on that end of the vehicle with it, from which some resulting force
> >is
> >transferred torsionally through the bar to the outside tire. With enough
> >tire grip and stiff enough bar the entire inner unsprung weight will be
> >carried; the inner wheel will lift off the ground if roll is great enough
> >relative to suspension travel. In reality, the end of the bar located on
> >the
> >outside of the turn is, for all purposes, fixed. However, that's really
> >immaterial to the argument.
> >
> >My argument states that the moments at the ends of the swaybar are equal,
> >not
> >the forces. What is torsion, but a moment, which is force applied a
> >specific
> >distance perpendicularly about a centerpoint of, in this case, a shaft.
> >Consider the moment equation derived from Newton's Law that for every
> >reaction there is an equal and opposite reaction. Assuming that the end
> >links connect to the suspension in identical locations, the body pivot
> >points
> >are in identical locations, the diameter and shape of the bar is
> >symetrical,
> >etc., etc. :
> >
> > (A1xB1) = (A2 x B2) ... the summation of the moments must equal zero
> >
> >where A1 = left force, B1 = left arm length, A2 = right force, B2 =right
> >arm
> >length
> >
> >which transcribes to A1 = A2 x (B2 / B1)
> >
> >when the arms are equal length B1 = B2 then A1 = A2, the forces are the
> >same
> >
> >however when the lever arms are a different length the applied forces side
> >to
> >side will vary based on the ratio of the lever arm lengths; the applied
> >forces are not equal. The end result is that you'll see a different amount
> >of roll in one turning direction than the other based on the bias of the
> >forces. It is the force from the outside bar end that will be transferred
> >to
> >the outside tire (indirectly via the various connection points), so it
> >should
> >be easy to see that outside tire loading will not be the same at the same
> >point of body roll side to side, hence handling balance side to side is
> >affected.
> >
> >Further, very few street bars are the straight, Speedway-style type. Most
> >street bars comprise multiple bends, which changes their character
> >dynamically when compared to a straight shaft. When the bar lever arms are
> >identical the bending torsion is distributed equally throughout the bar
> >about
> >it's center, resulting in an equal reaction from side to side. When the
> >lever arms are not equal the bending torsion will be biased more to one
> >side
> >of the bar than the other. With a multiple bend type of bar this will
> >result
> >in different dynamic reactions of the bar between the two turning
> >directions,
> >resulting in an added roll bias force factor; dependent on the particulars
> >of
> >the bar.
> >
> >The again, what the h*ll do I know; I've been wrong before. Im certainly
> >open to opposing arguments.
> >
> >M Sipe
> >
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