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
>
|